Kvant 1971, No. 11 · Pages

Popular science physics and mathematics
quantum
11
1971
magazine
Academy
Sciences of the USSR
And
Academy of Pedagogical
Sciences of the USSR

In the room:

Interaction of atoms and molecules 1 G. Ya. Myakishchev
On the best approximations 8 D. B. Fuks, M. B. Fuks
Gravity field of a spherically homogeneous body 16 I. I. Otsenevtsky
“Irreducible” case 20 V. G. Yankelevich
Laboratory “Quenta”
Demonstration of the orbits of bodies moving under the influence of central gravity 22 R. Wood
Problem “Quenta”
Problems 24
Problem solutions
M70-M73; F79-F82 26
N. V. Vasiliev, A. L. Toom, B. V. Bukhovtsev, I. Sh. Slobodetsky
V Weighing mathematical optimization 35 V. L. Gutenmacher
V Weighing physical optimization 42 I. A. Zaitsev
Applicant’s workshop
Graphs in problems on uniform processes 48 B. A. Kordemsky
Statics 54 L. G. Aslamazov
Information
Correspondence Optimization of the Asian Part of the USSR 58
Reviews, bibliography
From galaxies to elementary particles 63 V. A. Leitskovtsev
Answers, directions, solutions 64
Collector’s Corner
Badges of the Weighty Union Optimka, 3rd page, region.
Crossword 4th page region.

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Electromagnetic nature
intermolecular forces

It was almost impossible to begin a theoretical study of mechanical forces before the beginning of the 20th century. Simple and well-studied boundary forces during the interaction of bodies of columnar mass like a molecule could not play a noticeable role. It remained to propose that molecular forces are of an electromagnetic nature. Nothing was known about any other forces at that time.

Any atom, and especially molecules, is a complex system of a large number of loaded particles. It was not known how it worked. But the strength of their interaction undoubtedly depends on the structure of the molecules. However, the research began and began, naturally, with the simplest cases.

Orientation forces

For many molecules, such as water, the distribution of positive and negative charges is such that, on average, the centers of these charges do not coincide. Such a molecule can be approximately considered as an electric diopole: a combination of two exact charges +q and –q at a small distance l from each other (Fig. 1). The electrical properties of such a molecule are characterized by the diopole moment

$$p = ql$$

At first they were not able to calculate the values ​​of the diopole moments of molecules. Before the advent of quantum mechanics, this was generally impossible.

But, if we consider the diopole moments $p_1$ and $p_2$ of two molecules to be known, then the strength of interaction between the molecules depends. They will arise between two molecules, one of which has a diopole moment, and the second does not.

A diopole molecule creates an electric field that polarizes molecules with electric charges uniformly distributed throughout the volume.

to him. Positive charges will mix in the direction of the electric field, and negative charges will mix against it. As a result, the nonpolar molecule distances itself a little in this case as well. It is proportional to the dipole moment $p$ of a polar molecule, the coefficient $\alpha$, which characterizes the ability of a non-polar molecule to polarize and is again inversely proportional to the seventh power of the distance between the molecules:

$$F_n \sim \frac{p \alpha}{r^7}.$$

These tensile forces are called inductive or polarization forces, since they arise due to the polarization of molecules caused by electrostatic induction.

Dispersion forces

It is well known that tensile forces also exist between nonpolar molecules. For example, atoms of noble gases do not have a dipole moment, but nevertheless interact with each other. The nature of these forces was clarified only after the creation of quantum mechanics.

The qualitative and roughly approximate appearance of these forces can be explained as follows. In atoms and molecules, electrins move around the nuclei in some complex way. And on average over time, the dipole moments of molecules may turn out to be zero.

But at each moment of time, electrins occupy a specific position. Therefore, the instantaneous value of the dipole moment is from zero. An instantaneous dipole creates a field that polarizes neighboring nonpolar atoms. Because of this, the interaction of instantaneous dipoles appears. The total interaction between nonpolar molecules is the average result of the interaction of all possible instantaneous

dipoles with dipole moments that they induce in neighboring molecules due to induction.

Quantum calculation leads to the conclusion that the interaction force in this case is proportional to the product of molecules $\alpha_1$ and $\alpha_2$ and inversely proportional to the seventh power of the distance:

$$F_{\pi} \sim \frac{\alpha_1 \alpha_2}{r^7}.$$

These forces are called dispersive for the reason that the dispersion of a substance (the dependence of the refractive index of light on frequency) is determined by the same properties of molecules as these forces.

Dispersion forces act between all atoms and molecules, since the mechanism of their appearance does not depend on whether they have permanent dipole moments or not. Typically, these forces exceed in magnitude both orientational and inductive forces. Only during the interaction of molecules with large dipole moments, for example, water molecules, the orientational force is greater than the dispersion force (three times for water molecules). When polar molecules (CO, HCl, etc.) interact, the dispersion forces are ten and hundreds of times greater than all others.

But the main thing is that all three types of tension decrease in the same way with distance as $\frac{1}{r^7}$. However, at large distances compared to the size of molecules, the finite speed of propagation of electromagnetic interactions begins to affect. Because of this, at distances of the order of $10^{-5}$ cm, the gravitational forces begin to decrease as much as $\frac{1}{r^8}$. This was discovered in Deryagin’s experiments.

Pushback forces

Now let’s turn our attention to the repulsive forces acting between molecules at very short distances. On the one hand, there is a

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On the other hand, it’s easier, but on the other hand, it’s more difficult. It is simpler in the sense that these forces grow very quickly as the molecules approach each other, and therefore this or that rapid change in force with distance does not have a noticeable effect on the course of various processes.

The difficulty is that the pushing forces, to a much greater extent than the attractive forces, depend on the individual molecules. Knowing how molecule A pushes molecules B and C away, we are not yet able to judge what pushing forces will act between molecules B and C. When molecules come together in a non-presidential manner, the individuality of their structure is more pronounced than when there are large distances between them.

The assumption that the pushing forces return according to the law leads to a fairly good result

$$F_{от} \sim \frac{1}{r^{13}}.$$

Considering that the attractive forces with decreasing distance increase in proportion to $\frac{1}{r^7}$, and the pushing forces increase as $\frac{1}{r^{13}}$, we can draw an approximate dependence of the force on the distance. At figure 4, pushing forces are considered positive and attractive forces are considered negative. At a distance $r_0$, approximately equal to the sum of the radii of the molecules, the force is zero.

Potential energy of molecular interaction

When studying the behavior of a large set of molecules, it is more convenient to use not the interaction force of molecules, but potential energy.

It is necessary to calculate the average characteristics of the system, and the concept of the average force of interaction between molecules is meaningless, since the sum of all forces acting between molecules, according to Newton’s third law, is zero. The average potential energy, as we will soon see, largely determines the state and properties of matter.

Since the change in potential energy is equal to the work done by the force, then from the known dependence of the force on the distance one can find the dependence of the potential energy on the distance*).

It should be taken into account that if the force changes with distance as $\frac{1}{r^n}$, where $n$ is a certain number, then the potential energy will be less according to the law $U \sim \frac{1}{r^{n-1}}$. This can already be seen from the dimension messages. Energy has the dimension of work, that is, the product of force and distance. Therefore, the dimension of energy is equal to the dimension of force multiplied by the dimension of distance.

