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23.14. Prove that the product of the lengths of all sides and diagonals drawn from one vertex of a regular $n$-gon inscribed in a circle of radius 1 is equal to $n$.
untranslated text:
23.14. Докажите, что произведение длин всех сторон и диагоналей, проведённых из одной вершины правильного $n$-угольника, вписанного в... | 23.14. The vertices of a regular $n$-gon can be considered as the points $1, \varepsilon, \varepsilon^{2}, \ldots, \varepsilon^{n-1}$, where $\varepsilon$ is a primitive $n$-th root of unity. The lengths of the sides and diagonals drawn from the point 1 are $|1-\varepsilon|,\left|1-\varepsilon^{2}\right|, \ldots,\left|... | proof | Geometry | proof | Yes | Yes | olympiads | false | 38,732 |
23.15. a) Prove that the polynomial $P(x)=x^{4 n}+x^{3 n}+x^{2 n}+$ $+x^{n}+1$ is divisible by the polynomial $Q(x)=x^{4}+x^{3}+x^{2}+x+1$ if and only if $n$ is not divisible by 5.
b) Prove that if the numbers $m$ and $n$ are coprime, then $x^{(m-1) n}+\ldots+x^{2 n}+x^{n}+1$ is divisible by $x^{m-1}+\ldots+x^{2}+x+1$... | 23.15. a) The roots of the polynomial $P(x)$ are the primitive 5th roots of unity. Therefore, $Q(x)$ is divisible by $P(x)$ if and only if all the primitive 5th roots of unity are roots of the polynomial $Q(x)$. Let $\varepsilon$ be a primitive 5th root of unity. If $n$ is divisible by 5, then $Q(\varepsilon)=5$. If $n... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,733 |
23.17. Given $2 n+2$ numbers $a_{-n}, a_{-n+1}, \ldots, a_{n+1}$. Consider $2 n+2$ numbers
$$
b_{q}=\frac{1}{\sqrt{2 n}} \sum_{p=-n}^{n+1} a_{p} \varepsilon^{p q}, \quad q=-n,-n+1, \ldots, n+1
$$
where $\varepsilon=\cos \frac{\pi}{n}+i \sin \frac{\pi}{n}$. Prove that
$$
a_{p}=\frac{1}{\sqrt{2 n}} \sum_{q=-n}^{n+1} b... | 23.17. Consider the number
$$
s_{p}=\sqrt{2 n} \sum_{q=-n}^{n+1} b_{q} \varepsilon^{-p q}=\sum_{r=-n}^{n+1} a_{r} \sum_{q=-n}^{n+1} \varepsilon^{(r-p) q}
$$
If $r=p$, then $\sum_{q=-n}^{n+1} \varepsilon^{(r-p) q}=2 n$. If $r \neq p$, then this sum is zero. Indeed, $\varepsilon$ is a primitive $(2 n)$-th root of unity... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,734 |
23.18. Prove that if $z_{0}$ is a root of a polynomial with real coefficients, then $\bar{z}_{0}$ is also a root of this polynomial. | 23.18. Let $P(z)=a_{0} z^{n}+a_{1} z^{n-1}+\ldots+a_{n}$, where $a_{0}, a_{1}, \ldots, a_{n}$ are real numbers. Then $\overline{P(z)}=a_{0} \bar{z}^{n}+a_{1} \bar{z}^{n-1}+\ldots+a_{n}=P(\bar{z})$. Therefore, if $P\left(z_{0}\right)=0$, then $P\left(\bar{z}_{0}\right)=\overline{P\left(z_{0}\right)}=0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,735 |
23.19. Prove that for any natural numbers $a$, $b$, $c$ the polynomial $x^{3a} + x^{3b+1} + x^{3c+2}$ is divisible by $x^2 + x + 1$. | 23.19. If $\varepsilon$ is a root of the polynomial $x^{2}+x+1$, then $\varepsilon^{3}=1$. Therefore, $\varepsilon^{3 a}+\varepsilon^{3 b+1}+\varepsilon^{3 c+2}=1+\varepsilon+\varepsilon^{2}=0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,736 |
23.20. Given the polynomial $x^{n}+a_{1} x^{n-1}+\ldots+a_{n}$. Find the polynomial whose roots are: a) the squares of the roots of this polynomial; b) the cubes of the roots of this polynomial. | 23.20. a) Let's multiply the polynomials
$$
x^{n}+a_{1} x^{n-1}+\ldots+a_{n}=\left(x-x_{1}\right) \ldots\left(x-x_{n}\right)
$$
and
$$
x^{n}-a_{1} x^{n-1}+a_{2} x^{n-2}-a_{3} x^{n-3}+\ldots \pm a_{n}=\left(x+x_{1}\right) \ldots\left(x+x_{n}\right)
$$
The result is the polynomial
$$
\left(x^{n}+a_{2} x^{n-2}+\ldots... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,737 |
23.21. For which natural numbers $n$ is the expression $a^{n}(b-c)+$ $+b^{n}(c-a)+c^{n}(a-b)$ divisible by $a^{2}+b^{2}+c^{2}+a b+b c+c a$? | 23.21. Answer: only for $n=4$. It is easy to verify that for $n=4$ the result of the division is $(a-b)(b-c)(a-c)$. We will show that for all other natural $n$ the expression $a^{n}(b-c)+b^{n}(c-a)+c^{n}(a-b)$ does not divide $a^{2}+b^{2}+c^{2}+ab+bc+ca$. It is sufficient to check that the first expression does not div... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,738 |
23.22. a) Prove that the polynomial $P_{n}(x)=(x+1)^{n}-x^{n}-1$ is divisible by $x^{2}+x+1$ if and only if $n=6 k \pm 1$.
b) Prove that $P_{n}(x)$ is divisible by $\left(x^{2}+x+1\right)^{2}$ if and only if $n=6 k+1$.
c) Prove that $P_{n}(x)$ is not divisible by $\left(x^{2}+x+1\right)^{3}$. | 23.22. a) The polynomial $P_{n}(x)$ is divisible by $x^{2}+x+1$ if and only if $P_{n}(\varepsilon)=0$, where $\varepsilon$ is a primitive third root of unity. The equality $\varepsilon^{2}=-\varepsilon-1$ shows that the expression $P_{n}(\varepsilon)=\left(-\varepsilon^{2}\right)^{n}-\varepsilon^{n}-1$ depends only on ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,739 |
24.1. Prove that the equation $x^{n}+a_{1} x^{n-1}+\ldots+a_{n}=0$ can be transformed into the form $y^{n}+b_{2} y^{n-2}+\ldots+b_{n}=0$ using the substitution $y=x+c$, where $c$ is some number.
288 Chapter 24. Equations Solvable in Radicals
Problem 24.1 shows that it is sufficient to consider cubic equations of the ... | 24.1. The required substitution has the form $y=x+\frac{a_{1}}{n}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,740 |
24.2. Find the roots of the equation $x^{3}+p x+q=0$, representing them in the form $x=\sqrt[3]{\alpha}+\sqrt[3]{\beta}$ and finding expressions for $\alpha$ and $\beta$. | 24.2. The equality must hold
$$
x^{3}=\alpha+\beta+3 \sqrt[3]{\alpha \beta}(\sqrt[3]{\alpha}+\sqrt[3]{\beta})=\alpha+\beta+3 \sqrt[3]{\alpha \beta} x
$$
Therefore, the numbers $\alpha$ and $\beta$ must be chosen so that the equalities $3 \sqrt[3]{\alpha \beta}=-p$ and $\alpha+\beta=-q$ are satisfied. From these equal... | \sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,741 |
24.3. Solve the equation $x^{3}+p x+q=0$, using the identity
$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x+\omega y+\omega^{2} z\right)\left(x+\omega^{2} y+\omega z\right)$, where $\omega^{2}+\omega+1=0$. (Choose $y$ and $z$ such that $-3 y z=p$ and $\left.y^{3}+z^{3}=q.\right)$ | 24.3. If $-3 y z=p$, then $y^{3} z^{3}=-p^{3} / 27$. Therefore, $y^{3}$ and $z^{3}$ are the roots of the quadratic equation $t^{2}-q t-\frac{p^{3}}{27}=0$. The roots of this equation are $\frac{q}{2} \pm \sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}$. Choose one of the three values of the cube root $\sqrt[3]{\frac{q}{2}+\sqr... | x_{1}=-(y+z),x_{2}=-(\omegay+\omega^{2}z),x_{3}=-(\omega^{2}y+\omegaz) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,742 |
24.6. Let $x_{1}, x_{2}, x_{3}$ be the roots of the equation $x^{3} + p x + q = 0$. Let $\alpha = x_{1} + \omega x_{2} + \omega^{2} x_{3}$ and $\beta = x_{1} + \omega^{2} x_{2} + \omega x_{3}$, where $\omega^{2} + \omega + 1 = 0$ (these expressions are called Lagrange resolvents).
a) Prove that $3 x_{1} = \alpha + \be... | 24.6. a) The roots $x_{1}, x_{2}, x_{3}$ can be easily found from the system of linear equations $x_{1}+x_{2}+x_{3}=0, x_{1}+\omega x_{2}+\omega^{2} x_{3}=\alpha, x_{1}+\omega^{2} x_{2}+\omega x_{3}=\beta$. To find $x_{1}$, we need to add these equations and use the fact that $\omega^{2}+\omega+1=0$. To find $x_{2}$, w... | notfound | Algebra | proof | Yes | Yes | olympiads | false | 38,745 |
24.7. Prove that the polynomial $x^{3} + p x + q$ has multiple roots if and only if $\frac{q^{2}}{4} + \frac{p^{3}}{27} = 0$.
| The expression $\frac{q^{2}}{4} + \frac{p^{3}}{27}$ is called the discriminant of the cubic polynomial $x^{3} + p x + q$.
