Textbook "mathematical analysis"

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Tutorial
“mathematical analysis”*)

M. L. SMOLYANSKY

The rapid development of computer technology, the introduction of high-speed computers in all areas of the national economy, science and technology, has led to the emergence of new specialties. One of them is the specialty “computer programmer”. And as the fleet of computers increases, the need for programmers, computers and other specialists servicing these machines will constantly increase. By preparing the program and computing, both computing educational institutions and technical ones are remembered. They are also prepared in some schools. Now there are already several dozens, if not hundreds, of schools and classes in which mathematics is taught at an advanced level. However, complete teaching in these schools is greatly hampered by the lack of textbooks specifically designed for such schools. The schoolchild had to either use books written for calculating educational institutions, or limit himself to the material given in school during lessons.

Now two books have been published: “Algebra”**) and a solved book, the main goal of which is to correct this situation. Both of these books form an inseparable whole.

*) N. Ya. Vilenkin, S. I. Shvartsburg, Mathematical analysis, textbook for grades IX, X of secondary schools with special mathematics, “Prosveshchenie”, 1969.
**) N. Ya. Vilenkin, R. S. Guter, S. I. Schwartzburg, B. V. Ovchinsky and V. G. Ashkinez, Algebra, “Enlightenment”, 1968

Who are these books addressed to? Of course, first of all, the authors had in mind students from schools with a mathematical specialization. But not only they will benefit from these books. All students whose interests extend beyond the scope of the school curriculum will learn a lot from them. Both high school mathematics teachers and pedagogical students will find a lot of materials in it.

Books presenting an advanced course in mathematics can be written in different ways. You can remove special attention to complex algebraic transformations, confusing trionometric identities, and solving globe-breaking equations. On the other hand, it is possible to present without significant changes the mathematical disciplines that are covered in the calculations of educational information. The authors of the book “Mathematical Analysis” take a different path. Of course, after studying this book, the student will learn to solve trionometric and exponential equations, and prove various trionometric identities, and differentiate and interpret various functions. But, unlike most books on calculus for students, the solved book puts a lot of emphasis on clarifying the meaning of the basic concepts that students encounter.

The book begins with a presentation of the theory of active numbers. It is possible that the authors devote too much space to this issue. But without a clear understanding of what a real number is, it is impossible to fully understand what a limit is, or even what a limit is.

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falls out of the general style of the book. One must think that the inclusion of this chapter is due to the need, along with the main objectives of teaching, to solve the problem of preparing students for university entrance exams in mathematics. And at these exams, as you know, examiners often get carried away with all sorts of exponential, logarithmic and trigonometric equations, identities, etc. However, here too the authors not only give solutions to all kinds of equations, but also consider general questions related to these equations, questions about approximate methods for solving equations. The inclusion of the last question is due to the fact that in some schools with a mathematical specialization, computational mathematics is not studied, and familiarity with the methods of chords and tangents, with the method of successive approximations is useful for anyone finishing high school. Further, in schools where there is a special course in computational mathematics, it may be useful to study relevant issues in this book.

The eighth chapter treats the integral (both indefinite and definite), and the ninth deals with series. Unfortunately, the chapter devoted to intergrams turns out to be too heavy-handed. Here it was necessary to sacrifice the rigor of the presentation, not to define the concept of volume, but to consider it very important.

Exercises occupy a significant place in the book. Along with the usual tasks, it includes quite a few tasks of increased difficulty. The exercises are usually arranged in order of increasing difficulty. However, the rate of increase in the difficulty of tasks is quite high; it leads students to difficult exercises much faster than ordinary textbooks in a mass school. The exercises instill good skills in solving difficult problems and allow you to prepare for entrance exams to higher educational institutions with in-depth study of mathematics. Large exercises are placed with an asterisk and can serve as material for grit work.

Overall, the book was useful and interesting. The small circulation (30,000) is somewhat surprising. Let’s hope that the second edition of the book will not be long in coming.

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ANSWERS, INSTRUCTIONS, SOLUTIONS.

To the article “Geometry of collisions”

1. The kinematic diagram of the collision is shown at figure 1. The triangle $2C1'$ is rectangular and $tg\alpha = \frac{u_1}{u_2} = \frac{u_1}{u_2} = \frac{m_2}{m_1} = 3$. From the right angled equilateral triangle $2'2C$ we find $v'_2 = \frac{u_2}{\cos\beta}$, and since $u_2 = \frac{m_1}{m_1 + m_2} v_1$, then
   $$v'_2 = v_1 \frac{m_1}{m_1 + m_2} \cdot \frac{1}{\cos\beta} = v_1 \frac{\sqrt{2}}{4}.$$
   From the triangle $2C1'$ we find that
   $$v'_1 = \frac{u_2}{\cos\alpha} = v_1 \frac{m_1}{m_1 + m_2} \sqrt{1 + tg^2\alpha} = v_1 \frac{\sqrt{10}}{4}.$$