*) See the article by R. G. Mints “Graphs of potential energy”, «Kvant» No. 5, 1971, pp. 8-16.

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piya. This point is called the turning point. Subsequently, the molecule moves in the positive direction $r$ and goes to infinity.

A completely different picture will be observed at $E = E_2 < 0$ (see Fig. 9). In this case, the molecule is in a potential well and cannot leave it. At the turning points $r_2$ and $r_3$ the kinetic energy is zero. A so-called bound state arises. molecules vibrate to each other. The division of the system into two independent particles is impossible without increasing the total energy to $E > 0$.

Energy of interaction of molecules in gases, liquids and solids

Let us now establish the number of criteria for various gases, liquids and solids from the point of view of molecular-kinetic theory.

Gases

You can get a deeper understanding of the state of matter called a real gas if you picture the dependence of the potential energy of one of the molecules on the distance to its close neighbors (Fig. 10). When a molecule moves, its potential energy is almost exactly zero over most of its path, since the distance between molecules in a gas is, on average, much greater than their size. At points 1 and 2 there are close neighbors of the molecule in question. This molecule passes through a fairly significant

a considerable distance from neighbor 1 and closer from neighbor 2.

The average potential energy of a molecule is negative and very small. Its magnitude is equal to the area of ​​the figure limited by the potential curve between points 1 and 2 and the $r$ axis, divided by the length of the segment 1-2 (the average value of the segment 1-2). The total average energy is necessarily greater than zero (straight line at Fig. 10), since at $E < 0$ we would have a bound state of molecules. This is possible only if the average kinetic energy of a molecule in a gas is greater than the average value of its potential energy:

$$\bar{E}_{\kappa} > |\bar{U}|.$$

After all

$$\bar{E} = \bar{E}_{\kappa} + \bar{U},$$

and $\bar{U}$ is negative.

Liquids

In liquids and solids, molecules are located at a small distance from each other. Therefore, each molecule interacts with several of its molecules at once. Let us limit ourselves to taking into account the interaction of a given molecule with two close neighbors located approximately at a distance of $2r_0$ from each other.

The desired potential curve can be obtained superimposed by the curve depicted on figure 7 (pair interaction) on the same curve, shifted relative to the first by a distance slightly greater than $2r_0$. The potential energies add up, and therefore the depth of the potential well almost doubles, and the maximum

energy simuts decrease (Fig. 11). The course of the potential curve, taking into account interactions with other molecules, is shown in figure 12.

In order for a molecule to be unable to leave the liquid, its average energy must be negative ($\bar{E} < 0$). Only in this case the molecule remains inside the potential well formed by its neighbors. If $\bar{E} > 0$, then the molecule will not stay inside the liquid and will leave it.

Since $\bar{E} = \bar{E}_k + \bar{U}$ and $\bar{U} < 0$, then in a liquid the average kinetic energy is less than the absolute value of the average potential energy: $\bar{E}_k < |\bar{U}|$, and only slightly less ($\bar{E}_k < |\bar{U}|$), so $|\bar{E}| \leq |\bar{U}|_0$ is the minimum (in absolute value) value of the potential energy. That is why the oscillations of the molecule in the hole do not last long. Due to the randomness of the movement of molecules, their energy continuously changes and becomes too more, then less energy $\bar{E}$.

As soon as the energy of a molecule exceeds the height of the potential barrier separating one well from another, the molecule jumps from one equilibrium position to another. This is precisely what determines the nature of the thermal motion of molecules in a liquid and a flowing liquid. The “settled life” time of a molecule is the time it remains inside a certain potential well. With increasing temperature, the average energy increases, and jumps occur more frequently.

Solids

The potential energy of interaction of a solid molecule with its nearest neighbors is similar to the interaction energy of liquid molecules (Fig. 12). Only the depth of the potential well here is somewhat greater, since the molecules are located closer to each other. Condition $\bar{E}_k < |\bar{U}|$ is also satisfied for solid bodies. But the kinetic energy of solid molecules is significantly less than that of liquid molecules. After all, solids form liquids when cooled. Accordingly, in solids the average kinetic energy of molecular motion is significantly less than the absolute value of the average potential energy:

$$\bar{E}_k \ll |\bar{U}|.$$

At figure 12, the average energy of a molecule inside a potential well is depicted by a straight line segment. The particle oscillates in the potential well. The height of the barriers between adjacent wells is large, and the molecules hardly move from one equilibrium position to another. To move molecules must receive energy that significantly exceeds the average energy. This event is unlikely. That’s why solids, unlike liquids, retain their shape.

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we beam the point $A_4$. We denote the vector $\vec{OA}_4$ by $a_4$, from the point $A_3$ we put aside $a_4$, etc.

We get two broken lines: $A_1 A_3 A_5 \dots$ and $A_2 A_4 A_6 \dots$, volumetric axis $l$ (Fig. 3). It may happen that one of the points $A_k$ lands directly on $l$ - this is where the process stops. It may turn out that this will not happen - then the process will be endless.

All together is called the nose-pulling algorithm (ABH).

Let us now project all the constructions onto a straight line perpendicular to the $l$ axis. The axis $l$ itself will go into one point, the vectors $a_k$ into segments, which we denote by $b_k (k=1, 2, 3, \dots)$. Let us obtain the Euclidean algorithm already known to us for the segments $b_1$ and $b_2$. In particular,

$$\frac{b_1}{b_2} = \left[ m_1; m_2, m_3, m_4, \dots \right]$$

and, therefore,

$$\frac{b_2}{b_1} = \left[ 0; m_1, m_2, m_3, \dots \right].$$

Now we will show how to use ABH to solve our geometric problem (see § 1). Let us consider the lattice mentioned in § 1, the points of which correspond to the approximation of the number $\alpha$ (it is convenient for us to assume that $0 < \alpha < 1$; obviously, when proving the Hurwitz-Borel theorem we can limit ourselves to such numbers).

Let us take ABH to the vectors $a_1 = \vec{OA}_1$ and $a_2 = \vec{OA}_2$, on which our lattice is built, and to the axis $l = OY$. It turns out two broken lines $A_1 A_3 A_5 \dots$, $A_2 A_4 A_6 \dots$, solid axis $OY$ (Fig. 4).

The numbers $m_1, m_2, \dots$ appearing in ABH are nothing more than partial partial numbers $\alpha$. More precisely, the expansion of $\alpha$ into a continued fraction looks like this:

$$\alpha = \left[ 0; m_1, m_2, \dots \right]$$

(recall that $0 < \alpha < 1$).

Let us now prove several properties of our broken lines.

Theorem 1. The points $A_1, A_2, A_3, \dots$ are lattice points.

The validity of this assertion is clear from figure 4. Its complete proof is based on Theorem 2 of Part I. We leave the details to the reader.

Theorem 2. Between the broken lines $A_1, A_3 A_5 \dots$ and $A_2 A_4 A_6 \dots$ (maybe in the protected area in the drawing) the power of the lattice is mute.

Proof. ABH consists of sequential deposition of vectors: $A_1 B, BA_3; A_2 C, CA_4; A_3 D, DE; \dots$. For each such vector, we construct a parallelogram, the sides of which are this vector and the vector to which it is parallel. We obtain a sequence of parallelograms

$$OA_2 BA_1, OA_2 A_3 B, OA_2 CA_3, OCA_4 A_3, OA_4 DA_3, \dots$$

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Theorem 3. Point $A_k$ ($n \geq 2$) we answer the ($k-2$)th suitable fraction $\alpha_{k-2}$ of number $\alpha$.