## 24.3. Solving Equations of the 4th Degree | 24.7. First solution. Let the considered polynomial have roots $x_{1}, x_{2}$, and $x_{3}$, with $x_{1}=x_{2}$. The coefficient of $x^{2}$ is zero, so $x_{1}+x_{2}+x_{3}=0$. Therefore, $x_{3}=-2 x_{1}$. Then
$$
\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)=x^{3}-3 x_{1}^{2} x+2 x_{1}^{3},
$$
i.e., $p=-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,746 |
24.8. Solve the equation $x^{4}+a x^{2}+b x+c=0$ by representing the polynomial $x^{4}+a x^{2}+b x+c$ as the difference of squares of two polynomials. | 24.8. Let us use the fact that
$$
x^{4}+a x^{2}+b x+c=\left(x^{2}+\frac{a}{2}+t\right)^{2}-\left(2 t x^{2}-b x+\left(t^{2}+a t-c+\frac{a^{2}}{4}\right)\right)
$$
Choose $t$ such that the discriminant
$$
D=b^{2}-8 t\left(t^{2}+a t-c+\frac{a^{2}}{4}\right)
$$
is equal to zero. Then
$$
x^{4}+a x^{2}+b x+c=\left(x^{2}... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,747 |
24.9. Solve the equation $x^{4}+a x^{2}+b x+c=0$, by finding the equation that the sum of two roots of the given equation satisfies (Euler).
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 24.9. Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the roots of the equation $x^{4} + a x^{2} + b x + c = 0$. Let $u = x_{1} + x_{2} = - (x_{3} + x_{4})$. Then
$$
x^{4} + a x^{2} + b x + c = (x^{2} - u x + \alpha)(x^{2} + u x + \beta)
$$
i.e.,
$$
\alpha + \beta - u^{2} = a, \quad u(\alpha - \beta) = b, \quad \alpha \beta = c... | notfound | Number Theory | proof | Yes | Yes | olympiads | false | 38,748 |
24.10. Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the roots of the equation $x^{4} + a x^{2} + b x + c = 0$. Let $\alpha = -\left(x_{1} + x_{2}\right)\left(x_{3} + x_{4}\right)$, $\beta = -\left(x_{1} + x_{3}\right) \times$ $\times\left(x_{2} + x_{4}\right)$, and $\gamma = -\left(x_{1} + x_{4}\right)\left(x_{2} + x_{3}\right)... | 24.10. a) By the condition $x_{1}+x_{2}=-\left(x_{3}+x_{4}\right)$, therefore $x_{1}+x_{2}=\sqrt{\alpha}$. Similarly, $x_{1}+x_{3}=\sqrt{\beta}$ and $x_{1}+x_{4}=\sqrt{\gamma}$. Consequently,
$$
\begin{aligned}
& 2 x_{1}=\sqrt{\alpha}+\sqrt{\beta}+\sqrt{\gamma} \\
& 2 x_{2}=\sqrt{\alpha}-\sqrt{\beta}-\sqrt{\gamma} \\
... | y^{3}+2^{2}+(^{2}-4)y-b^{2}=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,749 |
24.11. Solve the equations
\[
\begin{gathered}
x^{5}-10 x^{3}\left(1-x^{2}\right)+5 x\left(1-x^{2}\right)^{2}=a \\
x^{7}-21 x^{5}\left(1-x^{2}\right)+35 x^{3}\left(1-x^{2}\right)^{2}-7 x\left(1-x^{2}\right)^{3}=a
\end{gathered}
\] | 24.11. Let $x=\cos \varphi$. From de Moivre's formula, it follows that the equations in question have the form $\cos 5 \varphi=a$ and $\cos 7 \varphi=a$. Therefore, the solutions to these equations are of the form
$$
x=\frac{1}{2}\left(\sqrt[n]{a+i \sqrt{1-a^{2}}}+\sqrt[n]{a-i \sqrt{1-a^{2}}}\right)
$$
where $n=5$ or... | \frac{1}{2}(\sqrt[n]{+i\sqrt{1-^{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,750 |
24.12. Solve the equation $x^{5}-5 a x^{3}+5 a^{2} x-b=0$. | 24.12. We will use the same method that was used in problem 24.2 to solve the cubic equation. Specifically, we will look for a solution in the form $x=\alpha+\beta$. Using the fact that $(\alpha+\beta)^{3}=\alpha^{3}+\beta^{3}+3 \alpha \beta(\alpha+\beta)$ and $(\alpha+\beta)^{5}=\alpha^{5}+\beta^{5}+5 \alpha \beta\lef... | \sqrt[5]{\frac{b}{2}+\sqrt{\frac{b^{2}}{4}-^{5}}}+\sqrt[5]{\frac{b}{2}-\sqrt{\frac{b^{2}}{4}-^{5}}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,751 |
24.13. What seventh-degree equations can be solved in the same way as was used in solving problem $24.12?$ | 24.13. Answer: $x^{7}-7 a x^{5}+14 a^{2} x^{3}-7 a^{3} x-b=0$.
Let $\alpha \beta=a$. Then
$$
\begin{array}{lr}
(\alpha+\beta)^{7}=\alpha^{7}+\beta^{7}+7 a\left(\alpha^{5}+\beta^{5}\right)+21 a^{2}\left(\alpha^{3}+\beta^{3}\right)+35 a^{3}(\alpha+\beta) \\
(\alpha+\beta)^{5}= & \alpha^{5}+\beta^{5}+5 a\left(\alpha^{3}... | x^{7}-7^{5}+14^{2}x^{3}-7^{3}b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,752 |
25.1. Prove that if the limit of a sequence exists, then it is unique. | 25.1. Suppose that $a$ and $b$ are limits of the sequence $\left\{a_{n}\right\}$, and $a \neq b$. Let $\varepsilon=\frac{1}{3}|a-b|$. According to the definition of a limit, we can choose numbers $N_{1}$ and $N_{2}$ such that $\left|a_{n}-a\right|<\varepsilon$ for $n>N_{1}$ and $\left|a_{n}-b\right|<\varepsilon$ for $n... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,753 |
25.2. Let $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ be two sequences, and $\lim _{n \rightarrow \infty} a_{n}=a$ and $\lim _{n \rightarrow \infty} b_{n}=b$. Prove that:
a) $\lim _{n \rightarrow \infty}\left(a_{n}+c\right)=a+c$ and $\lim _{n \rightarrow \infty}\left(c a_{n}\right)=c a$ for any number $c$;
b) $... | 25.2. a) The statement that $\lim _{n \rightarrow \infty}\left(a_{n}+c\right)=a+c$ is obvious: the same number $N$ for the sequence $\left\{a_{n}+c\right\}$ can be chosen as for the sequence $\left\{a_{n}\right\}$.
