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719. Find explicit formulas for simulating a continuous two-dimensional random variable $(X, Y)$, if the component $X$ is given by the probability density function $f_{1}(x) = x / 2$ in the interval $(0; 2)$; the component $Y$ is uniformly distributed in the interval $\left(x_{i}, x_{i}+3\right)$ with density $f_{2}(y)... | Solution. We will generate the component $X$ according to rule 2 (§ 3):
$$
\int_{0}^{x_{l}}(x / 2) \mathrm{d} x=r_{i}
$$
From this, we obtain an explicit formula for calculating possible values of $X$:
$$
x_{i}=2 \sqrt{r_{i}}
$$
We will find the conditional distribution function of the component $\gamma$, taking in... | x_{i}=2\sqrt{r_{i}},\quady_{i}=3r_{i}^{\}+x_{i} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,511 |
721. Find explicit formulas for the distribution of a continuous two-dimensional random variable ( $X, Y$ ), given by the probability density function $f(x, y)=6 y$ in the region bounded by the lines $y=0, y=x, x=1$. | Solution. Let's find the probability density of the component $\mathrm{X}$:
$$
f_{1}(x)=\int_{0}^{x} f(x, y) d y=6 \int_{0}^{x} y \mathrm{~d} y=3 x^{2}(0<x<1)
$$
Simulate $X$ according to rule 2 (3):
$$
3 \int_{0}^{x_{i}} x^{2} \mathrm{~d} x=r_{i}
$$
From this, we obtain the explicit formula for calculating possibl... | x_{i}=\sqrt[3]{r_{i}},\quady_{i}=x_{i}\sqrt{r_{i}^{\}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,512 |
724. The system consists of two blocks connected in series. The system fails if at least one block fails. The first block contains two elements: $\boldsymbol{A}, \boldsymbol{B}$ (they are connected in parallel) and fails when both elements fail simultaneously. The second block contains one element $C$ and fails when th... | The solution is as follows. a) We will select three random numbers from Table 9: 0.10, 0.09, and 0.73; according to the rule (if the random number is less than the probability of the event, the event occurs; if the random number is greater than or equal to the probability of the event, the event does not occur), we wil... | 0.022 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,513 |
727. The device consists of two nodes connected in series. The first node contains two elements: $A, B$, which are connected in parallel. The second node contains one element $C$. The time of failure-free operation of the elements is distributed according to the exponential law with parameters equal to 0.04, 0.05, and ... | Solution. a) Let's simulate the time (in hours) of failure-free operation of the elements using the formulas:
$$
\begin{aligned}
& t_{A}=-(1 / 0.04) \ln r_{1}=25\left(-\ln r_{1}\right) \\
& t_{B}=-(1 / 0.05) \ln r_{2}=20\left(-\ln r_{2}\right) \\
& t_{C}=-(1 / 0.10) \ln r_{3}=10\left(-\ln r_{3}\right)
\end{aligned}
$$... | P^{*}=0.36,\overline{i^{*}}=9 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,514 |
730. In a three-channel queuing system with rejections, requests arrive according to a Poisson process. The time between the arrivals of two consecutive requests is exponentially distributed with $f(\tau)=5 e^{-5 \tau}$. The service time for each request is 0.5 minutes.
Find the expected value $a$ of the number of req... | Solution. Let $T_{1}=0$ be the moment of the first request's arrival. The request will go to the first channel and be served by it. The moment of the end of the first request's service is $T_{1}+0.5=0+0.5=0.5$. We record one in the counter of served requests.
The moments of arrival of subsequent requests will be found... | 13.5 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,515 |
732. In a single-channel queuing system with rejections, a Poisson flow of requests arrives.
The time between the arrival of two consecutive requests is distributed according to the law \( f(\tau)=0.8 \mathrm{e}^{-0.8 \tau} \); the service time for requests is random and distributed according to the law \( f_{1}(t)=1.... | The solution is as follows. The time between the arrival of two consecutive requests is distributed according to the law $f(\tau)=0.8 \mathrm{e}^{-0.8 \tau}$, so the values $\tau_{i}$ will be generated using the formula
$$
\tau_{i}=-(1 / 0.8) \ln r_{i}=1.25\left(-\ln r_{i}\right)
$$
Random numbers $r_{i}$ are taken f... | 66 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,516 |
734. Output the formula
$$
I_{1}^{\hbar}=(b-a) \frac{\sum_{i=1}^{n} \varphi\left(x_{i}\right)}{n}
$$
where $x_{i}=a+(b-a) r_{i}$, for estimating the definite integral $l=\int_{a}^{b} \varphi(x) \mathrm{d} x$. | Solution. Let us introduce a random variable $X$, uniformly distributed over the integration interval $(a, b)$ with density $f(x)=1/(b-a)$. Then the mathematical expectation is
$$
M[\varphi(X)]=\int_{a}^{b} \varphi(x) f(x) \mathrm{d} x=\frac{1}{b-a} \int_{a}^{b} \varphi(x) \mathrm{d} x
$$
## Therefore
$$
\int_{a}^{b... | I_{i}^{*}=(b-)\cdot\frac{\sum_{i=1}^{n}\varphi(x_{i})}{n} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,517 |
741. Find the estimate $I_{1}^{*}$ of the definite integral $I=\int_{0}^{\frac{\pi}{2}} \cos x \mathrm{~d} x$ based on the data from ten trials. Take the random numbers $r_{i}$ from the first row of table 63. | Solution. We will generate 10 possible values of $X$ using the formula
$$
x_{i}=0+\frac{\pi}{2} r_{i}=1.571 r_{i}
$$
The results of the trials are presented in Table 64.
Table 64
| Trial Number $i$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :... | 1.01 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,519 |
742. As an approximate value of the definite integral $I=\int_{a}^{b} \varphi(x) \mathrm{d} x$, the estimate $I_{1}^{*}=(b-a) \frac{\Sigma \varphi\left(x_{i}\right)}{n}$ is taken. Prove that the variance $\sigma^{2}$ of the averaged function $(b-a) \varphi(X)$ is
$$
\sigma^{2}=(b-a) \int_{a}^{b} \varphi^{2}(x) \mathrm... | Solution. Given that $D(C X)=C^{2} D(X)$, we obtain
$$
\sigma^{2}=D[(b-a) \varphi(X)]=(b-a)^{2} D[\varphi(X)]
$$
Using formula (**) (Chapter VI, § 3)
$$
D[\varphi(X)]=\int_{a}^{b} \varphi^{2}(x) \mathrm{d} x-[M[\varphi(X)]]^{2}
$$
By formula (*) (Chapter VI, § 3)
$$
M[\varphi(X)]=\int_{a}^{b} \varphi(x) f(x) \math... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,520 |
743. Find the estimate $I^{*}$ of the integral $I=\int_{0}^{1} \mathrm{~d} x \int_{k}^{1}(x+y) \mathrm{d} y$. Conduct 10 trials. | Solution. The region of integration is bounded by the lines $g=x, y=1, x=0$ and, obviously, belongs to the unit square. The area of the region of integration (a right triangle) $S=(1 \cdot 1) / 2=0.5$.