Proof.
For $A_2$ and $A_3$ this is clear: $A_2$ corresponds to $p=0$, $q=1$, that is, an approximation equal to $0=\alpha_0$ (recall that $m_0=0$); $A_3$ corresponds to $p=1$, $q=m_1$ (Fig. 8), that is, an approximation of $1/m_1=\alpha_1$.

Let us already know that for $i \leq k-1$, where $k \geq 4$, the point $A_i$ corresponds to the approximation $\alpha_{i-2}$. Let us show that $A_k$ corresponds to $\alpha_{k-2}$.

We must prove that $A_kC=p_{k-2}$, $A_kH=q_{k-2}$ (Fig. 9). Triangle $A_{k-2}A_kE$ is similar to triangle $OA_{k-1}M$; triangle $A_{k-2}A_kB$ is similar to triangle $OA_{k-1}D$; The similarity coefficient in both cases is equal to $m_{k-2}$. We have:
$$A_kC = BC + A_kB = A_{k-2}F + m_{k-2}A_{k-1}D = p_{k-4} + m_{k-2}p_{k-3} = p_{k-2},$$
$$A_kH = EH + A_kE = A_{k-2}G + m_{k-2}A_{k-1}M = q_{k-4} + m_{k-2}q_{k-3} = q_{k-2}$$
(see article by N. M. Beskin).

The theorem has been proven.

Theorem 3, together with the corollary of Theorem 2, leads us to the following conclusion:
If for a given irrational number $\alpha$ there are an infinite number of approximations with a quality coefficient less than the given one, then there are an infinite number of them among the suitable fractions of the number $\alpha$.

The last question remains: how to find the quality coefficients of suitable fractions?

Fortunately, this question can be completely resolved, and the answer is written down in an elegant and convenient formula.

§ 3. Quality coefficients of suitable fractions.

Theorem 4. Let $\alpha=[0; m_1, m_2, \dots]$, $\mu_k$ be the quality coefficient of the $k-1$th suitable fraction. Then:

$$\frac{1}{\mu_k} = m_k + [0; m_{k+1}, m_{k+2}, \dots] + [0; m_{k-1}, m_{k-2}, \dots, m_1]$$

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Proof.
Consider the points $A_{k+2}$ and $A_{k+1}$, corresponding to the approximations $\alpha_k$ and $\alpha_{k-1}$ (Fig. 10). Through $q_k$, $r_k$; $q_{k-1}$, $r_{k-1}$ we denote their distances to the axes $OX$, $OY$. We know that $\mu_k = r_{k-1}q_{k-1}$ (see § 3). Let’s prove the formula:

$$q_k r_{k-1} + q_{k-1} r_k = 1$$ (3)
$$r_k / r_{k-1} = [0; m_{k+1}, m_{k+2}, \dots]$$ (4)
$$\frac{q_k}{q_{k-1}} = [m_k; m_{k-1}, m_{k-2}, \dots, m_1]$$ (5)

Formula (3) proves this. Divide the parallelogram $OA_{k+1}BA_{k+2}$ with straight lines parallel to the coordinate axes into triangles: $A_{k+1}QO$, $OPA_{k+2}$, $A_{k+2}MB$, $BRA_{k+1}$ and a straight line $MPQR$ (see Fig. 11). The area of ​​the parallelogram $OA_{k+1}BA_{k+2}$ is 1, since this is the main parallelogram of our lattice. Finding the areas of the parts:

$$1 = \frac{1}{2} r_{k-1} q_{k-1} + \frac{1}{2} r_k q_k + \frac{1}{2} r_{k-1} q_{k-1} + \frac{1}{2} r_k q_k + (r_{k-1} - r_k) (q_k - q_{k-1}) = q_k r_{k-1} + q_{k-1} r_k.$$

Formula (4), in essence, is already known to us: if we discard the part of our nose-pulling algorithm that precedes obtaining the point $A_{k+2}$, we obtain the same algorithm for the vectors $OA_{k+1}$, $OA_{k-2}$. Therefore, formula (4) follows directly from formula (2).

The most necessary proof is found in formula (5). Let us construct the broken line $A'_{k+2} \dots A'_2$ (Fig. 12), symmetrically to the broken line $A'_{k+2} \dots A'_2$ relative to the point $O$. This broken line, together with the broken line $A'_{k+1} \dots A'_1$, is, as can be seen, the result of applying $ABH$ to the vectors $\overrightarrow{OA}_{k+2}$, $\overrightarrow{OA}_{k+1}$ and negative

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direction of the $OX$ axis. This algorithm, as we see, is finite and ends at point $A_1$. It is structured like this: first, from the point $A'_{k+2}$ it lays down the vector $OA_{k+1}$ $m_k$ times, then from the point $A_{k+1}$ it lays down the vector $\vec{OA}'_k$ $m_{k-1}$ times, etc. The last laying down: the vector $\vec{OA}'_2$ lays down $m_1$ times.

By virtue of formula (4) we have:

$$q_k/q_{k-1} = [m_k; m_{k-1}, \dots, m_1].$$

So, all formulas have been proven. Let us now prove Theorem 4.

We have:

$$\frac{1}{\mu_k} = \frac{q_k r_{k-1} + q_{k-1} r_k}{r_{k-1} q_{k-1}} = \frac{r_k}{r_{k-1}} + \frac{q_k}{q_{k-1}} = [0; m_{k+1} \dots] + [m_k; m_{k-1}, \dots, m_1]$$

Q.E.D.

§ 4. Proof of the Hurwitz-Borel theorem.

We must show that the infinite set of suitable fractions of the number $\alpha = [0; m_1, m_2, \dots]$ has quality coefficients less than $\frac{1}{\sqrt{5}}$, that is, that for the infinite many $k$ the inequalities $\mu_k < \frac{1}{\sqrt{5}}$ or $\frac{1}{\mu_k} > \sqrt{5}$ hold. Let’s use Theorem 7.

There are three possible cases.

1°. Among the infinite numbers $m_1, m_2, \dots$ many different from 1 and 2. This case is the simplest. If $m_k > 2$, then $m_k \geq 3$ and $\frac{1}{\mu_k} = m_k + \frac{1}{m_{k+1} + \dots} + \frac{1}{m_{k-1} + \dots} > m_k \geq 3 > \sqrt{5}$.

2°. Among the numbers $m_1, m_2, \dots$ there are only a finite number different from 1 and 2, but two are an infinite number. In this case, for infinite many $k$ we have:

$$\frac{1}{\mu_k} = m_k + [0; m_{k+1}, \dots] + [0; m_{k-1}, \dots] > 2 + \frac{1}{3} + \frac{1}{3} = \frac{8}{3} > \sqrt{5}$$

3°. Starting at some point, all partial partials are equal to 1. Let, say, $m_{k_0} > 1$ and $m_k = 1$ for $k > k_0$. If the difference $k - k_0$ is even, then

$$\frac{1}{\mu_k} = 1 + [0; 1, 1, \dots, 1, m_{k_0}, \dots] > 1 + 2 [0; 1, 1, \dots] = 1 + \frac{4}{\sqrt{5} + 1} = \sqrt{5}$$

(We used here the fact that if $k$ is even and $m'_{k+1} > m'_{k+1}$, then

$$[m_0; m_1, \dots, m_k, m'_{k+1}, \dots] > [m_0, m_1, \dots, m_k, m'_{k+1}, \dots]$$

and equality $[1; 1, 1, \dots] = \frac{\sqrt{5} + 1}{2}$).

