problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
27.19. Prove that $3 / 10 < \lg 2 < 1 / 3$. | 27.19. The inequality $3 / 10 < \lg 2$ is equivalent to $10^{3} < 2^{10} = 1024$, and the inequality $\lg 2 < 1 / 3$ is equivalent to $8 = 2^{3} < 10$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,845 |
27.21. Compare the numbers $\log _{2} 3$ and $\log _{3} 5$. | 27.21. Let $a=\log _{2} 3$ and $b=\log _{3} 5$. Then $2^{a}=3$, so $8^{a}=3^{3}>5^{2}=9^{b}>8^{b}$. Therefore, $\log _{2} 3>\log _{3} 5$. | \log_{2}3>\log_{3}5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,847 |
27.22. Compare the numbers $\log _{20} 80$ and $\log _{80} 640$. | 27.22. It is clear that $\log _{20} 80=1+2 \log _{20} 2$ and $\log _{80} 640=1+3 \log _{80} 2$. Further, $\log _{20} 2=\frac{1}{\log _{2} 20}$ and $\log _{80} 2=\frac{1}{\log _{2} 80}$. Moreover, $3 \log _{2} 20=$ $=\log _{2} 8000>\log _{2} 6400=2 \log _{2} 80$. Therefore, $\log _{20} 80<\log _{80} 640$. | \log_{20}80<\log_{80}640 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,848 |
27.23. Compare the numbers $\log _{5} 7$ and $\log _{13} 17$. | 27.23. It is clear that $\log _{5} 7-1=\log _{5} \frac{7}{5}$ and $\log _{13} 17-1=\log _{13} \frac{17}{13}$. It is easy to verify that $\frac{7}{5}>\frac{17}{13}$. Therefore, $\log _{5} \frac{7}{5}>\log _{13} \frac{7}{5}>\log _{13} \frac{17}{13}$. In the end, we get that $\log _{5} 7>\log _{13} 17$. | \log_{5}7>\log_{13}17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,849 |
27.24. Compare the numbers $\log _{3} 7$ and $\log _{7} 27$.
## 27.5. Irrationality of logarithms | 27.24. It is clear that $\log _{7} 27=3 \log _{7} 3=\frac{3}{\log _{3} 7}$. Let's prove the inequality $\left(\log _{3} 7\right)^{2}>3$, i.e., $\log _{3} 7>\sqrt{3}$. It is easy to verify that $\sqrt{3}3^{\sqrt{3}}$ follows from the inequality $7^{4}>3^{7}$, i.e., $2401>$ $>2187$. In the end, we get that $\log _{3} 7>\... | \log_{3}7>\log_{7}27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,850 |
27.25. Prove that the following numbers are irrational:
a) $\log _{2} 3$; b) $\log _{\sqrt{2}} 3$; c) $\log _{5+3 \sqrt{2}}(3+5 \sqrt{2})$. | 27.25. a) Suppose that $\log _{2} 3=p / q$, where $p$ and $q$ are natural numbers. Then $2^{p / q}=3$, i.e., $2^{p}=3^{q}$. This cannot be.
b) Suppose that $\log _{\sqrt{2}} 3=p / q$, where $p$ and $q$ are natural numbers. Then $(\sqrt{2})^{p / q}=3$, i.e., $2^{p}=3^{2 q}$. This cannot be.
c) This problem is equivale... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,851 |
27.26. Provide an example of positive irrational numbers $a$ and $b$ such that the number $a^{b}$ is an integer.
## 27.6. Some Notable Limits | 27.26. Let $a=\sqrt{2}$ and $b=\log _{\sqrt{2}} 3$. The numbers $a$ and $b$ are irrational (problems 6.16 and 27.25). In this case, $a^{b}=(\sqrt{2})^{\log _{\sqrt{2}} 3}=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,852 |
27.27. Prove that $\lim _{x \rightarrow \pm \infty}(1+1 / x)^{x}=e$. | 27.27. We will use the fact that $\lim _{n \rightarrow \infty}(1+1 / n)^{n}=e$ (problem 25.37). For each $x \geqslant 1$, we can choose a natural number $n$ such that $n \leqslant x\left(1+\frac{1}{n+1}\right)^{n}=\left(1+\frac{1}{n+1}\right)^{n+1}\left(1+\frac{1}{n}\right)^{-1}
\end{gathered}
$
As $n \rightarrow \inf... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,853 |
27.28. Prove that $\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=1$. | 27.28. According to problem $27.27 \lim _{x \rightarrow 0}(1+x)^{1 / x}=e$. From this, using the continuity of the logarithm, we obtain $\lim _{x \rightarrow 0} \frac{1}{x} \ln (1+x)=1$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,854 |
27.29. Prove that $\lim _{x \rightarrow 0} \frac{(1+x)^{a}-1}{x}=a$ for any real $a$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 27.29. Let $(1+x)^{a}=1+y$. Then $y \rightarrow 0$ as $x \rightarrow 0$. Further, $a \ln (1+x)=\ln (1+y)$. Therefore,
$$
\frac{(1+x)^{a}-1}{x}=\frac{y}{x}=\frac{y}{\ln (1+y)} \cdot \frac{a \ln (1+x)}{x}
$$
Using the limit from problem 27.28, we obtain the required result. | Calculus | proof | Yes | Yes | olympiads | false | 38,855 | |
27.30. Prove that $\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}=\ln a$ for any positive $a$.
## 27.7. Hyperbolic Functions
| The same role that trigonometric functions play for the circle is played by hyperbolic functions for the hyperbola $x^{2}-y^{2}=1$:
$$
\begin{aligned}
& \left.\operatorname{sh} x=\frac{e^{x}-e^{-... | 27.30. Let $a^{x}-1=y$. Then $y \rightarrow 0$ as $x \rightarrow 0$. Moreover, $x=\frac{\ln (1+y)}{\ln a}$. Therefore, $\frac{a^{x}-1}{x}=\frac{y}{\ln (1+y)} \cdot \ln a$. Using the limit from problem 27.28, we obtain the required result. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,856 |
27.31. Prove that the point with coordinates $x=\operatorname{ch} t, y=$ $=\operatorname{sh} t$ lies on the hyperbola $x^{2}-y^{2}=1$. | 27.31. The formula for the difference of squares shows that
$$
\left(\frac{e^{t}+e^{-t}}{2}\right)^{2}-\left(\frac{e^{t}-e^{-t}}{2}\right)^{2}=e^{t} \cdot e^{-t}=1
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,857 |
27.32. Prove that
\[
\begin{aligned}
& \operatorname{sh}(x \pm y)=\operatorname{sh} x \operatorname{ch} y \pm \operatorname{ch} x \operatorname{sh} y \\
& \operatorname{ch}(x \pm y)=\operatorname{ch} x \operatorname{ch} y \pm \operatorname{sh} x \operatorname{sh} y
\end{aligned}
\]
The inverse hyperbolic functions ar... | 27.32. It is clear that
\[
\begin{aligned}
& 4 \operatorname{sh} x \operatorname{ch} y = e^{x+y} + e^{x-y} - e^{-x+y} - e^{-x-y} \\
& 4 \operatorname{ch} x \operatorname{sh} y = e^{x+y} - e^{x-y} + e^{-x+y} - e^{-x-y}
\end{aligned}
\]
From this, the first equality easily follows. The second equality is proved similar... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,858 |
27.33. Prove that
$\operatorname{Arsh} x=\ln \left(x+\sqrt{x^{2}+1}\right)$
$\operatorname{Arch} x=\ln \left(x \pm \sqrt{x^{2}-1}\right)= \pm \ln \left(x+\sqrt{x^{2}-1}\right)$;
$\operatorname{Arth} x=\frac{1}{2} \ln \frac{1+x}{1-x}$.