For $c=0$, the statement that $\lim _{n \rightarrow \infty}\left(c a_{n}\right)=0$ is obvious. If $c \ne... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,754 |
25.3. Prove that if $a_{n}<b_{n}<c_{n}$ for all $n$ and $\lim _{n \rightarrow \infty} a_{n}=$ $=a=\lim _{n \rightarrow \infty} c_{n}$, then $\lim _{n \rightarrow \infty} b_{n}=a$. | 25.3. For a given $\varepsilon>0$, choose numbers $N_{1}$ and $N_{2}$ such that $\left|a-a_{n}\right|N_{1}$ and $\left|a-c_{n}\right|N_{2}$. Then if $n>\max \left(N_{1}, N_{2}\right)$, we have $a_{n}>a-\varepsilon$ and $c_{n}<a+\varepsilon$. Therefore, $a-\varepsilon<a_{n} \leqslant b_{n} \leqslant c_{n}<a+\varepsilon$... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,755 |
25.5. Given a positive number $a$ and a sequence of positive numbers $x_{n}$. Prove that if $\lim _{n \rightarrow \infty} \frac{x_{n}-a}{x_{n}+a}=0$, then $\lim _{n \rightarrow \infty} x_{n}=a$. | 25.5. Let $\left|\frac{x_{n}-a}{x_{n}+a}\right|<\varepsilon<1$. Then
$$
\frac{-2 \varepsilon a}{1+\varepsilon}<x_{n}-a<\frac{2 \varepsilon a}{1-\varepsilon}
$$ | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,757 |
25.6. Prove that if $\lim _{n \rightarrow \infty} x_{n}=a$, then $\lim _{n \rightarrow \infty} \frac{x_{1}+\ldots+x_{n}}{n}=a$ (Cauchy). | 25.6. Instead of the sequence $\left\{x_{n}\right\}$, we can consider the sequence $\left\{x_{n}-a\right\}$, so we can assume that $a=0$. For any $\varepsilon>0$, we can choose $N$ such that if $n \geqslant N$, then $\left|x_{n}\right|N$
$$
\left|\frac{1}{n}\left(x_{1}+\ldots+x_{n}\right)\right| \leqslant \frac{C}{n}+... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,758 |
25.7. Let $\left\{a_{n}\right\}$ be a sequence of positive numbers, and $\lim _{n \rightarrow \infty} a_{n}=a$. Prove that $\lim _{n \rightarrow \infty} \sqrt[n]{a_{1} \ldots a_{n}}=a$. | 25.7. The function $\ln x$ is continuous, so $\lim _{n \rightarrow \infty} \ln a_{n}=\ln a$. Therefore, according to problem $25.6 \lim _{n \rightarrow \infty} \frac{\ln a_{1}+\ldots+\ln a_{n}}{n}=\ln a$. Hence, $\lim _{n \rightarrow \infty} \ln \sqrt[n]{a_{1} \ldots a_{n}}=\ln a$. Now, using the continuity of the func... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,759 |
25.8. Prove that if $\lim _{n \rightarrow \infty}\left(x_{n+1}-x_{n}\right)=a$, then $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=a$. | 25.8. Let $y_{1}=x_{1}$ and $y_{n}=x_{n}-x_{n-1}$ for $n \geqslant 2$. Then $x_{n}=y_{1}+y_{2}+\ldots$ $\ldots+y_{n}$. By the condition $\lim _{n \rightarrow \infty} y_{n}=a$. Therefore, according to problem 25.6 $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=\lim _{n \rightarrow \infty} \frac{y_{1}+y_{2}+\ldots+y_{n}}{... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,760 |
25.9. Prove that if $\lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=a$, then $\lim _{n \rightarrow \infty} \sqrt[n]{a_{n}}=a$. | 25.9. Consider the sequence $b_{1}=a_{1}, b_{n}=\frac{a_{n+1}}{a_{n}}$ for $n>1$. By the condition, $\lim _{n \rightarrow \infty} b_{n}=a$. Therefore, according to problem 25.7, $\lim _{n \rightarrow \infty} \sqrt[n]{b_{1} \ldots b_{n}}=a$. But $b_{1} \ldots b_{n}=a_{n}$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,761 |
25.10. Let $P(x)$ be a polynomial with integer coefficients. Prove that if $\lim _{n \rightarrow \infty}\{P(n) \alpha\}=0$, where $\{\cdot\}$ denotes the fractional part, then the number $\alpha$ is rational.
## 25.2. Weierstrass Theorem
- A sequence $\left\{a_{n}\right\}$ is called bounded if one can choose numbers ... | 25.10. It is more convenient to prove a more general statement: if $\{P(n) \alpha\}=a_{n}+\varepsilon_{n}$, where $\lim _{n \rightarrow \infty} \varepsilon_{n}=0$ and $a_{n}$ takes only a finite number of values (as $n$ runs through all natural numbers), then $\alpha$ is rational.
We apply induction on $m$ - the degre... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,762 |
25.11. Prove that any bounded sequence has at least one limit point (Bolzano-Weierstrass). | 25.11. It can be assumed that the numbers $c_{1}$ and $c_{2}$ from the definition of a bounded sequence of integers are integers. The interval $\left[c_{1}, c_{2}\right]$ is divided into a finite number of intervals of length 1. At least one of these intervals contains infinitely many terms of the sequence under consid... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,763 |
25.12. Prove that any bounded sequence $\left\{a_{n}\right\}$ contains a convergent subsequence $\left\{a_{n_{k}}\right\}$, i.e., one can choose a strictly increasing sequence of natural numbers $n_{1}, n_{2}, n_{3}, \ldots$ such that the sequence $\left\{b_{k}\right\}$, where $b_{k}=a_{n_{k}}$, converges.
A sequence ... | 25.12. Let $c$ be a limit point of the sequence $\left\{a_{n}\right\}$. Then the inequality $\left|a_{n}-c\right|<1 / k$ holds for an infinite number of terms $a_{n}$. Therefore, one can choose $a_{n_{k}}$ such that $\left|a_{n_{k}}-c\right|<1 / k$. Moreover, this can be done in such a way that $n_{1}<n_{2}<n_{3}<\ldot... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,764 |
25.13. Prove that any increasing (or at least non-decreasing) sequence $\left\{a_{n}\right\}$ that is bounded above has a limit (Weierstrass). | 25.13. A non-decreasing sequence bounded above is bounded, since $a_{n} \geqslant a_{1}$ for all $n$. Therefore, according to the Bolzano-Weierstrass theorem (problem 25.11), the sequence $\left\{a_{n}\right\}$ has a limit point $c$. First, we will prove that the limit point is unique. Suppose there are two limit point... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,765 |
25.14. Prove that the sequence $\left\{x_{n}\right\}$ converges if and only if for any $\varepsilon>0$ one can choose a number $N$ such that $\left|x_{n}-x_{m}\right|<\varepsilon$ for any $m, n \geqslant N$ (Cauchy criterion).
## 25.3. Calculation of Limits | 25.14. If the sequence $\left\{x_{n}\right\}$ has a limit $a$, then for any $\varepsilon>0$ we can choose $N$ such that $\left|x_{n}-a\right|<\varepsilon$ for all $n \geqslant N$. For any $\varepsilon>0$ we can choose a number $N$ such that $\left|x_{n}-x_{m}\right|<\varepsilon$ for any $m, n \geqslant N$. Let $\vareps... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,766 |
25.15. Calculate $\lim _{n \rightarrow \infty}(\sqrt{n+1}-\sqrt{n})$. | 25.15. Answer: 0. It is clear that
$$
0<\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{\sqrt{n}}
$$
Therefore, $\lim _{n \rightarrow \infty}(\sqrt{n+1}-\sqrt{n})=0$. | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,767 |
25.16. Calculate $\lim _{n \rightarrow \infty}\left(\sqrt{n^{2}+n}-n\right)$. | 25.16. Answer: 1/2. By multiplying and dividing $\sqrt{n^{2}+n}-n$ by $\sqrt{n^{2}+n}+n$, we get
$$
\sqrt{n^{2}+n}-n=\frac{n}{\sqrt{n^{2}+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1} \rightarrow \frac{1}{2}
$$ | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,768 |
25.17. a) Prove that $\lim _{n \rightarrow \infty} \frac{n}{10^{n}}=0$.
b) Prove that $\lim _{n \rightarrow \infty} \frac{\lg n}{n}=0$. | 25.17. a) It is clear that
$$
\frac{n}{10^{n}}=\frac{n}{n-1} \cdot \frac{n-1}{n-2} \cdot \ldots \cdot \frac{2}{1} \cdot \frac{1}{10^{n}}<2^{n-1} \frac{1}{10^{n}}=\frac{1}{2} \cdot \frac{1}{5^{n}} \rightarrow 0
$$
b) Let $10^{k} \leqslant n \leqslant 10^{k+1}$. Then $\frac{\lg n}{n} \leqslant \frac{k+1}{10^{k}} \right... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,769 |
25.18. a) Prove that $\lim _{n \rightarrow \infty} \frac{n^{k}}{a^{n}}=0$ for any natural $k$ and any $a>1$.