2. From the scattering diagram of a part with the same masses (Fig. 6 on page 23) we find that the recoil angle is $30^\circ$, the speed of the $\alpha$ particle after scattering is $v'_\alpha = v_\alpha \cos 60^\circ = \frac{1}{2} v_\alpha$, and the speed of the recoil nucleus is $v_\alpha = v_\alpha \cos 30^\circ = \frac{\sqrt{3}}{2} v_\alpha$ ($v_\alpha$ is the speed of the $\alpha$ particle before collision with the nucleus). Therefore, the energy of the $\alpha$-particle after scattering

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running_head:
folio:

E_{\alpha} = \frac{m v'_{\alpha}^2}{2} = \frac{1}{4} \frac{m v'_{\alpha}^2}{2} = \frac{1}{4} E_0, and the energy of the recoil core is $E_{\pi} = \frac{3}{4} E_0$.

3. \frac{\Delta E}{E} = \frac{v_n^2 - v'_n^2}{v_n^2} = \frac{5 - \sqrt{7}}{9}.

4. 0,24 $E_0$.

5. The scatterplot is shown at figure 2. $\varphi = \theta$ and $tg\varphi = tg\theta = \frac{v_1}{u_1 + u_2}$. From triangle 1’C2 it is found that
   v'_1 = \sqrt{u'_1^2 - u'_2^2} = u_2 \sqrt{\left(\frac{u'_1}{u_2}\right)^2 - 1} = u_2 \sqrt{\left(\frac{M}{m}\right)^2 - 1}.
   Therefore, the impulses of the part after the collision decreased: $mu'_1 = \alpha mu_1$ and $Mu'_2 = \alpha Mu_2$.
   The speeds of the part in the center of inertia system also decreased: $u'_1 = \alpha u_1$ and $u'_2 = \alpha u_2$.
   The scattering diagram will be as shown in figure 3. Moreover, point 2’ lies on a circle of radius $u'_2 = \alpha u_2$ with center at point $C$. The recoil angle (122’ angle) is maximum when the 22’ segment touches this circle.
   In this case, the angle $\omega$ of the part will be such that tg\omega = \frac{u'_2 + u'_1}{v'_2} = \frac{u'_2 + u'_1}{\sqrt{u'_2^2 - u'_2^2}} = \frac{\alpha \left(1 + \frac{u_1}{u_2}\right)}{\sqrt{1 - \alpha^2}} = \frac{\alpha}{\sqrt{1 - \alpha^2}} \left(1 + \frac{M}{m}\right).

6. Let’s draw a kinematic diagram. In the center of mass system, the momenta of the part after the collision are equal in magnitude and opposite in direction $mu'_1 = Mu'_2$ (or $\frac{u'_1}{u'_2} = \frac{M}{m} = \frac{u_1}{u_2}$), and the kinetic energy of the part is equal to before the collision
   $E_0 = \frac{(mu_1)^2}{2m_1} + \frac{(Mu_2)^2}{2M} = \frac{(mu_1)^2}{2} \left(\frac{1}{m} + \frac{1}{M}\right) = \frac{(Mu_2)^2}{2} \left(\frac{1}{m} + \frac{1}{M}\right)$,

(since $mu_1 = M_2 u_2$), after the collision has decreased and will become equal
$E_1 = \frac{(mu'_1)^2}{2m} + \frac{(Mu'_2)^2}{2M} = \frac{(m_1 u'_1)^2}{2} \left(\frac{1}{m} + \frac{1}{M}\right) = \frac{(Mu'_2)^2}{2} \left(\frac{1}{m} + \frac{1}{M}\right)$.
According to the scattering conditions, it will be as shown in figure 3. Moreover, point 2’ lies on a circle of radius $u'_2 = \alpha u_2$ with center at point $C$. The recoil angle (122’ angle) is maximum when the 22’ segment touches this circle.
In this case, the angle $\omega$ of the part will be such that tg\omega = \frac{u'_2 + u'_1}{v'_2} = \frac{u'_2 + u'_1}{\sqrt{u'_2^2 - u'_2^2}} = \frac{\alpha \left(1 + \frac{u_1}{u_2}\right)}{\sqrt{1 - \alpha^2}} \left(1 + \frac{M}{m}\right).

By the way
“The Theory of Comparison and Argument of Residues”

1. $-297 = (-10) \cdot 31 + 13$;

2. $-1005 = (-11) \cdot 98 + 73$.

3. Let $n_1$ be a divisor of $N$. Then $n_2 = \frac{N}{n_1}$ is also a divisor. Since $n_1 \cdot n_2 = N$,

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then at least one of the numbers $n_1, n_2$ is not greater than $\sqrt{N}$. The result easily follows from here.