The hyperbolic amplitude of a number $x$ is the angle $\alpha(-\pi / 2<\alpha<\pi... | 27.33. Let $y=\operatorname{Arsh} x$. Then $x=\operatorname{sh} y=\frac{e^{y}-e^{-y}}{2}$. Therefore, $e^{2 y}-2 x e^{y}-1=0$. Solving this quadratic equation with respect to $e^{y}$, we get $e^{y}=x \pm \sqrt{x^{2}+1}$. But $e^{y} \geqslant 0$, so $e^{y}=x+\sqrt{x^{2}+1}$,
i.e., $y=\ln \left(x+\sqrt{x^{2}+1}\right)$.... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,859 |
27.34. Prove the following properties of the hyperbolic amplitude:
a) $\operatorname{ch} x=1 / \cos \alpha$
b) $\operatorname{th}(x / 2)=\operatorname{tg}(\alpha / 2)$. | 27.34. a) It is clear that $1=\operatorname{ch}^{2} x-\operatorname{sh}^{2} x=\operatorname{ch}^{2} x-\operatorname{tg}^{2} \alpha$. Therefore, $\operatorname{ch}^{2} x=1+\operatorname{tg}^{2} \alpha=1 / \cos ^{2} \alpha$. Moreover, $\operatorname{ch} x$ and $\cos \alpha$ are positive.
b) It is clear that $e^{x}=\oper... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,860 |
28.1. Let the function $f(x)$ be differentiable at the point $x_{0}$. The line passing through the points $\left(x_{0}, y_{0}\right)$ and $\left(x_{1}, y_{1}\right)$, where $y_{0}=$ $=f\left(x_{0}\right)$ and $y_{1}=f\left(x_{1}\right)$, is given by the equation $y-y_{0}=k\left(x_{1}\right)\left(x-x_{0}\right)$. Prove ... | 28.1. The considered straight line is given by the equation $y-y_{0}=$ $=\frac{f\left(x_{1}\right)-f\left(x_{0}\right)}{x_{1}-x_{0}}\left(x-x_{0}\right)$, therefore $k\left(x_{1}\right)=\frac{f\left(x_{1}\right)-f\left(x_{0}\right)}{x_{1}-x_{0}}$. It is directly evident from the definition of the derivative that $\lim ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,861 |
28.2. Prove that if the function $f(x)$ is differentiable at the point $x_{0}$, then it is continuous at this point. | 28.2. Let's write the identity
$$
f(x)-f\left(x_{0}\right)=\frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}\left(x-x_{0}\right)
$$
In this case, $\lim _{x \rightarrow x_{0}} \frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}=f^{\prime}\left(x_{0}\right)$ and $\lim _{x \rightarrow x_{0}}\left(x-x_{0}\right)=0$. Therefore, $\lim _{x \ri... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,862 |
28.3. Prove that if the functions $f(x)$ and $g(x)$ are differentiable at the point $x_{0}$, then:
a) $(f+g)^{\prime}\left(x_{0}\right)=f^{\prime}\left(x_{0}\right)+g^{\prime}\left(x_{0}\right)$
b) $(f g)^{\prime}\left(x_{0}\right)=f^{\prime}\left(x_{0}\right) g\left(x_{0}\right)+f\left(x_{0}\right) g^{\prime}\left(x... | 28.3. a) It directly follows from the properties of the limit of the sum of two functions.
b) Let $h(x)=f(x) g(x)$. Then
$$
h(x)-h\left(x_{0}\right)=f(x)\left(g(x)-g\left(x_{0}\right)\right)+g\left(x_{0}\right)\left(f(x)-f\left(x_{0}\right)\right) .
$$
Divide both sides of this equation by $x-x_{0}$ and note that $f... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,863 |
28.4. Let the function $f$ be differentiable at the point $x_{0}$, and the function $g$ be differentiable at the point $y_{0}=f\left(x_{0}\right)$. Suppose that the point $x_{0}$ has a neighborhood $U\left(x_{0}\right)$ such that if $x$ belongs to $U\left(x_{0}\right)$ and $x \neq x_{0}$, then $f(x) \neq f\left(x_{0}\r... | 28.4. Identity
$$
\frac{g \circ f(x)-g \circ f\left(x_{0}\right)}{x-x_{0}}=\frac{g \circ f(x)-g \circ f\left(x_{0}\right)}{f(x)-f\left(x_{0}\right)} \cdot \frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}
$$
shows that the function $g \circ f$ is differentiable at the point $x_{0}$ and $(g \circ f)^{\prime}\left(x_{0}\right)=... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,864 |
28.6. Prove that $\left(x^{n}\right)^{\prime}=n x^{n-1}$ for any natural $n$.
| 28.6. It is clear that
$$
\frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}=\frac{x^{n}-x_{0}^{n}}{x-x_{0}}=x^{n-1}+x^{n-2} x_{0}+\ldots+x_{0}^{n-1}
$$
As $x \rightarrow x_{0}$, each of the $n$ terms tends to $x_{0}^{n-1}$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,866 |
28.7. Prove that $\left(x^{a}\right)^{\prime}=a x^{a-1}$ for any real $a$ and positive $x$. | 28.7. It is easy to check that
$$
\frac{x^{a}-x_{0}^{a}}{x-x_{0}}=x_{0}^{a} \frac{\left(x^{a} / x_{0}^{a}\right)-1}{x-x_{0}}=x_{0}^{a-1} \frac{(1+y)^{a}-1}{y}
$$
where $y=\frac{x-x_{0}}{x_{0}}$. Clearly, $y \rightarrow 0$ as $x \rightarrow x_{0}$. Moreover, $\lim _{y \rightarrow 0} \frac{(1+y)^{a}-1}{y}=$ $=a$ (probl... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,867 |
28.8. Prove that $(\sin x)^{\prime}=\cos x$ and $(\cos x)^{\prime}=-\sin x$. | 28.8. It is clear that
$$
\begin{aligned}
& \sin x - \sin x_{0} = 2 \sin \frac{x - x_{0}}{2} \cos \frac{x + x_{0}}{2} \\
& \cos x - \cos x_{0} = -2 \sin \frac{x - x_{0}}{2} \sin \frac{x + x_{0}}{2}
\end{aligned}
$$
It remains to note that $\frac{1}{x - x_{0}} \sin \frac{x - x_{0}}{2} \rightarrow \frac{1}{2}$ as $x \r... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,868 |
28.9. Prove that $(\operatorname{tg} x)^{\prime}=\frac{1}{\cos ^{2} x}$ for $x \neq \frac{(2 k+1) \pi}{2}$ and $(\operatorname{ctg} x)^{\prime}=-\frac{1}{\sin ^{2} x}$ for $x \neq k \pi$. | 28.9. According to problem 28.3 b)
$$
\left(\frac{\sin x}{\cos x}\right)^{\prime}=\frac{(\sin x)^{\prime} \cos x-\sin x(\cos x)^{\prime}}{\cos ^{2} x}=\frac{\cos ^{2} x+\sin ^{2} x}{\cos ^{2} x}=1
$$
For $(\operatorname{ctg} x)^{\prime}$ the calculations are similar. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,869 |
28.10. Prove that $\left(a^{x}\right)^{\prime}=a^{x} \ln a$ for $a>0$. | 28.10. It is clear that $a^{x}-a^{x_{0}}=a^{x_{0}}\left(a^{x-x_{0}}-1\right)$. It remains to note that $\lim _{x \rightarrow x_{0}} \frac{a^{x-x_{0}}-1}{x-x_{0}}=\ln a$ according to problem 27.30. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,870 |
28.11. Prove that $\left(\log _{a} x\right)^{\prime}=\frac{1}{x \ln a}$. | 28.11. Since $\log _{a} x=\frac{\ln x}{\ln a}$, it is sufficient to check that $(\ln x)^{\prime}=$ $=1 / x$. The equality $\ln x-\ln x_{0}=\ln \frac{x}{x_{0}}=\ln \left(1+\frac{x-x_{0}}{x_{0}}\right)$ shows that $\frac{\ln x-\ln x_{0}}{x-x_{0}}=\frac{1}{x_{0}} \frac{\ln (1+y)}{y}$, where $y=\frac{x-x_{0}}{x_{0}}$. It r... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,871 |
28.12. Prove that $(\arcsin x)^{\prime}=\frac{1}{\sqrt{1-x^{2}}}$ and $(\operatorname{arctg} x)^{\prime}=$ $=\frac{1}{1+x^{2}}$. | 28.12. If $x=\sin y$, then $\arcsin x=y$. Therefore, $\arcsin$ is the inverse function of $\sin$. Thus, according to problem 28.5,
$$
(\arcsin x)^{\prime}=\frac{1}{(\sin y)^{\prime}}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-x^{2}}}
$$
If $x=\operatorname{tg} y$, then $\operatorname{arctg} x=y$, therefore
$$
(\operatorname{... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,872 |
28.13. Let the functions $u(x)$ and $v(x)$ be differentiable, and the function $u(x)$ be positive. Prove that the function $u^{v}=u(x)^{v(x)}$ is differentiable and find its derivative. | 28.13. It is clear that if $y=u^{v}$, then $\ln y=v \ln u$. Differentiating this equality, we get $\frac{y^{\prime}}{y}=v^{\prime} \ln u+v \frac{u^{\prime}}{u}$. Therefore, $y^{\prime}=y\left(v^{\prime} \ln u+\frac{v u^{\prime}}{u}\right)$. | y^{\}=y(v^{\}\lnu+\frac{vu^{\}}{u}) | Calculus | proof | Yes | Yes | olympiads | false | 38,873 |
28.14. Calculate the derivative of the function $f(x)=x^{\sin x}$ (for $x>0$). | 28.14. According to problem $28.13 f^{\prime}(x)=x^{\sin x}\left(\frac{\sin x}{x}+\cos x \cdot \ln x\right)$. | f^{\}(x)=x^{\sinx}(\frac{\sinx}{x}+\cosx\cdot\lnx) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,874 |
