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Example 9. Find the residue of the function
$$
f(z)=e^{1 / z^{2}} \cos z
$$
at the point $z=0$.
|
Solution. Since the residue at the point $z=0$ is equal to the coefficient of $z^{-1}$, we immediately obtain that in this case the residue is zero, since the function $f(z)$ is even and its expansion in the neighborhood of the point $z=0$ cannot contain odd powers of $z$.
## Problems for Independent Solution
Find the residues at the singular points of the following functions:
324. $f(z)=\frac{\tan z}{z^{2}-\frac{\pi}{4} z}$.
325. $f(z)=z^{3} e^{1 / x}$.
326. $f(z)=\frac{\cosh z}{\left(z^{2}+1\right)(z-3)}$.
327. $f(z)=\frac{e^{z}}{\frac{1}{4}-\sin ^{2} z}$.
328. $f(z)=\frac{e^{z}}{z^{3}(z-1)}$.
329. $f(z)=\frac{z}{(z+1)^{3}(z-2)^{2}}$.
330. $f(z)=\frac{e^{-1 / z^{2}}}{1+z^{4}}$.
331. $f(z)=z^{2} \sin \frac{1}{z}$.
332. $f(z)=\cos \frac{1}{z}+z^{3}$.
333. $f(z)=\frac{\sin 2 z}{(z+i)\left(z-\frac{i}{2}\right)^{2}}$.
334. $f(z)=\frac{1-\cos z}{z^{3}(z-3)}$.
335. $f(z)=e^{z^{2}+1 / z^{2}}$.
336. $f(z)=\frac{e^{i z}}{\left(z^{2}-1\right)(z+3)}$.
337. $f(z)=\frac{\cos z}{z^{3}-\frac{\pi}{2} z^{2}}$.
338. $f(z)=\frac{e^{\pi z}}{z-i}$.
339. $f(z)=\frac{z^{2 n}}{(z-1)^{n}} \quad(n>0-$ integer).
340. $f(z)=\cot ^{2} z$.
341. $f(z)=\sin z \cos \frac{1}{z}$.
342. $f(z)=e^{z /(z-1)}$.
343. $f(z)=\frac{\sin \frac{1}{z}}{1-z}$.
344. $f(z)=\frac{e^{1 / z}}{1+z}$.
345. $f(z)=e^{\left(z^{2}+1\right) / z}$ 346. $f(z)=e^{z} \sin \frac{1}{z}$.
## § 11. The Residue Theorem of Cauchy. Application of Residues to the Evaluation of Definite Integrals. Summation of Some Series Using Residues
## $1^{\circ}$. The Residue Theorem of Cauchy.
Theorem. If the function $f(z)$ is analytic on the boundary $C$ of the region $D$ and everywhere inside the region, except for a finite number of singular points $z_{1}, z_{2}, \ldots, z_{n}$, then
$$
\int_{C} f(z) d z=2 \pi i \sum_{k=1}^{n} \operatorname{res} f\left(z_{k}\right)
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 4. Compute the integral
$$
\int_{|x|=2} \frac{1}{z-1} \sin \frac{1}{z} d z
$$
|
Solution. In the circle $|z| \leqslant 2$, the integrand has two singular points $z=1$ and $z=0$. It is easy to establish that $z=1$ is a simple pole, therefore
$$
\operatorname{res}\left(\frac{1}{z-1} \sin \frac{1}{z}\right)=\left.\frac{\sin \frac{1}{z}}{(z-1)^{\prime}}\right|_{z=1}=\sin 1
$$
To determine the nature of the singular point $z=0$, we write the Laurent series for the function $\frac{1}{z-1} \sin \frac{1}{z}$ in the neighborhood of this point. We have
$$
\begin{aligned}
\frac{1}{z-1} \sin \frac{1}{z}= & -\frac{1}{1-z} \sin \frac{1}{z}=-\left(1+z+z^{2}+\ldots\right)\left(\frac{1}{z}-\frac{1}{3!z^{3}}+\frac{1}{5!z^{5}}-\ldots\right)= \\
= & -\left(1-\frac{1}{31^{\prime}}+\frac{1}{5!}-\ldots\right) \frac{1}{z}+\frac{c_{-2}}{z^{2}}+\frac{c_{-3}}{z^{3}}+\ldots+\text { regular part } \\
& c_{-k} \neq 0, \quad k=2,3, \ldots
\end{aligned}
$$
Since the Laurent series contains an infinite number of terms with negative powers of $z$, the point $z=0$ is an essential singularity. The residue of the integrand at this point is
Therefore,
$$
\int_{|z| \neq 2} \frac{1}{z-1} \sin \frac{1}{z} d z=2 \pi i(\sin 1-\sin 1)=0
$$
## Problems for Independent Solution
Calculate the integrals:
347. $\int_{|z|=1} z \operatorname{tg} \pi z d z . \quad$ 348. $\int_{C} \frac{z d z}{(z-1)^{2}(z+2)}$, where $C: x^{2 / 3}+y^{2 / 3}=3^{2 / 3}$,
348. $\int_{\mid z==2} \frac{e^{z} d z}{z^{3}(z+1)}$
349. $\int_{|z-i|=3} \frac{e^{z^{2}}-1}{z^{3}-i z^{2}} d z . \quad 351 . \quad \int_{|z|=1 / 2} z^{2} \sin \frac{1}{z} d z$.
350. $\int_{|z|=\sqrt{3}} \frac{\sin \pi z}{z^{2}-z} d z$
351. $\int_{|z+1|=4} \frac{z d z}{e^{z}+3}$
352. $\int_{|z|=1} \frac{z^{2} d z}{\sin ^{3} z \cos z}$
353. $\int_{|z-i|=1} \frac{e^{z} d z}{z^{4}+2 z^{2}+1} \cdot 356 . \int_{|x|=-4} \frac{e^{i z} d z}{(z-n)^{3}}$.
359. $\int_{C} \frac{\sin \pi z}{\left(z^{2}-1\right)^{2}} d z, C: \frac{x^{2}}{4}+y^{2}=1 . \quad 360 . \int_{C} \frac{z+1}{z^{2}+2 z-3} d z, \quad C: x^{2}+y^{2}=16$.
361. $\int_{C} \frac{z \sin z}{(z-1)^{5}} d z, \quad C: \frac{x^{2}}{3}+\frac{y^{2}}{9}=1$. 362. $\int_{C} \frac{d z}{z^{4}+1} d z, \quad C: x^{2}+y^{2}=2 x$.
362. $\int_{|z|=1} z^{3} \sin \frac{1}{z} d z . \quad$ 364. $\int_{|z|=1 / 3}(z+1) e^{1 / z} d z$.
363. $\int_{|z|=2 / 3}\left(\sin \frac{1}{z^{2}}+e^{z^{2}} \cos z\right) d z$.
Residue of a function at the point at infinity
It is said that the function $f(z)$ is analytic at the point at infinity $z=\infty$, if the function
$$
\varphi(\zeta)=f\left(\frac{1}{\zeta}\right)
$$
is analytic at the point $\zeta=0$.
For example, the function $f(z)=\sin \frac{1}{z}$ is analytic at the point $z=\infty$, since the function
$$
\varphi(\zeta)=f\left(\frac{1}{\zeta}\right)=\sin \zeta
$$
is analytic at the point $\zeta=0$.
Let the function $f(z)$ be analytic in some neighborhood of the point at infinity (except the point $z=\infty$ itself).
The point $z=\infty$ is called an isolated singular point of the function $f(z)$ if there are no other singular points of the function $f(z)$ in some neighborhood of this point.
The function $f(z)=\frac{1}{\sin z}$ has a non-isolated singularity at infinity: the poles $z_{k}=k \pi$ of this function accumulate at infinity as $k \rightarrow \infty$.
It is said that $z=\infty$ is a removable singularity, a pole, or an essential singularity of the function $f(z)$ depending on whether $\lim _{z \rightarrow \infty} f(z)$ is finite, infinite, or does not exist.
The criteria for the type of singularity at infinity, related to the Laurent series expansion, differ from the criteria for finite singular points.
Theorem 1. If $z=\infty$ is a removable singularity of the function $f(z)$, then the Laurent series expansion of $f(z)$ in the neighborhood of this point does not contain positive powers of $z$; if $z=\infty$ is a pole, then this expansion contains a finite number of positive powers of $z$; in the case of an essential singularity - an infinite number of positive powers of $z$.
The Laurent series expansion of the function $f(z)$ in the neighborhood of the point at infinity is called the Laurent series expansion of $f(z)$ that converges everywhere outside a circle of sufficiently large radius $R$ centered at the point $z=0$ (except, perhaps, the point $z=\infty$ itself).
Let the function $f(z)$ be analytic in some neighborhood of the point $z=\infty$ (except, perhaps, the point $z=\infty$ itself).
The residue of the function $f(z)$ at infinity is defined as
$$
\operatorname{res} f(\infty)=\frac{1}{2 \pi i} \int_{\gamma^{-}} f(z) d z
$$
where $\gamma^{-}$ is a sufficiently large circle $|z|=\rho$, traversed clockwise (so that the neighborhood of the point $z=\infty$ remains on the left, as in the case of a finite point $z=a$).
From this definition, it follows that the residue of the function at infinity is equal to the coefficient of $z^{-1}$ in the Laurent series expansion of $f(z)$ in the neighborhood of $z=\infty$, taken with the opposite sign:
$$
\operatorname{res} f(\infty)=-c_{-1}
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 6. Compute the integral
$$
I=\int_{|z|=2} \frac{d z}{1+z^{4}}
$$
|
The poles (finite) of the integrand
$$
f(z)=\frac{1}{1+z^{4}}
$$
are the roots $z_{1}, z_{2}, z_{3}, z_{4}$ of the equation $z^{4}=-1$, which all lie inside the circle $|z|=2$. The function $f(z)=\frac{1}{1+z^{4}}$ has an expansion in the neighborhood of the infinitely distant point
$$
f(z)=\frac{1}{1+z^{4}}=\frac{1}{z^{4}} \frac{1}{1+\frac{1}{z^{4}}}=\frac{1}{z^{4}}-\frac{1}{z^{8}}+\frac{1}{z^{12}}-\cdots
$$
from which it is clear that res $f(\infty)=-c_{-1}=0$. By the equality (3)
$$
I=2 \pi i \sum_{k=1}^{4} \operatorname{res} f\left(z_{k}\right)=-2 \pi i \text { res } f(\infty)=0
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2. Find the logarithmic residue of the function
$$
f(z)=\frac{\operatorname{ch} z}{e^{i z}-1}
$$
with respect to the contour $C:|z|=8$.
|
Solution. We find the zeros $z_{k}$ of the function $f(z)$. For this, we solve the equation $\cosh z=0$ or $e^{z}+e^{-z}=0$. Writing the last equation as $e^{2 z}=-1$, we find
$2 z=\operatorname{Ln}(-1)=(2 k+1) \pi i$, so $z_{k}=\frac{2 k+1}{2} \pi i(k=0, \pm 1, \pm 2, \ldots)$ (all zeros are simple). To find the poles of the function $f(z)$, we solve the equation $e^{i z}-1=0$ or $e^{i z}=1$. We have $i z=\operatorname{Ln} 1=2 m \pi i, z_{m}=2 m \pi(m=0, \pm 1, \pm 2, \ldots)$. In the circle $|z|<8$ there are zeros
$$
z_{k}=\frac{2 k+1}{2} \pi i \quad(k=0, \pm 1, \pm 2,-3)
$$
and simple poles
$$
z_{m}=2 m \pi \quad(m=0, \pm 1)
$$
of the function $f(z)$. The number of zeros $N=6$, the number of poles $P=3$. Therefore,
$$
\frac{1}{2 \pi i} \int_{|z|=8} \frac{f^{\prime}(z)}{f(z)} d z=6-3=3
$$
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 3. Find the logarithmic residue of the function
$$
f(z)=\frac{1+z^{2}}{1-\cos 2 \pi z}
$$
with respect to the circle $|z|=\pi$.
|
Solution. Setting $1+z^{2}=0$, we find two simple zeros of the function $f(z): a_{1}=-i, a_{2}=i$. Setting $1-\cos 2 \pi z=0$, we find the poles of the function $f(z): z_{n}=n, n=0, \pm 1, \pm 2, \ldots$. The multiplicity of the poles is $k=2$.
In the circle $|z|<\pi$, the function has two simple zeros $a_{1}=-i, a_{2}=i$ and seven double poles
$$
z_{1}=-3, \quad z_{2}=-2, \quad z_{3}=-1, \quad z_{4}=0, \quad z_{5}=1, \quad z_{6}=2, \quad z_{7}=3
$$
Thus, $N=2$ and $P=7$. By the logarithmic residue theorem, the logarithmic residue of the function $f(z)$ with respect to the circle $|z|=\pi$ is
$$
\frac{1}{2 \pi i} \int_{|z|=\pi} \frac{f^{\prime}(z)}{f(z)} d z=2-7 \cdot 2=-12
$$
## Problems for Independent Solution
Find the logarithmic residues of the given functions with respect to the specified contours:
429. $f(z)=\frac{z}{1+z^{3}}$,
$C:|z|=2$.
430. $f(z)=\cos z+\sin z$,
$C:|z|=4$.
431. $f(z)=\left(e^{z}-2\right)^{2}$,
$C:|z|=8$.
432. $f(z)=\operatorname{th} z$,
$C:|z|=8$.
433. $f(z)=\operatorname{tg}^{3} z$,
$C:|z|=6$.
434. $f(z)=1-\operatorname{th}^{2} z$,
$C:|z|=2$.
Principle of the Argument. The logarithmic residue of the function $f(z)$ with respect to a closed contour $C$ is equal to the increment $\Delta_{C} \operatorname{Arg} f(z)$ of the argument of $f(z)$ when the contour $C$ is traversed, divided by $2 \pi$:
$$
\frac{1}{2 \pi i} \int_{C} \frac{f^{\prime}(z)}{f(z)} d z=\frac{1}{2 \pi} \Delta_{C} \operatorname{Arg} f(z)
$$
Therefore, the difference between the number of zeros and poles of the function $f(z)$ enclosed in the domain $D$ is
$$
N-P=\frac{1}{2 \pi} \Delta_{C} \operatorname{Arg} f(z)
$$
In other words, the difference $N-P$ is equal to the number of turns made by the vector in the $w$-plane from the point $w=0$ to the point $w=f(z)$ as the point $z$ traces the contour $C$ (the number of turns is positive if the vector rotates counterclockwise and negative otherwise).
In the special case where the function $w=f(z)$ is analytic in the domain $D$ and on its boundary $C$, where it does not vanish, the logarithmic residue of $f(z)$ with respect to $C$ gives the number of zeros of $f(z)$ in $D$, which is equal to the change in $\operatorname{Arg} f(z)$ when the contour $C$ is traversed, divided by $2 \pi$:
$$
\frac{1}{2 \pi i} \int_{C} \frac{f^{\prime}(z)}{f(z)} d z=\frac{1}{2 \pi} \Delta_{C} \operatorname{Arg} f(z)=N
$$
This is the case, for example, for a polynomial $Q_{n}(z)=\sum_{k=0}^{n} a_{k} z^{k}$.
|
-12
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 4. Find the number of roots in the right half-plane $\operatorname{Re} z>0$ of the equation
$$
Q_{5}(z) \equiv z^{5}+z^{4}+2 z^{3}-8 z-1=0
$$
|
Solution. By the argument principle, the number of zeros inside the contour $C$ is
$$
N=\frac{1}{2 \pi} \Delta_{C} \operatorname{Arg} Q_{5}(z)
$$
where the contour $C$ consists of the semicircle $C_{R}:|z|=R, \operatorname{Re} z>0$, and its diameter on the imaginary axis; the radius $R$ is taken to be so large that all the zeros of the polynomial $Q_{5}(z)$, located in the right half-plane, fall inside the semicircle $|z|>0$. We have
$$
Q_{5}(z)=z^{5}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)
$$
Hence,
$$
\operatorname{Arg} Q_{5}(z)=\operatorname{Arg}\left[z^{3}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)\right]=
$$
$$
\begin{aligned}
& =\operatorname{Arg} z^{5}+\operatorname{Arg}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)= \\
& =5 \operatorname{Arg} z+\operatorname{Arg}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)
\end{aligned}
$$
The increment of the argument of $Q_{5}(z)$ when traversing the semicircle $C_{R}$ in the positive direction will be
$$
\Delta_{C_{R}} \operatorname{Arg} Q_{5}(z)=5 \Delta_{C_{R}} \operatorname{Arg} z+\Delta_{C_{R}} \operatorname{Arg}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)
$$
In this equality, we take the limit as $R \rightarrow \infty$:
$$
\lim _{R \rightarrow \infty} \Delta_{C_{R}} \operatorname{Arg} Q_{5}(z)=5 \lim _{R \rightarrow \infty} \Delta_{C_{R}} \operatorname{Arg} z+\lim _{R \rightarrow \infty} \Delta_{C_{R}} \operatorname{Arg}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)
$$
Both limits on the right-hand side exist and are equal to
$$
\lim _{R \rightarrow \infty} \Delta_{C_{R}} \operatorname{Arg} z=\pi, \quad \lim _{R \rightarrow \infty} \Delta_{C_{R}} \operatorname{Arg}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)=0
$$
Thus,
$$
\lim _{R \rightarrow \infty} \Delta_{C_{R}} \operatorname{Arg} Q_{5}(z)=5 \pi
$$
Now let the point $z$ move along the imaginary axis from $z=i R$ to $z=-i R$. Let $z=i t, -R \leqslant t \leqslant R$. Then
$$
\dot{Q}_{5}(i t)=u(t)+i v(t)=t^{4}-1+i\left(t^{5}-2 t^{3}-8 t\right)
$$
from which
$$
\left\{\begin{array}{l}
u=t^{4}-1 \\
v=t^{5}-2 t^{3}-8 t
\end{array}\right.
$$
These are the parametric equations of the curve described by the point $w=Q_{5}(z)$ in the $(\boldsymbol{u}, \boldsymbol{v})$ plane as the point $z$ traverses the imaginary axis from top to bottom. To construct this curve, we find the points of its intersection with the coordinate axes $O u$ and $O v$. Setting $u$ and $v$ to zero, we get respectively
$$
\begin{aligned}
& t^{4}-1=0, \quad \text { or } t= \pm 1 \\
& t^{5}-2 t^{3}-8 t=0, \text { or } t= \pm 2, \quad t=0
\end{aligned}
$$
Note that equations (2) and (3) do not have common real roots, so the polynomial $Q_{5}(z)$ has no zeros on the imaginary axis. Therefore, the application of the argument principle to the contour is valid. The roots of equations (2) and (3) are arranged in descending order, i.e., in the order of traversal of the contour, and we find the corresponding values of $\boldsymbol{u}$ and $v$:
| № | $t$ | $u$ | $v$ |
| :---: | ---: | ---: | ---: |
| 1 | 2 | 15 | 0 |
| 2 | 1 | 0 | -9 |
| 3 | 0 | -1 | 0 |
| 4 | -1 | 0 | 9 |
| 5 | -2 | 15 | 0 |
Figure 12
Further,
$$
\begin{aligned}
& \lim _{t \rightarrow \pm \infty} u=+\infty \\
& \lim _{t \rightarrow \pm \infty} v= \pm \infty
\end{aligned}
$$
These data allow us to construct the curve of interest (Figure 12).