3. 6. $n^2+n+1=n(n+1)+1; n^2+n+1=(n-1)(n+2)+3$ (with $n=1$, remainder 0).
      b) $n^4+1=(n^3-3n^2+9n-27)(n+3)+82$.

4. At $n=-1, 0, 1, 2$. Take advantage of what
   $$\frac{n^2+1}{n-1}=n+1+\frac{2}{n-1}.$$

5. Of three consecutive numbers, one is divisible by 3, and one (or two) is divisible by 2.

6. $a^2+b^3+c^3-(a+b+c)=a(a-1)(a+1)+b(b-1)(b+1)+c(c-1)(c+1)$.

7. Consider prime numbers of the form $p, p+d, p+2d, \dots, p+14d$. It is easy to prove that $d$ must be divisible by 2, 3, 5, 7, 11, 13 (if $d$ is not divisible, for example, by 13, then the average of the numbers $p, p+d, \dots, p+14d$ there is a number divisible by 13), which means
   $$d > 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 = 30000.$$

8. $n \equiv 1$ (mod 3).

9. $n \equiv 5$ (mod 7).

10. Reason in the same way as in problem 3.

11. Let’s write the number $a=\overline{a_n a_{n-1} a_{n-2} \dots a_0}$ ($a_0, a_1, \dots, a_n$ – numbers) in the form $a=a_2 a_1 a_0+1000 a_5 a_4 a_3+1000^2 a_8 a_7 a_6+ \dots$. Since $1000 \equiv -1$ (mod 7) and $1000 \equiv 1$ (mod 37), then
    $$a \equiv \overline{a_2 a_1 a_0}-\overline{a_5 a_4 a_3}+\dots$$
    $$a \equiv \overline{a_2 a_1 a_0}+\overline{a_5 a_4 a_3}+\dots$$

12. If the number $5A$ and $A$ have the same sum of digits, then the number $5A-A=4A$ is divisible by 9.

13. Let $15x^2-7y^2=9$, then $y \equiv 0$ (mod 3), therefore $x \equiv 0$ (mod 3); dividing by 9, we get $15\left(\frac{x}{3}\right)^2-7\left(\frac{y}{3}\right)^2=1$, but the equality $15x^2-7y^2=1$ cannot be satisfied (make sure of this).

14. Consider the comparison
    $$n^2+3n+5 \equiv 0$$ (mod 11).

15. Consider what remainders the squares of natural numbers give when divided by 3, 5, 7.

16. Prove that in 9-arithmetic there is $x^3+y^3+z^3 \neq 5$ for no $x, y, z$.

17. If $ab$ is divisible by the prime $p$ and $a$ is not divisible by $p$, then $b$ is divisible by $p$, i.e. $b=0$.

18. This is the remainder $m-a$. It is clear that $a+(m-a)=0$.

Adding $-a$ to both sides of the equation $a+x=b$ gives $x=b+(-a)$.

21. Let $m=pq$, where $1<p<m$ and $1<q<m$. Then $pq=0$ is $m$-arithmetic.

22. Take advantage of the fact that in 31-arithmetic $30=-1$; $29=-2; \dots; 16=-15$, and also by the fact that for any remainder $a$ and odd $n$ there will be $(-a)^n=-a^n$.

To the article “On entrance exams in mathematics in technical schools in 1969”

Option 1.

1. 10 m.
2. $\frac{6 \pm \sqrt{100-\pi^2}}{4}$.
3. $4 < x \leq 7; -3 \leq x < -1; -17 < x < -16$.
4. $\frac{\pi}{4} + \pi n < x < \frac{\pi}{2} + \pi n$;
   $$\frac{\pi}{2} + \pi n < x < \frac{3\pi}{4} + \pi n.$$
5. $\frac{3}{2} h$.

Option 2.

1. 6 o’clock, 9 o’clock
2. $x = \frac{\pi}{4} + \pi n$.
3. $1 < x < 2^4; 2 < x < 2^4; 2^4 < x < 2^{15}$
4. $x = \pm \alpha + 2\pi n; \frac{\pi}{4} < \alpha < \frac{\pi}{3}$.
5. $\frac{7a \sqrt{3} \pi a^3}{576}$.

Option 3.

1. 5 o’clock 10 o’clock; 13 hours, 20 minutes; 4 hours 26 minutes 40 seconds.

2. Resolved no.

3. $\frac{1}{3} < x < 1; \frac{1}{3^7} < x < \frac{1}{3^{12}}$.

4. $2\pi n < x < \frac{\pi}{3} + 2\pi n$;
   $$2\pi n - \frac{\pi}{3} < x < 2\pi n.$$

5. $\frac{4\sqrt{3} \pi a^3}{27 \sin^3 2\alpha}$.

Option 4.