We use the formula
$$
f^{*}=S \cdot \frac{\sum_{i=1}^{N} f\left(x_{i}, y_{i}\right)}{N}=0.5 \frac{\sum_{i=1}^{N} f\l... | 0.435 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,521 |
745. Output the formula $I_{2}^{*}=\frac{1}{n} \sum_{i=1}^{n} \frac{\varphi\left(x_{i}\right)}{f\left(x_{i}\right)}$ for estimating the definite integral $I=\int_{a}^{b} \varphi(x) \mathrm{d} x$, where $f(x)$ is the density function of the "auxiliary" random variable $X$ in the integration interval $(a, b) ; x_{i}$ are... | Solution. Let $f(x)$ be the probability density function of some random variable $X$ over the interval of integration $(a, b)$, i.e.,
$$
\int_{a}^{b} f(x) d x=1.
$$
We can represent the integral $I$ as follows:
$$
I=\int_{a}^{b} \frac{\varphi(x)}{f(x)} \cdot f(x) d x
$$
Thus, the integral $I$ is represented as the ... | I_{2}^{*}=\frac{1}{n}\sum_{i=1}^{n}\frac{\varphi(x_{i})}{f(x_{i})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,522 |
746. Find the estimate $I_{2}^{*}$ of the integral $I=\int_{0}^{1} \mathrm{e}^{x} \mathrm{~d} x$. | Solution. Since $\mathbf{e}^{x}=1+x+\ldots$, we will take the function $f(x)=C(1+x)$ as the density of the "auxiliary" random variable $\boldsymbol{X}$. From the condition $C \int_{0}(1+x) d x=1$, we find $C=2 / 3$. Thus, $f(x)=(2 / 3)(x+1)$.
Let's write the required integral as follows:
$$
I=\int_{0}^{1} \frac{\math... | 1.713 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,523 |
748. Derive the formula $l_{3}^{*}=(b-a) c\left(n_{1} / n\right)$ for estimating the definite integral $I=\int_{a}^{b} \varphi(x) \mathrm{d} x$, where the integrand is non-negative and bounded ( $0 \leqslant \varphi(x) \leqslant c$ ), based on the interpretation of the integral as an area. | The problem is solved. Let's consider a two-dimensional random variable ( $X, Y$ ), uniformly distributed in a rectangle $D$ with base ( $b-a$ ) and height $c$, with probability density $f(x, y)=1 /(b-a) c$. The component $X$ is uniformly distributed in the interval ( $a, b$ ) with density $1 /(b-a)$; the component $Y$... | l_{3}^{*}=(b-)(n_{1}/n) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,524 |
752. Output the formula
$$
I_{i}^{*}=\frac{b-a}{n} \sum_{i=1}^{n}\left[\varphi\left(x_{i}\right)-g\left(x_{i}\right)\right]+\int_{a}^{b} g(x) \mathrm{d} x
$$
where $x_{i}=a+(b-a) r_{i}, g(x) \simeq \varphi(x)$, for estimating the integral
$$
I=\int_{a}^{b} \varphi(x) \mathrm{d} x
$$ | Solution. Let us consider a random variable $\boldsymbol{X}$, uniformly distributed over the integration interval $(a, b)$ with density $f(x)=1/(b-a)$.
Suppose we have found a function $g(x)$ that is "close" to $\varphi(x)$ and whose integral can be computed without resorting to the Monte Carlo method. Then the mathem... | I_{i}^{*}=\frac{b-}{n}\sum_{i=1}^{n}[\varphi(x_{i})-(x_{i})]+\int_{}^{b}(x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,526 |
753. Find the estimate $I_{4}^{*}$ of the integral $I=\int_{0}^{1} \sqrt{1+x^{2}} \mathrm{~d} x$.[^1] | Solution. Since $\sqrt{1+x^{2}}=1+(1 / 2) x^{2}+\ldots(|x| < 1)$
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | 0.100 | 0.973 | 0.253 | 0.376 | $\|0.520\|$ | $\|0.135\|$ | 0.863 | 0.467 | $\|0.354\|$ | 0.8 |
| $x_{i}^{2}$ | $0.0... | 1.145 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,527 |
764. Prove that adding a non-random function $\varphi(t)$ to a random function $X(t)$ does not change the correlation function: if $Y(t)=X(t)+\varphi(t)$, then $K_{y}\left(t_{1}, t_{\mathbf{z}}\right)=K_{x}\left(t_{1}, t_{\mathbf{s}}\right)$. | Solution. Let's find the mathematical expectation:
$$
m_{y}(t)=M[X(t)+\varphi(t)]=m_{x}(t)+\varphi(t) .
$$
Let's find the centered function:
$$
\begin{gathered}
\dot{Y}(t)=Y(t)-m_{y}(t)=[X(t)+\varphi(t)]-\left[m_{x}(t)+\varphi(t)\right]= \\
=X(t)-m_{x}(t)=X(t) .
\end{gathered}
$$
Thus, $\dot{Y}(t)=\dot{X}(t)$.
Let... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,528 |
768. Let $X(t)$ be a random function, $\varphi(t)$ be a non-random function. Prove: if $Y(t)=X(t)+\varphi(t)$, then $D_{y}(t)=D_{x}(t)$ | Solution. First method. For any fixed value of the argument, the section $X(t)$ is a random variable, $\varphi(t)$ is a constant number. It is known that adding a constant number to a random variable does not change its variance, therefore $D_{y}(t)=$ $=D[X(t)+\varphi(t)]=D_{x}(t)$.
Second method. Adding a non-random ... | D_{y}()=D_{x}() | Algebra | proof | Yes | Yes | olympiads | false | 32,529 |
773. Prove that the correlation function of the product of two centered uncorrelated random functions is equal to the product of the correlation functions of the factors. | Solution. Let $Z(t)=\dot{X}(t) \dot{Y}(t)$. The mathematical expectation of the product of uncorrelated functions is equal to the product of the expectations of the factors, so $m_{z}(t)=m_{i}(t) m_{i}(t)$. The mathematical expectation of any centered function is zero, so $m_{z}(t)=0 \cdot 0=0$ and, therefore, $\dot{Z}... | K_{z}=K_{x}K_{y} | Algebra | proof | Yes | Yes | olympiads | false | 32,530 |
775. Find: a) the mathematical expectation; b) the correlation function; c) the variance of the random function $X(t)=U \cos 2 t$, where $U$ is a random variable, and $M(U)=5, D(U)=6$. | Solution. a) Let's find the required mathematical expectation (the non-random factor $\cos 2 t$ is taken out of the sign of the mathematical expectation):
$$
M[X(t)]=M[U \cos 2 t]=\cos 2 t M(U)=5 \cos 2 t,
$$
b) Let's find the centered function:
$$
\dot{X}(t)=X(t)-m_{x}(t)=U \cos 2 t-5 \cos 2 t=(U-5) \cos 2 t
$$
Le... | )5\cos2,\quadb)6\cos2t_{1}\cos2t_{2},\quad)6\cos^{2}2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,531 |
779. Find the mutual correlation function of two random functions: $X(t)=t^{2} U$ and $Y(t)=t^{3} U$, where $U$ is a random variable, and $D(U)=5$. | Solution. Let's find the mathematical expectations:
$$
m_{x}(t)=M\left(t^{2} U\right)=t^{2} m_{z}, \quad m_{y}(t)=M\left(t^{3} U\right)=t^{3} m_{x}
$$
Let's find the centered functions:
$$
\begin{aligned}
& \dot{X}(t)=X(t)-m_{x}(t)=t^{2} U-t^{2} m_{a}=t^{2}\left(U-m_{a}\right) \\
& \dot{Y}(t)=Y(t)-m_{y}(t)=t^{2} U-t... | 5t_{1}^{2}t_{2}^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,532 |
801. Prove that the cross-correlation function of a random function $X(t)$ and its derivative is equal to the partial derivative of the correlation function with respect to the argument that "corresponds to the derivative" [if the index $\dot{x}$ is in the first (second) position, then one must differentiate with respe... | Solution. a) By the definition of the cross-correlation function,
$$
R_{x \dot{x}}\left(t_{1}, t_{2}\right)=M\left[\dot{X}\left(t_{1}\right) \dot{X}\left(t_{2}\right)\right]
$$
The product under the expectation sign can be represented as a partial derivative with respect to the argument $t_{9}$:
$$
\dot{X}\left(t_{1... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,533 |
805. Find the correlation function of the random function $Z(t)=X(t)+X^{\prime}(t)$, given the correlation function $K_{x}$. | Solution. By Theorem 2 ($\S 2$), $K_{z}=K_{x}+K_{\dot{x}}+R_{x \dot{x}}+R_{\dot{x} x}$. Considering that the correlation function of the derivative (Theorem 2)
$$
K_{\dot{x}}=\frac{\partial^{2} K_{x}}{\partial t_{1} \partial t_{2}}
$$
and the mutual correlation functions (Theorem 3)
$$
R_{x \dot{x}}=\frac{\partial K... | K_{z}=K_{x}+\frac{\partial^{2}K_{x}}{\partialt_{1}\partialt_{2}}+\frac{\partialK_{x}}{\partialt_{2}}+\frac{\partialK_{x}}{\partialt_{1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,534 |
811. Given the mathematical expectation $m_{x}(t)=3 t^{2}+1$ of the random function $X(t)$, find the mathematical expectation of the integral $Y(t)=\int_{0} X(s) \mathrm{d} s$. | Solution. The desired expected value
$$
m_{u}(t)=\int_{0}^{t} m_{x}(s) d s=\int_{0}^{t}\left(3 s^{2}+1\right) \cdot d s=t^{2}+t
$$ | ^{2}+ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,535 |
816. The correlation function $K_{x}\left(t_{1}, t_{3}\right)=$ $=\cos \omega t_{1} \cos \omega t_{2}$ of the random function $X(t)$ is given. Find: a) the correlation function; b) the variance of the integral $Y(t)=$ $=\int_{0}^{1} X(s) \mathrm{d} s$. | Solution. a) The correlation function of the integral $Y(t)=$ $=\int_{0}^{t} X(s)$ ds is equal to the double integral of the given correlation function:
$$
\begin{aligned}
& K_{y}\left(t_{1}, t_{2}\right)=\int_{0}^{t_{1}} \int_{0}^{t_{1}} K_{x}\left(s_{1}, s_{2}\right) d s_{1} d s_{2}=\int_{0}^{t_{1}} \int_{0}^{t_{1}}... | \frac{\sin^{2}\omega}{\omega^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,536 |
828. Prove that if the correlation function of a random function $X(t)$ is known, then the mutual correlation functions of the random functions $X(t)$ and $Y(t)=$
$=\int_{0}^{-} X(s) \mathrm{d} s$ are expressed by integrals: a) $\left.R_{x y}=\int_{0}^{t_{2}} K_{x}\left(t_{1}, s\right) \mathrm{d} s ; \sigma\right) R_{y... | Solution. By the definition of the cross-correlation function, $R_{x y}=M\left[\dot{X}\left(t_{1}\right) \mathcal{Y}\left(t_{2}\right)\right]$. Let's find the centered function:
$$
\begin{aligned}
Y(t)= & Y(t)-m_{y}(t)=\int_{0}^{t} X(s) d s-\int_{0}^{t} m_{x}(s) d s= \\
& =\int_{0}^{t}\left[X(s)-m_{x}(s)\right] d s=\i... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,537 |
830. Given a random function $X(t)=\cos (t+\varphi)$, where $\varphi$ is a random variable uniformly distributed in the interval $(0,2 \pi)$. Prove that $X(t)$ is a stationary function. | Solution. Let's find the mathematical expectation of $X(t)$:
$$
\begin{gathered}
m_{x}(t)=M[\cos (t+\varphi)]=M[\cos t \cos \varphi-\sin t \sin \varphi]= \\
=\cos t \cdot M[\cos \varphi]-\sin t \cdot M[\sin \varphi] .
\end{gathered}
$$
Considering that (see Chapter VI, § 3)
$M[\cos \varphi]=(1 / 2 \pi) \int_{0}^{2 \... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,538 |
840. Prove that the correlation function of a stationary random function is an even function. | Solution. The correlation function of any random function does not change when the arguments are permuted. In particular, for a stationary function, the correlation function of which depends only on the difference of the arguments, by swapping the arguments, we get $k_{x_{5}}\left(t_{2}-t_{1}\right)=k_{x}\left(t_{1}-t_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,539 |
843. Prove that the variance of a stationary random function $X(t)$ is constant and equal to the value of the correlation function at the origin: $D_{x}(t)=k_{x}(0)$. | Solution. The variance of any random function is equal to the value of its correlation function at equal argument values. In particular, for a stationary function, whose correlation function depends only on the difference of the arguments, we get
$$
D_{x}(t)=K_{x}(t, t)=k_{x}(t-t)=k_{x}(0)
$$ | D_{x}()=k_{x}(0) | Algebra | proof | Yes | Yes | olympiads | false | 32,540 |
852. Prove that if the correlation function $k_{x}(\tau)$ of a differentiable stationary random function $X(t)$ is known, then the correlation function of its derivative $k_{\dot{x}}(\tau)=-k_{x}^{\prime \prime}(\tau)$. | Solution. It is known that the correlation function of the derivative of any differentiable random function is equal to the second mixed derivative of its correlation function:
$$
K_{\dot{x}}\left(t_{i}, t_{2}\right)=\frac{\partial^{2} K_{x}\left(t_{1}, t_{2}\right)}{\partial t_{1} \partial t_{2}}
$$
By the condition... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,541 |
865. Prove that the mutual correlation function of a differentiable stationary random function $X(t)$ and its derivative $X^{\prime}(t)=\dot{x}$ is equal to the first derivative of the correlation function $k_{x}(\tau)$, taken with its (opposite) sign, if the index $\dot{x}$ is in the second (first) position: a) $r_{x ... | The solution is as follows. a) By the definition of the cross-correlation function,
$$
R_{x \dot{x}}=M\left[\dot{X}\left(t_{1}\right) X^{\prime}\left(t_{2}\right)\right]=M\left\{\frac{\left.\partial \mid X\left(t_{1}\right) \hat{X}\left(t_{2}\right)\right]}{\partial t_{2}}\right\}
$$
The operations of finding the mat... | r_{x\dot{x}}(\tau)=k_{x}^{\}(\tau) | Calculus | proof | Yes | Yes | olympiads | false | 32,543 |
867. Prove that the mutual correlation function of a stationary random function $X(t)$ and its derivative changes sign when the arguments $t_{i}$ and $t_{2}$ are interchanged. | Solution. Using problem 865:
$$
R_{x \dot{x}}\left(t_{1}, t_{2}\right)=k_{x}(\tau) .
$$
By changing the order of the arguments in the cross-correlation function, we get $R_{x \dot{x}}\left(t_{2}, t_{1}\right)$. The index $\dot{x}$ is in the second position; therefore, $k_{x}(\tau)$ needs to be differentiated with res... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,544 |
880. Prove that, knowing the spectral density of a differentiable stationary random function, one can find the spectral density of its derivative using the formula $s_{\dot{x}}(\omega)=\omega^{2} s_{x}(\omega)$. | Solution. The derivative of a stationary function is also stationary (see problem 853), therefore the spectral density of the derivative
$$
s_{\dot{x}}(\omega)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} k_{\dot{x}}(\tau) \mathrm{e}^{-i \omega \tau} \mathrm{d} \tau
$$
Considering that $\boldsymbol{k}_{\dot{x}}(\tau)=-k_{... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,545 |
887. Find the spectral density of a stationary random function, given its correlation function $k_{x}(\tau)=e^{-|\tau|} \cos \tau$. | Solution. Given that $|\tau|=-\tau$ when $\tau<0$ and $|\tau|=\tau$ when $\tau>0$.