From Figure 12, it is clear that the vector $w=Q_{5}(z)$ will rotate by an angle $\varphi=3 \pi$ in the negative direction. Therefore,
$\Delta_{C} \operatorname{Arg}_{5}(z)=5 \pi-3 \pi=2 \pi$,
from which the number of zeros in the right half-plane will be
$$
N=\frac{2 \pi}{2 \pi}=1
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 5. Find the number of roots of the equation
$$
Q_{7}(z) \equiv z^{7}-2 z-5=0
$$
in the right half-plane.
|
Solution. We choose the contour $C$ as indicated in Example 4. Then $\Delta_{C_{R}} \operatorname{Arg} Q_{7}(z)=\Delta_{C_{R}} \operatorname{Arg}\left(z^{7}-2 z-5\right)=$
$$
\begin{aligned}
& =\Delta_{C_{R}} \operatorname{Arg}\left[z^{7}\left(1-\frac{2}{z^{6}}-\frac{5}{z^{7}}\right)\right]=7 \Delta_{C_{R}} \operatorname{Arg} z+\Delta_{C_{R}} \operatorname{Arg}\left(1-\frac{2}{z^{6}}-\frac{5}{z^{7}}\right)= \\
& =7 \pi+\Delta_{C_{R}} \operatorname{Arg}\left(1-\frac{2}{z^{6}}-\frac{5}{z^{7}}\right) \rightarrow 7 \pi \quad \text { as } \quad R \rightarrow \infty
\end{aligned}
$$
Fig. 13
Let $z=$ it $(-R \leqslant t \leqslant R)$. Then
$$
Q_{7}(i t)=u(t)+i v(t)=-5+i\left(-t^{7}-2 t\right)
$$
from which
$$
\left\{\begin{array}{l}
u=-5 \\
v=-t\left(t^{6}+2\right)
\end{array}\right.
$$
Since $u \neq 0$, the application of the argument principle is valid ( $Q_{7}(z)$ has no zeros on the imaginary axis). This line is a straight line (Fig. 13). The vector $w=Q_{7}(z)$ makes a turn in the negative direction by $\pi$ radians. Therefore,
$$
\Delta_{\dot{c}_{R}} \operatorname{Arg} Q_{1}(z) \xrightarrow[R \rightarrow \infty]{\longrightarrow} 7 \pi-\pi=6 \pi
$$
and
$$
N=\frac{6 \pi}{2 \pi}=3
$$
## Problems for Independent Solution
Determine the number of roots in the right half-plane for the following equations:
435. $z^{4}+2 z^{3}+3 z^{2}+z+2=0$. 436. $z^{3}-2 z-5=0$.
436. $z^{3}-4 z^{2}+5=0$.
437. $2 z^{3}-z^{2}-7 z+5=0$.
438. $z^{5}+5 z^{4}-5=0$.
440: $z^{12}-z+1=0$.
Rouche's Theorem. Suppose functions $f(z)$ and $\varphi(z)$, analytic in the closed domain $\bar{D}$, bounded by the contour $C$, satisfy the inequality $|f(z)|>|\varphi(z)|$ at all points of this contour. Then their sum $F(z)=f(z)+\varphi(z)$ and the function $f(z)$ have the same number of zeros (counting multiplicities) in the domain $D$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 6. Find the number of zeros of the function
$$
F(z)=z^{8}-4 z^{5}+z^{2}-1
$$
inside the unit circle $|z|<1$.
|
Solution. Let us represent the function $F(z)$ as the sum of two functions $f(z)$ and $\varphi(z)$, which we choose, for example, as follows:
$$
f(z)=-4 z^{5}, \quad \varphi(z)=z^{8}+z^{2}-1
$$
Then on the circle $|z|=1$ we will have
$$
\begin{aligned}
& |f(z)|=\left|-4 z^{5}\right|=4 \\
& |\varphi(z)|=\left|z^{8}+z^{2}-1\right| \leqslant\left|z^{8}\right|+\left|z^{2}\right|+1=3
\end{aligned}
$$
Thus, on the boundary $|z|=1$ of the circle, the inequality $|f(z)|>|\varphi(z)|$ holds. The function $f(z)=-4 z^{5}$ has a fivefold zero at the origin. By Rouché's theorem, the function
$$
F(z)=f(z)+\varphi(z)=z^{8}-4 z^{5}+z^{2}-1
$$
has five zeros in the circle $|z|<1$. Note that a different choice of functions $f(z)$ and $\varphi(z)$ is also possible, for example, such as:
$$
f(z)=z^{8}-4 z^{5}, \quad \varphi(z)=z^{2}-1
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 7. Determine the number of roots of the equation
$$
z^{6}-6 z+10=0
$$
inside the circle $|z|<1$.
|
Solution. Let, for example, $f(z)=10$ and $\varphi(z)=z^{6}-6 z$. On the circle $|z|=1$ we have
$$
|f(z)|=10, \quad|\varphi(z)|=\left|z^{6}-6 z\right| \leqslant\left|z^{6}\right|+6|z|=7
$$
Thus, in all points of the circle $|z|=1$, the inequality $|f(z)|>|\varphi(z)|$ holds. The function $f(z)=10$ has no zeros inside the circle $|z|<1$, and therefore, by Rouché's theorem, the function $z^{6}-6 z+10$ also has no zeros.
## Problems for Independent Solution
Using Rouché's theorem, find the number of roots of the given equations in the specified domains:
441. $z^{4}-3 z^{3}-1=0, \quad|z|<2$.
442. $z^{3}+z+1=0, \quad|z|<\frac{1}{2}$.
443. $z^{5}+z^{2}+1=0, \quad|z|<2$.
444. $z^{8}+6 z+10=0,|z|<1$.
445. $27 z^{11}-18 z+10=0,|z|<1$.
446. $z^{8}-6 z^{6}-z^{3}+2=0,|z|<1$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 8. How many roots of the equation
$$
z^{4}-5 z+1=0
$$
lie in the annulus $1<|z|<2 ?$
|
Solution. Let $N$ be the number of roots of equation (4) in the ring $1<|\varphi(z)|$, since $|f(z)|=|-5 z|=5,|\varphi(z)|=\left|z^{4}+1\right| \leqslant$ $\left|z^{4}\right|+1=2$. The function $f(z)=-5 z$ has one root in the circle $|z|<1$, and thus $N_{1}=1$.
In the circle $|z|<2$, $|f(z)|>\left|\varphi(z)\right|$, since $|f(z)|=\left|z^{4}\right|=2^{4}=16,|\varphi(z)|=|1-5 z| \leqslant 1+5|z|=11$. The function $f(z)=z^{4}$ has four roots in the circle $|z|<2$, and therefore $N_{2}=4$.
The number of roots of equation (4) in the ring $1<|z|<2$ will be $N=4-1=3$.
## Problems for Independent Solution
Determine the number of roots of the given equations in the specified rings:
447. $4 z^{4}-29 z^{2}+25=0, \quad 2<|z|<3$.
448. $z^{7}-5 z^{4}+z^{2}-2=0,1<|z|<2$.
449. $z^{6}-8 z+10=0, \quad 1<|z|<3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 9. Find the number of roots of the equation
$$
z^{2}-a e^{z}=0, \quad \text { where } \quad 0<a<e^{-1}
$$
in the unit circle $|z|<1$.
|
Solution. Let $f(z)=z^{2}$ and $\varphi(z)=-a e^{z}$. On the circle $|z|=1$ we have
$$
\begin{aligned}
& |f(z)|=\left|z^{2}\right|=1 \\
& |\varphi(z)|=\left|-a e^{z}\right|=a\left|e^{z}\right|=a\left|e^{x+i y}\right|=a e^{x} \leqslant a e|\varphi(z)|$, if $|z|=1$. The function $f(z)=z^{2}$ in the circle $|z|0, \quad \text { since } \quad 00, \quad \text { since } \quad a1), \quad|z|\frac{e^{R}}{R^{n}}, \quad|z|<R$.
452. $z^{2}-\cos z=0, \quad|z|<2$.
453. $z^{4}-\sin z=0, \quad|z|<\pi$.
454. $z^{2}+\operatorname{ch} i z=0, \quad|z|<0.5$.
455. $\operatorname{ch} z=z^{2}-4 z,|z|<1$.
456. $2^{z}=4 z, \quad|z|<1$.
Note: There are some parts in the original text that seem to be incomplete or contain errors, such as the conditions and inequalities. Please review the original text for clarity.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 10. Find the number of roots of the equation
$$
\lambda-\boldsymbol{z}-e^{-z}=0, \quad \lambda>1
$$
in the right half-plane $\operatorname{Re} z>0$.
|
Solution. Consider the contour composed of the segment $[-i R, i R]$ and the right semicircle $|z|=R$. Let $f(z)=z-\lambda$ and $\varphi(z)=e^{-z}$. On the segment $[-i R, i R]$, where $z=i y$, we have
$$
\begin{aligned}
& |f(z)|=|i y-\lambda|=\sqrt{\lambda^{2}+y^{2}} \geqslant \sqrt{\lambda^{2}}=\lambda>1 \\
& |\varphi(z)|=\left|e^{-z}\right|=\left|e^{-i y}\right|=1
\end{aligned}
$$
and, consequently, $|f(z)|>|\varphi(z)|$.
On the semicircle $|z|=R$, where $\operatorname{Re} z=x>0$ for sufficiently large $R$ $(R>\lambda+1)$, we have $|f(z)|>|\varphi(z)|$, since
$$
\begin{aligned}
& |f(z)|=|z-\lambda| \geqslant|z|-\lambda=R-\lambda>1 \\
& |\varphi(z)|=\left|e^{-z}\right|=\left|e^{-x-i y}\right|=\left|e^{-x} e^{-i y}\right|=e^{-x}\left|e^{-i y}\right|=e^{-x} \leqslant 1 \quad(x>0)
\end{aligned}
$$
By Rouché's theorem, inside the specified contour for any sufficiently large $R$, the given equation has as many roots as the equation $f(z)=z-\lambda=0$, i.e., one root. Therefore, in the entire right half-plane, the given equation has a unique root.
## Problems for Independent Solution
457. Show that the equation $z e^{\lambda-z}=1$, where $\lambda>1$, has a unique real and positive root in the unit disk $|z| \leqslant 1$.
458. Show that the equation $1+z+\alpha z^{n}=0$, where $n$ is a natural number greater than one, has at least one root in the disk $|z| \leqslant 2$ for any $\alpha$.
459. Let $f(z)$ and $\varphi(z)$ be functions that are analytic in some neighborhood of the point $a$, and $C$ be a circle centered at the point $a$ such that along the circumference of this circle we have
$$
|\alpha f(z)|+|\beta \varphi(z)|<r .
$$
Show that the equation $F(z)=z-a-\alpha f(z)-\beta f(z)=0$ has one and only one root inside the circle $C$.
## CHAPTER
## Conformal Mappings
## § 13. Conformal Mappings
## $1^{\circ}$. Concept of Conformal Mapping.
Definition. A mapping of a neighborhood of a point $z_{0}$ onto a neighborhood of a point $w_{0}$, realized by the function $w=f(z)$, is called conformal if at the point $z_{0}$ it has the property of preserving angles between curves and the constancy of dilations (Fig. 14).
This means that:
1) if under the mapping $w=f(z)$ the curves $\gamma_{1}$ and $\gamma_{2}$ are transformed into the curves $\Gamma_{1}$ and $\Gamma_{2}$, respectively, then the angle $\varphi$ between the tangents $k_{1}$ and $k_{2}$ to the curves $\gamma_{1}$ and $\gamma_{2}$ at the point $z_{0}$ will be equal to the angle $\Phi$ between the corresponding tangents $K_{1}$ and $K_{2}$ to the curves $\Gamma_{1}$ and $\Gamma_{2}$ at the point $w_{0}$, i.e., $\Phi=\varphi$.
2) if in the complex plane $z$ we take an infinitesimally small circle centered at the point $z_{0}$, then in the plane $w$ it will correspond to an infinitesimally small circle centered at the point $w_{0}$.
Therefore, it is said that a conformal mapping has the property of angle conservation and similarity in the small.
If under the mapping $w=f(z)$ the angles between corresponding directions are equal not only in magnitude but also in the direction of measurement, then such a mapping is called a conformal mapping of the first kind.
A conformal mapping in which angles are preserved only in absolute magnitude but the direction of their measurement is reversed is called a conformal mapping of the second kind.
A simple example of a conformal mapping of the first kind is the mapping $w=z$, and of the second kind is the mapping $w=\bar{z}$.
In the following, we will consider only conformal mappings of the first kind.
A mapping $w=f(z)$ is called conformal in a domain $D$ if it is conformal at every point of this domain.
Criterion for Conformality. For the mapping $w=f(z)$ to be conformal in a domain $D$, it is necessary and sufficient that in this domain the function $w=f(z)$ is univalent and analytic, and $f^{\prime}(z) \neq 0$ for all $z \in D$.
If we do not assume the univalence of $f(z)$, then the mapping realized by this function will not be one-to-one, and thus will not be conformal. For example, the function $w=z^{4}$, defined in the half-ring $1 \leqslant|z| \leqslant 2,0 \leqslant \arg z \leqslant \pi$, is analytic in it, and, moreover, the condition $w^{\prime}=4 z^{3} \neq 0$ is satisfied everywhere in the half-ring. However, the function $w=z^{4}$ maps the given half-ring onto the region $1 \leqslant|w| \leqslant 16,0 \leqslant \arg w \leqslant 4 \pi$, i.e., a region that twice covers the corresponding ring on the $w$-plane, which violates the one-to-one correspondence.
Fig. 14
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 1. Find the approximate value of the smallest characteristic number of the kernel by the Ritz method
$$
K(x, t)=x t ; \quad a=0, b=1
$$
|
Solution. As the coordinate system of functions $\psi_{n}(x)$, we choose the system of Legendre polynomials: $\psi_{n}(x)=P_{n}(2 x-1)$. In formula (1), we limit ourselves to two terms, so that
$$
\varphi_{2}(x)=a_{1} \cdot P_{0}(2 x-1)+a_{2} \cdot P_{1}(2 x-1) .
$$
Noting that
$$
\psi_{1} \equiv P_{0}(2 x-1)=1 ; \quad \psi_{2} \equiv P_{1}(2 x-1)=2 x-1,
$$
we find
$$
\begin{gathered}
\left(\psi_{1}, \psi_{1}\right)=\int_{0}^{1} d x=1, \quad\left(\psi_{1}, \psi_{2}\right)=\left(\psi_{2}, \psi_{1}\right)=\int_{0}^{1}(2 x-1) d x=0 \\
\left(\psi_{2}, \psi_{2}\right)=\int_{0}^{1}(2 x-1)^{2} d x=\frac{1}{3}
\end{gathered}
$$
Further,
$$
\begin{aligned}
& \left(K \psi_{1}, \psi_{1}\right)=\int_{0}^{1}\left(\int_{0}^{1} K(x, t) \psi_{1}(t) d t\right) \psi_{1}(x) d x=\int_{0}^{1} \int_{0}^{1} x t d x d t=\frac{1}{4} \\
& \left(K \psi_{1}, \psi_{2}\right)=\int_{0}^{1} \int_{0}^{1} x t(2 x-1) d x d t=\frac{1}{12} \\
& \left(K \psi_{2}, \psi_{2}\right)=\int_{0}^{1} \int_{0}^{1} x t(2 t-1)(2 x-1) d x d t=\frac{1}{36}
\end{aligned}
$$
In this case, system (3) takes the form
$$
\left|\begin{array}{cc}
\frac{1}{4}-\sigma & \frac{1}{12} \\
\frac{1}{12} & \frac{1}{36}-\frac{1}{3} \sigma
\end{array}\right|=0
$$
or
$$
\sigma^{2}-\sigma\left(\frac{1}{12}+\frac{1}{4}\right)=0
$$
From this, $\sigma_{1}=0, \sigma_{2}=\frac{1}{3}$. The largest eigenvalue $\sigma_{2}=\frac{1}{3}$, so the smallest characteristic number $\lambda=\frac{1}{\sigma_{2}}=3$.