28.15. Compute the derivative of the function $f(x)=\cos ^{x} a - \sin ^{x} a$, where $0<a<\pi / 2$ is a constant angle.
## 28.3. Multiple root of a polynomial | 28.15. According to problem 28.13
$f^{\prime}(x)=\cos ^{x} a \cdot \ln \cos a-\sin ^{x} a \cdot \ln \sin a=\cos ^{x} a\left(\ln \cos a-\operatorname{tg}^{x} a \cdot \ln \sin a\right)$. | f^{\}(x)=\cos^{x}\cdot\ln\cos-\sin^{x}\cdot\ln\sin | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,875 |
28.16. Prove that a polynomial $f(x)$ of degree $n \geqslant 2$ has a multiple root if and only if $f(x)$ and $f^{\prime}(x)$ have a common root. | 28.16. Suppose that $f(x)=\left(x-x_{0}\right)^{m} g(x)$, where $m \geqslant 2$. Then the polynomial $f^{\prime}(x)=m\left(x-x_{0}\right)^{m-1} g(x)+\left(x-x_{0}\right)^{m} g^{\prime}(x)$ has the root $x_{0}$.
Suppose that $f(x)=\left(x-x_{0}\right) g(x)$, and $g\left(x_{0}\right) \neq 0$. Then $f^{\prime}(x)=g(x)+\l... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,876 |
28.17. Prove that if
$$
\begin{gathered}
x_{0}^{4}+a_{1} x_{0}^{3}+a_{2} x_{0}^{2}+a_{3} x_{0}+a_{4}=0 \\
4 x_{0}^{3}+3 a_{1} x_{0}^{2}+2 a_{2} x_{0}+a_{3}=0
\end{gathered}
$$
then $x^{4}+a_{1} x^{3}+a_{2} x^{2}+a_{3} x+a_{4}$ is divisible by $\left(x-x_{0}\right)^{2}$. | 28.17. Let $f(x)=x^{4}+a_{1} x^{3}+a_{2} x^{2}+a_{3} x+a_{4}$. By the condition $f\left(x_{0}\right)=f^{\prime}\left(x_{0}\right)=0$. Therefore, $x_{0}$ is a double root of the polynomial $f(x)$, i.e., the polynomial $f(x)$ is divisible by $\left(x-x_{0}\right)^{2}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,877 |
28.18. Prove that the polynomial
$$
f_{n}(x)=1+x+\frac{x^{2}}{2!}+\ldots+\frac{x^{n}}{n!}
$$
does not have multiple roots.
See also problems $23.22,24.7$.
## 28.4. Derivative of a polynomial | 28.18. Suppose that the polynomial $f_{n}(x)$ has a multiple root. Then the polynomials $f_{n}(x)$ and $f_{n}^{\prime}(x)=f_{n-1}(x)$ have a common root $x_{0}$. Therefore, $f_{n}\left(x_{0}\right)-f_{n-1}\left(x_{0}\right)=0$. But $f_{n}(x)-f_{n-1}(x)=\frac{x^{n}}{n!}$, so $x_{0}=0$. This leads to a contradiction, sin... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,878 |
28.19. Let $P(x)$ be a polynomial with integer coefficients. Prove that all coefficients of its $n$-th derivative $P^{(n)}(x)$ are divisible by $n!$ for any natural number $n$. | 28.19. It is sufficient to prove the required statement for the monomial $x^{m}$. It is clear that if $n \leqslant m$, then
$\left(x^{m}\right)^{(n)}=m(m-1) \ldots(m-n+1) x^{m-n}=n!\frac{m!}{n!(m-n)!} x^{m-n}=n!C_{m}^{n} x^{m-n}$.
The number $n!C_{m}^{n}$ obviously divides $n!$, since the number $C_{m}^{n}$ is an int... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,879 |
28.20. Prove that the arithmetic mean of the roots of a polynomial is equal to the arithmetic mean of the roots of its derivative. | 28.20. Let $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}$. Then, by Vieta's theorem, the sum of the roots of the polynomial $f$ is $-\frac{a_{n-1}}{a_{n}}$, so their arithmetic mean is $-\frac{a_{n-1}}{n a_{n}}$. The sum of the roots of the derivative $f^{\prime}(x)=n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_{1}$ ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,880 |
28.21. Let $f$ and $g$ be polynomials of degree $n$. Prove that $f g^{(n)}-f' g^{(n-1)}+f'' g^{(n-2)}-f^{(3)} g^{(n-3)}+\ldots+(-1)^{n} f^{(n)} g-$ is a constant. | 28.21. The derivative of the given expression is $f g^{(n+1)}+f' g^{(n)}-$ $-f' g^{(n)}-f'' g^{(n-2)}+f'' g^{(n-2)}+\ldots+(-1)^{n} f^{(n)} g'+(-1)^{n} f^{(n+1)} g$. All intermediate terms cancel each other out, leaving only $f g^{(n+1)} \pm f^{(n+1)} g$. But $f$ and $g$ are polynomials of degree $n$, so their $(n+1)$-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,881 |
28.22. Let $p$ and $q$ be real numbers. Determine how many real roots the cubic equation $x^{3} + p x + q = 0$ has depending on the signs of the numbers $p$ and $D = \frac{q^{2}}{4} + \frac{p^{3}}{27}$. | 28.22. Let $f(x)=x^{3}+p x+q$. Then $f^{\prime}(x)=3 x^{2}+p$. If $p \geqslant 0$, the function $f$ is monotonic and its graph intersects the $x$-axis at exactly one point. In this case, the equation has exactly one real root. Note that in this case $D \geqslant 0$.
Now suppose $p<0$. If $D>0$, the equation has one re... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,882 |
28.23. Let $f(x)=\left(x-x_{1}\right) \ldots\left(x-x_{n}\right)$, where the numbers $x_{1}, \ldots, x_{n}$ are pairwise distinct and non-zero. Prove that
$$
\sum_{i=1}^{n} \frac{x_{i}^{k}}{f^{\prime}\left(x_{i}\right)}= \begin{cases}0 & \text { for } 0 \leqslant k \leqslant n-2 \\ 1 & \text { for } k=n-1\end{cases}
$... | 28.23. The polynomial $\sum_{i=1}^{n} \frac{g\left(x_{i}\right) f(x)}{f^{\prime}\left(x_{i}\right)\left(x-x_{i}\right)}$ takes the values $g\left(x_{1}\right), \ldots, g\left(x_{n}\right)$ at the points $x_{1}, \ldots, x_{n}$. Therefore, if $g(x)$ is a polynomial of degree not higher than $n-1$, then $g(x)=\sum_{i=1}^{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,883 |
28.24. Let $P(x)=\left(x-x_{1}\right) \ldots\left(x-x_{n}\right)$, where $x_{1}, \ldots, x_{n}$ are real numbers. Prove that $\left(P^{\prime}(x)\right)^{2} \geqslant P(x) P^{\prime \prime}(x)$ for all real $x$. | 28.24. If $x=x_{i}$, then the inequality is obvious. Therefore, we will assume that $x$ is not a root of the polynomial $P$. Then
$$
\frac{P^{\prime}(x)}{P(x)}=\frac{1}{x-x_{1}}+\ldots+\frac{1}{x-x_{n}}
$$
Differentiating both sides of this equation, we get
$$
\frac{P^{\prime \prime}(x) P(x)-\left(P^{\prime}(x)\righ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,884 |
28.25. The four roots of a fourth-degree polynomial form an arithmetic progression. Prove that the roots of its derivative also form an arithmetic progression. | 28.25. After replacing $x$ with $x+c$, we can consider that the polynomial of degree 4 has roots $\pm a / 2, \pm 3 a / 2$. The derivative of the polynomial $\left(x^{2}-\frac{a^{2}}{4}\right)\left(x^{2}-\frac{9 a^{2}}{4}\right)$ is $4 x^{3}-5 a x$. Therefore, the roots of the derivative are $\pm \frac{\sqrt{5}}{2} a$ a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,885 |
28.26. Prove that any polynomial can be represented as the difference of two monotonically increasing polynomials.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 28.26. Let $f$ be a given polynomial. Consider polynomials $F$ and $G$ that have the following properties: $F'=\frac{\left(f'\right)^{2}+f'+1}{2}$, $G'=\frac{\left(f'\right)^{2}-f'+1}{2}$, $F(0)=f(0)$, and $G(0)=0$. Then $f=F-G$, and $F'>0$ and $G'>0$ (i.e., $F$ and $G$ are monotonically increasing), since $\left(f'\ri... | Number Theory | proof | Yes | Yes | olympiads | false | 38,886 | |
28.27. Prove that the polynomial $P(x)=a_{0}+a_{1} x^{k_{1}}+a_{2} x^{k_{2}}+$ $+\ldots+a_{n} x^{k_{n}}$ has no more than $n$ positive roots. | 28.27. We apply induction on $n$. For $n=1$, we get the polynomial $a_{0}+a_{1} x^{k}$, which has no more than one positive root. Suppose the polynomial $P(x)$ has more than $n$ positive roots. Between any two positive roots of the polynomial, there is at least one positive root of its derivative. Therefore, the polyno... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,887 |
28.28. The numbers $a_{1}, \ldots, a_{n}$ are pairwise distinct, and the number $a$ is positive. Prove that the sum of the lengths of the intervals on which the inequality $\sum_{k=1}^{n} \frac{1}{x-a_{k}}>a$ holds is $\frac{n}{a}$. | 28.28. We can assume that $a_{1}<\ldots<a_{n}$. Consider the polynomial $P(x)=\left(x-a_{1}\right) \ldots\left(x-a_{n}\right)$. It is easy to verify that $\frac{P^{\prime}(x)}{P(x)}=\sum_{k=1}^{n} \frac{1}{x-a_{k}}$. The function $f(x)=\frac{P^{\prime}(x)}{P(x)}$ is monotonically decreasing on each of the intervals whe... | \frac{n}{} | Algebra | proof | Yes | Yes | olympiads | false | 38,888 |
28.29. Prove that if all roots of the polynomial
$$
P(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n}
$$
with real coefficients are real and distinct, then
$$
a_{i}^{2}>\frac{n-i+1}{n-i} \cdot \frac{i+1}{i} a_{i-1} a_{i+1}, \quad i=1,2, \ldots, n-1
$$
(Newton).