889*. Find the spectral density of a stationary function, given its correlation function $k_{x}(\tau)=$ $=D \mathrm{e}^{-\alpha |\tau|}[\cos \beta \tau + (\alpha / \beta) \sin \beta |\tau|] (\alpha>0)$.
Hint. Expand the brackets and us... | \frac{2D\alpha^{3}}{\pi(\alpha^{2}+\omega^{2})^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,546 |
892. Find the spectral density of a stationary function, given its correlation function $k_{x}(\tau)=$ $=100 \mathrm{e}^{-0.1|\tau|}(1+0.1|\tau|)$.
Hint. Use problem 891.
893*. Find the spectral density of a stationary function, given its correlation function $k_{x}(\tau)=$ $=D \mathrm{e}^{-\alpha|\tau|}\left(1+\alph... | Solution. First method. We use the formula
$$
s_{x}(\omega)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} k_{x}(\tau) \mathrm{e}^{-i \omega \tau} \mathrm{d} \tau
$$
Substituting the given correlation function, we get
$$
s_{x}(\omega)=\frac{D}{2 \pi} \int_{-\infty}^{\infty} \mathrm{e}^{-\alpha|\tau|}\left(1+\alpha|\tau|+\f... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,547 |
900. Prove that for stationary and stationary connected random functions $X(t)$ and $Y(t)$, the following relationship holds between the mutual spectral densities: $s_{x y}(-\omega)=s_{y x}(\omega)$. | Solution. By the definition of the cross-spectral density,
$$
s_{x y}(\omega)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} r_{x y}(\tau) e^{-i \omega \tau} d \tau
$$
Therefore,
$$
\varepsilon_{x y}(-\infty)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} r_{x y}(\tau) e^{-(1-\infty) \tau} d \tau=\frac{1}{2 \pi} \int_{-\infty}^{\... | proof | Other | proof | Yes | Yes | olympiads | false | 32,548 |
802. Prove that, knowing the spectral density $s_{x}(\omega)$ of a differentiable stationary random function $X(t)$, one can find the cross-spectral density of the function $X(t)$ and its derivative using the formula $\delta_{x \dot{x}}(\omega)=i \omega s_{x}(\omega)$. | Solution. According to the definition of the spectral density,
$$
s_{x \dot{x}}(\omega)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} r_{x \dot{x}}(\tau) e^{-a / \tau} \mathrm{d} \tau
$$
It is known (see problem v65) that $r_{x i}(\tau)=\frac{d k_{x}(\tau)}{d \tau}, \quad$ Therefore, we have
$$
\begin{aligned}
& k_{x}(\ta... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,549 |
905. Find the correlation function of a stationary random function $X(t)$, given its spectral density $s_{x}(\omega)=s_{0}$ in the interval $-\omega_{0} \leqslant \omega \leqslant \omega_{0} ;$ outside this interval $s_{x}(\omega)=0$. | Solution. We use the formula
$$
k_{x}(\tau)=2 \int_{0}^{\infty} s_{x}(\omega) \cos \omega \tau \mathrm{d} \omega
$$
Considering that $s_{x}(\omega)=s_{0}$ in the interval ( $0 ; \omega_{0}$ ), we get
$$
k_{x}(\tau)=2 s_{0} \int_{0}^{\omega} \cos \omega \tau d \omega=2 s_{0}\left(\sin \omega_{0} \tau\right) / \tau
$$ | k_{x}(\tau)=2s_{0}\frac{\sin\omega_{0}\tau}{\tau} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,550 |
907. Find the correlation function of a stationary random function, given its spectral density $\delta_{x}(\omega)=D \alpha / \pi\left(\alpha^{2}+\omega^{2}\right)$. | Solution. First method. From problem 886, it follows that the correlation function $k_{x}(\tau)=D e^{-\alpha|\tau|}$ corresponds to the given spectral density. Since $k_{x}(\tau)$ and $s_{x}(\omega)$ are related by mutually inverse Fourier transforms, the sought correlation function is $k_{x}(\tau)=D \mathrm{e}^{-\alph... | k_{x}(\tau)=De^{-\alpha|\tau|} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,551 |
909. Find the correlation function of stationary white noise - a stationary random function with constant spectral density $s_{x}(\omega)=s_{0}$. | Solution. We use the formula
$$
k_{x}(\tau)=\int_{-\infty}^{\infty} s_{x}(\omega) \mathrm{e}^{i \tau \omega \omega} \mathrm{d} \omega=s_{0} \int_{-\infty}^{\infty} \mathrm{e}^{l \tau \omega \omega} \mathrm{d} \omega
$$
Taking into account that $\frac{1}{2 \pi} \int_{-\infty}^{\infty} \mathrm{e}^{\tau \tau \omega} \ma... | k_{x}(\tau)=2\pis_{0}\delta(\tau) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,552 |
910. A stationary random function $X(t)$ with mean $m_{x}=5$ is input to a linear stationary dynamic system described by the equation $Y^{\prime}(t)+2 Y(t)=5 X^{\prime}(t)+$ $+6 X(t)$. Find the mean of the random function $Y(t)$ at the output of the system in the steady state (after the transient process has decayed). | Solution. Let's equate the mathematical expectations of the left and right parts of the given differential equation:
$$
\begin{aligned}
M\left[Y^{\prime}(t)+2 Y(t)\right]= & M\left[5 X^{\prime}(t)+6 X(t)\right], \text{ or } M\left[Y^{\prime}(t)\right]+2 m_{y}= \\
& =5 M\left[X^{\prime}(t)\right]+6 m_{x} .
\end{aligned... | 15 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,553 |
912. A stationary random function $X(t)$ with the correlation function $\boldsymbol{k}_{x}(\tau)=6 \mathrm{e}^{-2|\tau|}$ is input into a linear stationary dynamic system described by the equation $3 Y^{\prime}(t)+Y(t)=4 X^{\prime}(t)+X(t)$. Find the variance of the random function $Y(t)$ at the output of the system in... | Solution. 1. Find the spectral density $s_{x}$ ( () ). Using the solution of problem 886, with $D_{x}=k_{x}(0)=6$ and $\alpha=2$, we get
$$
s_{x}(\omega)=D \alpha / \pi\left(\alpha^{2}+\omega^{2}\right)=12 / \pi\left(\omega^{2}+4\right)
$$ | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,554 |
Example 2. Find the range of $x$, if $|x-1|>2$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution. By the definition of absolute value
\[
\begin{gathered}
x-1>2 \text { when }(x-1)>0, \text { i.e., } x>1 \\
-(x-1)>2 \text { when }(x-1)<0, \text { i.e., } x<1
\end{gathered}
\]
From $x-1>2$ and $-x+1>2$ we get $x>3, x<-1$.
Therefore, the domain of the variable $x$ is the union of two infinite intervals: $... | (-\infty,-1),(3,+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,555 |
Example 4. Given the function $f(x)=x^{2}-10 x+16$.