## Problems for Independent Solution
Using the Ritz method, find the smallest characteristic numbers of the kernels ( $a=0$, $b=1):$
340. $K(x, t)=x^{2} t^{2} . \quad$ 341. $K(x, t)= \begin{cases}t, & x \geqslant t, \\ x, & x \leqslant t .\end{cases}$
341. $K(x, t)= \begin{cases}\frac{1}{2} x(2-t), & x \leqslant t, \\ \frac{1}{2} t(2-x), & x \geqslant t .\end{cases}$
$2^{\circ}$. Method of Traces. We call the $m$-th trace of the kernel $K(x, t)$ the number
$$
A_{m}=\int_{a}^{b} K_{m}(t, t) d t
$$
where $K_{m}(x, t)$ denotes the $m$-th iterated kernel.
For the smallest characteristic number $\lambda_{1}$, for sufficiently large $m$, the following approximate formula holds:
$$
\left|\lambda_{1}\right| \approx \sqrt{\frac{A_{2 m}}{A_{2 m+2}}}
$$
Formula (4) gives the value of $\left|\lambda_{1}\right|$ with a surplus.
Traces of even order for a symmetric kernel are computed by the formula
$$
A_{2 m}=\int_{a}^{b} \int_{a}^{b} K_{m}^{2}(x, t) d x d t=2 \int_{a}^{b} \int_{a}^{x} K_{m}^{2}(x, t) d t d x
$$
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 3. Using the Kellogg method, calculate the smallest characteristic number of the kernel $K(x, t)=x^{2} t^{2}, 0 \leqslant x, t \leqslant 1$.
|
Solution. Let $\omega(x)=x$. Then
$$
\begin{aligned}
& \omega_{1}(x)=\int_{0}^{1} x^{2} t^{2} t d t=\frac{x^{2}}{4} \\
& \omega_{2}(x)=\int_{0}^{1} x^{2} t^{4} \frac{1}{4} d t=\frac{1}{4} x^{2} \cdot \frac{1}{5} \\
& \omega_{3}(x)=\int_{0}^{1} \frac{1}{4 \cdot 5} x^{2} t^{4} d t=\frac{1}{4 \cdot 5^{2}} x^{2} \\
& \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
& \omega_{n}(x)=\frac{1}{4 \cdot 5^{n-1}} x^{2}
\end{aligned}
$$
Next,
$$
\left\|\omega_{n}(x)\right\|=\frac{1}{4 \cdot 5^{n-1}} \sqrt{\int_{0}^{1} x^{4} d x}=\frac{1}{4 \cdot 5^{n-1}} \cdot \frac{1}{\sqrt{5}}
$$
Thus, according to (7),
$$
\lambda_{1} \approx \frac{\frac{1}{4} \cdot \frac{1}{5^{n-2}} \cdot \frac{1}{\sqrt{5}}}{\frac{1}{4} \cdot \frac{1}{5^{n-1}} \cdot \frac{1}{\sqrt{5}}}=5
$$
## Problems for Independent Solution
Using the Kellogg method, find the smallest characteristic numbers of the following kernels:
347. $K(x, t)=x t$;
$0 \leqslant x, t \leqslant 1$.
348. $K(x, t)=\sin x \sin t$;
$-\pi \leqslant x, t \leqslant \pi$.
349. $K(x, t)=\left\{\begin{array}{ll}t, & x \geqslant t ; \\ x, & x \leqslant t ;\end{array} \quad 0 \leqslant x, t \leqslant 1\right.$.
350. $K(x, t)=\left\{\begin{array}{l}\frac{1}{2} x(2-t), \quad x \leqslant t, \\ \frac{1}{2} t(2-x), \quad x \geqslant t,\end{array} \quad 0 \leqslant x, t \leqslant 1\right.$.
Stability of a Compressed Rod (Longitudinal Bending of a Rod)
The equation of the deflected elastic line of the rod is
$$
\frac{d}{d x}\left(E I \frac{d y}{d x}\right)=M
$$
where $\boldsymbol{M}$ is the bending moment, $I$ is the moment of inertia of the cross-section of the rod with abscissa $x$, and $E$ is the Young's modulus.
Consider the case when the rod is compressed by forces applied to its ends. The magnitude of each of these forces is denoted by $P$. Then $M=-\boldsymbol{P} \cdot \boldsymbol{y}$ and equation (9) becomes
$$
\frac{d}{d x}\left(E I \frac{d y}{d x}\right)+P \cdot y=0
$$
The ends of the rod do not move in the direction perpendicular to the rod, so
$$
y(0)=y(l)=0,
$$
where $l$ is the length of the rod. Dividing both sides of equation (10) by $E$ and setting $\frac{\boldsymbol{P}}{\boldsymbol{E}}=\lambda$, we get
$$
\frac{d}{d x}\left(r \frac{d y}{d x}\right)+\lambda y=0
$$
Let $G(x, t)$ be the Green's function of the differential equation (12) with boundary conditions (11). Then the problem (12)-(11) is equivalent to the problem of solving the homogeneous Fredholm integral equation of the second kind
$$
y(x)=\lambda \int_{0}^{l} G(x, t) y(t) d t
$$
The kernel $G(x, t)$ of equation (13) is symmetric as the Green's function of a self-adjoint boundary value problem. Thus, the deflection $y(x)$ of the compressed rod satisfies the integral equation (13). For an arbitrarily chosen force $P$, the number $\lambda=\frac{P}{E}$ will not be a characteristic number, and the solution of equation (13) will be the function $y(x) \equiv 0$. In other words, an arbitrarily chosen compressive force will leave the rod straight. In the case when
$P=\lambda_{n} E$, where $\lambda_{n}$ is a characteristic number of the kernel $G(x, t)$, equation (13) will have a non-zero solution, which corresponds to the bending of the rod, i.e., the loss of its stability.
The smallest force $P$ at which the rod loses stability is called the critical force. It is obviously equal to $P_{\text {cr }}=\lambda_{1} E$, where $\lambda_{1}$ is the smallest characteristic number of equation (13). For practical purposes, a sufficiently good approximate formula is
$$
\lambda_{1} \approx \frac{1}{\sqrt{\int_{0}^{1} \int_{0}^{1} G^{2}(x, t) d x d t}}
$$
which gives $\lambda_{1}$ with a deficiency.
|
5
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 6. Find the length of the arc of the curve $y=\arcsin \sqrt{x}-\sqrt{x-x^{2}}$ from $x_{1}=0$ to $x_{2}=1$.
|
Solution. We have $y^{\prime}=\frac{1}{2 \sqrt{x} \sqrt{1-x}}-\frac{1-2 x}{2 \sqrt{x-x^{2}}}=\frac{x}{\sqrt{x-x^{2}}}=$ $=\sqrt{\frac{x}{1-x}}$ $L=\int_{0}^{1} \sqrt{1+y^{\prime 2}} d x=\int_{0}^{1} \sqrt{1+\frac{x}{1-x}} d x=\int_{0}^{1} \frac{1}{\sqrt{1-x}} d x=-\left.2 \sqrt{1-x}\right|_{0} ^{1}=2$.
Answer. $L=2$.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 6. Calculate the areas of figures bounded by the curves:
a) $\left\{\begin{array}{l}y=x \sqrt{9-x^{2}} \\ y=0(0 \leqslant x \leqslant 3)\end{array}\right.$
b) $\left\{\begin{array}{l}y=2 x-x^{2}+3 \\ y=x^{2}-4 x+3\end{array}\right.$
|
Solution. We will construct the corresponding regions (Fig. 3.8) and determine the appropriate limits of integration from them, omitting the systems of inequalities. Thus:
a) $S=\int_{D} \int d x d y=\int_{0}^{3} d x \int_{0}^{x \sqrt{9-x^{2}}} d y=$
$=\left.\int_{0}^{3} d x \cdot y\right|_{0} ^{x \sqrt{9-x^{2}}}=\int_{0}^{3} x \sqrt{9-x^{2}} d x=$
$=-\int_{0}^{3} \frac{1}{2}\left(9-x^{2}\right)^{1 / 2} d\left(9-x^{2}\right)=$
$=-\left.\frac{1}{3}\left(9-x^{2}\right)^{3 / 2}\right|_{0} ^{3}=9 ;$
b) $S=\iint_{D} d x d y=$
$$
=\int_{0}^{3}\left(2 x-x^{2}+3-x^{2}+4 x-3\right) d x=
$$
Fig. 3.9
$$
=\left.\left(-\frac{2 x^{3}}{3}+3 x^{2}\right)\right|_{0} ^{3}=9
$$
Answer. a) $S=9 ;$ b) $S=9$.
|
9
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 3. Calculate the mass of the surface $z=x y$, located inside the cylinder $x^{2}+\frac{y^{2}}{4}=1$, if the density is $\rho=\frac{|z|}{\sqrt{1+x^{2}+y^{2}}}$.
|
Solution. Given the symmetries of the integration region $\sigma$: $x^{2}+\frac{y^{2}}{4} \leqslant 1$, the equation of the surface, and the density function, it is sufficient to compute the integral over one quarter of the region and multiply the result by 4: $m=4 \int_{D} \rho(x, y, z) d \sigma$, where $D$ is the quarter of the region $\sigma$ lying in the first quadrant of the $O x y$ plane. Further, we have:
$$
\begin{gathered}
z=x y, z_{x}^{\prime}=y, z_{y}^{\prime}=x, \sqrt{1+{z_{x}^{\prime}}^{2}+z_{y}^{\prime 2}}=\sqrt{1+x^{2}+y^{2}} \\
\rho d \sigma=\frac{|z| \sqrt{1+x^{2}+y^{2}}}{\sqrt{1+x^{2}+y^{2}}} d x d y=x y d x d y
\end{gathered}
$$
$$
\begin{array}{ll}
m=4 \cdot \int_{\substack{x^{2}+y^{2} / 4 \leqslant 1 \\
x \geqslant 0, y \geqslant 0}} x y d x d y=4 \int_{0}^{1} x d x \int_{0}^{2 \sqrt{1-x^{2}}} y d y= & \left.2 \int_{0}^{1} x d x \cdot y^{2}\right|_{0} ^{2 \sqrt{1-x^{2}}}= \\
& =8 \int_{0}^{1}\left(1-x^{2}\right) x d x=2 .
\end{array}
$$
Answer. $m=2$.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 1. Calculate the mass of the tetrahedron bounded by the planes $x=0, y=0, z=0$ and $x / 10+y / 8+z / 3=1$, if the density distribution of mass at each point is given by the function $\rho=(1+x / 10+y / 8+z / 3)^{-6}$.
|
Solution. We have $m=\iint_{W} \int \rho d V$. The triple integral is reduced to a double and a definite one (see point $4^{\circ}$):
$$
m=\iint_{D} d x d y \int_{0}^{z} \frac{d z}{\left(1+\frac{x}{10}+\frac{y}{8}+\frac{z}{3}\right)^{6}}
$$
The upper limit, or the exit point from the region, is the ordinate of the point on the plane $\frac{x}{10}+\frac{y}{8}+\frac{z}{3}=1$ (Fig. 3.16), i.e., $z=3\left(1-\frac{x}{10}-\frac{y}{8}\right)$. The double integral is extended over the triangle, which is the base of the tetrahedron formed by
Fig. 3.16 the lines $x=0, y=0, \frac{x}{10}+\frac{y}{8}=1$; therefore (some steps are omitted):
$$
\begin{gathered}
m=\left.\iint_{D} d x d y\left[-\frac{3}{5}\left(1+\frac{x}{10}+\frac{y}{8}+\frac{z}{3}\right)^{-5}\right]\right|_{0} ^{3(1-x / 10-y / 8)}= \\
=\frac{3}{5} \int_{0}^{10} d x \int_{0}^{8(1-x / 10)}\left[2^{-5}-\frac{1}{\left(1+\frac{x}{10}+\frac{y}{8}\right)^{5}}\right] d y= \\
=-\left.\frac{3}{5} \int_{0}^{10} d x\left(\frac{y}{32}+\frac{8}{4}\left(1+\frac{x}{10}+\frac{y}{8}\right)^{-4}\right)\right|_{0} ^{8(1-x / 10)}= \\
=\frac{3}{5} \int_{0}^{10}\left(\frac{3}{8}-\frac{x}{40}-2\left(1+\frac{x}{10}\right)^{-4}\right) d x= \\
=-\left.\frac{3}{5}\left(\frac{3}{8} x-\frac{x^{2}}{80}+\frac{20}{3}\left(1+\frac{x}{10}\right)^{-3}\right)\right|_{0} ^{10}=2
\end{gathered}
$$
Answer. $m=2$.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 8. Check the conditions of Green's theorem for the line integral $\int_{L} 2 x y d x + x^{2} d y$ and compute this integral along the parabola $y=\frac{x^{2}}{4}$ from the origin to the point $A(2,1)$.
|
Solution. We have $P(x, y)=2 x y, Q(x, y)=x^{2}$. These functions are defined, continuous, and differentiable at any point $(x, y)$ in the plane. We have $\frac{\partial P}{\partial y}=2 x, \frac{\partial Q}{\partial x}=2 x$. The conditions of Green's theorem are satisfied. Therefore, the given integral is independent of the path of integration. Moreover, the expression $2 x y d x+x^{2} d y$ represents the total differential of some function $U(x, y)$. It is not difficult to guess that $U(x, y)=x^{2} y$. Therefore,
$$
\int_{O A} 2 x y d x+x^{2} d y=\left.x^{2} y\right|_{(0,0)} ^{(2, \mathrm{I})}=4
$$
Direct calculation gives $\left(y=\frac{x^{2}}{4}, d y=\frac{x}{2} d x\right)$ :
$$
\int_{O A} 2 x y d x+x^{2} d y=\int_{0}^{2}\left(\frac{x^{3}}{2}+\frac{x^{3}}{2}\right) d x=\left.\frac{x^{4}}{4}\right|_{0} ^{2}=4
$$
The answers match.
Answer. 4.
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 10. Calculate the integral $I=\oint_{L}\left(x^{2}-y^{2}\right) d x+2 x y d y$, where $L-$ is the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
|
Solution. We apply Green's formula and compute the double integral, transitioning to "generalized" polar coordinates. We have: $P=x^{2}-y^{2}, Q=2 x y, \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=2 y+2 y=4 y$. Therefore,
$$
\begin{aligned}
& \int\left(x^{2}-y^{2}\right) d x+2 x y d y=4 \iint_{D} y d x d y= \\
& =\left\{\begin{aligned}
x & =a r \cos \varphi \\
y & =b r \cos \varphi, 0 \leqslant \varphi \leqslant 2 \pi, 0 \leqslant r \leqslant 1 \\
d x d y & =a b r d r d \varphi
\end{aligned}\right\}= \\
& =4 \iint_{D} a b^{2} r^{2} \sin \varphi d r d \varphi=4 a b^{2} \int_{0}^{2 \pi} \sin \varphi d \varphi \int_{0}^{1} r^{2} d r= \\
& =\left.\left.4 a b^{2} \cos \varphi\right|_{0} ^{2 \pi} \cdot \frac{r^{3}}{3}\right|_{0} ^{1}=0
\end{aligned}
$$
Answer. $I=0$.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 1. Investigate the convergence of the series $\sum_{n=2}^{\infty} \frac{18}{n^{2}+n-2}$ and, if possible, find its sum.
|
Solution. We have $\frac{18}{n^{2}+n-2}=\frac{6}{n-1}-\frac{6}{n+2}$. Let's form the partial sum and find its limit. Notice which terms cancel each other out (!). We have:
$$
\begin{aligned}
& S_{n}=u_{2}+u_{3}+\ldots+u_{n}= \\
& =6\left[\left(\frac{1}{1}-\frac{1}{A}\right)+\left(\frac{1}{2}-\frac{1}{b}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{A}-\frac{1}{7}\right)+\ldots\right. \\
& +\left(\frac{1}{n-4}-\frac{1 /}{n+1}\right)+\left(\frac{1 /}{n+3}-\frac{1}{n}\right)+ \\
& \left.+\left(\frac{1 /}{n-2}-\frac{1}{n+1}\right)+\left(\frac{1}{n-1}+\frac{1}{n+2}\right)\right]= \\
& =6\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}\right)
\end{aligned}
$$
ll
$$
\lim _{n \rightarrow \infty} S_{n}=6 \lim _{n \rightarrow \infty}\left(\frac{11}{6}-\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}\right)=11
$$
Answer. The given series converges, and its sum is $S=11$.
|
11
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. Find the most probable number of hits in the ring in five throws, if the probability of hitting the ring with the ball in one throw is $p=0.6$.
|
Solution. We have $n=5 ; p=0.6 ; q=0.4$. For the number $k_{0}$, we obtain the estimate: $5 \cdot 0.6-0.4 \leqslant k_{0} \leqslant 5 \cdot 0.6+0.6$, i.e., $2.6 \leqslant k_{0} \leqslant 3.6$. Since $k_{0}$ is an integer, then $k_{0}=3$.