See also problem 10.34.
## 28.5. Identities | 28.29. Let $Q(y)=y^{n} P\left(y^{-1}\right)$. The roots of the polynomial $Q(y)$ are also real and distinct, so the roots of the quadratic trinomial
$Q^{(n-2)}(y)=(n-2)(n-3) \cdot \ldots \cdot 4 \cdot 3\left[n(n-1) a_{n} y^{2}+2(n-1) a_{n-1} y+2 a_{n-2}\right]$ are real and distinct. Therefore,
$$
(n-1)^{2} a_{n-1}^{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,889 |
28.30. Let $(x+a)^{n}=A_{0}+A_{1} x+\ldots+A_{n} x^{n}$. Find the coefficients $A_{0}, \ldots, A_{n}$ by successive differentiation. | 28.30. For $x=0$ we get $A_{0}=a^{n}$. Differentiating the original equality, we obtain
$$
n(x+a)^{n-1}=A_{1}+2 A_{2} x+\ldots+n A_{n} x^{n-1}
$$
For $x=0$ we get $A_{1}=n a^{n-1}$. Differentiating the new equality, we obtain
$$
n(n-1)(x+a)^{n-2}=2 \cdot 1 \cdot A_{2}+3 \cdot 2 \cdot A_{3} x+\ldots+n(n-1) A_{n} x^{n... | n(n-1)\ldots(n-+1)^{n-}=(-1)\cdot\ldots\cdot1\cdotA_{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,890 |
28.31. Compute the sum $1+2 x+3 x^{2}+\ldots+n x^{n-1}$ by differentiating the equality $1+x+x^{2}+\ldots+x^{n}=\frac{1-x^{n+1}}{1-x}$.
See also problems $14.16,30.14$.
## 28.6. Tangent and Normal | 28.31. Answer: $\left(\frac{1-x^{n+1}}{1-x}\right)^{\prime}=\frac{1-(n+1) x^{n}+n x^{n+1}}{(1-x)^{2}}$.
Remark. Another calculation of the required sum is given in the solution to problem 9.6. | \frac{1-(n+1)x^{n}+nx^{n+1}}{(1-x)^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,891 |
28.32. The tangent to the curve $y=e^{x}$ at the point $\left(x_{0}, y_{0}\right)$ intersects the $O x$ axis at the point $\left(x_{1}, 0\right)$. Prove that the difference $x_{1}-x_{0}$ is the same for all points on the curve. | 28.32. The tangent to the curve $y=f(x)$ at the point $\left(x_{0}, y_{0}\right)$ is given by the equation $y-y_{0}=f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)$. Therefore, for $x_{1}$ we obtain the equation $-y_{0}=e^{x_{0}}\left(x_{1}-x_{0}\right)$. Considering that $y_{0}=e^{x_{0}}$, we get $x_{0}-x_{1}=1$. | x_{0}-x_{1}=1 | Calculus | proof | Yes | Yes | olympiads | false | 38,892 |
28.33. On a parabola whose axis is parallel to the $O y$ axis, points $A_{1}, A_{2}$, and $A_{3}$ are taken. Let $k_{1}$ be the tangent of the angle of inclination of the tangent at point $A_{1}$, and $k_{i j}$ be the tangent of the angle of inclination of the secant $A_{i} A_{j}$. Prove that $k_{1}=k_{12}+k_{13}-k_{23... | 28.33. The numbers $k_{1}$ and $k_{ij}$ do not change under a parallel translation of the coordinate axes, so we can assume that we are dealing with the parabola $y = a x^{2}$. Let $A_{i} = \left(x_{i}, y_{i}\right)$. Then $k_{1} = 2 a x_{1}$ and
$$
k_{ij} = \frac{y_{i} - y_{j}}{x_{i} - x_{j}} = a \frac{x_{i}^{2} - x_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,893 |
28.34. Prove that the normal to the curve $y=f(x)$ at the point $\left(x_{0}, y_{0}\right)$ is given by the equation
$$
-f^{\prime}\left(x_{0}\right)\left(y-y_{0}\right)=x-x_{0}
$$ | 28.34. If a straight line is given by the equation $\frac{y-y_{0}}{x-x_{0}}=k$, then a line perpendicular to it, passing through the point ( $x_{0}, y_{0}$ ), is given by the equation $\frac{y-y_{0}}{x-x_{0}}=-\frac{1}{k}$, since $\operatorname{tg}\left(\varphi+90^{\circ}\right)=-\frac{1}{\operatorname{tg}(\varphi)}$. ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,894 |
28.35. The normal to the parabola $y=x^{2}$ at the point ( $x_{0}, y_{0}$ ) intersects the $O y$ axis at the point $\left(0, y_{1}\right)$. Prove that the difference $y_{1}-y_{0}$ is constant for all points on the parabola.
See also problem 8.17.
## 28.7. Functions differentiable on an interval | 28.35. The normal to the parabola $y=x^{2}$ at the point $\left(x_{0}, y_{0}\right)$ is given by the equation $\frac{y-y_{0}}{x-x_{0}}=-\frac{1}{2 x_{0}}$. To find $y_{1}$, we set $x=0$. As a result, we get $\frac{y_{1}-y_{0}}{-x_{0}}=-\frac{1}{2 x_{0}}$, i.e., $y_{1}-y_{0}=1 / 2$. | y_{1}-y_{0}=\frac{1}{2} | Calculus | proof | Yes | Yes | olympiads | false | 38,895 |
28.36. The function $f(x)$ is defined on the interval $[a, b]$ and at some interior point $x_{0}$ of this interval it attains a maximum or minimum value. Prove that if the derivative exists at the point $x_{0}$, then $f^{\prime}\left(x_{0}\right)=0$ (Fermat). | 28.36. Let for definiteness $f\left(x_{0}\right) \leqslant f(x)$ for all $x$ from the interval $[a, b]$. Consider the one-sided limits
$$
\lim _{x \rightarrow x_{0}^{+}} \frac{f(x)-f\left(x_{0}\right)}{x-x_{0}} \text { and } \lim _{x \rightarrow x_{0}-} \frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}
$$
In both limits, the ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,896 |
28.37. The function $f(x)$ is differentiable on the interval $[a, b]$, and $f(a)=f(b)$. Prove that there exists an interior point $x_{0}$ of this interval such that $f^{\prime}\left(x_{0}\right)=0$ (Rolle). | 28.37. The function $f(x)$ is continuous on the interval $[a, b]$, so by the Weierstrass theorem (problem 26.18) among its values there is a maximum $M$ and a minimum $m$. If $M=m$, then the function $f(x)$ is constant, so any interior point of the interval can be taken as $x_{0}$. If $M>m$, then one of these two value... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,897 |
28.38. The function $f(x)$ is differentiable on the interval $[a, b]$. Prove that there exists an interior point $x_{0}$ of this interval for which
$$
f^{\prime}\left(x_{0}\right)=\frac{f(a)-f(b)}{a-b} \quad(\text{Lagrange})
$$
The Lagrange theorem, written in the form
$$
f(b)-f(a)=f^{\prime}\left(x_{0}\right)(b-a)
... | 28.38. Consider the auxiliary function $F(x)=f(x)-$ $-\frac{f(a)-f(b)}{a-b}(x-a)$. This function is differentiable on the interval $[a, b]$ and $F(a)=F(b)=f(a)$. Therefore, we can apply Rolle's theorem (problem 28.37) to the function $F(x)$. As a result, we obtain that there exists an interior point $x_{0}$ of the inte... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,898 |
28.39. The function $f(x)$ is differentiable on the interval $[a, b]$, and $f^{\prime}(x)=0$ for all points $x$ in the interval $[a, b]$. Prove that the function $f(x)$ is constant on the interval $[a, b]$. | 28.39. Let $a < x \leqslant b$. Apply Lagrange's theorem (problem 28.38) to the function $f$ on the interval $[a, x]$. As a result, we get that $f(x) - f(a) = f'(x_0)(x - a)$ for some point $x_0$ in the interval $[a, x]$. By the condition $f'(x_0) = 0$, therefore $f(x) = f(a)$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,899 |