Find the values of the function for the argument values equal to the geometric mean and arithmetic mean
of its roots, respectively. | Solution. Let's find the roots, or zeros, of the function. By equating the function to zero, we get the quadratic equation $x^{2}-10 x+16=0$, the roots of which are $x_{1}=2, x_{2}=8$. The arithmetic mean of the roots is $x^{\prime}=\frac{x_{1}+x_{2}}{2}=5$ and the geometric mean is $x^{\prime \prime}=\sqrt{x_{1} x_{2}... | -8,-9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,556 |
Example 6. Prove that the function $f(x)=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}$ is odd. | Solution. Let's ensure that condition (1.9) is satisfied:
$$
f(-x)=(-x)-\frac{(-x)^{3}}{3}+\frac{(-x)^{5}}{5}=-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}=-\left(x-\frac{x^{3}}{3}+\frac{x^{5}}{5}\right)=-f(x)
$$
Prysher T. Show that the function $f(x)=\operatorname{tg} x \sin 3 x+\operatorname{ctg} 2 x$ is periodic and find it... | 2\pi | Algebra | proof | Yes | Yes | olympiads | false | 32,557 |
Example 8. Find the domain of the function
$$
f(x)=\frac{x-8}{x^{2}-7 x+12}
$$ | Solution. The given function is not defined for those values of $x$ for which the denominator is zero (since division by zero is meaningless). By equating the denominator to zero, we get $x^{2}-7 x+12=0$, from which $x_{1}=3, x_{2}=4$. Thus, the function is defined on the entire real line $-\infty<x<+\infty$, except fo... | (-\infty,3)\cup(3,4)\cup(4,+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,558 |
Example 9. Find the domain of the function
$$
f(x)=\sqrt{\left(9-x^{2}\right)\left(x^{2}-4\right)}+\sqrt[3]{5 x+7}
$$ | Solution. The function is the sum of two functions. The second one is defined for all $x$, since the cube root exists for any $x$. The first function $\sqrt{\left(9-x^{2}\right)\left(x^{2}-4\right)}$ is defined only for those values of $x$ for which the expression under the square root is non-negative (the square root ... | [-3,-2]\cup[2,3] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,559 |
Example 10. Find the domain of the function
$$
y=\arcsin \frac{2 x-1}{3}
$$ | Solution. The function $u=\arcsin t$ is defined for $-1 \leq t \leq 1$. Therefore, the given function is defined only for those values of $x$ for which
$$
-1 \leq \frac{2 x-1}{3} \leq 1 \text {, from which }-3 \leq 2 x-1 \leq 3
$$
Solving these inequalities, we get $-1 \leq x \leq 2$. Thus, the function is defined on... | [-1,2] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,560 |
Example 11. Find the domain of the function
\[
f(x)=\log \left(4-x^{2}\right)
\] | Solution. The logarithmic function is defined for positive values of its argument, i.e., for $4-x^{2}>0$, from which $x^{2}<4$ or $|x|<2$. Therefore, the given function is defined for $-2<x<2$, i.e., in the interval $(-2,2)$. | (-2,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,561 |
Example 14. Find the inverse functions for the given functions:
1) $y=3 x-5$
2) $y=\sqrt{1-x^{3}}$
3) $y=\arcsin 3 x$
4) $y=x^{2}+2$. | Solution. Solving each of the given equations for $x$, we obtain the following inverse functions:
1) $x=\frac{1}{3}(y+5)$
(defined for all $y$, i.e., for
$$
-\infty < y < +\infty
$$
7)
4) $x^{2} \leq 25$
5) $x^{2} \geq 16$.
2. Solve the equation $\left|x^{2}-9\right|+\left|x^{2}-4\right|=5$.
3. Given the function $f(... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,562 |
Example 2. Plot the graph of the function $y=\frac{1}{x^{2}}$. | Solution. This function is defined for all values of $x$, except for the point $x=0$. By assigning the argument the values specified below and calculating the corresponding function values, we construct the table:
| $x$ | -3 | -2 | -1 | $-\frac{1}{2}$ | $-\frac{1}{3}$ | $\frac{1}{3}$ | $\frac{1}{2}$ | 1 | 2 | 3 |
| :-... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,563 |
Example 3. Plot the graph of the function $y=\sqrt{x}$. | Solution. The function $y=\sqrt{x}$ is defined when the expression under the root is non-negative, i.e., $x \geq 0$. Therefore, the domain of the function is the infinite half-interval $[0, \infty)$.
By squaring both sides of the given equation, we find $y^{2}=x$. The obtained equation defines a parabola for which the... | \sqrt{x},fory\geq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,564 |
Example 7. Plot the graph of the function $f(x)=\operatorname{cosec} x$. | Solution. The function is defined for all real numbers except the points $x=k \pi(k=0, \pm 1, \pm 2, \pm 3, \ldots)$, i.e., its domain of definition is the union of intervals $\ldots(-\pi, 0)$, $(0, \pi),(\pi, 2 \pi) \ldots$
Since $\operatorname{cosec} x=\frac{1}{\sin x}$, the graph of the function can be constructed ... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,566 |
Example 8. Plot the graph of the function $y=\sin x+3$. | Solution. The graph of this function is a sine wave shifted upwards along the $O y$ axis by three units (Fig. 1.13) (see formula (1.18)).
Example g. Plot the graph of the function $y=\cos \left(x-\frac{\pi}{4}\right)$.
Solution. The graph of this function is a cosine wave shifted to the right along the $O x$ axis by ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,567 |
Example 10. Plot the graph of the function $y=3 \sin 2 x$. | Solution. Based on the simplest transformations of the graph (3 and
4), we conclude that the graph of this function is a sine wave, the abscissas of which are reduced by half, and the ordinates are increased by three times (Fig. 1.15). | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,568 |
Example 14. Plot the graph of the function $f(x)=|\sin x|$. | Solution. By the definition of absolute value (see § 1.1, equation (1.1)), we have $f(x)=\sin x$ if $\sin x \geq 0$ and $f(x)=-\sin x$ if $\sin x<0$.
Fig. 1.19
Therefore, the graph of the g... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,569 |
Example 1. Show that the sequence $x_{n}=\frac{1}{n}$ ( $n=1,2,3, \ldots$ ) has a limit of zero. Starting from which term do its values become and remain less than 0.001? | Solution. The sequence $x_{n}=\frac{1}{n}(n=1,2,3, \ldots)$ takes the values
$$
1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots
$$
Let $\varepsilon=0.001$. The inequality $\frac{1}{n}<\varepsilon$ is equivalent to $n>1000$. Therefore, $N=1000$.
Take an arbitrary number $\varepsilon>0$. We will show tha... | \lim_{narrow\infty}\frac{1}{n}=0 | Calculus | proof | Yes | Yes | olympiads | false | 32,572 |
Example 2. Prove that the limit of the sequence $y_{n}=4-\frac{1}{3^{n}}$ $(n=1,2,3, \ldots)$ is 4. | Solution. The given sequence takes the values
$$
3 \frac{2}{3}, 3 \frac{8}{9}, 3 \frac{26}{27}, 3 \frac{80}{81}, \ldots
$$
Let us define an arbitrarily small number $\varepsilon>0$ and form the difference
$$
y_{n}-a=\left(4-\frac{1}{3^{n}}\right)-4=-\frac{1}{3^{n}}
$$
We require that this difference in absolute val... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,573 |
Example 3. Prove that the limit of the sequence $z_{n}=-3+\frac{(-1)^{n}}{n^{2}}(n=1,2,3, \ldots)$ is -3. | Solution. The given sequence takes the values
$$
-4,-2 \frac{3}{4},-3 \frac{1}{9},-2 \frac{15}{16}, \ldots
$$
Let any number $\varepsilon>0$ be given. Consider the difference
$$
z_{n}-(-3)=\left[-3+\frac{(-1)^{n}}{n^{2}}\right]-(-3)=\frac{(-1)^{n}}{n^{2}}
$$
This difference will be less than $\varepsilon$ in absolu... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,574 |
Example 4. Prove that
$$
\lim _{x \rightarrow \infty} \frac{5 x+6}{6 x}=\frac{5}{6}
$$ | S o l u t i o n. To prove this, it is sufficient to show that the difference between the variable quantity $y=\frac{5 x+6}{6 x}$ and the constant $b=\frac{5}{6}$ as $n \rightarrow \infty$ is an infinitely small quantity. Transforming this difference, we get
$$
\frac{5 x+6}{6 x}-\frac{5}{6}=\frac{(5 x+6)-5 x}{6 x}=\fra... | \lim_{xarrow\infty}\frac{5x+6}{6x}=\frac{5}{6} | Calculus | proof | Yes | Yes | olympiads | false | 32,575 |
Example 5. Prove that
$$
\lim _{x \rightarrow 3}\left(2 x^{2}-5 x+4\right)=7
$$ | Solution. Since $x \rightarrow 3$, then $x=3+\alpha$, where $\alpha$ is infinitesimally small (see formula (1.38)). Substituting this expression for $x$ into the difference $\left(2 x^{2}-5 x+4\right)-7$ and transforming it, we get
$$
\begin{gathered}
{\left[2(3+\alpha)^{2}-5(3+\alpha)+4\right]-7=2\left(9+6 \alpha+\al... | \lim_{xarrow3}(2x^{2}-5x+4)=7 | Calculus | proof | Yes | Yes | olympiads | false | 32,576 |
Example 6. Find the one-sided limits of the function $f(x)=\frac{6}{x-3}$ as $x \rightarrow 3$ from the left and from the right. | Solution. The problem reduces to finding two limits
$$
\lim _{x \rightarrow 3-0} \frac{6}{x-3} \text { and } \lim _{x \rightarrow 3+0} \frac{6}{x-3}
$$
If $x \rightarrow 3-0$, i.e., $x$ approaches 3 while remaining less than 3, then the quantity $x-3$ is an infinitesimally small negative value. Its reciprocal will be... | \lim_{xarrow3-0}\frac{6}{x-3}=-\infty\lim_{xarrow3+0}\frac{6}{x-3}=+\infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,577 |
Example 7. Find the one-sided limits of the function $f(x)=3^{\frac{1}{x}}$ as $x \rightarrow 0$ from the left and right. | Solution.