b) Prove that $\lim _{n \rightarrow \infty} \frac{\left(\log _{a} n\right)^{k}}{n}=0$ for any natural $k$ and any $a>1$. | 25.18. a) If $n$ is sufficiently large, then $\sqrt[k]{a}>\frac{n+1}{n}=1+\frac{1}{n}$. Choose a number $q>1$ and a natural number $N$ such that $q1$, then $\lim _{m \rightarrow \infty} q^{-m}=0$.
b) Let $a^{m-1} \leqslant n \leqslant a^{m}$. Then $\frac{\left(\log _{a} n\right)^{k}}{n} \leqslant \frac{m^{k}}{a^{m-1}}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,770 |
25.19. Prove that $\lim _{n \rightarrow \infty} \frac{x^{n}}{n!}=0$. | 25.19. Choose a natural number $N$ such that $\frac{|x|}{N+1} \leqslant \frac{1}{2}$. If $n>N$, then $\frac{|x|^{n}}{n!} \leqslant \frac{|x|^{N}}{N!}\left(\frac{1}{2}\right)^{n-N}=\frac{C}{2^{n}}$, where $C=\frac{2^{N}|x|^{N}}{N!}$ is a constant. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,771 |
25.20. Prove that $\lim _{n \rightarrow \infty} \sqrt[n]{x}=1$ for any positive number $x$. | 25.20. Let $x>1$. Consider the auxiliary sequence $a_{n}=x^{1 / n}-1$. Clearly, $a_{n}>0$, so according to problem 13.9 a) $1+n a_{n} \leqslant\left(1+a_{n}\right)^{n}=x$. Therefore, $01$. Then $\lim _{n \rightarrow \infty} y^{1 / n}=1$, so $\lim _{n \rightarrow \infty} x^{1 / n}=1$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,772 |
25.21. Prove that $\lim _{n \rightarrow \infty} \sqrt[n]{n}=1$. | 25.21. Here is the solution. Let $a_{n}=\sqrt[n]{n}-1$. Then $a_{n} \geqslant 0$, so $n=\left(1+a_{n}\right)^{n} \geqslant 1+n a_{n}+\frac{n(n-1)}{2} a_{n}^{2} \geqslant \frac{n(n-1)}{2} a_{n}^{2}$. Therefore, $0 \leqslant a_{n} \leqslant \sqrt{\frac{2}{n-1}}$.
Second solution. Using the inequality between the arithme... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,773 |
25.22. Prove that $\lim _{n \rightarrow \infty} \sqrt[n]{n!}=\infty$. | 25.22. From the inequality $(2 n)!\geqslant n(n+1) \ldots(2 n)>n^{n+1}$ it follows that $\sqrt[2 n]{(2 n)!}>\sqrt[2 n]{n^{n+1}}>\sqrt{n}$ and $\sqrt[2 n+1]{(2 n+1)!}>\sqrt[2 n+1]{n^{n+1}}>\sqrt{n}$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,774 |
25.23. Let $a_{0}=a, b_{0}=b$, where $0<a<b$. Set $a_{n+1}=$ $=\frac{2 a_{n} b_{n}}{a_{n}+b_{n}}$ and $b_{n+1}=\frac{a_{n}+b_{n}}{2}$ for $n \geqslant 0$. Prove that $\lim _{n \rightarrow \infty} a_{n}=$ $=\lim _{n \rightarrow \infty} b_{n}=\sqrt{a b}$. | 25.23. If $0<a_{n}$ and $b_{n+1}<b_{n}$. Moreover, $4 a_{n} b_{n}<\left(a_{n}+b_{n}\right)^{2}$, i.e., $a_{n+1}<b_{n+1}$. Therefore, $a_{0}<a_{1}<a_{2}<\ldots<b_{2}<$ $<b_{1}<b_{0}$. This means that the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ converge. Let $\lim _{n \rightarrow \infty} a_{n}=\alpha$... | \alpha=\sqrt{} | Algebra | proof | Yes | Yes | olympiads | false | 38,775 |
25.24. Let among the numbers $2^{1}, 2^{2}, \ldots, 2^{n}$ exactly $a_{n}$ numbers start with the digit 1. Compute the limit $\lim _{n \rightarrow \infty} \frac{a_{n}}{n}$. | 25.24. Consider all natural numbers $a$ for which $10^{k-1} \leqslant a10^{k}$. The smallest power of two within these limits starts with a one, since otherwise we could divide the number $a=2^{m}$ by 2 and obtain a number within the same limits. The power of two following the smallest one starts with the digit 2 or 3.... | \lg2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,776 |
25.25. Let $x_{1}=\sqrt{a}$, where $a$ is some positive number, and $x_{n+1}=\sqrt{a+x_{n}}$ for $n \geqslant 1$. Prove that the limit $\lim _{n \rightarrow \infty} x_{n}$ exists and compute it. | 25.25. If we knew that the limit $\lim _{n \rightarrow \infty} x_{n}=c$ exists, finding $c$ would be easy. Indeed, $c=\sqrt{a+c}$, so $c^{2}-c-a=0$. Therefore, $c=1 / 2 \pm \sqrt{a+1 / 4}$. It is also clear that $c \geqslant 0$, so $c=1 / 2+\sqrt{a+1 / 4}$.
Let us now prove that the considered limit exists. First, not... | \frac{1}{2}+\sqrt{+\frac{1}{4}} | Algebra | proof | Yes | Yes | olympiads | false | 38,777 |
25.26. Let $x_{1}$ and $a$ be positive numbers, $x_{n+1}=\frac{a}{1+x_{n}}$ for $n \geqslant 1$. Prove that the limit $\lim _{n \rightarrow \infty} x_{n}$ exists, and compute it. | 25.26. If we knew that the limit $\lim _{n \rightarrow \infty} x_{n}=c$ exists, it would be easy to find it. Indeed, $c=\frac{a^{n \rightarrow \infty}}{1+c}$, so $c^{2}+c-a=0$. It is also clear that $c \geqslant 0$. Therefore, $c$ is the positive root of the equation $x^{2}+x-a=0$. Such a root is unique.
Let us now pr... | \frac{-1+\sqrt{1+4a}}{2} | Algebra | proof | Yes | Yes | olympiads | false | 38,778 |
25.28. Let $x_{1} \stackrel{n \rightarrow \infty}{=}, x_{2}=b$ and $x_{n+1}=\frac{1}{2}\left(x_{n}+x_{n-1}\right)$ for $n \geqslant 1$. Compute $\lim _{n \rightarrow \infty} x_{n}$. | 25.28. Answer: $\lim _{n \rightarrow \infty} x_{n}=\frac{a+2 b}{3}$. We will prove that $x_{n}=\frac{a+2 b}{3}+$ $+\left(\frac{-1}{2}\right)^{n-2} \frac{b-a}{3}$. For $n=1$ and 2, the required equality is easily verified. The relation $x_{n+1}=\frac{1}{2}\left(x_{n}+x_{n-1}\right)$ is also easily verified. | \frac{+2b}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,780 |
25.29. Let $p_{1}, \ldots, p_{k}$ be prime numbers, $a_{n}$ be the number of natural numbers not exceeding $n$ and divisible only by these prime numbers. Prove that $\lim _{n \rightarrow \infty} \frac{a_{n}}{n}=0$. | 25.29. Let $p_{1}^{\alpha_{1}} \ldots p_{k}^{\alpha_{k}} \leqslant n$. Then $p_{i}^{\alpha_{i}} \leqslant n$, so $\alpha_{i} \leqslant \frac{\lg n}{\lg p_{i}}$. Therefore, $a_{n} \leqslant\left(\frac{\lg n}{\lg p_{1}}+1\right) \ldots\left(\frac{\lg n}{\lg p_{k}}+1\right)$. If $n \geqslant p_{1}, \ldots, p_{k}$, then $a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,781 |
25.30. Let $x_{0}$ and $y_{0}$ be some positive numbers, with $x_{0}>y_{0}, x_{n+1}=\frac{x_{n}+y_{n}}{2}$ and $y_{n+1}=\frac{2 x_{n} y_{n}}{x_{n}+y_{n}}$ for $n \geqslant 0$. Prove that $\lim _{n \rightarrow \infty} x_{n}=\lim _{n \rightarrow \infty} y_{n}=\sqrt{x_{0} y_{0}}$. | 25.30. It is easy to check that $x_{0}>\frac{x_{0}+y_{0}}{2}>\frac{2 x_{0} y_{0}}{x_{0}+y_{0}}>y_{0}$, i.e., $x_{0}>$ $>x_{1}>y_{1}>y_{0}$. Similarly, $x_{n}>x_{n+1}>y_{n+1}>y_{n}$ for any $n$. Thus, the sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ are monotonic and bounded, so they converge to some numb... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,782 |
25.31. Let $x_{0}$ and $y_{0}$ be some positive numbers, with $x_{0}>y_{0}, x_{n+1}=\frac{x_{n}+y_{n}}{2}$ and $y_{n+1}=\sqrt{x_{n} y_{n}}$ for $n \geqslant 0$. Prove that both sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ converge to the same limit, called the arithmetic-geometric mean of the numbers $x_... | 25.31. It is easy to check that $x_{0}>\frac{x_{0}+y_{0}}{2}>\sqrt{x_{0} y_{0}}>y_{0}$, i.e., $x_{0}>$ $>x_{1}>y_{1}>y_{0}$. Similarly, $x_{n}>x_{n+1}>y_{n+1}>y_{n}$ for any $n$. Thus, the sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ are monotonic and bounded, so they converge to some numbers $x$ and $y$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,783 |
25.32. Let \(a_{0}=2 \sqrt{3}, b_{0}=3\) and \(a_{n+1}=\frac{2 a_{n} b_{n}}{a_{n}+b_{n}}, b_{n+1}=\sqrt{a_{n+1} b_{n}}\) for \(n \geqslant 0\). Prove that \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n}=\pi\). | 25.32. Let \( A_{n}=6 \cdot 2^{n} \operatorname{tg} \frac{\pi}{6 \cdot 2^{n}} \) and \( B_{n}=6 \cdot 2^{n} \sin \frac{\pi}{6 \cdot 2^{n}} \), i.e., \( A_{n} \) is the semiperimeter of a regular \( 6 \cdot 2^{n} \)-gon circumscribed around a circle of radius 1, and \( B_{n} \) is the semiperimeter of a regular \( 6 \cd... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,784 |
25.33. Let $x_{0}$ and $y_{0}$ be some non-negative numbers, $x_{n+1}=\frac{x_{n}+y_{n}}{2}$ and $y_{n+1}=\sqrt{x_{n+1} y_{n}}$ for $n \geqslant 0$.