Direct calculations of $p_{5}(k)$ lead to the values: $p_{5}(0)=$ $=0.01024 ; p_{5}(1)=0.0768 ; \quad p_{5}(2)=0.2304 ; p_{5}(3)=0.3456 ; p_{5}(4)=$ $=0.2592 ; p_{5}(5)=0.07776$. The largest of the numbers $p_{5}(k)-$ is $p_{5}(3)=$ $=0.3456$; it corresponds to the value $k_{0}=3$.
Answer. 3 .
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. Given a statistical series - the number of days missed due to illness by employees of a laboratory.
| Number of days | 0 | 2 | 3 | 4 | 5 | 7 | 10 | Total |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Number of employees | 7 | 3 | 5 | 2 | 5 | 6 | 2 | 30 |
Determine the average number of days missed due to illness per person and the dispersion of this number around the average.
|
Solution. Let's determine the sample mean of size $n=30(k=7)$ using formula (1):
$$
\bar{x}_{3}=\frac{1}{30}(7 \cdot 0+3 \cdot 2+5 \cdot 3+2 \cdot 4+5 \cdot 5+6 \cdot 7+2 \cdot 10)=\frac{116}{30}=3.87
$$
We will calculate the variance and standard deviation using formula (2):
$$
\begin{aligned}
\overline{x_{\mathrm{B}}^{2}}=\frac{1}{30}\left(7 \cdot 0^{2}+3 \cdot 2^{2}+5 \cdot 3^{2}+2 \cdot 4^{2}+5 \cdot 5^{2}+6 \cdot 7^{2}+2 \cdot\right. & \left.10^{2}\right)= \\
= & \frac{692}{30}=23.1
\end{aligned}
$$
We have $D_{\mathrm{B}}=\overline{x_{\mathrm{B}}^{2}}-\left(\bar{x}_{\mathrm{B}}\right)^{2}=23.1-3.87^{2}=8.3 ; \sigma=\sqrt{D}=2.8$. Rounding to the nearest whole number, we get that the average number (expected value) of absences in a year per person is 4 days, with a spread of 3 days.
Problem 2. The sample provided in the following table corresponds to the service time (in minutes) for a customer in a store:
| Time Interval | $34 \div 38$ | $38 \div 42$ | $42 \div 46$ | $46 \div 50$ | $50 \div 54$ | $54 \div 58$ | $\Sigma$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Number of Customers | 2 | 3 | 15 | 18 | 10 | 2 | 50 |
Provide a forecast of the service time by the store and the standard deviation of this time. Find the mode of the variation series.
Solution. We transition from the interval table to the variation series by replacing the time interval with the center of this interval:
$$
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline x_{i} & 36 & 40 & 44 & 48 & 52 & 56 & \Sigma \\
\hline n_{i} & 2 & 3 & 15 & 18 & 10 & 2 & 50 \\
\hline
\end{array}
$$
The mode is the variant with the maximum frequency; $M o=48$.
The numbers in the variation series are not convenient for calculations because they are large, so we transition to conditional variants with a new center $c=48$ and step (difference) $h=4$. Let $u_{i}=\frac{x_{i}-c}{h}=\frac{x_{i}-48}{4}$. We will calculate the sample mean and variance for the new variant. For convenience, we provide a table with intermediate calculations.
| $i$ | $x_{2}$ | $n_{2}$ | $u_{i}$ | $u_{2} n_{2}$ | $n_{i} u_{2}^{2}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 36 | 2 | -3 | -6 | 18 |
| 2 | 40 | 3 | -2 | -6 | 12 |
| 3 | 44 | 15 | -1 | -15 | 15 |
| 4 | 48 | 18 | 0 | 0 | 0 |
| 5 | 52 | 10 | 1 | 10 | 10 |
| 6 | 56 | 2 | 2 | 4 | 8 |
| $\Sigma$ | - | 50 | -3 | -13 | 63 |
Using the sums in the last row of the table, we calculate: $\bar{u}_{\mathrm{B}}=-\frac{13}{50}=-0.26, \overline{u_{\mathrm{B}}^{2}}=\frac{63}{50}=1.26, \quad D_{\mathrm{B}}=1.26-0.0676=1.1924$, $\sigma(x)=1.1$.
The reverse transition to the variants $x$ is carried out using formulas that follow from the formulas in point $8^{\circ}$:
$$
\bar{x}_{\mathrm{B}}=c+\bar{u}_{\mathrm{B}} \cdot h, \quad D_{\mathrm{B}}(x)=D_{\mathrm{B}}(u) h^{2}, \quad \sigma(x)=\sigma(u) h
$$
We get: $\bar{x}_{\mathrm{B}}=48-4 \cdot 0.26=46.96 ; D_{\mathrm{B}}(x)=19.08 ; \sigma(x)=4.4$.
Forecast of service time: $\approx 47$ min, spread $\approx 4.4$ min.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example. Compute the limit
$$
\lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}
$$
|
Solution. Here $(2 n+1)^{2}-(n+1)^{2}=3 n^{2}+2 n-$ is a polynomial of the second degree (an infinitely large sequence of order $n^{2}$) and $n^{2}+n+1$ is a polynomial of the second degree (an infinitely large sequence of order $n^{2}$).
1. Factor out $n^{2}$ in the numerator, we get
$$
(2 n+1)^{2}-(n+1)^{2}=n^{2}\left(3+\frac{2}{n}\right)
$$
2. Factor out $n^{2}$ in the denominator, we get
$$
n^{2}+n+1=n^{2}\left(1+\frac{1}{n}+\frac{1}{n^{2}}\right)
$$
3. We have
$$
\lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}=\lim _{n \rightarrow \infty} \frac{n^{2}(3+2 / n)}{n^{2}\left(1+1 / n+1 / n^{2}\right)}
$$
4. Canceling $n^{2}$ and using the theorem on the limit of a quotient, we get
$$
\lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}=\frac{\lim _{n \rightarrow \infty}(3+2 / n)}{\lim _{n \rightarrow \infty}\left(1+1 / n+1 / n^{2}\right)}=3
$$
Answer. $\lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}=3$.
Conditions of the Problems. Calculate the limit.
1. $\lim _{n \rightarrow \infty} \frac{(5-n)^{2}+(5+n)^{2}}{(5-n)^{2}-(5+n)^{2}}$. 2. $\quad \lim _{n \rightarrow \infty} \frac{(4-n)^{3}-(2-n)^{3}}{(1-n)^{2}-(2+n)^{4}}$.
2. $\lim _{n \rightarrow \infty} \frac{(3-n)^{3}-(2-n)^{3}}{(1-n)^{3}-(1+n)^{3}}$.
3. $\lim _{n \rightarrow \infty} \frac{(2-n)^{2}-(1+n)^{2}}{(1+n)^{2}-(2-n)^{2}}$.
4. $\lim _{n \rightarrow \infty} \frac{(3+n)^{2}-(2+n)^{2}}{(2+n)^{2}-(1-n)^{2}}$.
5. $\lim _{n \rightarrow \infty} \frac{(n+2)^{3}-(n+2)^{2}}{(n-2)^{3}-(n+2)^{3}}$.
6. $\lim _{n \rightarrow \infty} \frac{(1+3 n)^{3}-27 n^{3}}{(1+4 n)^{2}+2 n^{2}}$.
7. $\lim _{n \rightarrow \infty} \frac{(3-2 n)^{2}}{(n-3)^{3}-(n+3)^{3}}$.
8. $\lim _{n \rightarrow \infty} \frac{(2+n)^{3}}{(n+2)^{2}-(n+1)^{3}}$.
9. $\lim _{n \rightarrow \infty} \frac{(n+2)^{2}-(n+5)^{3}}{(3-n)^{3}}$.
Answers. 1. $-\infty$. 2.0.3.0. 4. $-1.5 .1 / 3.6 .-\infty .7 .9 .8 .-2 / 9$. 9. -1.10 .1 .
## 3.3. Calculation of $\lim _{n \rightarrow \infty}[f(n) / g(n)]$
Problem Statement. Calculate the limit
$$
\lim _{n \rightarrow \infty} \frac{f(n)}{g(n)}
$$
where $f(n)$ is an infinitely large sequence of order $n^{\alpha}$ and $g(n)$ is an infinitely large sequence of order $n^{\beta}$ $(\alpha, \beta \in \mathbb{R})$.
Plan of Solution.
1. Factor out $n^{\alpha}$ in the numerator, we get $f(n)=n^{\alpha} \varphi(n)$, where $\lim _{n \rightarrow \infty} \varphi(n)=a, a \neq 0$.
2. Factor out $n^{\beta}$ in the denominator, we get $g(n)=n^{\beta} \psi(n)$, where $\lim _{n \rightarrow \infty} \psi(n)=b, b \neq 0$.
3. We have
$$
\lim _{n \rightarrow \infty} \frac{f(n)}{g(n)}=\lim _{n \rightarrow \infty} \frac{n^{\alpha} \varphi(n)}{n^{\beta} \psi(n)}
$$
4. We get:
if $\alpha>\beta$, then $\lim _{n \rightarrow \infty} \frac{f(n)}{g(n)}=\infty$;
if $\alpha<\beta$, then $\lim _{n \rightarrow \infty} \frac{f(n)}{g(n)}=0$;
if $\alpha=\beta$, then by the theorem on the limit of a quotient
$$
\lim _{n \rightarrow \infty} \frac{f(n)}{g(n)}=\frac{\lim _{n \rightarrow \infty} \varphi(n)}{\lim _{n \rightarrow \infty} \psi(n)}=\frac{a}{b}
$$
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example. Compute the limit
$$
\lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}
$$
|
Solution. The numerator $n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}$ is an infinitely large sequence of order $n^{2}$, and the denominator $(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}$ is an infinitely large sequence of order $n^{2}$.
1. Factor out $n^{2}$ in the numerator, we get
$$
n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}=n^{2}\left(\frac{1}{n^{5 / 6}}+2 \sqrt[5]{1+\frac{1}{n^{10}}}\right)
$$
2. Factor out $n^{2}$ in the denominator, we get
$$
(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}=n^{2}\left(1+\frac{1}{n^{3 / 4}}\right) \sqrt[3]{1-\frac{1}{n^{3}}}
$$
3. We have
$$
\lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}=\lim _{n \rightarrow \infty} \frac{n^{2}\left(1 / n^{5 / 6}+2 \sqrt[5]{1+1 / n^{10}}\right)}{n^{2}\left(1+1 / n^{3 / 4}\right) \sqrt[3]{1-1 / n^{3}}}
$$
4. Canceling $n^{2}$ and using theorems about limits, we finally get
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}=\lim _{n \rightarrow \infty} \frac{1 / n^{5 / 6}+2 \sqrt[5]{1+1 / n^{10}}}{\left(1+1 / n^{3 / 4}\right) \sqrt[3]{1-1 / n^{3}}}= \\
&=\frac{\lim _{n \rightarrow \infty}\left(1 / n^{5 / 6}+2 \sqrt[5]{1+1 / n^{10}}\right)}{\lim _{n \rightarrow \infty}\left(1+1 / n^{3 / 4}\right) \sqrt[3]{1-1 / n^{3}}}=2
\end{aligned}
$$
NOTE. In this case, the property of the root was used, according to which $\lim _{n \rightarrow \infty} \sqrt[5]{1+1 / n^{10}}=1$ and $\lim _{n \rightarrow \infty} \sqrt[3]{1-1 / n^{3}}=1$.
$$
\text { Answer. } \lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}=2
$$
Conditions of the Problems. Calculate the limit.
1. $\lim _{n \rightarrow \infty} \frac{n \sqrt[3]{3 n^{2}}+\sqrt[4]{4 n^{8}+1}}{(n+\sqrt{n}) \sqrt{7-n+n^{2}}}$.
2. $\lim _{n \rightarrow \infty} \frac{\sqrt{n-1}-\sqrt{2 n^{2}+3}}{\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{5}+2}}$.
3. $\lim _{n \rightarrow \infty} \frac{\sqrt{2 n^{3}+3}-\sqrt{n+5}}{\sqrt[3]{n^{3}+2}-\sqrt{n-1}}$.
4. $\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2}+3}+3 n^{3}}{\sqrt[4]{n^{12}+2 n+1}-n^{2}}$.
5. $\lim _{n \rightarrow \infty} \frac{\sqrt{3 n+2}-\sqrt[3]{125 n^{3}+n}}{\sqrt[5]{n}+n^{2}}$.
6. $\lim _{n \rightarrow \infty} \frac{n \sqrt{n}-\sqrt[3]{27 n^{6}+n^{4}}}{(n+\sqrt[4]{n}) \sqrt{4+n^{2}}}$.
7. $\lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt{n^{2}+2}}{\sqrt[4]{n^{4}+1}-\sqrt[3]{n^{2}-1}}$.
8. $\lim _{n \rightarrow \infty} \frac{\sqrt{n^{5}+3}+\sqrt{n-2}}{\sqrt[4]{n^{4}+2}-\sqrt{n-2}}$.
9. $\lim _{n \rightarrow \infty} \frac{10 n^{3}-\sqrt{n^{3}+2}}{\sqrt{4 n^{6}+3}-n}$.
10. $\lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt[3]{8 n^{3}+3}}{\sqrt[4]{n+5}+n}$.
Answers. 1. $\sqrt{2}$. 2.0.3. $+\infty .4$. 3. 5.5. 6. -3. 7. -1 . 8. $+\infty$. 9.5. 10. -2 .
## 3.4. Calculation of $\lim _{n \rightarrow \infty}\left[u(n)^{v(n)}\right]$
Problem Statement. Calculate the limit of the sequence
$$
\lim _{n \rightarrow \infty}\left[u(n)^{v(n)}\right]
$$
where $\lim _{n \rightarrow \infty} u(n)=1$ and $\lim _{n \rightarrow \infty} v(n)=\infty$.
## Plan of Solution.
1. Transform the expression under the limit so that we can use the second remarkable limit, i.e., factor out the one:
$$
\lim _{n \rightarrow \infty}\left[u(n)^{v(n)}\right]=\lim _{n \rightarrow \infty}\left((1+\alpha(n))^{1 / \alpha(n)}\right)^{\alpha(n) v(n)}
$$
where $\alpha(n)=u(n)-1$ is an infinitesimal sequence as $n \rightarrow \infty$. Since $\alpha(n) \rightarrow 0$ as $n \rightarrow \infty$, we have
$$
\lim _{n \rightarrow \infty}(1+\alpha(n))^{1 / \alpha(n)}=e
$$
2. If $\lim _{n \rightarrow \infty} a_{n}=a\left(a_{n}>0, a>0\right)$ and $\lim _{n \rightarrow \infty} b_{n}=b$, then
$$
\lim _{n \rightarrow \infty} a_{n}^{b_{n}}=a^{b}
$$
Therefore, if the limit
$$
\lim _{n \rightarrow \infty} \alpha(n) v(n)=\lim _{n \rightarrow \infty}(u(n)-1) v(n)
$$
exists, then we finally have
$$
\lim _{n \rightarrow \infty}\left[u(n)^{v(n)}\right]=e^{\lim _{n \rightarrow \infty}(u(n)-1) v(n)}
$$
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Example. Compute the limit
$$
\lim _{x \rightarrow 0} \frac{2 x \sin x}{1-\cos x}
$$
|
Solution. The expression under the limit sign is the ratio of two infinitesimals at the point $x=0$, since
$$
\lim _{x \rightarrow 0}(2 x \sin x)=0, \quad \lim _{x \rightarrow 0}(1-\cos x)=0
$$
The infinitesimals in the numerator and denominator are replaced by equivalent ones:
$$
\begin{array}{ll}
2 x \sin x \sim 2 x \cdot x, & x \rightarrow 0 \\
1-\cos x \sim \frac{x^{2}}{2}, & x \rightarrow 0
\end{array}
$$
Thus,
$$
\lim _{x \rightarrow 0} \frac{2 x \sin x}{1-\cos x}=\lim _{x \rightarrow 0} \frac{2 x \cdot x}{x^{2} / 2}=4
$$
Answer. $\lim _{x \rightarrow 0} \frac{2 x \sin x}{1-\cos x}=4$.
Conditions of the Problems. Calculate the limit.
1. $\lim _{x \rightarrow 0} \frac{\ln (1+\sin 2 x)}{\sin 3 x}$.
2. $\lim _{x \rightarrow 0} \frac{3 x^{2}+6 x}{\sin 3 x}$.
3. $\lim _{x \rightarrow 0} \frac{5^{x}-1}{\ln (1+x)}$.
4. $\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{\cos 5 x-\cos 3 x}$.
5. $\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{e^{2 x^{2}}-1}$.
6. $\lim _{x \rightarrow 0} \frac{\sqrt{9+x}-3}{3 \operatorname{arctg} 2 x}$.
7. $\lim _{x \rightarrow 0} \frac{\operatorname{tg} 2 x}{e^{2 x}-1}$.
8. $\lim _{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{\sin ^{2} x}$.
9. $\lim _{x \rightarrow 0} \frac{\sin 2 x}{\ln (1-2 x)}$.
10. $\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{\ln (e-2 x)-1}$.
Answers. 1. 2/3. 2. 2. 3. $\ln 5$.
4. $-1 / 4$.
5. 1 .
6. $1 / 12$. 7. 1. 8. $1 / 4$. 9. -1. 10. $-e$.
3.9. Calculation of $\lim _{x \rightarrow a}[f(x) / g(x)]$
PROBLEM STATEMENT. Calculate the limit
$$
\lim _{x \rightarrow a} \frac{f(x)}{g(x)}
$$
where $f(x)$ and $g(x)$ are infinitesimal functions at the point $x=a$.