28.40. The function $f(x)$ is differentiable on the interval $[a, b]$.
a) Prove that this function is non-decreasing (on this interval) if and only if $f^{\prime}(x) \geqslant 0$ for any point $x$ in the interval $(a, b)$.
b) Prove that if $f^{\prime}(x) \geqslant 0$ for any point $x$ in the interval $(a, b)$ and the... | 28.40. a) It is evident from the definition of the derivative that \( f'(x) \geqslant 0 \) for a non-decreasing function \( f(x) \).
Suppose now that \( f'(x) \geqslant 0 \) for any point \( x \) in the interval \((a, b)\). Take two points \( x \) and \( y \) on the segment \([a, b]\) such that \( x < y \). Apply Lagr... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,900 |
28.42. Prove that if $x>0$, then $\cos x>1-\frac{x^{2}}{2}$ and $\sin x>$
$x-\frac{x^{3}}{6}$. | 28.42. Let $f(x)=\cos x$ and $g(x)=1-\frac{x^{2}}{2}$. Then $f(0)=g(0)$ and $f^{\prime}(x)=-\sin x>-x=g^{\prime}(x)$ for $x>0$, since $\sin x < x$ for $x > 0$. Therefore, $f(x) > g(x)$ for $x > 0$ according to problem 28.41.
Now let $f(x)=\sin x$ and $g(x)=x-\frac{x^{3}}{6}$. Then $f^{\prime}(x)=\cos x$ and $g^{\prime... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,902 |
28.43. Prove that if $0<x<\frac{\pi}{2}$, then $\sin x < x - \frac{1}{3} x^{3}$.
Note. The proofs of the inequalities in problems 28.42 and 28.43 are based on the fact that an inequality for derivatives implies an inequality for functions. In other words, an inequality for functions implies an inequality for antideriv... | 28.43. It is sufficient to prove that $(\operatorname{tg} x)^{\prime}=\frac{1}{\cos ^{2} x}>1+x^{2}$, i.e., $\frac{1}{\cos ^{2} x}-1=\operatorname{tg}^{2} x>x^{2}$. The inequality $\operatorname{tg} x>x$ is proven in the solution to problem 11.1. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,903 |
28.44. a) Let $0<\alpha<1$ and $x \geqslant 0$. Prove that $x^{\alpha}-\alpha x \leqslant 1-\alpha$.
b) Let $a, b, p$ and $q$ be positive numbers, with $\frac{1}{p}+\frac{1}{q}=1$. Prove that $a b \leqslant \frac{1}{p} a^{p}+\frac{1}{q} b^{q}$. | 28.44. a) Let $f(x)=x^{\alpha}-\alpha x$. Then $f^{\prime}(x)=\alpha\left(x^{\alpha-1}-1\right)$. Therefore, $f^{\prime}(x)>0$ for $0<x<1$ and $f^{\prime}(x)<0$ for $x>1$. Hence, $f(x)$ is increasing for $0<x<1$ and decreasing for $x>1$. Thus, for non-negative $x$, the maximum value of $f(x)$ is $f(1)=1-\alpha$.
b) Th... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,904 |
28.45. The functions $f(x)$ and $g(x)$ are differentiable on the interval $[a, b]$, and the derivative $g'(x)$ does not vanish at any interior point of this interval. Prove that there exists an interior point $x_0$ of the interval $[a, b]$ such that
$$
\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'\left(x_{0}\right)}{g'\left(x_... | 28.45. Consider the auxiliary function
$$
F(x)=(f(b)-f(a))(g(x)-g(a))-(f(x)-f(a))(g(b)-g(a))
$$
It is clear that
$$
F^{\prime}(x)=(f(b)-f(a)) g^{\prime}(x)-f^{\prime}(x)(g(b)-g(a))
$$
and $F(a)=F(b)=0$. Therefore, we can apply Rolle's theorem (problem 28.37) to the function $F(x)$. As a result, we obtain that there... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,905 |
28.51. a) Prove that $\ln x < n\left(x^{1 / n} - 1\right) < x^{1 / n} \ln x$ for any positive number $x \neq 1$.
b) Prove that $\lim _{n \rightarrow \infty} n\left(x^{1 / n} - 1\right) = \ln x$ for any positive number $x \neq 1$. | 28.51. a) Let $y=x^{1 / n}$. Then, after canceling by $n$, the required inequalities can be written as $\ln y<y-1<y \ln y$. Such inequalities have been proven in the solution to problem 28.50.
b) This follows directly from a), since $\lim _{n \rightarrow \infty} x^{1 / n}=1$ (problem 25.20). | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,911 |
28.52. Let $x>0$.
a) Prove that $\lim _{n \rightarrow \infty} n \ln (1+x / n)=x$.
b) Prove that $\lim _{n \rightarrow \infty}(1+x / n)^{n}=e^{x}$. | 28.52. a) According to problem 28.50
$$
\frac{x / n}{1+x / n}<\ln \left(1+\frac{x}{n}\right)<\frac{x}{n}
$$
therefore $\frac{x}{1+x / n}<n \ln \left(1+\frac{x}{n}\right)<x$.
b) It is clear that $(1+x / n)^{n}=e^{n \ln (1+x / n)}$. According to part a), the equality $\lim _{n \rightarrow \infty} n \ln (1+x / n)=x$ ho... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,912 |
28.53. Prove that $e^{x}>x^{e}$ for any positive $x \neq e$. | 28.53. The required inequality is equivalent to the inequality $x > e \ln x$, i.e., $\ln x / x < 1 / e$. Consider the function $f(x) = \ln x / x$. It is easy to verify that $f'(x) = (1 - \ln x) / x^2$. The derivative of the function $f$ is zero only at the point $x = e$. Moreover, $\lim _{x \rightarrow +0} (\ln x / x) ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,913 |
28.54. Let $a$ and $b$ be positive numbers. Prove that $b \cdot 2^{a} + a \cdot 2^{-b} \geqslant a + b$. | 28.54. Consider the function $f(x)=b \cdot x^{a}+a \cdot x^{-b}$, where $x \geqslant 1$. It is clear that $f^{\prime}(x)=a b\left(x^{a-1}-x^{-b-1}\right)>0$ for $x>1$, since $x^{a-1} / x^{-b-1}=$ $=x^{a+b}>1$. Therefore, the function $f$ is increasing and $f(2) \geqslant f(1)$, i.e., $b \cdot 2^{a}+a \cdot 2^{-b} \geqs... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,914 |
28.55. Let $a>b>0$. Prove that
$$
\sqrt{a b}<\frac{a-b}{\ln a-\ln b}<\frac{a+b}{2}
$$ | 28.55. Let $a=b x$, where $x>1$. The required inequality can be rewritten as
$$
\sqrt{x}1$, then $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{\prime}=\frac{1}{2} x^{-1}\left(x^{1 / 2}+x^{-1 / 2}\right)>x^{-1}=(\ln x)^{\prime}$ and $\left(2 \frac{x-1}{x+1}\right)^{\prime}=\frac{4}{(x+1)^{2}}g^{\prime}(x)$ for $x>1$, then... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,915 |
28.58. Prove that if $x>-1$ and $x \neq 0$, then
$$
\frac{2|x|}{2+x}<|\ln (1+x)|<\frac{|x|}{\sqrt{1+x}}
$$
See also problem 8.17.