a) Prove that if $0 \leqslant x_{0}<y_{0}$, then
$$
\lim _{n \rightarrow \infty} x_{n}=\lim _{n \rightarrow \infty} y_{n}=\frac{\sqrt{y_{0}^{2}-x_{0}^{2}}}{\arccos \left... | 25.33. a) Let $x_{0}=y_{0} \cos \alpha$, i.e., $\alpha=\arccos \left(x_{0} / y_{0}\right)$. Then $x_{1}=y_{0} \frac{1+\cos \alpha}{2}=y_{0} \cos ^{2} \frac{\alpha}{2}$ and $y_{1}=\sqrt{y_{0} \cos ^{2} \frac{\alpha}{2} y_{0}}=y_{0} \cos \frac{\alpha}{2}$. Therefore, $x_{1}=y_{1} \cos \frac{\alpha}{2}$. Continuing this r... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,785 |
25.34. a) Prove that for any natural $n$ the inequality $2 \leqslant(1+1 / n)^{n}<3$ holds.
b) Prove that for any natural $n$ the inequality $(n / 3)^{n}<n!$ holds. | 25.34. a) First solution. According to problem 13.9, if $0<\alpha \leqslant 1 / n$, then the inequality $1+n \alpha \leqslant(1+1 / n)^{n}<1+n \alpha + n^{2} \alpha^{2}$ holds. For $\alpha=1 / n$, we obtain the required result.
Second solution. It is clear that
$$
\begin{gathered}
\left(1+\frac{1}{n}\right)^{n}=1+n \... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,786 |
25.35. Find the first digit of the number $2^{400}$. | 25.35. Answer: 2. The first digit of the number $2^{400}=\left(2^{10}\right)^{40}=1024^{40}$ coincides with the first digit of the number $1,024^{40}$. On the one hand, according to problem 25.34, we get $1,024^{40}1+40 \cdot 0,024+\frac{40 \cdot 39}{2} 0,024^{2}=2,40928>2$. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,787 |
25.36. Prove that if \( m \) and \( n \) are natural numbers, and \( m > n \), then
\[
\left(1+\frac{1}{m}\right)^{m}>\left(1+\frac{1}{n}\right)^{n} \quad \text { and } \quad\left(1+\frac{1}{m}\right)^{m+1}<\left(1+\frac{1}{n}\right)^{n+1}
\] | 25.36. It is sufficient to prove the required inequalities in the case when $m=n+1$.
First solution. Let $0 < x < 1$
$$
\left(1+\frac{1}{n+1}\right)^{n+1} > \left(1+\frac{1}{n}\right)^{n}
$$
Second solution. The inequality $\left(1+\frac{1}{n+1}\right)^{n+1}>\left(1+\frac{1}{n}\right)^{n}$ is equivalent to the inequa... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,788 |
25.37. Prove that there exists a limit $e=\lim _{n \rightarrow \infty}(1+1 / n)^{n}$. | 25.37. According to problem 25.36, the sequence $a_{n}=(1+1 / n)^{n}$ is monotonically increasing, and according to problem 25.34, this sequence is bounded above. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,789 |
25.38. a) Prove that
$$
\left(1+\frac{1}{n}\right)^{n}<2+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!}
$$
b) Prove that
$$
e \geqslant 2+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{k!}
$$
for any $k$.
c) Prove that $e=\lim _{n \rightarrow \infty}\left(2+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!}\right)$. | 25.38. a) By the binomial formula of Newton
$$
\left(1+\frac{1}{n}\right)^{n}=1+n \frac{1}{n}+C_{n}^{2} \frac{1}{n^{2}}+\ldots+C_{n}^{k} \frac{1}{n^{k}}+\ldots+C_{n}^{n} \frac{1}{n^{n}}
$$
Moreover, $C_{n}^{k}=\frac{n(n-1) \ldots(n-k-1)}{k!}<\frac{n^{k}}{k!}$.
b) If $k \leqslant n$, then
$$
\begin{aligned}
&\left(1... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,790 |
25.39. Prove that
$$
0<e-\left(1+1+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!}\right)<\frac{1}{n!n}
$$ | 25.39. According to problem 25.38
$$
\begin{array}{r}
e-\left(1+1+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!}\right)=\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}+\ldots= \\
=\frac{1}{(n+1)!}\left(1+\frac{1}{n+2}+\frac{1}{(n+2)(n+3)}+\ldots\right)< \\
<\frac{1}{(n+1)!}\left(1+\frac{1}{n+2}+\frac{1}{(n+2)^{2}}+\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,791 |
25.40. Does the sequence $a_{n}=\sin (2 \pi n!e)$ converge? | 25.40. A n s w e r: yes, converges to 0. According to problem 25.39, the fractional part of the number $n!e$ is between 0 and $1 / n$. Therefore, $0<\sin (2 \pi n!e)<$ $<\sin (2 \pi / n)$ for $n \geqslant 4$. | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,792 |
25.41. Prove that the number $e$ is irrational. | 25.41. Suppose that $e=m / n$, where $m$ and $n$ are natural numbers. Then according to problem 25.39
$$
0<\frac{m}{n}-\left(2+\frac{1}{2!}+\ldots+\frac{1}{n!}\right)<\frac{1}{n!n}
$$
After multiplying by $n!$ we get that $0<a<1 / n$, where $a$ is an integer. This cannot be | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,793 |