SOLUTION PLAN.
1. We need to replace $f(x)$ and $g(x)$ with equivalent infinitesimal functions. However, the table of equivalent infinitesimal functions is compiled for the point $x=0$. Therefore, we first make a variable substitution $x-a=t$ and will find the limit as $t \rightarrow 0$.
2. Transform the expression under the limit sign using algebraic and trigonometric formulas, and replace the infinitesimal functions in the product and quotient with equivalent ones.
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example. Calculate the limit
$$
\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\tan^{2} 2 x}
$$
|
Solution.
1. Since
$$
\lim _{x \rightarrow \pi}[\cos 3 x-\cos x]=0, \quad \lim _{x \rightarrow \pi} \operatorname{tg}^{2} 2 x=0
$$
the expression under the limit sign is a ratio of two infinitesimal functions as $x \rightarrow \pi$. We need to replace these infinitesimal functions with equivalent ones. For this, we first make the substitution $x-\pi=t$:
$$
\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\operatorname{tg}^{2} 2 x}=\lim _{t \rightarrow 0} \frac{\cos 3(\pi+t)-\cos (\pi+t)}{\operatorname{tg}^{2} 2(\pi+t)}
$$
2. Using trigonometric formulas and replacing the product and quotient of infinitesimal functions with equivalent ones, we get
$$
\begin{aligned}
& \lim _{t \rightarrow 0} \frac{\cos 3(\pi+t)-\cos (\pi+t)}{\operatorname{tg}^{2} 2(\pi+t)}= \\
& \quad=\lim _{t \rightarrow 0} \frac{\cos t-\cos 3 t}{\operatorname{tg}^{2} 2 t}=\lim _{t \rightarrow 0} \frac{-2 \sin 2 t \sin (-t)}{\operatorname{tg}^{2} 2 t}=\lim _{t \rightarrow 0} \frac{2 \cdot 2 t \cdot t}{4 t^{2}}=1
\end{aligned}
$$
Answer. $\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\operatorname{tg}^{2} 2 x}=1$.
Conditions of the Problems. Calculate the limits.
1. $\lim _{x \rightarrow 1} \frac{x^{3}-1}{\ln x}$.
2. $\lim _{x \rightarrow \pi} \frac{1+\cos 5 x}{\sin ^{2} 3 x}$.
3. $\lim _{x \rightarrow 1 / 2} \frac{1+\cos 2 \pi x}{\operatorname{tg}^{2} 2 \pi x}$.
4. $\lim _{x \rightarrow 2} \frac{\sin 3 \pi x}{\sin 8 \pi x}$.
5. $\lim _{x \rightarrow 2} \frac{\sqrt{x^{2}-x-1}-1}{\ln (x-1)}$.
6. $\lim _{x \rightarrow \pi / 2} \frac{\operatorname{tg} 5 x}{\operatorname{tg} 3 x}$.
7. $\lim _{x \rightarrow 1} \frac{1-x^{3}}{\sin \pi x}$.
8. $\lim _{x \rightarrow 1} \frac{2-\sqrt{5-x}}{\sin \pi x}$.
9. $\lim _{x \rightarrow \pi} \frac{\operatorname{tg} 5 x}{\sin 3 x}$.
10. $\lim _{x \rightarrow 2} \frac{2^{x}-4}{\sin \pi x}$.
Answers. 1. 3. 2. 5/18. 3. $1 / 2$. 4. 3/8. 5. 3/2. 6. 3/5. 7. $3 / \pi$. 8. $-1 /(4 \pi)$. 9. $-5 / 3$. 10. $(4 \ln 2) / \pi$.
### 3.10. Calculation of $\lim _{x \rightarrow 0}\left[u(x)^{v(x)}\right]$
Problem Statement. Calculate the limit
$$
\lim _{x \rightarrow 0}\left[u(x)^{v(x)}\right]
$$
where $\lim _{x \rightarrow 0} u(x)=1$ and $\lim _{x \rightarrow 0} v(x)=\infty$.
## PLAN OF SOLUTION.
1. Transform the expression under the limit sign:
$$
u(x)^{v(x)}=e^{v(x) \ln u(x)}
$$
2. Since the exponential function $e^{x}$ is continuous, we can take the limit under the sign of this function. We have
$$
\lim _{x \rightarrow 0}\left[u(x)^{v(x)}\right]=\lim _{x \rightarrow 0} e^{v(x) \ln u(x)}=e^{\lim _{x \rightarrow 0}[v(x) \ln u(x)]}
$$
3. Calculate the limit of the exponent
$$
\lim _{x \rightarrow 0}[v(x) \ln u(x)]
$$
replacing infinitesimal functions with equivalent ones.
4. Write the final answer.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example. Compute the limit
$$
\lim _{x \rightarrow 0} \sqrt[3]{x\left(2+\sin \frac{1}{x}\right)+8 \cos x}
$$
|
Solution.
1. Since the function $y=\sqrt[3]{x}$ is continuous for all $x$, by passing to the limit under the sign of a continuous function, we get
$$
\lim _{x \rightarrow 0} \sqrt[3]{x\left(2+\sin \frac{1}{x}\right)+8 \cos x}=\sqrt[3]{\lim _{x \rightarrow 0}\left[x\left(2+\sin \frac{1}{x}\right)+8 \cos x\right]}
$$
2. Since $x$ is an infinitesimal function at the point $x=0$, and $2+\sin (1 / x)$ is a bounded function in the neighborhood of the point $x=0$, then $x(2+\sin (1 / x))$ is an infinitesimal function at the point $x=0$, i.e.,
$$
\lim _{x \rightarrow 0} x\left(2+\sin \frac{1}{x}\right)=0
$$
3. Since $\cos x$ is continuous at the point $x=0$, then
$$
\lim _{x \rightarrow 0} \cos x=1
$$
and, using the properties of the limit of a function at a point, we get
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \sqrt[3]{x\left(2+\sin \frac{1}{x}\right)+8 \cos x}=\sqrt[3]{\lim _{x \rightarrow 0} x\left(2+\sin \frac{1}{x}\right)+8 \lim _{x \rightarrow 0} \cos x}=2 \\
& \text { Answer. } \lim _{x \rightarrow 0} \sqrt[3]{x\left(2+\sin \left(\frac{1}{x}\right)\right)+8 \cos x}=2
\end{aligned}
$$
Conditions of the Problems. Calculate the limit.
1. $\lim _{x \rightarrow 0} \sqrt{9 \cos 2 x+2 x \operatorname{arctg} \frac{1}{x}}$.
2. $\lim _{x \rightarrow \pi / 2} \sqrt{4 \sin x+(2 x-\pi) \sin \frac{x^{2}}{2 x-\pi}}$.
3. $\lim _{x \rightarrow 0} \sqrt{\cos x+\operatorname{arctg} x \cos ^{2} \frac{1}{x^{2}}}$.
4. $\lim _{x \rightarrow 0} \sqrt{4 \cos x+\ln (1+2 x) \sin \frac{1}{x}}$.
5. $\lim _{x \rightarrow 0} \sqrt{4 \cos ^{2} x+\left(e^{2 x}-1\right) \operatorname{arctg} \frac{1}{x^{2}}}$.
6. $\lim _{x \rightarrow 0} \ln \left[\left(e^{x^{3}}-\cos x\right) \cos \frac{2}{x}+\operatorname{tg}\left(x+\frac{\pi}{4}\right)\right]$.
7. $\lim _{x \rightarrow 0} \sqrt{4 \sin x+\left(e^{\sin ^{2} x}-1\right) \cos \frac{1}{x}}$.
8. $\lim _{x \rightarrow 2} \sqrt{\ln (x+2)+\sin \left(4-x^{2}\right) \cos \frac{x+2}{x-2}}$.
9. $\lim _{x \rightarrow 1} \operatorname{tg}\left(\arccos x+\sin \frac{x-1}{x+1} \cos \frac{x+1}{x-1}\right)$.
10. $\lim _{x \rightarrow 0} \ln \left(3+\operatorname{arctg} x \sin \frac{1}{x}\right)$.
Answers. 1. 3.
11. 2 .
12. 1 .
13. 2 .
14. 2 .
15. $0 . \quad 7.2$.
16. $\ln 2$.
17. $0 . \quad 10 . \ln 3$.
Chapter 4
## DIFFERENTIATION
In the study of the topic DIFFERENTIATION, you will become familiar with the concepts of the derivative and differential of a function of one variable through examples, learn to compute derivatives using the rules of differentiation for sums, products, quotients, and composite functions, learn to differentiate functions given parametrically, compute higher-order derivatives, and apply derivatives and differentials in approximate calculations and in solving geometric problems.
Using the package SOLVER.CA, you can compute limits, perform numerical calculations, and compute derivatives of any order to verify the correctness of the results you obtain.
## 4.1. The Concept of a Derivative
Problem Statement. Find the derivative of the function $f(x)$ at the point $x=0$ based on the definition.
Plan of Solution.
1. By definition,
$$
f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}
$$
(Recall that when computing the limit $x \rightarrow 0$, but $x \neq 0$.)
2. Compute the limit
$$
\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}
$$
3. If the limit exists and equals $A$, then $f^{\prime}(0)=A$; if the limit does not exist, then $f^{\prime}(0)$ does not exist.
|
2
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Calculus
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math-word-problem
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Yes
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Yes
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olympiads
| false
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Example. According to the definition, find the derivative of the function
$$
f(x)=\left[\begin{array}{ll}
1-\cos \left(x \sin \frac{1}{x}\right), & x \neq 0 \\
0, & x=0
\end{array}\right.
$$
at the point $x=0$.
|
Solution.
1. By definition
$$
f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}=\lim _{x \rightarrow 0} \frac{1-\cos (x \sin (1 / x))-0}{x}
$$
2. Since $\sin (1 / x)$ is a bounded function and $x$ is an infinitesimal function as $x \rightarrow 0$, by the theorem on the product of an infinitesimal function and a bounded function, $x \sin (1 / x) \rightarrow 0$ as $x \rightarrow 0$. Replacing the infinitesimal function in the numerator with an equivalent one and again using the mentioned theorem, we get
$$
\lim _{x \rightarrow 0} \frac{1-\cos (x \sin (1 / x))-0}{x} \lim _{x \rightarrow 0} \frac{x^{2} \sin ^{2}(1 / x)}{2 x}=0
$$
3. Thus, the limit exists and is equal to zero. Therefore, $f^{\prime}(0)=0$.
Answer. $f^{\prime}(0)=0$.
1. $f(x)=\left[\begin{array}{l}\sin \left(x^{3}+x^{2} \sin \frac{2}{x}\right), x \neq 0, \\ 0, x=0 .\end{array}\right.$
2. $f(x)=\left[\begin{array}{l}\operatorname{tg}\left(x^{2} \cos \frac{1}{9 x}\right)+2 x, x \neq 0, \\ 0, x=0 .\end{array}\right.$
3. $f(x)=\left[\begin{array}{l}\arcsin \left(x \cos \frac{1}{5 x}\right) \\ 0, x=0 .\end{array}\right.$
4. $f(x)=\left[\begin{array}{l}\ln \left(1-\operatorname{tg}\left(x^{2} \sin \frac{1}{x}\right)\right), x \neq 0, \\ 0, x=0 .\end{array}\right.$
5. $f(x)=\left[\begin{array}{l}\operatorname{tg}\left(x \sin \frac{3}{x}\right), x \neq 0, \\ 0, x=0 .\end{array}\right.$
6. $f(x)=\left[\begin{array}{l}\sqrt{1+\ln \left(1+x^{2} \sin \frac{1}{x}\right)}-1, x \neq 0, \\ 0, x=0 .\end{array}\right.$
7. $f(x)=\left[\begin{array}{l}\sin \left(e^{x^{2} \sin (5 / x)}-1\right), x \neq 0, \\ 0, \quad x=0 .\end{array}\right.$
8. $f(x)=\left[\begin{array}{l}x^{3} \cos \frac{4}{3 x}+3 x, \quad x \neq 0, \\ 0, \quad x=0 .\end{array}\right.$
9. $f(x)=\left[\begin{array}{l}\operatorname{arctg}\left(x-x^{2} \sin \frac{1}{3 x}\right), x \neq 0, \\ 0, x=0 .\end{array}\right.$
10. $f(x)=\left[\begin{array}{l}\sin ^{2} x \cos \frac{5}{x}+2 x, x \neq 0, \\ 0, x=0 .\end{array}\right.$
Answers. 1. 0. 2. 2. 3. Does not exist. 4. 0. 5. Does not exist. 6. 0. 7. 0. 8. 3. 9. 1. 10. 2.
## 4.2. Calculation of Derivatives
Problem Statement. Find the derivative of the function $y=f(x)$.
Solution Plan. The problem is solved in several stages. At each stage, it is necessary to recognize the type of function and apply the corresponding differentiation rule.
The following types of functions are possible.
- The function has the form $C_{1} u_{1}+C_{2} u_{2}+\ldots+C_{n} u_{n}$, where $u_{1}(x)$, $u_{2}(x), \ldots, u_{n}(x)$ are some functions and $C_{1}, C_{2}, \ldots, C_{n}$ are some constants. Use the formula for the derivative of a linear combination
$$
\left(C_{1} u_{1}+C_{2} u_{2}+\ldots+C_{n} u_{n}\right)^{\prime}=C_{1} u_{1}^{\prime}+C_{2} u_{2}^{\prime}+\ldots+C_{n} u_{n}^{\prime}
$$
- The function has the form $u \cdot v$. Use the formula for the derivative of a product
$$
(u \cdot v)^{\prime}=u^{\prime} \cdot v+u \cdot v^{\prime}
$$
- The function has the form $\frac{u}{v}$. Use the formula for the derivative of a quotient:
$$
\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} \cdot v-u \cdot v^{\prime}}{v^{2}}
$$
- The function has the form $u(v(x))$. Use the formula for the derivative of a composite function
$$
u(v(x))^{\prime}=u^{\prime}(v) \cdot v^{\prime}(x)
$$
- The function has the form $u(x)^{v(x)}$. The derivative of such a function is calculated using the formula
$$
u(x)^{v(x)}=e^{v \ln u}
$$
The transition from one stage to the next is made until each sign of the derivative contains a tabular function.
|
0
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Calculus
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math-word-problem
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Yes
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Yes
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olympiads
| false
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Example. Find the area of the region bounded by the graphs of the functions
$$
y=x^{2}-4 x+3, \quad y=-x^{2}+2 x+3
$$
|
## Solution.
1. Find the abscissas $a$ and $b$ of the points of intersection of the graphs. For this, solve the equation
$$
x^{2}-4 x+3=-x^{2}+2 x+3
$$
We get $a=0, \quad b=3$.
2. Investigate the sign of the function $\varphi=x^{2}-4 x+3-\left(-x^{2}+2 x+3\right)$ on the interval $[a, b]=[0,3]$. For this, assign $x$ any value from $(0,3)$, for example $x=1$. We get that $\varphi(1)=-4$. Therefore, $\varphi<0$ when $x \in(0,3)$. Hence, $x^{2}-4 x+3 \leq-x^{2}+2 x+3$ when $x \in[0,3]$ and the region $D$ is defined by the system of inequalities
$$
\left\{\begin{aligned}
0 & \leq x \leq 3 \\
x^{2}-4 x+3 & \leq y \leq-x^{2}+2 x+3
\end{aligned}\right.
$$
3. Apply the formula (1) with $v(x)=-x^{2}+2 x+3, \quad u(x)=x^{2}-4 x+3, a=0$ and $b=3$:
$$
S_{D}=\int_{0}^{3}\left(-x^{2}+2 x+3-x^{2}+4 x-3\right) d x=\int_{0}^{3}\left(-2 x^{2}+6 x\right) d x=9
$$
Answer. $S=9(\text{units of length})^{2}$.
Conditions of the Problems. Calculate the areas of regions bounded by the graphs of the given functions.
1. $y=32-x^{2}, \quad y=-4 x$.
2. $y=3 \sqrt{x}, \quad y=3 / x, \quad x=4$.
3. $x=5-y^{2}, \quad x=-4 y$.
4. $y=\sqrt{e^{x}-1}, \quad y=0, \quad x=\ln 4$.
5. $y=\sin x, \quad y=\cos x, \quad x=0(x \geq 0)$.
6. $y=\sqrt{x}, \quad y=1 / x, \quad x=16$.
7. $x=27-y^{2}, \quad x=-6 y$.
8. $y=\sin x, \quad y=\cos x, \quad x=0(x \leq 0)$.
9. $y=\sqrt{9-x^{2}}, \quad y=0, \quad x=0, \quad x=3 / 2$.
10. $y=2 / x, \quad y=5 e^{x}, \quad y=2, \quad y=5$.
Answers. 1. 288. $2.14-3 \ln 4$. 3. 36. 4. $(6 \sqrt{3}-2 \pi) / 3$. 5. $\sqrt{2}-1$.
6. $42-\ln 16$.
7. 288.
8. $1+\sqrt{2}$.
9. $(3 \pi+\sqrt{3}) / 4$. 10. 3.
8.8. Calculation of arc lengths $y=f(x)$
PROBLEM STATEMENT. Calculate the length of the curve given by the equation
$$
y=f(x)
$$
and bounded by points with abscissas $x=a$ and $x=b$.