Prove these facts yourself).
Strictly speaking, the case when all incomplete partials are equal to 1 has not been considered. In this case we have

$$\frac{1}{\mu_k} = 1 + [0; 1, 1, \dots] + [0; 1, 1, \dots, 1]$$

and for even $k$ we get

$$\frac{1}{\mu_k} < 1 + 2 [0; 1, 1, \dots] = \sqrt{5}$$

due to the properties of continued fractions.

In all cases, the existence of an infinite multi-approximation with a quality coefficient less than $1/\sqrt{5}$ is proven.

It is clear that at the same time in the case $3^\circ$ and only in this case the number $\sqrt{5}$ cannot be notified. Strictly speaking, in this case

$$\lim_{k \to \infty} \frac{1}{\mu_k} = \sqrt{5},$$

that is, for very large $k$, the number $1/\mu_k$ is very close to $\sqrt{5}$.

Thus, we have proven all statements of the Hurwitz-Borel theorem.

In fact, we have proven even more. It is clear that Theorem 7 allows us to solve the question of the best approximations of the individual for each $\alpha$. For example, if the partial partials of a given number $\alpha$ are not all 1, starting from some number, that is, if the case $3^\circ$ does not occur, then, as can be seen from our reasoning, $\sqrt{5}$ can be replaced by $2^2/3$ (and in fact even by $2\sqrt{2}$; prove). From Theorem 7 it follows that

$$m_k < \frac{1}{\mu_k} < m_k + 2.$$

From this it is clear that if incomplete partial numbers $\alpha$ grow unlimitedly, then the number $\frac{1}{\mu_k}$ can be arbitrarily large, that is, the number $\mu_k$ can be arbitrarily small, that is, for a given $\alpha$ there are approximations with an arbitrarily small quality factor.

If the numbers $m_k$ are all smaller than a certain number $N$, then the quality coefficient of approximations of this number cannot be less than $\frac{1}{N+2}$. For example, the quality coefficient of rational approximations of the number $\sqrt{2} = [1; 2, 2, \dots]$ cannot be less than $\frac{1}{4}$ (in fact, the smallest quality coefficient among the approximations of the number $\sqrt{2}$ has the fraction $\frac{3}{2}$; this coefficient is equal to $6 - 4\sqrt{2} \approx 0,34$).

So, surprisingly, it turns out that the “best” approximate numbers, like $\sqrt{2}$, are the worst approximations by rational numbers.

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Field of softness of a spherical-homogeneous body

I. I. Ogievetsky

Newton’s law of universal gravitation equally determines the fall of an apple to the ground, the movement of the Moon around the Earth, and the orbits of the planets. It is based on the fact that the tension force between material points with masses $m_1$ and $m_2$, located at a distance $l$ from each other, is equal to

$$F = \gamma \frac{m_1 m_2}{l^2},$$

(1)

where $\gamma = 6,67 \cdot 10^{-11} \frac{H \cdot m^2}{\kappa^2}$ is the gravitational constant (in the SI system). But the Earth, Moon and other real bodies are not exact, they have finite sizes. How to determine the tension force between stranded bodies?

The solution to this problem led Newton to the creation of the most important branch of mathematics - integral calculus. To find the force of constriction of a material point to a conjugate body, you should mentally divide this body into small (in the limit - infinitesimal) parts, calculate the contribution to the constriction force from each of these parts using formula (1) and then add (interpret) these contributions. Celestial bodies can be approximately considered spherical-homogeneous, that is, such that at any given point their density is entirely determined by the distance of this point from the center of the body. Newton discovered a remarkable fact:

$$F = \gamma \frac{4\pi R^2 \rho m}{d^2} = \gamma \frac{M \cdot m}{d^2},$$

(2)

where $M$ is the surface mass. In other words, the tensile force outside the spherical surface is the same as if the entire mass of the surface were concentrated at its center. This immediately implies a similar statement for a spherical homogeneous body.

*) This property is often used as very important, without proof. Use it in some articles «Kvant». Therefore, it is very useful to understand how it proves.

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Fig. 3.

Let us now use inversion and replace $l$ in (6) with $\bar{l} \frac{d}{R}$ according to (4):

$$\Delta F_{OQ} = \gamma \frac{m\rho R^2 \cos \alpha \Delta S}{d^2 \bar{l}^2} =$$
$$= \gamma \frac{m\rho R^2 \Delta \omega}{d^2},$$

(7)

where the values ​​are entered
$$\Delta \omega = \frac{\cos \alpha \Delta S}{\bar{l}^2}.$$

Let’s make an important observation. Note that $\alpha$ is also the angle between the normal (radius $OP$) to the surface element $\Delta S$ and the direction $\bar{Q}P^*$, and cosa is set for any point $P$.

It is therefore easy to verify that $\Delta \omega$ is the solid angle at which the surface element $\Delta S$ is visible from point $\bar{Q}$.

If you introduce a cone with the vertex at point $\bar{Q}$ and a running contour $\Delta S$ is formed, then the solid angle at the vertex of this cone will be $\Delta \omega$ (Fig. 3).

Numerically, the angle $\Delta \omega$ is equal to the area that this cone cuts out from a sphere of unit radius with center at point $\bar{Q}$.

*) This is where the inversion operation is essential. With its help, we introduced the distance $\bar{l}$ into the denominator (6) instead of the distance $l$, thanks to which the solid angle $\Delta \omega$ appeared in the formula.

Full acceptance of all surfaces $\Delta S$:

$$F = \sum_{s} \Delta F_{OQ} = \gamma \frac{m\rho R^2}{d^2} \sum_{s} \Delta \omega =$$
$$= \gamma \frac{m\rho 4\pi R^2}{d^2} = \gamma \frac{mM}{d^2},$$

(8)

since $\sum_{s} \Delta \omega = 4\pi$ is the total solid angle at which the entire sphere is visible from the interior point $\bar{Q}$, and $4\pi R^2\rho = M$ is the total mass of the spherical surface. Thus, Statement 1 is proven.

Let us now consider a material point $\bar{Q}$ of mass $m$ inside a hollow spherical surface. Again from the messages of symmetry it is very clear that the force generated by the spherical surface as a whole can only be directed along the line $O\bar{Q}$.

Therefore, we need to add the components of the force $\Delta F_{OQ}$ from the elements of the sphere $\Delta S$, equal to

$$\Delta F_{O\bar{Q}} = \gamma \frac{m\rho \Delta S \cos \beta}{\bar{l}^2} =$$
$$= \gamma \frac{m\rho R^2 \Delta S \cos \beta}{d^2 l^2},$$

(9)

if we use it again, we take it from inversion and replace $\bar{l}$ with $l \frac{d}{R}$ according to correspondence (4).