1. 3 hours, 2 hours.
2. $x = \pi n \pm \frac{\pi}{3}$.
3. $\frac{5}{2} < x < 3; x > 4$.
4. $\frac{\pi}{3} + 4\pi k < x < 3\pi + 4\pi k$.

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5. $a^2 \sin \alpha \left[ 1 + \frac{1}{\cos \beta} \right]$.

To the article
“Someone loses, someone finds…”

1. $x = 13$.

2. $x = 2$.

3. $x = -1$.

4. $x = 9$.

5. $x = 1$.

6. There are no solutions.

7. $x = \frac{\pi n}{3}$.

8. $x = \arctg 2 + \pi n; \quad x = \frac{\pi}{2} + \pi k$.

To the article
“TV is your helper in preparing for the day”

Test No. 1

1. $\frac{1}{1 - 3x}$, if $x \leqslant 0; \quad \frac{x + 1}{3x^2 - 4x + 1}$, if $0 < x < 1 \left( x \neq \frac{1}{3} \right); \quad \frac{1}{x - 1}$, if $x > 1$.

2. $\sqrt{2}$. 3. $-(\sqrt{3} a + 1)$. 4. $m$ if $\frac{a}{m^2 - 2a^2} > 0$, and $(-m)$ if $\frac{a}{m^2 - 2a^2} < 0$.

3. $S_7 = b_5 - b_1 = 560$.

Test No. 2

1. $-2; 3; \frac{1}{2}; \frac{2}{3}$. 2. $3 \pm \sqrt{2}, 3 \pm 3 \sqrt{2}$.

2. $\begin{cases} x = 3 \\ y = 5 \end{cases} \begin{cases} x = 5 \\ y = 3 \end{cases}$

3. 90 rub., 135 rub., 5. 60 km/h.

Test No. 3

1. $x_1 = 1; x_2 = -\frac{1}{3}$. 2. $x_{1,2} = \pm 1$.

2. $x_1 = 1; x_2 = -6$. 4. $8 \mathrm{dm}, 25 \mathrm{dm}, 15 \mathrm{dm}$.

3. $\frac{R^2}{4} (8 \sqrt{3} - 9)$.

To the task
“Calendar from cubes”
(«Kvant» №3)

The task contains two (not very tricky) traps.

It is easy to understand that both dice must have the numbers 1 and 2, since there are numbers 11 and 22. It is also necessary that both dice have the number 0, since otherwise it is impossible to make up all the numbers from 01 to 09 (you cannot place all the numbers from 1 to 9 on the same die: it has only six faces).

By writing the numbers 0, 1, 2 on the faces of each cube, we leave 6 faces free, on which we need to place 7 numbers: from 3 to 9. This can be done, since the number 9 is obtained by turning over the number 6.

Before calculating how many numbers to place on two cubes, let’s clarify which two “calendars” (two pairs of cubes) we will consider different. Firstly, we will assume that the cubes on which we write the numbers are exactly the same, and we do not distinguish them; secondly, we, of course, consider the cubes to be identical, from which one is obtained from the other by rotation in space; and, finally, we will not take into account how the number is written on the edge - where is the top, where is the bottom, etc., we are only interested in what number is written.

Let’s put both cubes on the table as zeros. Let there be numbers on one die: 0, 1, 2, 3. On two empty faces you need to place two numbers from the remaining heels (4, 5, 6, 7, 8). This can be done in $C_5^2 = 10$ ways. Leaving the zero in place (at the bottom), the heels of the digits can be placed in 5!=120 ways. One of them can be reduced to the others by rotating the cube in order in the vertical axis. There will be four times fewer different methods. So, the total number of ways to arrange the numbers for the first die will be $\frac{10 \cdot 5!}{4} = 300$. On the second die you can arrange the remaining numbers in $\frac{5!}{4} = 30$ ways. Thus, the total number of ways will be $300 \times 30 = 9000$.

Solution to the “encryption” placed in #4 «Kvant»

Here each letter is protected by a letter. Punctuation marks and spaces between words are left in place.

Let’s write the letters of the Russian alphabet in a circle clockwise (or counterclockwise) in order of their order (the letters I and E are skipped). In the same way, we write down the first 15 prime numbers in a circle.

Teler, you can proceed directly to decryption.

Let’s take the first letter of the encryption - K. It corresponds to the first number on the “numeric” circle - 2. You need to count the second letter from K on the “letter” circle counterclockwise (respectively) clockwise (e). This is letter 3. The first letter has been deciphered.

The second letter of the encryption, P, is answered by the third letter, counted on the letter circle from P counterclockwise. This is the letter H. The third letter E of the encryption in the same way (missing not 5 letters) will turn into A, etc.

The final result is:

“Knowledge is only knowledge when it is acquired through the efforts of one’s thoughts, and not memory alone.”