SOLUTION PLAN. The length $l$ of a piecewise smooth curve $y=f(x)$, bounded by points with abscissas $x=a$ and $x=b$, is equal to
$$
l=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x
$$
1. Find $y^{\prime}=f^{\prime}(x)$.
2. Compute the differential of the arc length
$$
d l=\sqrt{1+\left(y^{\prime}\right)^{2}} d x
$$
3. Find the arc length by evaluating the definite integral (1). Write the answer, not forgetting the dimension.
|
9
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Calculus
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math-word-problem
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Yes
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Yes
|
olympiads
| false
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Example. Evaluate the double integral
$$
\iint_{D}\left(54 x^{2} y^{2}+150 x^{4} y^{4}\right) d x d y
$$
where the region $D$ is bounded by the lines $x=1, y=x^{3}$ and $y=-\sqrt{x}$.
|
Solution.
1. Let's define the region $D$ by inequalities. It is obvious that $-\sqrt{x} \leq x^{3}$. Therefore, $-\sqrt{x} \leq y \leq x^{3}$. Since $x$ appears under the square root, $x \geq 0$. For $x$, the possible inequalities are $0 \leq x \leq 1$ or $1 \leq x$. In the second case, the region is unbounded, which is unacceptable.
Thus,
$$
D=\left\{\begin{array}{cc}
(x, y): & 0 \leq x \leq 1 \\
-\sqrt{x} \leq y \leq x^{3}
\end{array}\right\}
$$
2. We transition from the double integral to the iterated integral:
$$
\iint_{D}\left(54 x^{2} y^{2}+150 x^{4} y^{4}\right) d x d y=\int_{0}^{1} d x \int_{-\sqrt{x}}^{x^{3}}\left(54 x^{2} y^{2}+150 x^{4} y^{4}\right) d y
$$
3. Using the properties of the definite integral, we sequentially integrate first with respect to $y$ (considering $x$ as a constant), and then with respect to $x$:
$$
\begin{aligned}
& \int_{0}^{1} d x \int_{-\sqrt{x}}^{x^{3}}\left(54 x^{2} y^{2}+150 x^{4} y^{4}\right) d y= \\
& =\int_{0}^{1}\left[54 x^{2} \int_{-\sqrt{x}}^{x^{3}} y^{2} d y+150 x^{4} \int_{-\sqrt{x}}^{x^{3}} y^{4} d y\right] d x= \\
& =\int_{0}^{1}\left[\left.54 x^{2} \frac{y^{3}}{3}\right|_{-\sqrt{x}} ^{x^{3}}+\left.150 x^{4} \frac{y^{5}}{5}\right|_{-\sqrt{x}} ^{x^{3}}\right] d x= \\
& =\int_{0}^{1}\left[18 x^{11}+18 x^{7 / 2}+30 x^{19}+30 x^{13 / 2}\right] d x=11
\end{aligned}
$$
Answer. $\iint_{D}\left(54 x^{2} y^{2}+150 x^{4} y^{4}\right) d x d y=11$.
Conditions of the Problems. Calculate the double integrals over the regions $D$ bounded by the given lines.
1. $\iint_{D}(2 x-y) d x d y, \quad y=x^{2}, y=\sqrt{x}$.
2. $\iint_{D}(x-y) d x d y, \quad y=2-x^{2}, y=2 x-1$.
3. $\quad \iint_{D}(y \ln x) d x d y, \quad y=\frac{1}{x}, y=\sqrt{x}, x=2$.
4. $\quad \iint_{D}(\cos 2 x+\sin y) d x d y, \quad y=\frac{\pi}{4}-x, y=0, x=0$.
5. $\quad \iint_{D} \sin (x+y) d x d y, \quad y=x, y=\frac{\pi}{2}, x=0$.
6. $\iint_{D} \frac{x^{2}}{y^{2}} d x d y, \quad y=\frac{1}{x}, y=x, x=2$.
7. $\iint_{D}\left(x^{2}+y\right) d x d y, \quad y=x^{2}, y=\sqrt{x}$.
8. $\quad \iint_{D}(2 x-y) d x d y, \quad y=x^{2}, y=x, x=1, x=2$.
9. $\quad \int_{D} \sqrt{1-x^{2}-y^{2}} d x d y, \quad y=\sqrt{1-x^{2}}, \quad y=0, x=0$.
10. $\iint_{D} x^{2} y^{2} \sqrt{1-x^{3}-y^{3}} d x d y, \quad y=\sqrt[3]{1-x^{3}}, \quad y=0, \quad x=0$.
Answers. 1. $I=1 / 10 . \quad 2 . \quad I=64 / 15 . \quad$ 3. $I=5(2 \ln 2-1) / 8$.
4. $I=(\pi+1-2 \sqrt{2}) / 4 . \quad$ 5. $I=1 . \quad 6 . \quad I=9 / 4 . \quad$ 7. $I=33 / 140$.
5. $I=9 / 10$. $\quad 9 . \quad I=\pi / 6$. $\quad 10 . \quad I=4 / 135$.
## 12.3. Double Integral
## in Polar Coordinates
Problem Statement. Calculate the double integral
$$
\iint_{D} f(x, y) d x d y
$$
where the region $D$ is bounded by two circles
$$
y^{2}+a_{1} y+b_{1} x+x^{2}=0, \quad y^{2}+a_{2} y+b_{2} x+x^{2}=0
$$
$$
\left(a_{1}=0, a_{2}=0, \quad b_{1} b_{2}>0 \quad \text { or } \quad b_{1}=0, b_{2}=0, \quad a_{1} a_{2}>0\right)
$$
and two lines
$$
m_{1} y+k_{1} x=0, \quad\left(m_{1}^{2}+k_{1}^{2} \neq 0\right), \quad m_{2} y+k_{2} x=0, \quad\left(m_{2}^{2}+k_{2}^{2} \neq 0\right)
$$
## Plan of Solution.
1. Define the region $D$ by inequalities in the Cartesian coordinate system.
Notice that the circles $y^{2}+a_{1} y+b_{1} x+x^{2}=0$ and $y^{2}+a_{2} y+b_{2} x+x^{2}=0$ pass through the origin and their centers are located on the $O X$ axis (when $a_{1}=0, a_{2}=0$) or on the $O Y$ axis (when $b_{1}=0, b_{2}=0$) on the same side of the origin (since $b_{1} b_{2}>0$ or $a_{1} a_{2}>0$). Therefore, the circle with the smaller radius is located inside the other. Suppose, for example, this is the circle $y^{2}+a_{1} y+b_{1} x+x^{2}=0$. The region $D$ is between the circles, so the coordinates of the points in the region $D$ satisfy the inequalities
$$
y^{2}+a_{1} y+b_{1} x+x^{2} \geq 0, \quad y^{2}+a_{2} y+b_{2} x+x^{2} \leq 0
$$
The lines $m_{1} y+k_{1} x=0$ and $m_{2} y+k_{2} x=0$ pass through the origin. The region $D$ is located between them. Considering in which half-plane the circles and, consequently, the region $D$ are located, we determine which of the following pairs of inequalities the coordinates of the points in the region $D$ satisfy:
$$
\begin{array}{ll}
m_{1} y+k_{1} x \geq 0, & m_{2} y+k_{2} x \geq 0 \\
m_{1} y+k_{1} x \leq 0, & m_{2} y+k_{2} x \geq 0 \\
m_{1} y+k_{1} x \geq 0, & m_{2} y+k_{2} x \leq 0 \\
m_{1} y+k_{1} x \leq 0, & m_{2} y+k_{2} x \leq 0
\end{array}
$$
2. Since the region $D$ is bounded by circles and lines passing through the origin, it is easier to solve the given problem in polar coordinates
$$
\left\{\begin{array}{l}
x=\varrho \cos \varphi \\
y=\varrho \sin \varphi
\end{array}\right.
$$
In this case, $(\varrho, \varphi) \in D^{\prime}$, and the desired integral is defined by the formula
$$
\iint_{D} f(x, y) d x d y=\iint_{D^{\prime}} f(\varrho \cos \varphi, \varrho \sin \varphi) \varrho d \varrho d \varphi
$$
3. To find the region $D^{\prime}$, we replace $x$ with $\varrho \cos \varphi$ and $y$ with $\varrho \sin \varphi$ in the inequalities defining the region $D$. Then we solve the obtained inequalities for $\varrho$ and $\varphi$. Thus, we get
$$
D^{\prime}=\left\{\begin{array}{rlrl}
& =\varphi_{1} & \leq \varphi \leq \varphi_{2} \\
\varrho_{1}(\varphi) & \leq \varrho \leq \varrho_{2}(\varphi)
\end{array}\right\}
$$
4. We transition from the double integral to the iterated integral:
$$
S=\int_{\varphi_{1}}^{\varphi_{2}} d \varphi \int_{\varrho_{1}(\varphi)}^{\varrho_{2}(\varphi)} f(\varrho \cos \varphi, \varrho \
|
11
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math-word-problem
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Yes
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Yes
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olympiads
| false
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Example. Evaluate the double integral
$$
\iint_{D} \frac{x}{y^{5}} d x d y
$$
where the region $D$ is defined by the inequalities
$$
1 \leq \frac{x^{2}}{16}+y^{2} \leq 3, \quad y \geq \frac{x}{4}, \quad x \geq 0
$$
|
SOLUTION.
1. The region $D$ is defined by inequalities in the Cartesian coordinate system:
$$
D=\left\{(x, y): \begin{array}{c}
1 \leq \frac{x^{2}}{16}+y^{2} \leq 3 \\
\\
y \geq \frac{x}{4}, \quad x \geq 0
\end{array}\right\}
$$
2. Since the region $D$ is bounded by ellipses and lines passing through the origin, it is easier to solve the problem using generalized polar coordinates
$$
\left\{\begin{array}{l}
x=4 \varrho \cos \varphi \\
y=\varrho \sin \varphi
\end{array}\right.
$$
In this case, $(\varrho, \varphi) \in D^{\prime}$, and the desired integral is given by the formula
$$
\iint_{D} \frac{x}{y^{5}} d x d y=\iint_{D^{\prime}} \frac{4 \varrho \cos \varphi}{\varrho^{5} \sin ^{5} \varphi} 4 \varrho d \varrho d \varphi
$$
3. To find the region $D^{\prime}$, we replace $x$ with $a \varrho \cos \varphi$ and $y$ with $b \varrho \sin \varphi$ in the inequalities defining the region $D$:
$$
\left\{\begin{array}{l}
1 \leq \frac{16 \varrho^{2} \cos ^{2} \varphi}{16}+\varrho^{2} \sin ^{2} \varphi \leq 3 \\
\varrho \sin \varphi \geq \frac{4 \varrho \cos \varphi}{4}, \varrho \cos \varphi \geq 0
\end{array}\right.
$$
Solving these inequalities with respect to $\varrho$ and $\varphi$, we get
4. Transitioning from the double integral to the repeated integral and integrating sequentially, we obtain
$$
\iint_{D} \frac{x}{y^{5}} d x d y=\iint_{D^{\prime}} \frac{4 \varrho \cos \varphi}{\varrho^{5} \sin ^{5} \varphi} 4 \varrho d \varrho d \varphi=\int_{\pi / 4}^{\pi / 2} d \varphi \int_{1}^{\sqrt{3}} 16 \varrho^{-3} \cdot \frac{\cos \varphi}{\sin ^{5} \varphi} d \varrho=
$$
$$
=16 \int_{\pi / 4}^{\pi / 2} \frac{d \sin \varphi}{\sin ^{5} \varphi} \int_{1}^{\sqrt{3}} \varrho^{-3} d \varrho=\left.\left.16\left(-\frac{1}{4 \sin ^{4} \varphi}\right)\right|_{\pi / 4} ^{\pi / 2}\left(-\frac{1}{2 \varrho^{2}}\right)\right|_{1} ^{\sqrt{3}}=4
$$
Answer. $\iint_{D} \frac{x}{y^{5}} d x d y=4$.
PROBLEM CONDITIONS. Compute the double integrals.
1. $\iint_{D} \frac{y}{x} d x d y$
$$
D=\left\{1 \leq \frac{x^{2}}{9}+\frac{y^{2}}{4} \leq 2, y \geq 0, y \leq \frac{2 x}{3}\right\}
$$
2. $\int_{D}^{D} \int^{x} \frac{x}{y} d x d y$ $D=\left\{1 \leq \frac{x^{2}}{4}+\frac{y^{2}}{9} \leq 4, x \geq 0, y \geq \frac{3 x}{2}\right\}$.
3. $\iint_{D} \frac{x}{y} d x d y$ $D=\left\{1 \leq \frac{x^{2}}{16}+\frac{y^{2}}{4} \leq 4, x \geq 0, y \geq \frac{x}{2}\right\}$.
4. $\int_{D}^{D} \int^{3} y d x d y$ $D=\left\{1 \leq \frac{x^{2}}{4}+\frac{y^{2}}{9} \leq 1, x \geq 0, y \geq 0\right\}$.
5. $\iint_{D} \frac{x}{y} d x d y$
$D=\left\{1 \leq \frac{x^{2}}{4}+\frac{y^{2}}{16} \leq 5, x \geq 0, y \geq 2 x\right\}$.
6. $\iint_{D} \frac{x}{y} d x d y$
$D=\left\{1 \leq \frac{x^{2}}{9}+\frac{y^{2}}{4} \leq 5, x \geq 0, y \geq \frac{2 x}{3}\right\}$.
7. $\iint_{D} \frac{x}{y^{3}} d x d y$
$D=\left\{1 \leq \frac{x^{2}}{4}+y^{2} \leq 25, x \geq 0, y \geq \frac{x}{2}\right\}$.
8. $\int_{D}^{D} \int^{y} \frac{y}{x^{3}} d x d y$
$D=\left\{1 \leq x^{2}+\frac{y^{2}}{16} \leq 9, y \geq 0, y \leq 4 x\right\}$.
9. $\iint_{D} \frac{8 y}{x^{3}} d x d y$
$D=\left\{1 \leq \frac{x^{2}}{4}+y^{2} \leq 4, y \geq 0, y \leq \frac{x}{2}\right\}$.
10. $\iint_{D} \frac{9 x}{y^{3}} d x d y$
$D=\left\{1 \leq \frac{x^{2}}{4}+\frac{y^{2}}{9} \leq 36, x \geq 0, y \geq \frac{3 x}{2}\right\}$.
Answers. 1. $I=\ln 2$.
2. $I=3 \ln 2$.
3. $I=12 \ln 2$.
4. $I=6$.
5. $I=4 \ln 2$.
6. $I=9 \ln 2$.
7. $I=2 \ln 5$.
8. $I=8 \ln 3$.
9. $I=\ln 2$.
10. $I=2 \ln 6$.
## 12.5. Calculation of Volumes Using Double Integrals
PROBLEM STATEMENT. Find the volume of the body bounded by the surfaces
$g_{i}(x, y)=0(i=1,2, \ldots), \quad z=f_{1}(x, y), z=f_{2}(x, y)\left(f_{2}(x, y) \geq f_{1}(x, y)\right)$.
## PLAN OF SOLUTION.
1. The volume of the cylindrical prism bounded by the given surfaces is determined by the formula
$$
V=\iint_{D}\left[f_{2}(x, y)-f_{1}(x, y)\right] d x d y
$$
where $D$ is the projection of the body onto the $X O Y$ plane.
2. To find $D$, we define the body using inequalities and eliminate $z$.
For example, suppose the coordinates of the points of the body satisfy the inequalities $0 \leq z \leq f(x, y), g_{1}(x, y) \geq 0$ and $g_{2}(x, y) \leq 0$. Then the body is defined by the system of inequalities
$$
\left\{\begin{array}{l}
g_{1}(x, y) \geq 0 \\
g_{2}(x, y) \leq 0 \\
0 \leq z \leq f(x, y)
\end{array}\right.
$$
Eliminating $z$, we get
$$
D=\left\{\begin{array}{ll}
& g_{1}(x, y) \geq 0 \\
(x, y): & g_{2}(x, y) \leq 0 \\
0 \leq f(x, y)
\end{array}\right\}
$$
3. We compute the double integral using the formula (1) with $f_{2}=f(x, y)$ and $f_{1}=0$.
We write the answer, not forgetting the dimension.
|
4
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Calculus
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math-word-problem
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Yes
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Yes
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olympiads
| false
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Example 1. Find the volume of the body bounded by the surfaces
$$
x=17 \sqrt{2 y}, \quad x=2 \sqrt{2 y}, \quad z=1 / 2-y, \quad z=0
$$
|
## Solution.
1. By formula (1) with $f_{2}=1 / 2-y$ and $f_{1}=0$, the desired volume is
$$
V=\iint_{D}\left(\frac{1}{2}-y\right) d x d y
$$
where $D$ is the projection of the body onto the $X O Y$ plane.
2. To find $D$, we define the body using inequalities and eliminate $z$ from them. In this case, the body is defined by the system of inequalities
$$
\left\{\begin{array}{l}
x \leq 17 \sqrt{2 y} \\
x \geq 2 \sqrt{2 y} \\
0 \leq z \leq 1 / 2-y
\end{array}\right.
$$
Therefore,
$$
D=\left\{\begin{array}{ll}
(x, y): \begin{array}{l}
2 \sqrt{2 y} \leq x \leq 17 \sqrt{2 y} \\
0 \leq 1 / 2-y, \quad y \geq 0
\end{array}
\end{array}\right\}
$$
The inequality $y \geq 0$ is necessary because $y$ is under the square root.