Calculation of the sphere $S$ into two parts $S_\Pi$ and $S_\lambda$ by a plane passing through the point $\bar{Q}$ perpendicular to the line

O$\bar{Q}$ and the images at the intersection with the disk sphere (Fig. 4). It is obvious that parts of the spherical surface $S_\pi$ and $S_\pi$, lying to the right and left of this disk, will attract the material point $\bar{Q}$ in different directions, to the right and to the left. Matching (see Fig.4), $\cos \beta > 0$ for points $P$ from $S_\pi$ and $\cos \beta < 0$ for points from $S_\pi$. Let us now take into account that $\frac{\Delta S | \cos \beta |}{l^2} = \Delta \omega - \text{телесному углу}, под которым виден элемент$\Delta S$из точки$Q^*$). Складывая силы$\Delta F_{o\bar{Q}}$, находим полную силу$$F = \sum_S \Delta F_{o\bar{Q}} = \frac{\gamma m\rho R^2}{d^2} \left( \sum_S \Delta \omega - \sum_S \Delta \omega \right),$$(10) где суммы распространены, соответственно, на части поверхности$S_\pi$и$S_\pi$. Но$\sum_S \Delta \omega = \sum_S \Delta \omega = \omega_D - \text{solid angle} at which element $\Delta S$ is visible from point $Q$. Indeed (see Fig. 4), for a cone with vertex Fig. 4.

$Q$ formed by the tangents to the sphere $S$, the line of tangency is precisely the circle $D$ (check this). Thus,

$$F = 0$$

(*) As for a point outside the sphere, when calculating the force at a point inside the sphere we find that the formula includes a solid angle, this time for an inverted point lying outside the sphere.

for any point $\bar{Q}$ lying inside $S$. The proof of statement 2 is complete; inside a homogeneous spherical surface, the gravitational force of the real becomes zero.

It is very noteworthy that Newton attached great importance to the above-proven assumption about the gravitational properties of extended spherical-homogeneous bodies and surfaces. Only after proving these assumptions did Newton publish the derivation he had obtained many years earlier of Kepler’s laws of planetary motion from formula (1) for exact bodies (see J. Orir, “Popular Physics”, Moscow, 1964, pp. 87-98).

Exercises

1. A remarkable property of mathematical methods is that the same “mathematical model” can describe facts from seemingly completely different areas of physics. Refer to the fact that Coulomb’s law; describing the interaction of two precise electric charges, written mathematically in the same way as Newton’s gravitational law, formulate the “electrostatic analogues” of assumptions 1 and 2.

2. Prove that the force of attraction of an exact mass by a homogeneous unbounded plate is inversely proportional to the distance of this mass from the plate.

3. Find the force of attraction exerted by a homogeneous material hemisphere on an exact mass located on a straight line connecting the center of the sphere with its pole.

4. In Hinton’s imaginary flat world (see «Kvant» No. 4, 1970), the force of interaction between exact masses is inversely proportional to the distance between them. Keeping in mind that the “spheres” of a flat world are circles, formulate and prove propositions for the Hinton world similar to propositions 1 and 2. Show that the force of attraction experienced by an imaginary inhabitant of one of the planets of the Hinton system is proportional to the distance of the inhabitant from the center of the planet. If there are collebose planets in the system, then there is no gravitational field in the interior of the collebose (it is assumed that the planets are made of homogeneous matter).

20

V. G. Yankelevich

Our reader, seventh grader Misha Balkin, asks: “I saw in the reference book the Cardan formula for solving the cubic equation $x^3 + px + q = 0$:

$$x = \sqrt[3]{\frac{q}{2} + \sqrt{D}} + \sqrt[3]{\frac{q}{2} - \sqrt{D}},$$

where $D = \left(\frac{p}{3}\right)^3 + \left(\frac{q}{2}\right)^2$.

But it seems to me that this formula cannot always be used. For example, if we want to find the roots of the equation $x^3 - x = 0$ ($p = -1, q = 0$) using this formula, we get

$$x = \sqrt[3]{\sqrt{-\frac{1}{27}}} + \sqrt[3]{-\sqrt{-\frac{1}{27}}}.$$

This is some kind of minimum number, and the equation has three crowns: 0, 1 and -1. How to establish this contradiction?>

The editors believe that the answer to this question will be of interest to many of our readers. We are publishing an article by V. G. Yankelevch, a 10th grade student from the city of Kustanay, who tells how cubic equations can be solved in the case when the Kardanov formula produces negative numbers under the square root sign.

Back in the 16th century, a general formula was found that made it possible to express the solution to a third-degree equation

$$x^3 + px^2 + qx + r = 0$$

(1)

through its coefficients using arithmetic operations and knowledge of the roots. This formula (it is called the Del Ferro formula or the Tartaglia-Cardano formula) can be written as follows:

$$x = \sqrt[3]{\frac{c}{54} + \sqrt{\frac{D}{108}}} + \sqrt[3]{\frac{c}{54} - \sqrt{\frac{D}{108}}} - \frac{p}{3},$$

(2)

where $c = 2p^3 - 9pq + 27r$; $D = p^2q^2 - 4q^3 - 4p^3r + 18pqr - 27r^2$. Afterwards, as soon as the concept of a complex number took shape, it became clear that this formula allows us to obtain all solutions to equation (1) with any (complex) coefficients $p, q, r$. But, unfortunately, even if all the coefficients of the equation are general action numbers, then to find (hereinafter action!) solutions using formula (2) you have to extract the numbers; this difficulty arises when $D > 0$ (this is precisely the case when the equation has three different real roots).

Consider the cubic equation (1) with real and positive discriminants $D^*$).

We will show that any equation (1) can be reduced to an equation of the form

$$ay^3 - 3by^2 - 3ay + b = 0,$$

(3)

where $a$ and $b$ are real numbers and $a \neq 0$.

First we prove that equation (3) has roots

$$y_1 = \tg \frac{\varphi}{3},$$

$$y_2 = \tg \frac{\varphi + 2\pi}{3},$$

$$y_3 = \tg \frac{\varphi + 4\pi}{3},$$

*) $D$ is called the discriminant of the polynomial $x^3 + px^2 + qx + r$. It can be shown that $D = (x_1 - x_2)^2 (x_2 - x_3)^2 (x_3 - x_1)^2$, where $x_1, x_2, x_3$ are the roots of equation (1). If the coefficients $p, q, r$ are real and $D < 0$, then the equation has two conjugate complex roots and one real one, which is immediately found by formula (2).

22

Demonstration of the orbits of bodies moving under the influence of central gravity

R. Wood

In the last issue of our magazine, an article by the remarkable American physicist R. Wood about artificial mirages and tornadoes was published. Below we place another article by Wood, devoted to the study of the orbits of bodies in the central field. This article was published in the Physical Review in 1897.

After Wood published an article on the LABORATORY Demonstration of the orbits of planets and comets around the Sun using a steel ball and an electromagnet, the journal Nature published a letter from a professor at the University of London, which stated that this experiment does not demonstrate orbits in a gravitational field, since the force of attraction of the ball by a magnet varies inversely to the fifth power, and not to the square of the distance, as in the case of gravity.

Wood set to work always, drew a detailed diagram and realized that the ball intersects the magnetic lines at an angle and that its orbit depends only on the horizontal component of the tension. After careful measurements of the magnetic field strength in the stack plane, it was found to be almost exactly inversely proportional to the square of the distance.

Meanwhile, critical reports appeared in several other English technical journals, and Wood was pleased to send confirmation of the validity of his experiment, citing the results of measurements.