Let $x$ tend to 0, remaining less than it, i.e., $x \rightarrow -0$, then $x$ will be an infinitely small quantity, taking negative values, and the quantity $\frac{1}{x}$ will be a negative infinitely large quantity (see formula (1.41)). This function can be represented as follows:
$$
3^{\frac{1}{x}} = 3^{-... | \lim_{xarrow-0}3^{\frac{1}{x}}=0\lim_{xarrow+0}3^{\frac{1}{x}}=+\infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,578 |
Example 8. Find the limits
\[
\begin{gathered}
\lim _{x \rightarrow 2-0} \operatorname{arctg} \frac{1}{2-x} \text { and } \\
\lim _{x \rightarrow 2+0} \operatorname{arctg} \frac{1}{2-x}
\end{gathered}
\] | Solution. Let $x \rightarrow 2$ from the left, i.e., remaining less than 2, then the quantity $(2-x)$ will be infinitely small, taking positive values, and the quantity $\frac{1}{2-x}$ will be infinitely large, also taking positive values. Considering the definition of the function $z=\operatorname{arctg} y$ and the fa... | \frac{\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,579 |
Example 1. Find $\lim _{x \rightarrow 2}\left(4 x^{2}-6 x+3\right)$. | Solution. Since the limit of the algebraic sum of variables is equal to the same algebraic sum of the limits of these variables (formula (1.42)), a constant multiplier can be factored out of the limit sign (formula (1.44)), the limit of an integer positive power is equal to the same power of the limit (formula (1.45)),... | 7 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,580 |
Example 3. Find $\lim _{x \rightarrow 4} \frac{x^{2}-6 x+8}{x-4}$. | Solution. When $x=4$, both the numerator and the denominator of the given function become zero. This results in an indeterminate form $\frac{0}{0}$, which needs to be resolved. We will transform the given function by factoring the numerator using the formula
$$
x^{2}+p x+q=\left(x-x_{1}\right)\left(x-x_{2}\right)
$$
... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,581 |
Example 4. Find $\lim _{x \rightarrow 1} \frac{3 x^{2}-x-2}{4 x^{2}-5 x+1}$. | Solution. When $x=1$, both the numerator and the denominator become zero, resulting in an indeterminate form of $\frac{0}{0}$. We will transform the given function by factoring the numerator and the denominator using the formula
$$
a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)
$$
where $x_{1}$ and $x_{2}$ a... | \frac{5}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,582 |
Example 5. Find $\lim _{x \rightarrow-1} \frac{3 x^{4}+2 x^{3}-x^{2}+5 x+5}{x^{3}+1}$. | Solution. When $x=-1$, both the numerator and the denominator become zero. Factoring them, we find
\[
\begin{gathered}
x^{3}+1=(x+1)\left(x^{2}-x+1\right) \\
3 x^{4}+2 x^{3}-x^{2}+5 x+5=(x+1)\left(3 x^{3}-x^{2}+5\right)
\end{gathered}
\]
The second equality is obtained by directly dividing $3 x^{4}+2 x^{3}-x^{2}+5 x+... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,583 |
Example 6. Find $\lim _{x \rightarrow+\infty} \frac{6 x^{2}+5 x+4}{3 x^{2}+7 x-2}$. | Solution. As $x \rightarrow+\infty$, the numerator and denominator increase without bound (we get an indeterminate form of $\frac{\infty}{\infty}$). To find the limit, we transform the given fraction by dividing its numerator and denominator by $x^{2}$, i.e., by the highest power of $x$. Using the properties of limits,... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,584 |
Example 7. Find $\lim _{x \rightarrow+\infty} \frac{7 x^{2}+6 x-3}{9 x^{3}+8 x^{2}-2}$. | Solution. By dividing the numerator and the denominator of the fraction by $x^{3}$, i.e., by the highest degree, we get
$$
\begin{aligned}
\lim _{x \rightarrow+\infty} \frac{7 x^{2}+6 x-3}{9 x^{3}+8 x^{2}-2}= & \lim _{x \rightarrow+\infty} \frac{\frac{7}{x}+\frac{6}{x^{2}}-\frac{3}{x^{3}}}{9+\frac{8}{x}-\frac{2}{x^{3}... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,585 |
Example 9. Find $\lim _{x \rightarrow 3} \frac{x^{2}-9}{\sqrt{x+1}-2}$. | Solution. When $x=3$, both the numerator and the denominator of the fraction become zero. The denominator contains an irrational expression $\sqrt{x+1}$. To eliminate the irrationality in the denominator, we multiply the numerator and the denominator by $(\sqrt{x+1}+2)$. We get
$$
\begin{gathered}
\lim _{x \rightarrow... | 24 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,586 |
Example 10. Find $\lim _{x \rightarrow \pi} \frac{\sin ^{2} x}{1+\cos ^{3} x}$. | Solution. Factoring the numerator and the denominator and canceling the factor $1+\cos x \neq 0$, we get
$$
\begin{gathered}
\lim _{x \rightarrow \pi} \frac{\sin ^{2} x}{1+\cos ^{3} x}=\lim _{x \rightarrow \pi} \frac{1-\cos ^{2} x}{1+\cos ^{3} x}=\lim _{x \rightarrow \pi} \frac{(1+\cos x)(1-\cos x)}{(1+\cos x)\left(1-... | \frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,587 |
Example 11. Find $\lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}$. | Solution. The numerator and denominator of the fraction are the sums of $n$ terms of the corresponding arithmetic progressions. Finding these sums using the known formula, we get
\[
\begin{gathered}
\lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}=\lim _{n \rightarrow \infty} \frac{\frac{1+(2 n... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,588 |
Example 12. Find $\lim _{n \rightarrow \infty} \frac{1^{2}+2^{2}+3^{2}+\ldots+n^{2}}{n^{3}}$. | Solution. Using the formula for the sum of squares of a natural number series
$$
1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}
$$
we get
$$
\lim _{n \rightarrow \infty} \frac{1^{2}+2^{2}+3^{2}+\ldots+n^{2}}{n^{3}}=\lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 n^{3}}=
$$
$$
\begin{gathered}
=\frac{1}{... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,589 |
Example 13. Find $\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+x}-1}{\sqrt{1+x}-1}$. | Solution. Introduce a new variable $z$ by the formula
$$
1+x=z^{6}
$$
(the exponent is chosen so that it is possible to extract both a square and a cube root).