3. We compute the double integral:
$$
\begin{aligned}
& V=\iint_{D}\left(\frac{1}{2}-y\right) d x d y=\int_{0}^{1 / 2} d y \int_{2 \sqrt{2 y}}^{17 \sqrt{2 y}}\left(\frac{1}{2}-y\right) d x= \\
&= 15 \sqrt{2} \int_{0}^{1 / 2}\left(\frac{1}{2}-y\right) \sqrt{y} d y=1
\end{aligned}
$$
Answer. $V=1$ unit of volume.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 1. Find the mass of the plate $D$ with surface density $\mu=16 x+9 y^{2} / 2$, bounded by the curves
$$
x=\frac{1}{4}, \quad y=0, \quad y^{2}=16 x \quad(y \geq 0)
$$
|
Solution.
1. The mass of the plate $D$ with surface density $\mu=16 x+9 y^{2} / 2$ is determined by the formula
$$
m=\iint_{D}\left(16 x+\frac{9 y^{2}}{2}\right) d x d y
$$
2. We compute the obtained double integral in Cartesian coordinates:
a) define the region $D$ by a system of inequalities:
$$
\left\{\begin{array}{l}
0 \leq x \leq 1 / 4 \\
0 \leq y \leq 4 \sqrt{x}
\end{array}\right.
$$
The inequality $0 \leq x$ follows from $y^{2}=16 x$, i.e., $x$ is non-negative
b) convert the double integral to an iterated integral:
$$
m=\iint_{D}\left(16 x+\frac{9 y^{2}}{2}\right) d x d y=\int_{0}^{1 / 4} d x \int_{0}^{4 \sqrt{x}}\left(16 x+\frac{9 y^{2}}{2}\right) d y
$$
c) integrate sequentially, using the properties of the definite integral:
$$
\begin{aligned}
m=\int_{0}^{1 / 4} d x \int_{0}^{4 \sqrt{x}}(16 x & \left.+\frac{9 y^{2}}{2}\right) d y= \\
& =\left.\int_{0}^{1 / 4}\left(16 x y+\frac{3 y^{3}}{2}\right)\right|_{0} ^{4 \sqrt{x}} d x=160 \int_{0}^{1 / 4} x^{3 / 2} d x=2
\end{aligned}
$$
Answer. $m=2$ units of mass.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 3. Find the mass of the plate $D$ with surface density $\mu=x / y^{5}$, bounded by the curves
$$
\frac{x^{2}}{16}+y^{2}=1, \quad \frac{x^{2}}{16}+y^{2}=3, \quad y=\frac{x}{4}, \quad x=0 \quad\left(y \geq \frac{x}{4}, x \geq 0\right)
$$
|
Solution.
1. The mass of the plate $D$ with surface density $\mu=x / y^{5}$ is determined by the formula
$$
m=\iint_{D} \frac{x}{y^{5}} d x d y
$$
2. We calculate the obtained double integral:
a) define the region $D$ by inequalities in Cartesian coordinates
$$
D=\left\{(x, y): \begin{array}{c}
1 \leq \frac{x^{2}}{16}+y^{2} \leq 3 \\
\\
y \geq x / 4, \quad x \geq 0
\end{array}\right\}
$$
Since the region $D$ is bounded by ellipses and lines passing through the origin, it is easier to solve the problem in generalized polar coordinates
$$
\left\{\begin{array}{l}
x=4 \varrho \cos \varphi \\
y=\varrho \sin \varphi
\end{array}\right.
$$
In this case, $(\varrho, \varphi) \in D^{\prime}$, and the sought mass is determined by the formula
$$
m=\iint_{D} \frac{x}{y^{5}} d x d y=\iint_{D^{\prime}} \frac{4 \varrho \cos \varphi}{\varrho^{5} \sin ^{5} \varphi} 4 \varrho d \varrho d \varphi
$$
To find the region $D^{\prime}$, we replace $x$ with $4 \varrho \cos \varphi$ and $y$ with $\varrho \sin \varphi$ in the inequalities defining the region $D$:
$$
\left\{\begin{array}{l}
1 \leq \frac{16 \varrho^{2} \cos ^{2} \varphi}{16}+\varrho^{2} \sin ^{2} \varphi \leq 3 \\
\varrho \sin \varphi \geq 4 \varrho \cos \varphi / 4, \quad \varrho \cos \varphi \geq 0
\end{array}\right.
$$
Solving these inequalities with respect to $\varrho$ and $\varphi$, we get
$$
D^{\prime}=\left\{\begin{array}{lc}
& (\varrho, \varphi): \begin{array}{c}
1 \leq \varrho \leq \sqrt{3} \\
\frac{\pi}{4} \leq \varphi \leq \frac{\pi}{2}
\end{array}
\end{array}\right\}
$$
b) we transition from the double integral to the repeated integral:
$$
m=\iint_{D^{\prime}} \frac{4 \varrho \cos \varphi}{\varrho^{5} \sin ^{5} \varphi} 4 \varrho d \varrho d \varphi=\int_{\pi / 4}^{\pi / 2} d \varphi \int_{1}^{\sqrt{3}} 16 \varrho^{-3} \cdot \frac{\cos \varphi}{\sin ^{5} \varphi} d \varrho
$$
c) integrating sequentially, we get
$$
m=16 \int_{\pi / 4}^{\pi / 2} \frac{d \sin \varphi}{\sin ^{5} \varphi} \int_{1}^{\sqrt{3}} \varrho^{-3} d \varrho=\left.\left.16\left(-\frac{1}{4 \sin ^{4} \varphi}\right)\right|_{\pi / 4} ^{\pi / 2}\left(-\frac{1}{2 \varrho^{2}}\right)\right|_{1} ^{\sqrt{3}}=4
$$
Answer. $m=4$ units of mass.
Conditions of the Problem. Find the mass of the plate $D$ with surface density $\mu$, where $D$ is bounded by the given lines.
1. $\mu=2 x+y^{2}, \quad x=4, y=0, y=\sqrt{x}$.
2. $\mu=x^{2}+y, \quad x=1, y=0, y=2 \sqrt{x}$.
3. $\mu=x^{2}+2 y, \quad x=0, y=4, y=x^{2}(x \geq 0)$.
4. $\mu=x+y^{2}, \quad x=0, y=1, y=x^{2} / 4(x \geq 0)$.
5. $\quad \mu=\frac{x-y}{x^{2}+y^{2}}, \quad x=0, \quad y=0, \quad x^{2}+y^{2}=4, \quad x^{2}+y^{2}=9$
$$
(x \geq 0, y \leq 0)
$$
6. $\quad \mu=\frac{2 y-x}{x^{2}+y^{2}}$
$x=0, \quad y=0, \quad x^{2}+y^{2}=3, \quad x^{2}+y^{2}=5$ $(x \leq 0, y \geq 0)$.
7. $\quad \mu=\frac{y-x}{x^{2}+y^{2}}, \quad x=0, \quad y=0, \quad x^{2}+y^{2}=4, \quad x^{2}+y^{2}=16$
$$
(x \leq 0, \quad y \geq 0)
$$
8. $\quad \mu(x, y)=y, \quad y=0, \quad y=x \sqrt{3}, \quad x^{2}+\frac{y^{2}}{4}=1, \quad x^{2}+\frac{y^{2}}{4}=9$
$$
(y \geq 0, \quad y \leq x \sqrt{3})
$$
9. $\quad \mu(x, y)=\frac{y}{x^{2}}, \quad y=0, \quad y=x, \quad \frac{x^{2}}{4}+y^{2}=1, \quad \frac{x^{2}}{4}+y^{2}=4$ $(y \geq 0, \quad y \leq x)$.
10. $\mu(x, y)=\frac{x}{y^{2}}, \quad x=0, \quad y=x, \quad \frac{x^{2}}{9}+\frac{y^{2}}{4}=1, \quad \frac{x^{2}}{9}+\frac{y^{2}}{4}=25$
$(x \geq 0, y \geq x)$.
Answers. 1. $m=448 / 15$. 2. $m=11 / 7$. 3. $m=448 / 15$. 4. $m=11 / 7$. 5. $m=2$. 6. $m=3(\sqrt{5}-\sqrt{3})$. 7. $m=4$. 8. $m=52 / 3$. 9. $m=(\sqrt{2}-1) / 2$. 10. $m=18(\sqrt{2}-1)$.
## 12.9. Triple Integral
in Cartesian coordinates
Problem Statement. Calculate the triple integral
$$
\iiint_{\Omega} f(x, y, z) d x d y d z
$$
where the region $\Omega$ is bounded by certain surfaces.
Plan of Solution.
1. Define the region $\Omega$ by a system of inequalities, for example,
$$
\left\{\begin{aligned}
a & \leq x \leq b \\
y_{1}(x) & \leq y \leq y_{2}(x) \\
z_{1}(x, y) & \leq z \leq z_{2}(x, y)
\end{aligned}\right.
$$
2. Transition from the triple integral to the repeated integral:
$$
\iiint_{\Omega} f(x, y, z) d x d y d z=\int_{a}^{b} d x \int_{y_{1}(x)}^{y_{2}(x)} d y \int_{z_{1}(x, y)}^{z_{2}(x, y)} f(x, y, z) d z
$$
3. Using the properties of the definite integral, integrate sequentially first with respect to $z$ (considering $x$ and $y$ as constants), then with respect to $y$ (considering $x$ as a constant), and finally with respect to $x$.
Write the answer.
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 1. Find the volume of the body $\Omega$, bounded by the surfaces
$$
x=17 \sqrt{2 y}, \quad x=2 \sqrt{2 y}, \quad z=\frac{1}{2}-y, \quad z=0
$$
|
Solution.
1. Define the region $\Omega$ by inequalities. Since $17 \sqrt{2 y} \geq 2 \sqrt{2 y}$, for $x$ we have the inequalities $2 \sqrt{2 y} \leq x \leq 17 \sqrt{2 y}$. Since $y$ appears under the square root, $y \geq 0$. For $z$, the possible inequalities are $0 \leq z \leq 1 / 2-y$ or $1 / 2-y \leq z \leq 0$. In the first case, $0 \leq y \leq 1 / 2$. In the second case, $y \geq 1 / 2$, i.e., the region is unbounded, which is unacceptable.
Thus,
$$
\Omega=\left\{\begin{array}{cc}
& 2 \sqrt{2 y} \leq x \leq 17 \sqrt{2 y} \\
(x, y, z): & 0 \leq y \leq 1 / 2 \\
& 0 \leq z \leq 1 / 2-y
\end{array}\right\}
$$
2. We compute the volume using formula (1), reducing the triple integral to an iterated integral:
$$
V=\iiint_{\Omega} 1 \cdot d x d y d z=\int_{0}^{1 / 2} d y \int_{2 \sqrt{2 y}}^{17 \sqrt{2 y}} d x \int_{0}^{1 / 2-y} d z=1
$$
Answer. $V=1$ unit of volume.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 1. Find the mass of the body $\Omega$ with density $\mu=2 x$, bounded by the surfaces
$$
x=2 \sqrt{2 y}, \quad x=\sqrt{2 y}, \quad z=1-y, \quad z=0
$$
|
Solution.
1. The mass of the body $\Omega$ with density $\mu=2 x$ is determined by the formula
$$
m=\iiint_{\Omega} 2 x d x d y d z
$$
2. Let's define the region $\Omega$ using inequalities. Since $2 \sqrt{2 y} \geq \sqrt{2 y}$, for $x$ we have the inequalities $\sqrt{2 y} \leq x \leq 2 \sqrt{2 y}$. Since $y$ appears under the square root, $y \geq 0$. For $z$, the possible inequalities are $0 \leq z \leq 1-y$ or $1-y \leq z \leq 0$. In the first case, $0 \leq y \leq 1$. In the second case, $y \geq 1$, i.e., the region is unbounded, which is unacceptable.
Thus,
$$
\Omega=\left\{\begin{array}{cc}
& \sqrt{2 y} \leq x \leq 2 \sqrt{2 y} \\
(x, y, z): & 0 \leq y \leq 1 \\
& 0 \leq z \leq 1-y
\end{array}\right\}
$$
3. We compute $m$ by reducing the triple integral to a repeated integral:
$$
m=\iiint_{\Omega} 2 x d x d y d z=\int_{0}^{1} d y \int_{\sqrt{2 y}}^{2 \sqrt{2 y}} 2 x d x \int_{0}^{1-y} d z=1
$$
Answer. $m=1$ unit of mass.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 3. Find the derivative of the scalar field $u=x z^{2}+2 y z$ at the point $M_{0}(1,0,2)$ along the circle
$$
\left\{\begin{array}{l}
x=1+\cos t \\
y=\sin t-1 \\
z=2
\end{array}\right.
$$
|
Solution. The vector equation of the circle has the form
$$
\mathbf{r}(t)=(1+\cos t) \mathbf{i}+(\sin t-1) \mathbf{j}+2 \mathbf{k} .
$$
We find the vector $T$, tangent to it at any point $M$. We have
$$
\left.T=\frac{d r}{d t}=-\sin t \mathbf{i}+\cos t \mathbf{j}\right]
$$
The given point $M_{0}(1,0,2)$ lies in the $\boldsymbol{O} \boldsymbol{z}$ plane in the first octant, and corresponds to the parameter value $t=\frac{\pi}{2}$. At this point we will have
$$
\left.\tau\right|_{M_{0}}=-\sin \frac{\pi}{2} \mathbf{i}+\cos \frac{\pi}{2} \mathbf{j}=-\mathbf{i} .
$$
From this we obtain that the direction cosines of the tangent to the circle are $\cos \alpha=-1, \cos \beta=0, \cos \gamma=0$. The values of the partial derivatives of the given scalar field at the point $M_{0}(1,0,2)$ are
$$
\left.\frac{\partial u}{\partial x}\right|_{M_{0}}=\left.z^{2}\right|_{M_{0}}=4,\left.\quad \frac{\partial u}{\partial y}\right|_{M_{0}}=\left.2 z\right|_{M_{0}}=4,\left.\quad \frac{\partial u}{\partial z}\right|_{M_{0}}=\left.(2 x z+2 y)\right|_{M_{0}}=4
$$
Therefore, the sought derivative is
$$
\left.\frac{\partial u}{\partial l}\right|_{M_{0}}=\left.\frac{\partial u}{\partial r}\right|_{M_{0}}=4 \cdot(-1)+4 \cdot 0+4 \cdot 0=-4
$$
## Problems for Independent Solution
In the following problems, find the derivative of the given functions at the point $M_{0}\left(x_{0}, y_{0}, z_{0}\right)$ in the direction of the point $M_{1}\left(x_{1}, y_{1}, z_{1}\right)$.
58. $u=\sqrt{x^{2}+y^{2}+z^{2}}, \quad M_{0}(1,1,1), M_{1}(3,2,1)$
59. $u=x^{2} y+x z^{2}-2, \quad M_{0}(1,1,-1), M_{1}(2,-1,3)$.
60. $u=x e^{y}+y e^{x}-z^{2}, \quad M_{0}(3,0,2), M_{1}(4,1,3)$
61. $u=\frac{x}{y}-\frac{y}{x}, \quad M_{0}(1,1), M_{1}(4,5)$.
62. Find the derivative of the scalar field $u=\ln \left(x^{2}+y^{2}\right)$ at the point $M_{0}(1,2)$ of the parabola $y^{2}=4 x$ in the direction of this curve.
63. Find the derivative of the scalar field $u=\operatorname{arctg} \frac{y}{x}$ at the point $M_{0}(2,-2)$ of the circle $x^{2}+y^{2}-4 x=0$ along the arc of this circle.
64. Find the derivative of the scalar field $u=x^{2}+y^{2}$ at the point $M_{0}\left(x_{0}, y_{0}\right)$ of the circle $x^{2}+y^{2}=\boldsymbol{R}^{2}$ in the direction of this circle.
65. Find the derivative of the scalar field $u=2 x y+y^{2}$ at the point $(\sqrt{2}, 1)$ of the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ in the direction of the external normal to the ellipse at this point.
66. Find the derivative of the scalar field $u=x^{2}-y^{2}$ at the point $(5,4)$ of the hyperbola $x^{2}-y^{2}=9$ in the direction of this curve.
67. Find the derivative of the scalar field $u=\ln (x y+y z+x z)$ at the point $M_{0}(0,1,1)$ in the direction of the circle $x=\cos t, y=\sin t, z=1$.
68. Find the derivative of the scalar field $u=x^{2}+y^{2}+z^{2}$ at the point $M_{0}$ corresponding to the value of the parameter $t=\frac{\pi}{2}$ in the direction of the helical line $x=R \cos t$, $y=R \sin t, z=a t$.
## §9. Gradient of a Scalar Field
Let us have a scalar field defined by the scalar function
$$
u=f(x, y, z)
$$
the function $f$ is assumed to be differentiable.