## 28.9. L'Hôpital's Rule | 28.58. Consider the functions
\[
\begin{aligned}
& f(x)=\ln (1+x)-\frac{2 x}{2+x} \\
& g(x)=\ln (1+x)-\frac{x}{\sqrt{1+x}}
\end{aligned}
\]
It is easy to verify that \( f(0)=g(0)=0 \) and \( g'(0) \leqslant 0 \leqslant f'(0) \), and the inequalities are strict for \( x \neq 0 \). Indeed,
\[
\begin{aligned}
& f'(x)=\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,918 |
28.59. Let the functions $f(x)$ and $g(x)$ satisfy the conditions of Cauchy's theorem (problem 28.45) and, in addition, $f(a)=g(a)=$ =0. Prove that if $\lim _{x \rightarrow a+} \frac{f^{\prime}(x)}{g^{\prime}(x)}=A$, then $\lim _{x \rightarrow a+} \frac{f(x)}{g(x)}=A$ (L'Hôpital's rule). | 28.59. Fix a point $x$, where $a < x \leqslant b$, and apply Cauchy's theorem to the interval $[a, x]$. As a result, we obtain that there is a point $x_{1}$ inside this interval for which
$$
\frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f^{\prime}\left(x_{1}\right)}{g^{\prime}\left(x_{1}\right)}
$$
If $x \right... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,919 |
28.60. Calculate the limit $\lim _{x \rightarrow 0} \frac{x-\sin x}{x^{3}}$ using L'Hôpital's rule. | 28.60. Applying L'Hôpital's rule twice, we get
$$
\lim _{x \rightarrow 0} \frac{x-\sin x}{x^{3}}=\lim _{x \rightarrow 0} \frac{1-\cos x}{3 x^{2}}=\lim _{x \rightarrow 0} \frac{\sin x}{6 x}=\frac{1}{6}
$$ | \frac{1}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,920 |
28.61. Calculate the limit $\lim _{x \rightarrow a} \frac{a^{x}-x^{a}}{x-a}$. | 28.61. By L'Hôpital's rule
$$
\begin{aligned}
& \lim _{x \rightarrow a} \frac{a^{x}-x^{a}}{x-a}=\lim _{x \rightarrow a}\left(a^{x}-x^{a}\right)^{\prime}=\lim _{x \rightarrow a}\left(\ln a \cdot a^{x}-a x^{a-1}\right)= \\
&=\ln a \cdot a^{a}-a^{a}=a^{a}(\ln a-1)
\end{aligned}
$$ | ^{}(\ln-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,921 |
28.62. Calculate the limit $\lim _{x \rightarrow 0} \frac{\tan x - x}{x - \sin x}$.
### 28.10. The number of roots of the equation | 28.62. Answer: 2. Transform the ratio of derivatives:
$$
\frac{\frac{1}{\cos ^{2} x}-1}{1-\cos x}=\frac{1}{\cos ^{2} x} \cdot \frac{1-\cos x^{2}}{1-\cos x}=\frac{1+\cos x}{\cos ^{2} x}
$$
The last expression tends to 2 as $x \rightarrow 0$. | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,922 |
28.68. How many solutions does the equation $\left(\frac{1}{16}\right)^{x}=\log _{\frac{1}{16}} x$ have? | 28.68. Answer: three. The roots $x_{1}=1 / 2$ and $x_{2}=$ $=1 / 4$ are easily guessed. In addition, there is a root of the equation $x=\left(\frac{1}{16}\right)^{x}$. And this equation cannot have more than three roots according to problem 28.67.
Remark. If you try to solve this equation graphically, it may seem that... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,928 |
28.69. Prove that for even $n$ the polynomial
$$
f_{n}(x)=1+x+\frac{x^{2}}{2!}+\ldots+\frac{x^{n}}{n!}
$$
has no real roots, and for odd $n$ it has exactly one real root.
See also problem 24.4.
### 28.11. Periodic functions | 28.69. First, note that the polynomial $f_{n}(x)$ does not have multiple roots (Problem 28.18). Suppose that the required statement is proved for $n=0,1, \ldots, 2 k$. (We start with $k=0$: for the polynomial $f_{0}(x)=1$, the statement is obvious.) We will prove that the polynomial $f_{2 k+1}(x)$ has exactly one real ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,929 |
28.70. Prove that if $f(x)$ is a periodic function, then $f^{\prime}(x)$ is also a periodic function. | 28.70. Let $g(x)=f(x+T)$, where $T$ is a constant. Then $g^{\prime}(x)=(x+T)^{\prime} f^{\prime}(x+T)=f^{\prime}(x+T)$. Therefore, if $f(x+T)=f(x)$, then $g^{\prime}(x)=f^{\prime}(x)$ and $f^{\prime}(x)=f^{\prime}(x+T)$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,930 |
28.72. Let $x_{1}, \ldots, x_{n}$ be the roots of a polynomial, and $y_{1}, \ldots, y_{n-1}$ be the roots of its derivative. Prove that $\Sigma_{k}\left(x_{1}, \ldots, x_{n}\right)=$ $=\Sigma_{k}\left(y_{1}, \ldots, y_{n-1}\right)$ | 28.72. Let
$$
\begin{aligned}
& P(x)=\left(x-x_{1}\right) \ldots\left(x-x_{n}\right)= \\
& \quad=x^{n}-C_{n}^{1} \Sigma_{1}\left(x_{1}, \ldots, x_{n}\right) x^{n-1}+C_{n}^{2} \Sigma_{2}\left(x_{1}, \ldots, x_{n}\right) x^{n-2}-\ldots
\end{aligned}
$$
Then
$$
\begin{aligned}
\frac{1}{n} P^{\prime}(x)=x^{n-1}-\frac{n-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,932 |
28.73. Let $x_{1}, \ldots, x_{n}$ be positive numbers and $\Sigma_{k}=$ $=\Sigma_{k}\left(x_{1}, \ldots, x_{n}\right)$. Prove that
$$
\Sigma_{1} \geqslant \sqrt{\Sigma_{2}} \geqslant \sqrt[3]{\Sigma_{3}} \geqslant \ldots \geqslant \sqrt[n]{\Sigma_{n}}
$$
(Maclaurin). | 28.73. We apply induction on $n$. For $n=2$, we need to prove the obvious inequality $\frac{x_{1}+x_{2}}{2} \geqslant \sqrt{x_{1} x_{2}}$. Now let $n \geqslant 3$. It is sufficient to consider the case when $x_{1}<x_{2}<\ldots<x_{n}$. In this case, the derivative of the polynomial $\left(x-x_{1}\right) \ldots\left(x-x_... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,933 |
28.74. Let $x_{1}, \ldots, x_{n}$ be real numbers. Prove that for all $k=1,2, \ldots, n-1$ the inequalities
$$
\Sigma_{k}^{2}\left(x_{1}, \ldots, x_{n}\right) \geqslant \Sigma_{k-1}\left(x_{1}, \ldots, x_{n}\right) \Sigma_{k+1}\left(x_{1}, \ldots, x_{n}\right)
$$
(Newton).
### 28.13. Algebraic and Transcendental Fun... | 28.74. We apply induction on $n$. For $n=2$, we need to prove the obvious inequality $\left(x_{1}+x_{2}\right)^{2} \geqslant 4 x_{1} x_{2}$. Now let $n \geqslant 3$. It is sufficient to consider the case when $x_{1}<x_{2}<\ldots<x_{n}$. In this case, the derivative of the polynomial $\left(x-x_{1}\right) \ldots\left(x-... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,934 |
28.77. a) Let $a$ be a fixed number. Prove that any polynomial $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{n} x^{n}$ can be written in the form
$$
f(x)=A_{0}+A_{1}(x-a)+A_{2}(x-a)^{2}+\ldots+A_{n}(x-a)^{n}
$$
where $A_{0}, A_{1}, \ldots, A_{n}$ are constants.