Do six pins on his sleeve point to the laboratory’s surface, the movement tells the planet no eliptical轨道, parabolic轨道 and hyperbolic轨道 under the influence of a central force, the magnitude of which would vary inversely with the square of the distance. I conducted several experiments examining the motion of a small steel ball around a magnetic field. Obits of length give good results, and I believe that they can be useful when giving lectures.

The device for demonstration is very simple. It consists of a glass disk about 40 cm in diameter, with a small hole in the center. The cone of a large electromagnet is passed into this hole. The surface of the disk is covered with smoke (Fig. 1). You need to try to give the plate a horizontal position and the magnet axis a vertical position as accurately as possible.

A small polished steel ball about 5 mm in diameter (for example, from a wave bearing)

24

TASKS

M111. The square on side 1 contains a figure whose distance between any two points is not equal to 0.001. Prove that the area of ​​this figure does not exceed 0.34. (We can assume that the boundary of the figure referred to in the conditions consists of straight line segments and circular arcs.)
Try to get a more accurate estimate and prove a similar theorem in space.
G. V. Rosenblum

M112. In the table $m \times n$ the numbers are written so that for any two rows and any two columns, the sum of the numbers in the two opposite vertices of the rectangle they form is equal to the sum of the numbers in its two other vertices. Some of the numbers were erased, but let’s say you can restore the erased ones. Prove that there are at least $(n+m-1)$ numbers left.

M113. Prove that for any natural number $n$ there is a number made up of the digits 1 and 2 that is divisible by $2^n$.
B. M. Ivlev

M114. Several numbers are written in a circle. If for some four consecutive numbers $a, b, c, d$ it turns out that $(a-d)(b-c) < 0$, then the numbers $b$ and $c$ can be swapped. Prove that this operation can be performed only a finite number of times.
V. B. Alekseev

M115. A whole number of liters of water are poured into three vessels. It is allowed to pour a column of water into any vessel, as much as it already contains, from any other vessel. Prove,

Fig. 1.
a voltmeter, a current source and connected wires, is it possible to change the value of one of the resistances without breaking a single contact in the circuit?

A. R. Zilberman

F119. Light from source $S$ on its way to the screen passes through a resting glass cube with edge $l$ (Fig. 2). How much faster is light?

will reach the screen if the cube is moved at speed $v$? The speed of light in air is $c$, the refractive index of the stack is $n$ ($v \ll c, n \gg 1$).

G. V. Meledin

F120. There is water in a cylinder with a piston, inside of which at the initial moment there is a cavity of volume $V$. The gas pressure in the cavity is negligible. The piston exerts constant pressure $P$ on the water. What kinetic energy will the water acquire at the moment when the cavity disappears? The initial speed of water is zero. The force of gravity can be ignored.

G. L. Komkin

F121. Estimate the maximum force that the diameter will show when attached to the planes covering the Magdeburg hemispheres (hemispheres) with a radius

Fig. 3.

$R=20$ cm (Fig. 3). The hemispheres stretch in opposite directions as $F$. Atmospheric pressure is 1 atm.

F122. A ball is thrown into a glass of water, rotating around its axis, and floats on the surface of the water. Where on the surface will the ball be located?

Amendment

In «Kvant» No. 8 for 1971 on page 3, due to the fault of the editors, “a portrait from the bronze medal of Euclid” was mistakenly placed. In fact, images of Euclid’s mathematics have not reached our time. The image placed appears to be a portrait of the philosopher Euclid of Megara.

Find the error

As is known, by a point, the sum of the distances from the bark to the vertices of a given quadrilateral is minimal,

Fig. 1.

is the point of intersection of its diagonals (Fig. 1). We will reduce one of the sides of the quadrilateral, for example, $BC$ (Fig. 2), until the side $BC$

Fig. 2.

will not turn into a point. In this case, the point $O$ will approach $BC$ and at figure 3 will merge with it.

Fig. 3.

This means that the point whose sum of distances from the crust to the vertices of a given trigon is minimal is one of its vertices. But there is a theorem that states that such a point is the point from which all three sides of the triangle are visible at an angle $120^\circ$, and not one of the vertices.

Where is the mistake?

A. Skovlin

26

In this issue we publish the solution to problems M70–M73

M70

Let $l_1, l_2, \dots, l_n$ be several lines in the plane, among which there are two intersecting lines. Prove that it is possible, and in a unique way, to choose on each of these lines at the point $X_1, X_2, \dots, X_n$ so that the redistribution restored at the point $X_i$ to the line $l_i$ passes through the point $X_{i+1}$ (for all $i=1, 2, \dots, n-1$), and the redistribution restored to $l_n$ at the point $X_n$ passes through the point $X_1$.

Try to formulate and prove a similar theorem in space.

This problem can also be formulated as follows. Let $p_i$ be a mapping that assigns to each point $M$ (on the plane) its projection onto the line $l_i$, that is, $p_i(M)$ is the base of the redistribution dropped from the point $M$ to the line $l_i$. It is required to prove that on the line $l_1$ there is a unique point $X_1$ such that after $n$ mappings

$$X_1 \xrightarrow{p_n} X_n \xrightarrow{p_{n-1}} X_{n-1} \xrightarrow{p_{n-2}} \dots$$

turns into itself.

For an arbitrary point $X$ of the line $l_1$ denotes through $f(X)$ the point on the line $l_1$ at which it passes

$$X$$ after $n$ mappings $p_n, p_{n-1}, \dots, p_1$ (Fig. 1); it is written like this:

$$f = p_1 \circ p_2 \circ \dots \circ p_n.$$

Let’s choose the origin and direction on each of our lines $l_i$, that is, we’ll turn each of them into a numerical axis (we choose the same scale unit on all lines). (We identify a point (on a line) and its corresponding number—its coordinate—so that the maps $p_1, p_2, \dots, p_n$ and $f$ can now be considered functions that

27

Each number is assigned a new number.

Let us prove that the mapping $f$ is linear: $f(x) = ax + b$.

To do this, it is enough to prove the following three obvious statements.

(1) If $l$ and $l'$ are two straight lines and $p : l \rightarrow l'$ is a map that associates each point with its projection onto $l'$, then $p$ is a linear map.

(2) If $p_1$ and $p_2$ are linear maps, then so is $p_1^\circ p_2$.

(3) If $p_1, p_2, \dots, p_n$ are linear maps, then $f = p_1^\circ p_2^\circ \dots p_n^\circ$ is also a linear map.

It is clear that (3) is derived from (2) by induction. Check (2): if $p_1(x) = a_1x + b_1$ and $p_2(x) = a_2x + b_2$, then $(p_1^\circ p_2)(x) = p_1(p_2(x)) = a_1a_2x + (a_1b_2 + b_1)$. To prove (1), check this formula:

$$p(x) = x \cos \alpha + c,$$

where $\alpha$ is the angle between the positive direction of the axes $Ox$ and $O'x'$, $c$ is the coordinate of the point $p(O)$ in the system $Ox'$ (Fig. 2; consider also the case when the lines $l$ and $l_1$ are parallel).