25.42. Prove that $\lim _{n \rightarrow \infty} \frac{\sqrt[n]{n!}}{n}=\frac{1}{e}$.
## 25.5. Conjugate numbers | 25.42. First solution. Let $a_{n}=\frac{n!}{n^{n}}$. Then $\frac{a_{n+1}}{a_{n}}=$ $=\frac{(n+1)!}{(n+1)^{n}} \cdot \frac{n^{n}}{n!}=\frac{1}{(1+1 / n)^{n}} \rightarrow \frac{1}{e}$. Therefore, according to problem 25.9, $\lim _{n \rightarrow \infty} \sqrt[n]{a_{n}}=\frac{1}{e}$, i.e., $\lim _{n \rightarrow \infty} \fr... | \frac{1}{e} | Calculus | proof | Yes | Yes | olympiads | false | 38,794 |
25.43. Let $(1+\sqrt{2}+\sqrt{3})^{n}=q_{n}+r_{n} \sqrt{2}+s_{n} \sqrt{3}+t_{n} \sqrt{6}$, where $q_{n}, r_{n}, s_{n}, t_{n}$ are natural numbers. Compute the limits $\lim _{n \rightarrow \infty} \frac{r_{n}}{q_{n}}, \lim _{n \rightarrow \infty} \frac{s_{n}}{q_{n}}$ and $\lim _{n \rightarrow \infty} \frac{t_{n}}{q_{n}}... | 25.43. Let $\lambda_{1}=1+\sqrt{2}+\sqrt{3}, \lambda_{2}=1-\sqrt{2}+\sqrt{3}, \lambda_{3}=1+\sqrt{2}-\sqrt{3}$, $\lambda_{4}=1-\sqrt{2}-\sqrt{3}$. Then
$$
\begin{aligned}
& \lambda_{1}^{n}=q_{n}+r_{n} \sqrt{2}+s_{n} \sqrt{3}+t_{n} \sqrt{6}, \\
& \lambda_{2}^{n}=q_{n}-r_{n} \sqrt{2}+s_{n} \sqrt{3}-t_{n} \sqrt{6}, \\
& ... | \lim_{narrow\infty}\frac{r_{n}}{q_{n}}=\frac{1}{\sqrt{2}},\lim_{narrow\infty}\frac{s_{n}}{q_{n}}=\frac{1}{\sqrt{3}},\lim_{narrow\infty}\frac{t_{n}}{q_{n}}=\frac{1}{\sqrt{6}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,795 |
25.45. Prove that any (non-empty) bounded above set of real numbers has a unique least upper bound, and a set bounded below has a unique greatest lower bound. | 25.45. Let's conduct the proof only for the exact upper bound. Let $x_{0}$ and $x_{1}$ be two exact upper bounds, with $x_{1} > x_{0}$. Then for $\varepsilon = \frac{x_{1} - x_{0}}{2}$, there exists $x$ (a number from the considered set) such that $x + \frac{x_{1} - x_{0}}{2} > x_{1}$, i.e., $x > \frac{x_{1} + x_{0}}{2... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,797 |
26.1. Real numbers $x$ and $y$ satisfy the equations $x^{3}-3 x^{2}+5 x=1, y^{3}-3 y^{2}+5 y=5$. Find $x+y$.
## 26.2. Periodic functions | 26.1. Answer: $x+y=2$. The function $f(x)=x^{3}-3 x^{2}+5 x=(x-1)^{3}+$ $+2(x-1)+3$ is monotonically increasing, so for each real number $c$ the equation $f(x)=c$ has exactly one real solution.
The function $f(x)-3$ has the following property: if $x-1=$ $=1-y$, then $f(x)-3=-(f(y)-3)$. From the monotonicity of the fun... | x+2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,798 |
26.3. Prove that if the limit of the function $f(x)$ at the point $x_{0}$ exists, then it is unique. | 26.3. Suppose that the function $f(x)$ has two limits, $a$ and $b$, at the point $x_{0}$, and $a \neq b$. For $\varepsilon=\frac{1}{2}|a-b|$, choose $\delta_{1}$ and $\delta_{2}$ such that $|f(x)-a|<\varepsilon$ when $\left|x-x_{0}\right|<\delta_{1}$ and $|f(x)-b|<\varepsilon$ when $\left|x-x_{0}\right|<\delta_{2}$. Th... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,800 |
26.4. Prove that $\lim _{x \rightarrow x_{0}} f(x)=a$ if and only if $\lim _{n \rightarrow \infty} f\left(a_{n}\right)=a$ for any sequence $\left\{a_{n}\right\}$, for which $\lim _{n \rightarrow \infty} a_{n}=x_{0}$ and $a_{n} \neq x_{0}$ for all $n$ (it is also assumed that all points $a_{n}$ belong to the domain of t... | 26.4. Suppose first that $\lim _{x \rightarrow x_{0}} f(x)=a$. For a given $\varepsilon>0$, choose $\delta>0$ such that from the inequality $\left|x-x_{0}\right|N$. Then $\left|f\left(a_{n}\right)-a\right|0$, having the following property: for any $\delta>0$ there exists $x \neq x_{0}$, for which $\left|x-x_{0}\right|\... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,801 |
26.6. Prove that $\lim _{\alpha \rightarrow 0} \frac{\sin \alpha}{\alpha}=1$.
## 26.4. Continuity
A function $f(x)$ is called continuous at a point $x_{0}$ if $\lim _{x \rightarrow x_{0}} f(x)=$ $=f\left(x_{0}\right)$. If the function $f(x)$ is continuous at every point (in its domain), it is simply called continuous... | 26.6. According to problem $11.1 \cos \alpha<\frac{\sin \alpha}{\alpha}<1$ for $0<\alpha<\pi / 2$. The same inequalities are true for $-\pi / 2<\alpha<0$, since $\cos (-\alpha)=\cos \alpha$ and $\frac{\sin (-\alpha)}{-\alpha}=\frac{\sin \alpha}{\alpha}$. It remains to note that $\lim _{\alpha \rightarrow 0} \cos \alpha... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,803 |
26.7. a) Let $P(x)$ be a polynomial. Prove that the function $P(x)$ is continuous.
b) Let $P(x)$ and $Q(x)$ be polynomials, and suppose $Q\left(x_{0}\right) \neq 0$. Prove that the function $P(x) / Q(x)$ is continuous at the point $x_{0}$. | 26.7. Let's use problem 26.5. It is clear that $\lim _{x \rightarrow x_{0}} x=x_{0}$, therefore, applying induction on $n$, we get $\lim _{x \rightarrow x_{0}} x^{n}=x_{0}^{n}$. Hence,
$$
\lim _{x \rightarrow x_{0}}\left(a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}\right)=a_{n} x_{0}^{n}+a_{n-1} x_{0}^{n-1}+\ldots+a_{0}
$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,804 |
26.9. Prove that the function $f(x)=\sin x$ is continuous.
314 Chapter 26. Continuous and Discontinuous Functions | 26.9. It is clear that
$$
\sin (x+\varepsilon)-\sin x=2 \sin \frac{\varepsilon}{2} \cos \left(x+\frac{\varepsilon}{2}\right)
$$
Therefore, it is sufficient to prove that $\lim _{t \rightarrow 0} \sin t=0$. But if $|t|<\pi / 2$, then $|\sin t|<|t|$ according to problem 11.1. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,806 |
26.10. Let \( f(x) = x \sin \frac{1}{x} \) for \( x \neq 0 \) and \( f(0) = 0 \). Prove that the function \( f(x) \) is continuous (at all points \( x \)).
## 26.5. Intermediate Value Theorem | 26.10. It is clear that the function $f(x)$ is continuous at all points $x \neq 0$. Let's check that the function $f(x)$ is continuous at the point $x=0$, i.e., for any $\varepsilon>0$ we can choose $\delta>0$ such that if $|x|<\delta$, then $|f(x)|<\varepsilon$. But $|f(x)|=\left|x \sin \frac{1}{x}\right| \leqslant|x|... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,807 |
26.11. The function $f(x)$ is continuous on the interval $[a, b]$ and takes values of different signs at its endpoints. Prove that $f\left(x_{0}\right)=0$ for some point $x_{0}$ in this interval (Intermediate Value Theorem). | 26.11. Let for definiteness $f(a)>0$ and $f(b)<0$, then if $f(c) \leq 0$, we set $a_{1}=a$ and $b_{1}=c$; if $f(c)>0$, we set $a_{2}=c$ and $b_{2}=b_{1}$ (we assume that $f(c) \neq 0$, since otherwise the proof is immediately complete). Then we take the midpoint of the interval $\left[a_{2}, b_{2}\right]$ and repeat th... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,808 |
26.12. a) Let $f$ be a continuous function for which the equation $f(x)=x$ has no real solutions. Prove that the equation $f(f(x))=x$ also has no real solutions.
b) Let $f$ and $g$ be continuous functions satisfying the identity $f(g(x))=g(f(x))$. Prove that if the equation $f(x)=g(x)$ has no real solutions, then the ... | 26.12. a) If $f$ is a continuous function and the equation $f(x)=x$ has no real solutions, then either $f(x)>x$ for all $x$, or $f(x)<x$ for all $x$.
b) If $f$ and $g$ are continuous functions and the equation $f(x)=g(x)$ has no real solutions, then either $f(x)>g(x)$ for all $x$, or $f(x)<g(x)$ for all $x$. In the fi... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,809 |
26.13. The function $f(x)$ is continuous on the interval $[0,1]$ and takes values from the same interval. Prove that $f(x)=x$ for some point $x$ in this interval. | 26.13. Consider the auxiliary function $\varphi(x)=f(x)-x$. It is clear that $\varphi(0)=f(0) \geqslant 0$ and $\varphi(1)=f(1)-1 \leqslant 0$. Therefore, $\varphi(x)=0$ for some point $x$ in the interval $[0,1]$. In this case, $f(x)=x$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,810 |
26.14. On the segment $[0,1]$, $n$ numbers $x_{1}, \ldots, x_{n}$ are chosen. Prove that on the same segment, a number $x$ can be chosen such that $\frac{1}{n} \sum_{i=1}^{n}\left|x-x_{i}\right|=\frac{1}{2}$. | 26.14. Consider the function $f(x)=\frac{1}{n} \sum_{i=1}^{n}\left|x-x_{i}\right|$. All numbers $x_{i}$ are non-negative, so $f(0)=\frac{1}{n} \sum_{i=1}^{n}\left|-x_{i}\right|=\frac{1}{n} \sum_{i=1}^{n} x_{i}$. All numbers $x_{i}$ do not exceed 1, so $f(1)=\frac{1}{n} \sum_{i=1}^{n}\left|1-x_{i}\right|=\frac{1}{n} \su... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,811 |
26.15. Does there exist a function that is continuous on the interval $[0,1]$, takes irrational values at rational points, and rational values at irrational points, with all values belonging to the interval $[0,1]$?