b) Prove that $A_{0}=f(a), A_{1}=f^{\prime}(a), A_{2}=\frac... | 28.77. a) It is sufficient to note that
$$
x^{k}=(a+(x-a))^{k}=\sum_{i=0}^{k} C_{k}^{i} a^{i}(x-a)^{k-i}
$$
b) To find $A_{0}$, set $x=a$. As a result, we get $f(a)=A_{0}$. Differentiating the expression for $f(x)$, we get
$$
f^{\prime}(x)=A_{1}+2 A_{2}(x-a)+\ldots+n A_{n}(x-a)^{n-1}
$$
Setting $x=a$, we obtain $f^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,937 |
28.78. Let $a$ be a fixed number, and $f(x)$ be a function having derivatives up to order $n+1$ inclusive for any $x$ between $a$ and $b$ (for some $b$). Prove that if the number $x$ is between $a$ and $b$ and if
$T(x)=f(a)+\frac{f^{\prime}(a)}{1!}(x-a)+\frac{f^{\prime \prime}(a)}{2!}(x-a)^{2}+\ldots+\frac{f^{(n)}(a)}... | 28.78. It is clear that $T(a)=f(a)$. Further,
$$
\begin{aligned}
& T^{\prime}(x)=f^{\prime}(a)+\frac{f^{\prime \prime}(a)}{1!}(x-a)+\ldots+\frac{f^{(n)}(a)}{(n-1)!}(x-a)^{n-1} \\
& T^{\prime \prime}(x)=f^{\prime \prime}(a)+\frac{f^{\prime \prime \prime}(a)}{1!}(x-a)+\ldots+\frac{f^{(n)}(a)}{(n-2)!}(x-a)^{n-2}
\end{ali... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,938 |
29.1. Prove that $\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \ln \frac{x-a}{x+a}+C$. | 29.1. It is clear that $\left(\ln \frac{x-a}{x+a}\right)^{\prime}=(\ln (x-a)-\ln (x+a))^{\prime}=\frac{1}{x-a}-\frac{1}{x+a}=$ $=\frac{2 a}{x^{2}-a^{2}}$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,940 |
29.2. Let $F(x)$ be an antiderivative of the function $f(x)$ on the interval $[a, b]$, $\varphi(y)$ be a differentiable function on the interval $[p, q]$, such that $a \leqslant \varphi(y) \leqslant b$ for all $y$ from the interval $[p, q]$, and for any point $y_{0}$ from the interval $[p, q]$ there exists a neighborho... | 29.2. The formula for the derivative of a composition of functions (problem 28.4) shows that $(F(\varphi(y)))^{\prime}=F^{\prime}(\varphi(y)) \varphi^{\prime}(y)$. It remains to note that $F^{\prime}(\varphi(y))=f(\varphi(y))$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,941 |
29.3. Compute $\int \frac{d x}{x\left(1+\ln ^{2} x\right)} ; \int \frac{e^{x} d x}{1+e^{2 x}} ; \int \frac{\cos x d x}{1+\sin ^{2} x}$. | 29.3. According to problems 29.2 and 28.12
\[
\int \frac{d x}{x\left(1+\ln ^{2} x\right)}=\int \frac{d \ln x}{1+\ln ^{2} x}=\int \frac{d y}{1+y^{2}}=\operatorname{arctg} y+C=\operatorname{arctg} \ln x+C ;
\]
\[
\int \frac{e^{x} d x}{1+e^{2 x}}=\int \frac{d e^{x}}{1+e^{2 x}}=\int \frac{d y}{1+y^{2}}=\operatorname{arct... | \operatorname{arctg}\lnx+C,\operatorname{arctg}e^{x}+C,\operatorname{arctg}\sinx+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,942 |
29.4. Calculate $\int \operatorname{tg} x d x$. | 29.4. According to problem 29.2
$$
\int \operatorname{tg} x d x=\int-\frac{d \cos x}{\cos x}=-\ln \cos x+C
$$ | -\ln\cosx+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,943 |
29.5. Prove that $\int f(x) d x$ can be calculated as follows. Let $x=\varphi(y)$ be a differentiable function with an inverse function $y=\psi(x)$. Compute $\int f(\varphi(y)) \varphi^{\prime}(y) d y=F(y)$ and set $y=\psi(x)$. Then $F(\psi(x))$ is the desired integral. | 29.5. According to problem $29.2 \int f(\varphi(y)) \varphi^{\prime}(y) d y=\int f(\varphi(y)) d \varphi(y)=$ $=\int f(x) d x$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,944 |
29.6. Calculate $\int \sqrt{1-x^{2}} d x$. | 29.6. Let $x=\sin y$. Then, according to problem 29.5
$$
\begin{aligned}
\int \sqrt{1-x^{2}} d x & =\int \cos ^{2} y d y=\int \frac{1+\cos 2 y}{2} d y=\frac{y}{2}+\frac{1}{4} \int \cos 2 y d(2 y)= \\
& =\frac{y}{2}+\frac{\sin 2 y}{4}+C=\frac{\arcsin x}{2}+\frac{x \sqrt{1-x^{2}}}{2}+C
\end{aligned}
$$ | \frac{\arcsinx}{2}+\frac{x\sqrt{1-x^{2}}}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,945 |
29.7. Calculate $\int \sqrt{1+x^{2}} d x$. | 29.7. Let $x=\operatorname{sh} y$. Then, according to problem 29.5
$$
\begin{aligned}
\int \sqrt{1+x^{2}} d x & =\int \operatorname{ch}^{2} y d y=\int \frac{1+\operatorname{ch} 2 y}{2} d y=\frac{y}{2}+\frac{1}{4} \int \operatorname{ch} 2 y d(2 y)= \\
& =\frac{y}{2}+\frac{\operatorname{sh} 2 y}{4}+C=\frac{\operatorname... | \frac{1}{2}\ln(x+\sqrt{1+x^{2}})+\frac{x\sqrt{1+x^{2}}}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,946 |
29.8. Compute $\int \frac{d x}{1+\sqrt{x}}$. | 29.8. Let $x=y^{2}$. Then, according to problem 29.5
$$
\begin{aligned}
& \int \frac{d x}{1+\sqrt{x}}=\int \frac{2 y d y}{1+y}=2 \int\left(1-\frac{1}{1+y}\right) d y= \\
& \quad=2 y-2 \ln (1+y)+C=2 \sqrt{x}-2 \ln (1+\sqrt{x})+C
\end{aligned}
$$ | 2\sqrt{x}-2\ln(1+\sqrt{x})+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,947 |
29.9. Prove that
$$
\int f(x) g^{\prime}(x) d x=f(x) g(x)-\int f^{\prime}(x) g(x) d x
$$
(integration by parts formula). | 29.9. It is required to prove that
$$
f(x) g^{\prime}(x) d x=(f(x) g(x))^{\prime}-f^{\prime}(x) g(x)
$$
This immediately follows from the formula for the derivative of the product of two functions. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,948 |
29.10. Compute using integration by parts:
a) $\int x^{3} \ln x d x$; b) $\int \operatorname{arctg} x d x$; c) $\int x \cos x d x$; d) $\int x e^{x} d x$. | 29.10. а) $\int x^{3} \ln x d x=\int \ln x d\left(\frac{x^{4}}{4}\right)=\ln x \cdot \frac{x^{4}}{4}-\int \frac{x^{4}}{4} d(\ln x)=$ $=\ln x \cdot \frac{x^{4}}{4}-\int \frac{x^{4}}{4} \cdot \frac{1}{x} d x=\frac{x^{4}}{4} \ln x-\frac{x^{4}}{16}+C$.
б) $\int \operatorname{arctg} x d x=x \operatorname{arctg} x-\int \fra... | xe^{x}-e^{x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,949 |
29.11. Calculate $\int \frac{\ln x}{x} d x$. | 29.11. Apply the integration by parts formula for \( f(x) = g(x) = \ln x \). As a result, we get
\[
\int \ln x \frac{1}{x} d x = (\ln x)^{2} - \int \frac{1}{x} \ln x d x
\]
The antiderivative is defined up to a constant, so this equality should be understood as: \( I = (\ln x)^{2} - (I + C) \), where \( I = \int \fra... | \frac{(\lnx)^{2}}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,950 |
29.12. Calculate: a) $\int e^{a x} \sin b x d x$; b) $\int e^{a x} \cos b x d x$. | 29.12. a) Integrating by parts, we get
$$
\begin{aligned}
\int e^{a x} \sin b x d x & =-\frac{1}{b} e^{a x} \cos b x+\frac{a}{b} \int e^{a x} \cos b x d x= \\
& =-\frac{1}{b} e^{a x} \cos b x+\frac{a}{b^{2}} e^{a x} \sin b x-\frac{a^{2}}{b^{2}} \int e^{a x} \sin b x d x
\end{aligned}
$$
For \( I=\int e^{a x} \sin b x... | \inte^{}\sin=\frac{e^{}}{^{2}+b^{2}}(\sin-b\cos)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,951 |
29.13. Prove that
\[
\begin{gathered}
\int f g^{(n+1)} d x=f g^{(n)}-f^{\prime} g^{(n-1)}+f^{\prime \prime} g^{(n-2)}-\ldots \\
\ldots+(-1)^{n} f^{(n)} g+(-1)^{n+1} \int f^{(n+1)} g d x
\end{gathered}
\] | 29.13. Let's use the formula from problem 29.9, replacing $g$ with $g^{(n)}$:
$$
\int f g^{(n+1)} d x=f g^{(n)}-\int f^{\prime} g^{(n)} d x
$$
Similarly, we obtain
$$
\begin{aligned}
& \int f^{\prime} g^{(n)} d x=f^{\prime} g^{(n-1)}-\int f^{\prime \prime} g^{(n-1)} d x \\
& \int f^{\prime \prime} g^{(n-1)} d x=f^{\... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,952 |
29.14. Let $P(x)$ be a polynomial. Compute the integrals: a) $\int P(x) e^{a x} d x$; b) $\int P(x) \sin a x d x$; c) $\int P(x) \cos a x d x$.