From equality (A), it follows that $z \rightarrow 1$ as $x \rightarrow 0$. Substituting the expression of the old variable in terms of the new one, factoring... | \frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,590 |
Example 14. Find $\lim _{n \rightarrow \infty} \frac{12 n+5}{\sqrt[3]{27 n^{3}+6 n^{2}+8}}$. | Solution. As $n \rightarrow \infty$, the numerator and denominator also tend to infinity, resulting in an indeterminate form $\frac{\infty}{\infty}$. To find the limit, we divide the numerator and denominator by $\boldsymbol{n}$ and bring $\boldsymbol{n}$ under the root sign:
$$
\lim _{n \rightarrow \infty} \frac{12 n... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,591 |
Example 1. Find $\lim _{x \rightarrow \infty}\left(1+\frac{k}{x}\right)^{x}$. | Solution. When $\quad x \rightarrow \infty$, the expression $\quad\left(1+\frac{k}{x}\right)^{x} \rightarrow 1$, resulting in the indeterminate form $1^{\infty}$.
Introduce a new variable $\alpha$ by the formula
$$
\frac{k}{x}=\alpha
$$
from which $x=\frac{k}{\alpha}$. If $x \rightarrow \infty$, then $\alpha \righta... | e^{k} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,592 |
Example 2. Find $\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}$. | Solution. Since
$$
\frac{\ln (1+x)}{x}=\frac{1}{x} \ln (1+x)=\ln (1+x)^{\frac{1}{x}}
$$
then based on formula (1.58) we find
$$
\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=\lim _{x \rightarrow 0}\left[\ln (1+x)^{\frac{1}{x}}\right]=\ln \left[\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}\right]=\ln e=1
$$
Therefore,
$... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,593 |
Example 3. Find $\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}$. | Solution. Introduce a new variable $z$ by the formula
$$
a^{x}-1=z
$$
from which
$$
a^{x}=1+z
$$
Taking the logarithm of this equality to the base $e$, we obtain the formula
$$
x \ln a=\ln (1+z)
$$
from which
$$
x=\frac{\ln (1+z)}{\ln a}
$$
Obviously, as $x \rightarrow 0$ and $z \rightarrow 0$, therefore
$$
\l... | \ln | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,594 |
Example 4. Find $\lim _{x \rightarrow \infty}\left(\frac{x+2}{x-3}\right)^{x}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the text is already in English, there's no need for translation. If you meant to tr... | Solution. Dividing the numerator and the denominator of the fraction by $x$ and using formula (1.64) for $k=+2$ and $k=-3$, we get
$$
\lim _{x \rightarrow \infty}\left(\frac{x+2}{x-3}\right)^{x}=\lim _{x \rightarrow \infty}\left(\frac{1+\frac{2}{x}}{1+\frac{-3}{x}}\right)^{x}=\frac{e^{2}}{e^{-3}}=e^{5}
$$ | e^5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,595 |
Example 5. Find $\lim _{x \rightarrow 0} \frac{\sin a x}{x}$. | Solution. Let $a x=\alpha$, from which $x=\frac{\alpha}{a}$. If $x \rightarrow 0$, then $\alpha \rightarrow 0$, therefore
$$
\lim _{x \rightarrow 0} \frac{\sin a x}{x}=\lim _{\alpha \rightarrow 0} \frac{\sin \alpha}{\frac{\alpha}{a}}=\lim _{\alpha \rightarrow 0} a \frac{\sin \alpha}{\alpha}=a \lim _{\alpha \rightarrow... | \lim_{xarrow0}\frac{\sin}{x}= | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,596 |
Example 6. Find $\lim _{x \rightarrow 0} \frac{\tan x}{x}$. | Solution. Taking into account that $\operatorname{tg} x=\frac{\sin x}{\cos x}$ and $\lim _{x \rightarrow 0} \cos x=\cos 0=1$, based on the properties of limits (1.43) and (1.46) and formula (1.59), we obtain
$$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\operatorname{tg} x}{x} & =\lim _{x \rightarrow 0}\left(\frac{... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,597 |
Example 7. Find $\lim _{x \rightarrow 0} \frac{\sin ^{2} \frac{x}{3}}{x^{2}}$. | Solution. Taking into account formula (1.67), based on the properties of limits, we obtain
$$
\lim _{x \rightarrow 0} \frac{\sin ^{2} \frac{x}{3}}{x^{2}}=\lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{3}}{x}\right)^{2}=\left[\lim _{x \rightarrow 0} \frac{\sin \frac{x}{3}}{x}\right]^{2}=\left(\frac{1}{3}\right)^{2}=\... | \frac{1}{9} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,598 |
Example 8. Find $\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}$. | Solution. By dividing the numerator and the denominator by $x$, based on formula (1.67) we obtain
$$
\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}=\lim _{x \rightarrow 0} \frac{\frac{\sin a x}{x}}{\frac{\sin b x}{x}}=\frac{\lim _{x \rightarrow 0} \frac{\sin a x}{x}}{\lim _{x \rightarrow 0} \frac{\sin b x}{x}}=\fra... | \frac{}{b} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,599 |
Example 9. Find $\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+9}-3}$. | Solution. When $x=0$, both the numerator and the denominator become zero. The denominator contains an irrationality. We will eliminate the irrationality and use formula (1.59)
$$
\begin{gathered}
\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+9}-3}=\lim _{x \rightarrow 0} \frac{\sin x(\sqrt{x+9}+3)}{(\sqrt{x+9}-3)(\sqr... | 6 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,600 |
Example 10. Find $\lim _{x \rightarrow 0}\left(\frac{\sin 3 x}{x}\right)^{x+2}$. | Solution. This is a limit of the form (1.60), where $\varphi(x)=\frac{\sin 3 x}{x}, \psi(x)=x+2$.