## 26.6. Properties of functions continuous on an interval | 26.15. Answer: no, it does not exist. Indeed, if $f$ is the required function, then $f(x) \neq x$ for any point $x$ in the interval $[0,1]$. But according to problem 26.13, such continuous functions do not exist. | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,812 |
26.16. Prove that the function $f(x)$, continuous on the interval $[a, b]$, is bounded on this interval, i.e., the set of all values $f(x)$ for $x$ in the interval $[a, b]$ is a bounded set. | 26.16. We will prove that the function $f(x)$ is bounded above (boundedness from below is proved similarly). Suppose that for any natural number $n$ there is a point $x_{n}$ on the interval $[a, b]$ such that $f\left(x_{n}\right)>n$. According to problem 25.12, from the bounded sequence $\left\{x_{n}\right\}$, one can ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,813 |
26.18. Prove that a function $f(x)$, continuous on the interval $[a, b]$, attains a maximum and a minimum at some points of this interval (Weierstrass).
## 26.7. Convex Functions
A function $f(x)$ defined on the interval $[a, b]$ is called convex if
$$
f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant \frac{f\left(x_{1}... | 26.18. We will prove that on the interval $[a, b]$, there exists a point $x_{0}$ such that $f\left(x_{0}\right)=M$, where $M$ is the least upper bound of the set of values of $f(x)$ for all points $x$ in the interval $[a, b]$. (The proof for the greatest lower bound is analogous.) First, note that the set of values of ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,815 |
26.19. Prove that the function $f(x)$ is convex on the interval $[a, b]$ if and only if for any $n$ and any points $x_{1}, \ldots, x_{n}$ from this interval, the inequality
$$
f\left(\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}\right) \leqslant \frac{f\left(x_{1}\right)+f\left(x_{2}\right)+\ldots+f\left(x_{n}\right)}{n}
$$
hol... | 26.19. It is sufficient to prove that inequality (1) implies inequality (2) for any $n$. If inequality (1) holds, then
$$
\begin{aligned}
4 f\left(\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}\right) \leqslant 2 f\left(\frac{x_{1}+x_{2}}{2}\right)+ & 2 f\left(\frac{x_{3}+x_{4}}{2}\right) \leqslant \\
& \leqslant f\left(x_{1}\righ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,816 |
26.20. Let $f(x)$ be a convex function, and $p$ and $q$ be positive numbers whose sum is 1.
a) Prove that if the numbers $p$ and $q$ are rational, then $f\left(p x_{1}+q x_{2}\right) \leqslant p f\left(x_{1}\right)+q f\left(x_{2}\right)$.
b) Prove that if the function $f(x)$ is continuous, then $f\left(p x_{1}+q x_{2... | 26.20. a) If the numbers $p$ and $q$ are rational, then $p=\frac{m}{n}$ and $q=\frac{n-m}{n}$, where $m, n$, and $n-m$ are natural numbers. Apply the inequality from problem 26.19 to the numbers $x_{1}, \ldots, x_{1}, x_{2}, \ldots, x_{2}$ ( $m$ numbers $x_{1}$ and $n-m$ numbers $x_{2}$ ). As a result, we obtain the re... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,817 |
26.22. Prove that the function $f(x)=\ln x$ is concave on the interval $(0,+\infty)$. | 26.22. It is required to prove that if $x_{1}, x_{2}>0$, then $\ln \left(\frac{x_{1}+x_{2}}{2}\right) \geqslant$ $\geqslant \frac{\ln x_{1}+\ln x_{2}}{2}$. This inequality is equivalent to the inequality $\frac{x_{1}+x_{2}}{2} \geqslant$ $\geqslant \exp \left(\frac{\ln x_{1}+\ln x_{2}}{2}\right)=\sqrt{x_{1} x_{2}}$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,819 |
26.23. Prove the inequality between the arithmetic mean and the geometric mean for positive numbers $x_{1}, \ldots, x_{n}:$
$$
\frac{x_{1}+\ldots+x_{n}}{n} \geqslant \sqrt[n]{x_{1} \ldots x_{n}}
$$ | 26.23. The function $f(x)=\ln x$ is concave (Problem 26.22). Therefore, according to Jensen's inequality
$$
\ln \left(\frac{x_{1}+\ldots+x_{n}}{n}\right) \geqslant \frac{\ln x_{1}+\ldots+\ln x_{n}}{n}
$$
This inequality is equivalent to the required one.
Remark. For other proofs of the inequality between the arithme... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,820 |
26.24. Prove that if $A, B, p$ and $q$ are positive numbers, and $1 / p + 1 / q = 1$, then $A^{1 / p} B^{1 / q} \leqslant A / p + B / q$.
## 26.8. Uniform Continuity
Let the set $D$ be contained in the domain of the function $f(x)$. The function $f(x)$ is called uniformly continuous on the set $D$ if for any $\vareps... | 26.24. The function $f(x)=\ln x$ is concave (Problem 26.22). Therefore, according to Jensen's inequality,
$$
\ln \left(\frac{A}{p}+\frac{B}{q}\right) \geqslant \frac{1}{p} \ln A+\frac{1}{q} \ln B
$$
This inequality is equivalent to the required one.
Note. Other proofs are provided in the solutions to Problems 8.46 a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,821 |
26.25. a) Provide an example of a continuous function on the set of all real numbers that is not uniformly continuous.
b) Provide an example of a continuous function on the interval $(0,1)$ that is not uniformly continuous. | 26.25. a) The function $f(x)=x^{2}$ on the set of all real numbers is not uniformly continuous. Indeed, if this function were uniformly continuous, then for any $\varepsilon>0$ we could choose $\delta>0$ such that for any $x$ the inequality $\left|\left(x+\frac{\delta}{2}\right)^{2}-x^{2}\right|0$ we chose $\delta>0$ s... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,822 |
26.27. Let \( f(x) = x \sin (1 / x) \) for \( x \neq 0 \), and \( f(0) = 0 \). Prove that on the interval \([0,1]\), the function \( f \) is continuous but not of bounded variation. | 26.27. The continuity of the function $f(x)$ is proven in the solution to problem 26.10. Let $x_{1}=\frac{2}{\pi}, x_{2}=\frac{1}{\pi}, x_{3}=\frac{2}{3 \pi}, x_{4}=\frac{1}{2 \pi}, x_{5}=\frac{2}{5 \pi}, x_{6}=\frac{1}{3 \pi}$, $x_{7}=\frac{2}{7 \pi}, x_{8}=\frac{1}{4 \pi}, \ldots$ Then $f\left(x_{2 k}\right)=0$, so
... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,824 |
26.28. Prove that a function $f$, defined on the interval $[a, b]$, is of bounded variation if and only if it can be represented as the difference of two non-decreasing functions. | 26.28. It is clear that if the function $g$ is non-decreasing or non-increasing, then $\operatorname{Var}_{a}^{b}(g)=|g(a)-g(b)|$; in particular, $g$ is a function of bounded variation. It is also clear that if $g_{1}$ and $g_{2}$ are functions of bounded variation, then $f=g_{1}+g_{2}$ is also a function of bounded va... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,825 |
27.1. a) Let $a>1$. Prove that if $x_{1}>x_{2}$, then $a^{x_{1}}>a^{x_{2}}$.