## 29.2. Definite Integral
Let a function $f(x)$ be defined on the interval $[a, b]$. Divide this interval into parts by points $x_{0}=a<x_{1}<x_{2}<\ldots<x_{n-1}<$ $<b=x_{n... | 29.14. a) Let's use the formula from problem 29.13. If $g^{(n+1)}=e^{a x}$, then $g^{(n)}=\frac{e^{a x}}{a}, g^{(n-1)}=\frac{e^{a x}}{a^{2}}, \ldots$ Therefore,
$$
\int P(x) e^{a x} d x=e^{a x}\left(\frac{P}{a}-\frac{P^{\prime}}{a^{2}}+\frac{P^{\prime \prime}}{a^{3}}-\ldots\right)+C
$$
b) If $g^{(n+1)}=\sin a x$, the... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,953 |
29.17. Let $I$ be the exact upper bound of the lower integral sums, $\sigma$ be an arbitrary integral sum, and $s$ and $S$ be the lower and upper integral sums, respectively, corresponding to the same partition as $\sigma$. Prove that
$$
|\sigma-I| \leqslant S-s
$$ | 29.17. First, note that the exact upper bound of the lower integral sums exists, since according to problem 29.16, any lower integral sum \( s \) does not exceed some fixed integral sum \( S_{0} \). From the inequality \( s \leqslant S_{0} \), it follows that \( I \leqslant S_{0} \). This inequality holds for any upper... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,956 |
29.18. Prove that any continuous function $f(x)$ on the interval $[a, b]$ is integrable. | 29.18. According to problem 29.17, it is sufficient to prove that if the largest length of the subinterval of the partition tends to zero, then the difference $S-s$ tends to zero.
The function $f(x)$ is uniformly continuous on the interval $[a, b]$ (problem 26.26), so for any $\varepsilon>0$ we can choose $\delta>0$ s... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,957 |
29.19. Let $f(x)$ be a continuous function on the interval $[a, b]$. Prove that there exists a point $\xi$ in the interval $[a, b]$ such that $\int_{a}^{b} f(x) d x=f(\xi)(b-a)$. | 29.19. Let $M$ and $m$ be the maximum and minimum values of the function $f(x)$ on the interval $[a, b]$. Then each integral sum is bounded between $m(b-a)$ and $M(b-a)$. Therefore, $\int_{a}^{b} f(x) d x=c(b-a)$, where the number $c$ is between $m$ and $M$. A continuous function on an interval takes all values between... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,958 |
29.20. Prove that if a continuous function $f(x)$ is non-negative on the interval $[a, b]$, then $\int_{a}^{b} f(x) d x \geqslant 0$. Moreover, if $f\left(x_{0}\right)>0$ for some point $x_{0}$ in the interval $[a, b]$, then $\int_{a}^{b} f(x) d x>0$.
| It is often convenient to consider definite integrals $\int_{b}^{... | 29.20. If $f(x) \geqslant 0$ for all points of the segment $[a, b]$, then all integral sums are non-negative, so their limit is also non-negative.
Suppose now that $f\left(x_{0}\right)=c>0$. We can assume that $x_{0}$ is an interior point of the segment $[a, b]$. Choose $\delta>0$ such that $f(x)>c / 2$ when $\left|x-... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,959 |
29.21. Let $f(x)$ be a continuous function on the interval $[a, b]$ and $F(x)=\int_{a}^{x} f(t) d t$. Prove that $F^{\prime}(x)=f(x)$, i.e., $F(x)$ is an antiderivative of $f(x)$. | 29.21. Let points $x$ and $x+\Delta x$ lie on the interval $[a, b]$. Then $F(x+\Delta x)-F(x)=\int_{x}^{x+\Delta x} f(t) d t$. According to problem 29.19, this integral is equal to $f(\xi) \Delta x$, where $\xi$ is some point between $x$ and $x+\Delta x$. Therefore, $\lim _{\Delta x \rightarrow 0} \frac{F(x+\Delta x)-F... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,960 |
29.22. Let $f(x)$ be a continuous function on the interval $[a, b]$. Prove that
$$
\int_{a}^{b} f(x) d x=F(b)-F(a)
$$
where $F(x)$ is the antiderivative of the function $f(x)$ (the Newton-Leibniz formula).
| $\left.F(x)\right|_{a} ^{b}$.
## 29.3. Calculation of Integrals | 29.22. Consider the function $\Phi(x)=\int_{a}^{x} f(t) d t$. It is clear that $\Phi(b)=$ $=\int_{a}^{b} f(x) d x$. Moreover, according to problem $29.21 \Phi^{\prime}(x)=f(x)$. Therefore, $\Phi(x)$ is also an antiderivative, which means $\Phi(x)=F(x)+C$, where $C$ is some constant. Note that $\Phi(a)=0$. Therefore, $F... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,961 |
29.23. Let \( U_{n}=\int_{0}^{\pi / 2} \cos ^{n} x d x \).
a) Prove that if \( n \) is odd, then \( U_{n}=\frac{(n-1)!!}{n!!} \), and if \( n \) is even, then \( U_{n}=\frac{(n-1)!!}{n!!} \frac{\pi}{2} \), where \( n!! \) denotes the product of all natural numbers not exceeding \( n \) and having the same parity as \(... | 29.23. a) It is clear that $U_{0}=\frac{\pi}{2}$ and $U_{1}=\left.\sin x\right|_{0} ^{\pi / 2}=1$. Now let $n>1$. Then, integrating by parts, we get
$$
U_{n}=\int_{0}^{\pi / 2} \cos ^{n-1} x d \sin x=\left.\sin x \cos ^{n-1} x\right|_{0} ^{\pi / 2}-\int_{0}^{\pi / 2} \sin x d \cos ^{n-1} x
$$
The first term is zero, ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,962 |
29.24. Prove that $\int_{0}^{\pi / 2} \ln (\sin x) d x=-\frac{\pi}{2} \ln 2$. | 29.24. First solution. Let us recall that
$$
\sin \frac{\pi}{2 n} \sin \frac{2 \pi}{2 n} \ldots \sin \frac{(n-1) \pi}{2 n}=\frac{\sqrt{n}}{2^{n-1}}
$$
(problem 23.8 b). Therefore,
$$
\sum_{k=1}^{n-1} \frac{\pi}{2 n} \ln \sin \frac{k \pi}{2 n}=\frac{\pi}{2} \cdot \frac{-(n-1) \ln 2+\frac{1}{2} \ln n}{n} \rightarrow-\... | -\frac{\pi}{2}\ln2 | Calculus | proof | Yes | Yes | olympiads | false | 38,963 |
29.25. Let $f_{1}(x)=\int_{a}^{x} f(t) d x, f_{2}(x)=\int_{a}^{x} f_{1}(t) d x, \ldots, f_{n}(x)=$ $=\int_{a}^{x} f_{n-1}(t) d x$. Prove that
$$
f_{n}(x)=\frac{(-1)^{n-1}}{(n-1)!} \int_{a}^{x} f(t)(t-x)^{n-1} d t
$$
## 29.4. Calculation of Areas
Let the function $f(x)$ on the interval $[a, b]$ be non-negative. The t... | 29.25. According to problem 29.13
\[
\begin{aligned}
f_{n}(x)= & \int_{a}^{x} f_{n-1}(t) \cdot 1 d x= \\
= & \left(f_{n-1}(t)(t-x)-f_{n-2}(t) \frac{(t-x)^{2}}{2!}+f_{n-3}(t) \frac{(t-x)^{3}}{3!}-\ldots\right. \\
& \left.\ldots+(-1)^{n-2} f_{1}(t) \frac{(t-x)^{n-1}}{(n-1)!}\right)\left.\right|_{a} ^{x}-(-1)^{n-2} \int_... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,964 |
29.27. Calculate the area under the graph of the function $y=$ $=\sin x$ on the interval from 0 to $\pi$.
## 29.5. Calculation of Volumes
Let a body be located in space with rectangular coordinates $O x y z$, such that the projection of this body onto the $O x$ axis is the segment $[a, b]$. Suppose that a plane passi... | 29.27. The desired area is equal to $\int_{0}^{\pi} \sin x d x=-\left.\cos x\right|_{0} ^{\pi}=2$. | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,966 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.