Now, to solve the problem, it is enough to show that the linear equation $x = ax + b$ has a solution, and only one. If this is not the case (that is, if there are no solutions or more than one), then $a = 1$. Let us prove that this cannot be. Note that with each projection $p_k$: $l_{k+1} \rightarrow l_k$ the distance between any two points does not increase:

$$|p_k(x) - p_k(x')| \leq |x - x'|,$$

and if the lines $l_{k+1}$ and $l_k$ are not parallel, then strictly decreases. It follows that for any two points $x$ and $x'$

$$|f(x) - f(x')| < |x - x'|.$$

But if $a = 1$, that is, if $f(x) = x + b$, then $f$ preserves the distances between any two points (it simply shifts them all by the distance $b$). We got a contradiction. The proof is legal.

Problem M70 and our solution have been re-established without any changes for the case when the lines are located not on a plane, but in space (they can also be intersecting). In this case, $p_k$ means projection onto the line $l_k$ in space: a mapping that corresponds to each point the base of the perpendicular dropped from this point onto the line $l_k$.

Fig. 2. A mapping that assigns to each point on one line its projection onto another line is written as a linear function.

Fig. 3. Point $X_1$, for which $f(X_1) = X_1$, it is practically easy to find a method of successive approximations: starting from an arbitrary point $X$, lower the perpendiculars to the given lines and find the successive after each “cycle” of the point $X' = f(X)$ $X'' = f(X')$ and so on. It is not difficult to prove that the distances $|X - X_1|, |X' - X_1|, |X'' - X_1|$, … form an infinite decreasing geometric progression with the value $|a| < 1$, so that by repeating the construction several times, we can find the position of the point $X_1$ with any ironclad accuracy. A similar method is very used to find a solution to an equation of the form $f(X) = X$; it will be applied when the function $f$ decreases the distance between any two points. In our case, when $f$ is a linear function, labeled, the solution to equation $f(X) = X$ can be found exactly (think about how to find the point $X_1$ if $X, X'$ and $X''$ are known).

28

There is another possibility of generalization:
look at straight planes together. Can
prove, for example, the following fact: if the data $n$
planes in space, among which
there are three intersecting at one point,
then it is possible and in the only way
select on each of these planes
point $X_k$ ($k=1, \dots, n$) so that the
ndikulier, restored at point $X_k$ to co-
corresponding plane, passed through
point $X_{k+1}$ (and $X_{n+1}=X_1$). Now it’s already
$f=p_1 \cdot p_2 \dots p_n$ will be a linear mapping
plane to plane and coordinates will be written as follows: $f(x, y)=$ $(x', y')$, where $x'=a_1x+b_1y+c_1$, $y'=a_2x+b_2y+c_2$.

Prove that if the system of equations
$$\begin{cases} x=a_1x+b_1y+c_1, \\ y=a_2x+b_2y+c_2 \end{cases}$$
has no solutions or has more than one
solutions (this happens if $(a_1-1)(b_2-1)−a_2b_1=0$), then the system $x=a_1x+b_1y$, $y=a_2x+b_2y$ has a solution $(x_0, y_0)$, different from $(0,0)$, and that the mapping $f$ does not change the dis-
standing between points $(0,0)$ and $(x_0, y_0)$, which cannot happen in our problem.

Nanmore complete solutions to the problem
Slania F. Schmidel from Moscow, Yu. Obolonkov from Voronezh, M. Rozov from Minsk.

M71

A rectangular table of $m$ rows and $n$ columns is filled with numbers. Let’s rearrange the numbers in each line in ascending order. Prove that if you then rearrange the numbers in each column in ascending order, then in each row they will still appear in ascending order.

Think about what will happen if you act in a different order: in the original table, first rearrange the numbers in ascending order in the columns, and then in the rows: will the result be the same table as in the first case or a different one?

Let’s consider a table in which in each row the numbers are in ascending order, and prove that if $l<r$, then the largest number in the $r$th column (we denote it by $b$) is not less than the largest number in the $l$th column (we denote it by $a$). Indeed, the number at the intersection of the line where $a$ is and the $r$ column is no less than $a$ (because the numbers in the line are in ascending order).

Similarly, one can prove that the $k$th largest number (counting from the largest number in the $r$-column $(b_k)$) is no less than the $k$th largest number in the $l$th column

29

Different tables are read. To verify this, it is enough to give an example (Fig. 5).

The problem was solved correctly by P. Ayulov, A. Brevdo, L. Brauzinsky, G. Vysotskaya, S. Grigoryan, A. Gordenko, I. Jabbarov, A. Kleyin, A. Kmizhnerman, I. Kochubeevsky, G. Levin, M. Preezer, E. Skryt, A. Slinkin, A. Chernyak and others. In addition to the solution, there are many others close to ours in which induction is carried out; In some works it has been proven that it is possible to rearrange each column in succession so that the numbers will always appear in the rows in ascending order.

N. B. Vasiliev

M72

Solve the equation
$$\sqrt[3]{1-x} + \sqrt[3]{1+x} = p,$$
(1)
where $p$ is an arbitrary real number.

A naive attempt at a solution: cube both sides of the equation
$$2 + 3\sqrt[3]{1-x^2}\left(\sqrt[3]{1-x} + \sqrt[3]{1+x}\right) = p^3$$
(2)
and substitute here the original equality (1):
$$2 + 3p\sqrt[3]{1-x^2} = p^3.$$ (3)
Now it’s easy to find $x$:
$$x = \pm \sqrt{1 - \left(\frac{p^3 - 2}{3p}\right)^3}.$$ (4)
We were sent many letters where this answer was received (in this and other ways). There were also quite a few letters in which it was correct to find a set of values ​​of $p$ for which formula (4) makes sense:
$$p = -1, \quad 0 < p \leq 2.$$
Some even noticed that the value $p = -1$ should be discarded, since the value $x = 0$ given by the formula $p$ is not suitable. But it was not proven that for all $p$, $0 < p \leq 2$ the value $x$ given by formula (4) are solutions. (Usually such a proof is a test, but here it is very difficult to do.) And this must be proven, since not all transformations performed above remain equivalent: equation (2) is not equivalent to (3). To see this, substitute $x = 0$, $p = -1$ in (2) and (3). Equality (3) will return to the same place, but (2) will not. Therefore, the statements found in the letters that this transformation is equivalent are incorrect. (Indeed: try to derive (2) from (3). You will have to use equality (1), the conditions of the problem, and this cannot be done, since this is precisely what needs to be proven.) If this is so, then other roots may be constructed.

Let us examine in more detail the transformations we made to the equation. Let us denote $\sqrt[3]{1-x} = u$ and $\sqrt[3]{1+x} = v$ for brevity.

From Eq.
$$u + v = p$$
we moved on to the equation
$$u^3 + v^3 + 3uv(u + v) = p^3$$
(2)
using an equivalent transformation (raising the object into a cube). The next step is to substitute $p$ together with $u + v$ and go to the equation
$$u^3 + v^3 + 3uvp = p^3,$$
(3)
which we will write like this:
$$u^3 + v^3 - p^3 + 3uvp = 0,$$
$$(u + v)^3 - p^3 - 3uv(u + v - p) = 0.$$
Let’s apply the formula for the difference of two cubes and then put $(u + v - p)$ out of brackets:
$$(u + v - p)(u^2 + v^2 + p^2 + up + pv - uv) = 0.$$ (3’)
So, we managed to write equation (3) in this form (3’), from which it is easy to study why it is not equivalent to (2). It is clear that the left side of (3’) is equal to zero when at least one parenthesis is equal to zero:

1. $u + v - p = 0,$
2. $u^2 + v^2 + p^2 + up + pv - uv = 0.$

30

The first equality is our original equation. This means that the entire construction of the solution is related to the second one and we need to find out when it is executed. Multiplying it by 2, you can write it like this:

$$(u+p)^2+(v+p)^2+(u-v)^2=0.$$

This equality holds only for $u=v=-p$.