From (1.67)
$$
\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}=3, \lim _{x \rightarrow 0}(x+2)=2
$$
According to formula (1.62), we get
$$
\lim _{x \rightarrow 0}\left(\frac{\sin 3 x}{x}\right)^{x+2}=3^{2}=9
$$ | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,601 |
Example 11. Find $\lim _{x \rightarrow \infty}\left(\frac{2 x-1}{3 x+4}\right)^{x^{2}}$. | Solution. This is also a limit of the form $(1.60)$, where $\varphi(x)=\frac{2 x-1}{3 x+4}, \psi(x)=x^{2}$
Since
$$
\lim _{x \rightarrow \infty} \frac{2 x-1}{3 x+4}=\lim _{x \rightarrow \infty} \frac{2-\frac{1}{x}}{3+\frac{4}{x}}=\frac{2}{3}, \lim _{x \rightarrow \infty} x^{2}=\infty
$$
then according to formula (1.... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,602 |
Example 13. Find $\lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}$. | Solution. By adding and subtracting 1 from $\cos x$ and applying the corresponding formula, we get
$$
\begin{aligned}
& \lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}=\lim _{x \rightarrow 0}[1-(1-\cos x)]^{\frac{1}{x}}=\lim _{x \rightarrow 0}\left(1-2 \sin ^{2} \frac{x}{2}\right)^{\frac{1}{x}}= \\
& =\lim _{x \rightarr... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,603 |
Example 14. Find $\lim _{x \rightarrow 0}\left(1+\operatorname{tg}^{2} x\right)^{2 \operatorname{ctg}^{2} x}$. | Solution. Since $\operatorname{ctg} x=\frac{1}{\operatorname{tg} x}$, then
$$
\begin{gathered}
\lim _{x \rightarrow 0}\left(1+\operatorname{tg}^{2} x\right)^{2 \operatorname{ctg}^{2} x}=\lim _{x \rightarrow 0}\left(1+\operatorname{tg}^{2} x\right)^{\frac{2}{\operatorname{tg}^{2} x}}= \\
=\lim _{x \rightarrow 0}\left[\... | e^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,604 |
Example 1. Find $\lim _{x \rightarrow+\infty}\left(\sqrt{x^{2}+6 x+5}-x\right)$. | Solution. When $x \rightarrow+\infty$, the given function represents the difference of two infinitely large quantities taking positive values (case $\infty-\infty$). By multiplying and dividing the given function by $\left(\sqrt{x^{2}+6 x+5}+x\right)$, we get
$$
\begin{gathered}
\lim _{x \rightarrow+\infty}\left(\sqrt... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,605 |
Example 2. Find $\lim _{x \rightarrow 3}\left(\frac{1}{x-3}-\frac{6}{x^{2}-9}\right)$. | Solution. Here we also have an indeterminacy of the form $(\infty-\infty)$ as $x \rightarrow 3$. We perform the subtraction of fractions and cancel the factor $x-3 \neq 0$; passing to the limit, we find
$$
\begin{gathered}
\lim _{x \rightarrow 3}\left(\frac{1}{x-3}-\frac{6}{x^{2}-9}\right)=\lim _{x \rightarrow 3}\left... | \frac{1}{9} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,606 |
Example 3. Find $\lim _{x \rightarrow 0}\left(\frac{1}{4 \sin ^{2} x}-\frac{1}{\sin ^{2} 2 x}\right)$. | Solution. In this case, we also have an indeterminate form of the type $(\infty-\infty)$. By applying the appropriate trigonometric formula, performing the subtraction of fractions, and taking the limit, we get
$$
\begin{gathered}
\lim _{x \rightarrow 0}\left(\frac{1}{4 \sin ^{2} x}-\frac{1}{\sin ^{2} 2 x}\right)=\lim... | -\frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,607 |
Example 4. Find $\lim _{x \rightarrow 0} x \operatorname{ctg} \frac{x}{3}$. | Solution. When $x \rightarrow 0$, we get an indeterminate form of $0 \cdot \infty$. Rewriting the given function in another form and applying formula (1.59), we find
$$
\lim _{x \rightarrow 0} x \operatorname{ctg} \frac{x}{3}=\lim _{x \rightarrow 0} x \frac{\cos \frac{x}{3}}{\sin \frac{x}{3}}=\lim _{x \rightarrow 0} \... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,608 |
Example 5. Find $\lim _{x \rightarrow 2}(2-x) \tan \frac{\pi}{4} x$. | Solution. When $x \rightarrow 2$, we have an indeterminate form of $0 \cdot \infty$. Setting $x=2-\alpha$ and taking the limit, we find
$$
\begin{aligned}
& \lim _{x \rightarrow 2}(2-x) \operatorname{tg} \frac{\pi}{4} x=\lim _{\alpha \rightarrow 0} \alpha \operatorname{tg} \frac{\pi}{4}(2-\alpha)=\lim _{\alpha \righta... | \frac{4}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,609 |
Example 6. Find $\lim _{x \rightarrow+\infty} x\left(\operatorname{arctg} x-\frac{\pi}{2}\right)$. | Solution. Let $\operatorname{arctg} x=\alpha$, then $x=\operatorname{tg} \alpha$, if $x \rightarrow+\infty$, then $\alpha \rightarrow \frac{\pi}{2}$.
Consequently,
$$
\begin{gathered}
\lim _{x \rightarrow+\infty} x\left(\operatorname{arctg} x-\frac{\pi}{2}\right)=\lim _{\alpha \rightarrow \frac{\pi}{2}} \operatorname... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,610 |
Example 1. If $x \rightarrow 0$, which of the infinitesimal quantities $3 x, x^{2}, \sqrt{x}, x^{3}, \frac{1}{2} x$ are of the same order as $x$, higher order, and lower order compared to $x$? | S o l u t i o n. Let us consider the limits of the ratios of these quantities to $\boldsymbol{x}$.
Since
$$
\lim _{x \rightarrow 0} \frac{3 x}{x}=3, \lim _{x \rightarrow 0} \frac{\frac{1}{2} x}{x}=\frac{1}{2}
$$
the quantities $3 x$ and $\frac{1}{2} x$ are infinitesimals of the same order as the quantity $x$ (see fo... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,611 |
Example 4. Prove that as $x \rightarrow 0$ the infinitesimals $\operatorname{arctg} c x$ and $c x$ are equivalent.
62 | Solution. Let
$$
\operatorname{arctg} c x=z
$$
then
$$
\operatorname{tg} z=c x
$$
and $z \rightarrow 0$ as $x \rightarrow 0$.
Therefore,
$$
\lim _{x \rightarrow 0} \frac{\operatorname{arctg} c x}{c x}=\lim _{z \rightarrow 0} \frac{z}{\operatorname{tg} z}=1
$$
which is what we needed to prove. | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,612 |
Example 5. Find $\lim _{x \rightarrow 0} \frac{x \sin 6 x}{(\operatorname{arctg} 2 x)^{2}}$. | Solution. Taking into account that
$$
\sin c x \sim c x \text { and } \operatorname{arctg} c x \sim c x
$$
(see examples 3 and 4), we get
$$
\lim _{x \rightarrow 0} \frac{x \sin c x}{(\operatorname{arctg} 2 x)^{2}}=\lim _{x \rightarrow 0} \frac{x}{2 x} \cdot \frac{6 x}{2 x}=\frac{1}{2} \cdot 3=\frac{3}{2}
$$ | \frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,613 |
Example 6. Find $\lim _{x \rightarrow 3} \frac{\ln \left(x^{2}-5 x+7\right)}{x-3}$. | Solution. When $x \rightarrow 3$, we get an indeterminate form of $\frac{0}{0}$, since $x^{2}-5 x+7 \rightarrow 1$ and $\ln \left(x^{2}-5 x+7\right) \rightarrow 0$. The expression $x^{2}-5 x+7$ can be represented as:
$$
x^{2}-5 x+7=1+\left(x^{2}-5 x+6\right)=1+z
$$
where $\left(x^{2}-5 x+6\right)=z \rightarrow 0$ as ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,614 |
Example 7. Find $\lim _{x \rightarrow 0} \frac{(1+x)^{a}-1}{x}$. | Solution. When $x \rightarrow \mathbf{0}$, the quantity $(1+x)^{a}-1$ is an infinitesimal. Let
$$
(1+x)^{a}-1=z
$$
Since
$$
z \sim \ln (1+z)
$$
then
$$
(1+x)^{a}-1 \sim \ln \left[1+(1+x)^{a}-1\right]=\ln (1+x)^{a}=a \ln (1+x)
$$
Therefore,
$$
\lim _{x \rightarrow 0} \frac{(1+x)^{a}-1}{x}=\lim _{x \rightarrow 0} ... | a | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,615 |
Example 1. Prove that the function $y=x^{3}$ is continuous in the interval $(-\infty, \infty)$. | Solution. The function $y=x^{3}$ is defined for all $x$, i.e., in the infinite interval $(-\infty, \infty)$.
We will show that an infinitely small increment in the argument corresponds to an infinitely small increment in the function. Fix an arbitrary point $x$, and give the argument $x$ an increment $\Delta x$. The i... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,616 |
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