b) Let $a<1$. Prove that if $x_{1}>x_{2}$, then $a^{x_{1}}<a^{x_{2}}$. | 27.1. a) We can consider that $x_{1}=p_{1} / q$ and $x_{2}=p_{2} / q$. Then $a^{p_{1}}>a^{p_{2}}$, since $a>1$. For positive numbers $\alpha$ and $\beta$, the inequality $\alpha>\beta$ is equivalent to the inequality $\alpha^{q}>\beta^{q}$. Therefore, $a^{p_{1} / q}>a^{p_{2} / q}$.[^5]
328 Chapter 27. Logarithm and Ex... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,827 |
27.2. Let $a>1$. Prove that $a^{x}$ can be arbitrarily large if $x$ is sufficiently large. | 27.2. Let $a=1+\delta$, where $\delta>0$. Then
$$
(1+\delta)^{n}=1+n \delta+C_{n}^{2} \delta^{2}+\ldots>n \delta
$$
Therefore, if $n>y / \delta$, then $a^{n}>y$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,828 |
27.3. Prove that $a^{x_{1}+x_{2}}=a^{x_{1}} a^{x_{2}}$. | 27.3. We can consider that $x_{1}=p_{1} / q$ and $x_{2}=p_{2} / q$. It is clear that $a^{p_{1}+p_{2}}=a^{p_{1}} a^{p_{2}} \cdot$ Therefore, $a^{x_{1}+x_{2}}=\sqrt[q]{a^{p_{1}+p_{2}}}=\sqrt[q]{a^{p_{1}}} \cdot \sqrt[q]{a^{p_{2}}}=a^{x_{1}} a^{x_{2}}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,829 |
27.4. Prove that $a^{\lambda x}=\left(a^{x}\right)^{\lambda}$ for any rational $\lambda$. | 27.4. For a natural $\lambda$ this follows from problem 27.3. If $\lambda=p / q$, then it is sufficient to take the $q$-th root of both sides of the equality $a^{p x}=\left(a^{x}\right)^{p}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,830 |
27.6. Prove that if there exists a limit $\lim _{n \rightarrow \infty} x_{n}=x$, where $\left\{x_{n}\right\}$ is a sequence of rational numbers, then there exists a limit $\lim _{n \rightarrow \infty} a^{x_{n}}$, and this limit depends only on $x$.
Let $a$ be a positive number. Define $a^{x}$ for any $x$ as follows. L... | 27.6. Let us assume that $a>1$; the case $a<1$ is treated similarly. Consider auxiliary sequences of rational numbers $\left\{x_{n}^{\prime}\right\}$ and $\left\{x_{n}^{\prime \prime}\right\}$, converging to $x$, where $\left\{x_{n}^{\prime}\right\}$ is monotonically increasing and $\left\{x_{n}^{\prime \prime}\right\}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,832 |
27.7. a) Let $a>1$. Prove that if $x>y$, then $a^{x}>a^{y}$. b) Let $a<1$. Prove that if $x>y$, then $a^{x}<a^{y}$. | 27.7. a) Let us choose rational numbers $p$ and $q$ such that $x > p > q > y$. Then we can choose sequences of rational numbers $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ such that they converge to $x$ and $y$ and at the same time $x_{n} \geqslant p$ and $y_{n} \leqslant q$ for all $n$. According to problem 27.1... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,833 |
27.8. Prove that if $\lim _{n \rightarrow \infty} x_{n}=x$, where $\left\{x_{n}\right\}$ is a sequence of arbitrary (not necessarily rational) numbers, then $\lim _{n \rightarrow \infty} a^{x_{n}}=a^{x}$. | 27.8. The solution is similar to the solution of problem 27.6. We again choose the same sequences $\left\{x_{n}^{\prime}\right\}$ and $\left\{x_{n}^{\prime \prime}\right\}$. According to problem 27.7, from the inequalities $x_{k_{n}}^{\prime}<x_{n}<x_{k_{n}}^{\prime \prime}$ follow the inequalities $a^{x_{k_{n}}^{\prim... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,834 |
27.9. Prove that $a^{x_{1}+x_{2}}=a^{x_{1}} a^{x_{2}}$ for arbitrary (not necessarily rational) numbers $x_{1}$ and $x_{2}$. | 27.9. The relation $a^{x_{1}+x_{2}}=a^{x_{1}} a^{x_{2}}$ for arbitrary numbers follows from the analogous relation for rational numbers (Problem 27.3), since the function $f(x)=a^{x}$ is continuous (Problem 27.8). | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,835 |
27.10. Let \( a \) be a positive number, and \( a \neq 1 \). Prove that for any positive number \( x \), there exists a unique number \( y \) such that \( a^{y} = x \).
Let \( a \) and \( x \) be positive numbers, and \( a \neq 1 \). The logarithm of \( x \) to the base \( a \) is the number \( y = \log_{a} x \), for ... | 27.10. The function $f(x)=a^{x}$ is continuous (Problem 27.8) and monotonic (this can easily be derived from the statement of Problem 27.1, using the continuity of the function $f$). Moreover, according to Problem 27.2, the number $a^{y}$ can be arbitrarily large, as well as arbitrarily close to zero (to prove the latt... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,836 |
27.11. Prove that the function $f(x)=\log _{a} x$ is continuous. | 27.11. Let's prove the continuity at the point $x_{0}=a^{y_{0}}$, where $y_{0}=\log _{a} x_{0}$. For a given $\varepsilon>0$, take $\delta$ to be the smallest of the two positive numbers $\left|a^{y_{0}}-a^{y_{0}+\varepsilon}\right|$ and $\left|a^{y_{0}}-a^{y_{0}-\varepsilon}\right|$. The function $g(y)=a^{y}$ is monot... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,837 |
27.12. Prove that $\log _{a}\left(x_{1} x_{2}\right)=\log _{a} x_{1}+\log _{a} x_{2}$.
## 27.2. Exponential Function | 27.12. This follows from the corresponding property of the exponential function: $a^{y_{1}+y_{2}}=a^{y_{1}} a^{y_{2}}$ (problem 27.9). | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,838 |
27.13. Solve the equation $5^{2 x-1}+5^{x+1}=250$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 27.13. Answer: $x=2$. The function $f(x)=5^{2 x-1}+5^{x+1}$ is monotonically increasing, so it takes the value 250 at only one value of $x$. It is also clear that $f(2)=5^{3}+5^{3}=250$. | 2 | Algebra | proof | Yes | Yes | olympiads | false | 38,839 |
27.14. Solve the equation $6^{x}-2^{x}=32$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 27.14. A n s: $x=2$. Dividing both sides of the equation by $2^{x} \neq 0$, we get the equation $3^{x}-2=32 \cdot 2^{-x}$. The function $f(x)=3^{x}-2$ is monotonically increasing, while the function $g(x)=32 \cdot 2^{-x}$ is monotonically decreasing. Therefore, the equation $f(x)=g(x)$ cannot have more than one solutio... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,840 | |
27.15. How many digits does the number $2^{100}$ have?
## 27.3. Identities for logarithms | 27.15. Answer: 31 digits. It is clear that $2^{100}=1024^{10}>1000^{10}$, so the number $2^{100}$ has no fewer than 31 digits. On the other hand,
$$
\frac{1024^{10}}{1000^{10}}<\left(\frac{1025}{1000}\right)^{10}=\left(\frac{41}{40}\right)^{10}<\frac{41}{40} \cdot \frac{40}{39} \cdot \frac{39}{38} \cdot \ldots \cdot \... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,841 |
27.16. a) Prove that $\log _{a} x=\frac{\log _{b} x}{\log _{b} a}$.
b) Prove that $\log _{a} b=\frac{1}{\log _{b} a}$. | 27.16. a) By definition, $a^{\log _{a} x}=x$. Taking the logarithm of both sides of this equation to the base $b$, we get $\log _{a} x \cdot \log _{b} a=\log _{b} x$.
b) Write the identity from part a) for $x=b$ and note that $\log _{b} b=1$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,842 |
27.17. Suppose that $a^{2}+b^{2}=7 a b$ and $a b \neq 0$. Prove that
$$
\lg \left|\frac{a+b}{3}\right|=\frac{1}{2}(\lg |a|+\lg |b|)
$$
326 Chapter 27. Logarithm and Exponential Function
## 27.4. Inequalities and Comparison of Numbers | 27.17. The required equality can be rewritten as $\lg \frac{(a+b)^{2}}{9}=$ $=\lg a b$ (we used the fact that $a b>0$). From the condition, it follows that $(a+b)^{2}=a^{2}+b^{2}+2 a b=7 a b+2 a b=9 a b$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,843 |
27.18. Prove that
$$
2<\frac{1}{\log _{2} \pi}+\frac{1}{\log _{5} \pi}
$$ | 27.18. The right-hand side of the required inequality is $\frac{\lg 2}{\lg \pi}+\frac{\lg 5}{\lg \pi}$. Therefore, we need to prove that $2 \lg \pi < \lg 2 + \lg 5$, i.e., $\lg \left(\pi^{2}\right) < \lg 10$. It remains to note that $\pi^{2} < 9.87$ (Problem 33.4). | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,844 |
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