In application to our equation:

$$\sqrt[3]{1-x}=\sqrt[3]{1+x}=-p.$$

This means that $x=0$ for $p=-1$ is the only constructed root. All other roots given by formula (4) for $0<p\leqslant2$ satisfy equation (1).

Let us write down again the identities that we have successfully established (for symmetrina we replace $p$ with $-w$):

$$u^3+v^3+w^3-3uvw=(u+v+w)(u^2+v^2+w^2-uw-vw-uv),$$

$$2(u^2+v^2+w^2-uw-vw-uv)=$$

$$=(u-v)^2+(v-w)^2+(w-u)^2.$$

They are often useful in proving inequalities and in other problems.

M73

On a lottery ticket you need to mark 8 cells out of 64. What is the probability that after a drawing in which 8 cells out of 64 are also chosen (and we consider all such possibilities to be equally probable), it turns out that exactly 4 cells are guessed? 5 cells?…all 8 cells?

As is known, 8 cells out of 64 can be chosen in $C_{64}^8=\frac{64!}{8!56!}$ ways ($n!$ is the product of all integers from 1 to $n$ inclusive) $^*$). All these methods are equally reliable. Therefore, the probability of getting any one particular way is $1/C_{64}^8$. Now we need to calculate what time the case turned out to be exactly $k$ cells ($0\leqslant k\leqslant8$). The number of such cases is equal to the number of methods

$^*$) See, for example, the article by N. Ya. Vilenkin “Combinatorics” in «Kvant» No. 1, 1971.

Fig. 6. Red cells are winners; The 8 cells we have selected are marked with a cross. On this figure $k=2$, $8-k=6$.

methods in which $k$ cells are selected from 8 marked and $8-k$ cells from 56 unmarked. Therefore it is equal to $C_{8}^k\cdot C_{56}^{8-k}$. To find the desired time $f_k$ - the probability of guessing exactly $k$ cells, you need to multiply the probability of one method by this number. We get

$$f_k=\frac{C_{8}^k\cdot C_{56}^{8-k}}{C_{8}^4}.$$

Thus, the probabilities of guessing 0,1, …, 8 cells are equal to the corresponding ones.

$$f_0=\frac{C_{8}^0\cdot C_{56}^8}{C_{8}^4},$$

$$f_1=\frac{C_{8}^1\cdot C_{56}^7}{C_{8}^4}, \dots, f_8=\frac{C_{8}^8\cdot C_{56}^0}{C_{8}^4}.$$

The sum of these nine numbers is, of course, equal to one.

Formally, the problem is a solution, but from a practical point of view this is not yet the case. I would like to imagine how great the chances are of guessing a particular number of cells if we buy and fill, say, ten lottery tickets. To do this, you need to approximately represent these numbers in the form of decimal fractions. This can be done using a slide rule; for example, directly use formulas (1). This is what A. Zlomnik from the city of Zhukovsky did. We will describe another, as it seems to us, simpler method.

31

Fig. 7. Shown here are the numbers $f_k$ – the probabilities of guessing $k$ cells ($k = 0, 1, 2, \dots, 8$). At $k \geq 4$ the values ​​of $f_k$ are very close to zero.

First, let’s calculate the ratio of the $(k+1)$th of these numbers to the $k$th one:

$$\frac{f_{k+1}}{f_k} = \frac{C_8^{k+1} \cdot C_{56}^{8-k-1}}{C_{64}^8} : \frac{C_8^k \cdot C_{56}^{8-k}}{C_{64}^8} = \frac{(8-k)^2}{(k+1)(48+k+1)}$$

in part,

$$\frac{f_1}{f_0} = \frac{64}{49} \approx 1.31; \quad \frac{f_2}{f_1} = \frac{49}{100};$$

$$\frac{f_3}{f_2} = \frac{12}{51} \approx 0.235; \quad \frac{f_4}{f_3} = \frac{25}{208} \approx 0.12;$$

$$\frac{f_5}{f_4} \approx 0.06; \dots$$

By calculating the ratio of $f_2, f_3, f_4$ to $f_0$ with two decimal places:

$$\frac{f_2}{f_0} = 0.64; \quad \frac{f_3}{f_0} \approx 0.15; \quad \frac{f_4}{f_0} = 0.02.$$

It is clear that $f_5/f_0 < 0.02 \cdot 0.12$ and, for the chosen degree, exactly $f_5, f_6, f_7, f_8$ are equal to zero. We can make an equation

$$f_0(1+1,31+0,64+0,15+0,02)=1,$$

whence $f_0 \approx 0.32$. Now finding $f_1, f_2, f_3, f_4$:

$$f_1 \approx f_0 \cdot 1.31 \approx 0.42; \quad f_2 \approx f_0 \cdot 0.64 \approx 0.20;$$

$$f_3 \approx f_2 \cdot 0.24 \approx 0.047; \quad f_4 \approx f_3 \cdot 0.12 \approx 0.0056.$$

Thus, if you fill out ten tickets, then the average ones will most likely all be those on which 0, 1, 2 and, perhaps, 3 cells are guessed; The probability of guessing 4 or more cells is very low. By multiplying $f_4$ by $\frac{f_5}{f_4} \approx 0.06$ and so on, approximations of $f_5, f_6, f_7, f_8$ can be obtained. In part, $f_8 \approx 2 \cdot 10^{-10}$. This means that in order to have a real chance of guessing all 8 cells, you need to fill out several tens of billions of tickets.

Using the approximate calculation method that we applied, it can be shown that if the number of $N$ cells on a lottery ticket is much greater than the number of $m$ cells that need to be eliminated, and $m$ is sufficiently large, then the probability of guessing $k$ cells of the approximate is equal to

$$f_k = \frac{C_m^k C_{N-m}^m}{C_N^m} \approx \frac{\mu^k \cdot e^{-\mu}}{k!},$$

where $\mu = \frac{m^2}{N-2m}$ and $e \approx 2.718\dots$ is a constant number ($\pi$), so we need to use the fact that $1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\dots = e^x$ for any $x$).

A. L. Toom

In this issue we publish solutions to problems F79—F82

F79

The level of water entering the boat coincides with the water level in the lake. Where will the water level be higher if you throw a log into the boat?

When lowering a log into a boat, the boat will plunge into the lake, you will receive an additional volume of water $\Delta V = \frac{m}{\rho}$ ($m$ is the mass of the log, $\rho$ is the density of water). On the other hand, since the log floats in water, the water level in the boat should change in the same way as it would change if water with a mass equal to the mass of the log and a volume equal to the volume of the immersed part of the log, that is, $V = \frac{m}{\rho}$, was added to the boat. (We can mentally remove the log and fill the resulting water with water, but the water level in the boat will not change.) The same volume of water additionally flows out from the boat when the log is thrown into it. Therefore, the water level in the boat will be the same as the water level in the boat.