Definition. The gradient of a scalar field at a given point $M$ is a vector denoted by the symbol grad $u$ and defined by the formula
$$
\operatorname{grad} u=\frac{\partial u}{\partial x} \mathbf{i}+\frac{\partial u}{\partial y} \mathbf{j}+\frac{\partial u}{\partial z} \mathbf{k}
$$
We have
Using the formula (1) from §8 for the directional derivative,
$$
\frac{\partial u}{\partial l}=\left(\operatorname{grad} u, \mathbf{l}^{0}\right)
$$
where $\mathbf{l}^{0}$ is the unit vector in the direction of $\mathbf{l}$, i.e.
$$
\mathbf{l}^{0}=\frac{\mathbf{l}}{|\mathbf{l}|}=\mathbf{i} \cos \alpha+\mathbf{j} \cos \beta+\mathbf{k} \cos \gamma
$$
## Properties of the Gradient
1. The gradient is directed along the normal to the level surface (or level line, if the field is planar).
2. The gradient is directed in the direction of the increase of the field function.
3. The magnitude of the gradient is equal to the maximum directional derivative at the given point of the field:
$$
\left.\max \frac{\partial u}{\partial l}=|\operatorname{grad} u|=\sqrt{\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}+\left(\frac{\partial u}{\partial z}\right)^{2}}\right)
$$
These properties provide an invariant characteristic of the gradient. They indicate that the vector grad $u$ points in the direction and magnitude of the greatest change in the scalar field at the given point.
|
-4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 1. Using the invariant definition, calculate the divergence of the vector $a=x \mathbf{i}$ at the point $O(0,0,0)$, choosing as the surface $\sigma$ surrounding the point $O$, a sphere $\sigma_{\varepsilon}$ of radius $\varepsilon$ centered at this point.
|
Solution. By the definition of divergence at the given point, we have
$$
\operatorname{div} a(0)=\lim _{\left(\sigma_{k}\right) \rightarrow 0} \frac{\int\left(a, n^{0}\right) d \sigma}{v_{\varepsilon}}
$$
where $v_{\varepsilon}$ is the volume of the ball bounded by the sphere $\sigma_{\varepsilon}$, or
$$
\operatorname{div} a(0)=\lim _{k \rightarrow 0} \frac{\prod_{0}\left(a, n^{0}\right) d \theta}{v_{\varepsilon}}
$$
But since the volume of the ball is $v_{\varepsilon}=\frac{4}{3} \pi \varepsilon^{3}$, then
$$
\operatorname{div} a(0)=\lim _{\varepsilon \rightarrow 0} \frac{\sigma_{\varepsilon}\left(a, n^{0}\right) d \sigma}{(4 / 3) \pi \varepsilon^{3}}
$$
We will compute the flux $\oiint_{\sigma_{\varepsilon}}\left(a, n^{\dagger}\right) d \sigma$ of the given vector through the sphere $\sigma_{\varepsilon}$. The normal vector $n^{\dagger}$ to the sphere $\sigma_{\varepsilon}$ is directed along the radius of the sphere, so we can set
$$
\mathbf{n}^{\mathbf{n}}=\mathbf{r}^{\mathbf{0}}=\frac{\mathbf{r}}{|\mathbf{r}|}=\frac{\mathbf{r}}{\varepsilon}
$$
where $\mathbf{r}^{\mathbf{0}}$ is the unit vector of the radius vector $\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}$, or
$$
\mathbf{n}^{0}=\frac{x \mathbf{i}+y \mathbf{j}+z \mathbf{k}}{\varepsilon}
$$
The desired flux will be equal to
$$
\oiint_{\sigma_{\varepsilon}}\left(a, n^{0}\right) d \sigma=\oiint_{\sigma_{\varepsilon}} \frac{x^{2}}{\varepsilon} d \sigma
$$
Switching to coordinates on the sphere $\sigma_{\varepsilon}$
$$
x=\varepsilon \cos \varphi \sin \theta, \quad y=\varepsilon \sin \varphi \sin \theta, \quad z=\varepsilon \cos \theta .
$$
we get
$$
\begin{aligned}
\oiint_{\sigma_{\varepsilon}}\left(a, n^{0}\right) d \sigma=\iint_{\sigma_{\varepsilon}} \frac{\varepsilon^{2} \cos ^{2} \varphi \sin ^{2} \theta \varepsilon^{2} \sin \theta d \varphi d \theta}{\varepsilon}= \\
=\varepsilon^{3} \int_{0}^{2 \pi} \cos ^{2} \varphi d \varphi \int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3} \pi \varepsilon^{3} .
\end{aligned}
$$
Therefore,
$$
\operatorname{div} a(0)=\lim _{\varepsilon \rightarrow 0} \frac{(4 / 3) \pi \varepsilon^{3}}{(4 / 3) \pi \varepsilon^{3}}=1
$$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 5. Show that the vector field $a=\frac{2 \cos \theta}{r^{3}} \mathbf{e}_{r}+\frac{\sin \theta}{r^{3}} \mathbf{e}_{\theta}$
is solenoidal.
|
Solution. Using formula (5), we will have
$$
\begin{aligned}
& \operatorname{div}=\frac{1}{r^{2}} \frac{\theta}{\partial r}\left(r^{2} \frac{2 \cos \theta}{r^{3}}\right)+\frac{1}{r \sin \theta} \frac{\theta}{\partial \theta}\left(\sin \theta \frac{\sin \theta}{r^{3}}\right)+0= \\
&=\frac{1}{r^{2}}\left(-\frac{2 \cos \theta}{r^{2}}\right)+\frac{1}{r^{4} \sin \theta} 2 \sin \theta \cos \theta=0
\end{aligned}
$$
everywhere where $\boldsymbol{r} \neq 0$. This means that the vector field is solenoidal everywhere except at the point $\boldsymbol{r}=0$.
## Problems for Independent Solution
267. Find the equation of the vector lines of the following fields
a) $\mathbf{a}=e_{p}+\rho e_{4}+e_{z} ; 6$ 6) $=\rho e_{p}+\varphi e_{p}+z e_{z} ;$ c) $a=\frac{2 \alpha \cos \theta}{r^{3}} c+\frac{\alpha \sin \theta}{r^{3}} e_{\theta}, \alpha=$ const. Find the gradients of scalar fields:
a) In cylindrical coordinates
268. $u=\rho^{2}+2 \rho \cos \varphi-e^{8} \sin \varphi$.
269. $u=\rho \cos \varphi+2 \sin ^{2} \varphi-3^{\mu}$.
b) In spherical coordinates
270. $u=r^{2} \cos \theta$. 271. $u=3 r^{2} \sin \theta+e^{r} \cos \varphi-r$. 272. $u=\frac{\mu \cos \theta}{r^{2}}, \mu=$ const.
Calculate the divergence of the vector fields:
a) In cylindrical coordinates
273. $a=\rho e_{n}+x \sin \varphi e_{\varphi}+e^{\psi} \cos x \mathrm{e}_{2} . \quad 274.2=\varphi \operatorname{arctg} \rho e_{\rho}+2 e_{p}-x^{2} e^{2} e_{z}$.
b) In spherical coordinates
275. $a=r^{2} e_{s}-2 \cos ^{2} \varphi e_{\theta}+\frac{\varphi}{r^{2}+1} e_{\varphi}$.
Calculate the rotor of the following vector fields:
276. $a=(2 r+\alpha \cos \varphi) \mathrm{e}_{r}-\alpha \sin \theta \mathrm{e}_{\theta}+r \cos \theta \mathrm{e}_{\varphi}, \alpha=$ const.
277. $a=r^{2} \mathrm{e}_{r}+2 \cos \theta \mathrm{e}_{\theta}-\varphi \mathrm{e}_{\text {, }}$.
278. $\mathrm{a}=\cos \varphi \mathrm{e}_{5}-\frac{\sin \varphi}{\rho} \mathrm{e}_{5}+\rho^{2} \mathrm{e}_{2}$.
279. Show that the vector field $\mathrm{a}=\frac{2 \cos \theta}{\boldsymbol{r}^{3}} \mathbf{r}_{r}+\frac{\sin \theta}{\boldsymbol{r}^{3}} \mathbf{e}_{\text {, }}$ is potential.
280. Show that the vector field a $=f(r) e_{\text {, }}$, where $f(r)-$ any differentiable function, is potential.
$5^{\circ}$. Calculation of flux in curvilinear coordinates. Let $S$ be a part of the coordinate surface $q_{1}=C$, where $C=$ const, bounded by coordinate lines
$$
\begin{aligned}
& Q_{2}=\alpha_{1}, \quad Q_{2}=\alpha_{2} \quad\left(\alpha_{1}<\alpha_{2}\right) \\
& q_{3}=\beta_{1}, \quad q_{3}=\beta_{2} \quad\left(\beta_{1}<\beta_{2}\right)
\end{aligned}
$$
Then the flux of the vector field
$$
a=a_{1}\left(q_{1}, q_{2}, q_{3}\right) \mathbf{e}_{1}+a_{2}\left(q_{1}, q_{2}, q_{3}\right) e_{2}+a_{3}\left(q_{1}, q_{2}, q_{3}\right) e_{3}
$$
through the surface $S$ in the direction of the vector $\mathbf{e}_{1}$ is calculated by the formula
$$
\Pi=\int_{a_{1}}^{a_{1}} \int_{\beta_{1}}^{\beta_{2}} a_{1}\left(C, q_{2}, q_{3}\right) H_{2}\left(C, q_{2}, q_{3}\right) H_{3}\left(C, q_{2}, q_{3}\right) d q_{3} d q_{2}
$$
Similarly, the flux through a part of the surface $q_{2}=C$ or through a part of the surface $\varphi_{3}=C$, where $C=$ const, is calculated.
Example v. Calculate the flux of the vector field given in cylindrical coordinates: $a=\rho \mathrm{e}_{p}+\boldsymbol{z e} \mathbf{e}_{p}$ through the outer part of the lateral surface of the cylinder $\rho=1$, bounded by the planes $z=0, z=1$.
Solution. The cylinder is a coordinate surface $\rho=C=$ const, so the sought flux is
$$
\Pi=\int_{0}^{2 z} \int_{0}^{1} C^{2} d z d \varphi=2 \pi C^{2}
$$
From this, for the surface $\rho=1$, we get
$$
\Pi=2 \pi
$$
|
0
|
Calculus
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
Example 12. Calculate the circulation of the vector field given in cylindrical coordinates: $2=\rho \sin \varphi \mathrm{e}_{\rho}+\rho z \mathrm{e}_{\varphi}+\rho^{3} \mathrm{e}_{z}$, along the curve L: $\{\rho=\sin \varphi, z=0,0 \leqslant \varphi \leqslant \pi\}$ directly and using Stokes' theorem.
|
Solution. Coordinates of the given vector
$$
a_{p}=\rho \sin \varphi, \quad a_{v}=\rho z, \quad a_{n}=\rho^{3}
$$
The contour $L$ represents a closed curve located in the plane $z=0$ (Fig. 41).
1) Direct calculation of circulation.
Substituting the coordinates of the vector into formula (13), we get
$$
\boldsymbol{\mu}=\oint_{\mathbf{L}} \rho \sin \varphi d \rho+\rho^{2} z d \varphi+\rho^{3} d z
$$
On the curve $L$ we have:
$$
z=0, d z=0 ; \quad \rho=\sin \varphi, d \rho=\cos \varphi d \varphi ; \quad 0 \leqslant \varphi \leqslant \pi
$$
Therefore, the sought circulation will be equal to
$$
\boldsymbol{\mu}=\oint_{L} \rho \sin \varphi d \rho=\int_{0}^{\pi} \sin ^{2} \varphi \cos \varphi d \varphi=0
$$
2) Calculation of circulation using Stokes' theorem. According to Stokes' theorem, the sought circulation is equal to
$$
\mu=\oint_{\mathbf{L}}(\mathrm{a}, d r)=\iint_{s}\left(\mathrm{rot} \mathrm{a}, \mathrm{n}^{0}\right) d S
$$
where $S$ is the surface stretched over the contour $L$.
We find the rotor of the given field
$$
\operatorname{rot} \alpha=\frac{1}{\rho}\left|\begin{array}{ccc}
e_{\phi} & \rho e_{\phi} & e_{z} \\
\frac{\theta}{\partial \rho} & \frac{\theta}{\partial \varphi} & \frac{\theta}{\partial z} \\
\rho \sin ^{2} \varphi & \rho^{2} z & \rho^{3}
\end{array}\right|=-\rho e_{\rho}-3 \rho^{2} e_{\varphi}+(2 z-\cos \varphi) e_{z} .
$$
At points where $\rho=0$, the value of the rotor is determined by continuity, assuming
$$
\text { rot } a(0, \varphi, z)=(2 z-\cos \varphi) e_{z} \text {. }
$$
Thus, the rotor is defined throughout the entire three-dimensional space. Since the curve $L$ lies in the plane $z=0$, we can take as the surface $S$ stretched over this curve, the part of the plane $z=0$ bounded by the curve $L$. Then, for the normal vector $\mathbf{n}^{\prime \prime}$ to the surface $S$, we can take the unit vector $\mathbf{e}_{z}$, i.e., $\mathbf{n}^{\prime \prime}=\mathbf{e}_{z}$. We find the scalar product:
$\left(\operatorname{rot} \mathrm{A}, \mathrm{n}^{0}\right)=\left(-\rho e_{\rho}-3 \rho^{2} \mathrm{e}_{\varphi}+(2 z-\cos \varphi) e_{z}, \mathrm{e}_{z}\right)=2 z-\cos \varphi_{.}$.
Since, due to the orthonormality of the basis vectors $e_{\rho}, \mathbf{e}_{\varphi}, \mathbf{e}_{z}$, we have
$$
\left(e_{\rho}, e_{z}\right)=\left(e_{\varphi}, e_{z}\right)=0, \quad\left(e_{z}, e_{z}\right)=1
$$
The sought circulation is equal to
$$
\zeta=\iint_{S}(2 z-\cos \varphi) d S
$$
Considering that $z=0$ on $S$ and the element of area $d S$ of the coordinate surface $z=0$ is equal to
$$
d S=\rho d \rho d \varphi
$$
we finally get
$$
\begin{aligned}
& \zeta=-\iint_{S} \cos \varphi d S=-\iint_{S} \cos \varphi \cdot \rho d \rho d \varphi= \\
& =-\int_{0}^{\pi} \cos \varphi d \varphi \int_{0}^{\sin \varphi} \rho d \rho=-\frac{1}{2} \int_{0}^{\pi} \sin ^{2} \varphi \cos \varphi d \varphi=0
\end{aligned}
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
45. How many numbers need to be selected from a table of random numbers to be confident with a probability of at least 0.9 that among them there is at least one even number?
|
Solution. Let $n$ be the required number of random numbers. Consider the events:
$A_{k}$ - one randomly selected number is even $(k=1,2, \ldots, n)$;
$\overline{A_{k}}$ - one randomly selected number is odd;
$B$ - among $n$ random numbers there will be at least one even;
$\bar{B}$ - among $n$ random numbers there will be no even numbers.
The probabilities of all events $\overline{A_{k}}$ are equal. Let $P\left(A_{k}\right)=p$, $P\left(\bar{A}_{k}\right)=1-p=q$. Events $A_{k}$ and $\overline{A_{k}}$ are equally probable, so $p=q=0.5$. The probability of event $B$ is
$$
P(B)=1-P(\bar{B})
$$
Event $\bar{B}$ consists in all $n$ random numbers being odd. This means that $\bar{B}$ is the product of $\boldsymbol{n}$ independent events $\overline{A_{k}}: \bar{B}=\overline{A_{1}} \cdot \overline{A_{2}} \cdot \ldots \cdot \overline{A_{n}}$.
Applying the multiplication theorem of probabilities of independent events and equality $(*)$, we find $P(B)=1-q^{n}=1-0.5^{n}$.
By the condition $P(B) \geq 0.9$, we get: $1-0.5^{n} \geq 0.9$.
Solving this inequality, we sequentially find:
$$
\begin{gathered}
-0.5^{n} \geq-0.1 \\
0.5^{n} \leq 0.1 \\
n \cdot \lg 0.5 \leq \lg 0.1
\end{gathered}
$$
Since $\lg 0.5(\lg 0.1 / \lg 0.5) \approx 3.32$. Since $n$ is an integer, then $\boldsymbol{n} \geq 4$.
## 2.2. The Formula of Total Probability
The probability of an event $A$, which can occur given the occurrence of one of $n$ mutually exclusive events (hypotheses) $B_{i}(i=1,2, \ldots, n)$, forming a complete group, is found using the formula of total probability:
$$
P(A)=\sum_{i=1}^{n} P\left(B_{i}\right) P_{B_{i}}(A)
$$
In this formula, $P(A)$ is the probability of event $A$; $P\left(B_{i}\right)$ is the probability of event $B_{i}$, $P_{B_{i}}(A)$ is the conditional probability of event $A$, calculated given that event $B_{i}$ has occurred; $\sum_{i=1}^{n} P\left(B_{i}\right) P_{B_{i}}(A)$ is the sum of the products of the probabilities $P\left(B_{i}\right)$ of each of the events $B_{i}$ and the corresponding conditional probability $P_{B_{l}}(A)$.
The sum of the probabilities of the hypotheses $\sum_{i=1}^{n} P\left(B_{i}\right)=1$.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
118. The random variable $X$ is given by the distribution function:
$$
F(x)=\left\{\begin{array}{ccc}
0 & \text { if } & x \leq -c \\
\frac{1}{2}+\frac{1}{\pi} \arcsin \frac{x}{c} & \text { if } & -c < x \leq c \\
1 & \text { if } & x > c
\end{array}\right.
$$
( arcsine law ).
Find the mathematical expectation of this variable.
|
Solution. Let's find the probability density function of the random variable $X: f(x)=F^{\prime}(x)$. For $x \leq -c$ and for $x > c$, the function $f(x)=0$. For $-c < x \leq c$, we have
$$
f(x)=\left(\frac{1}{2}+\frac{1}{\pi} \arcsin \frac{x}{c}\right)^{\prime}=\frac{1}{\pi \sqrt{c^{2}-x^{2}}}
$$
The expected value of the random variable $X$ will be found using formula (54) with $a=-c$ and $b=c$:
$$
M(X)=\int_{-c}^{c} \frac{x d x}{\pi \sqrt{c^{2}-x^{2}}}
$$
Considering that the integrand is an odd function and the limits of integration are symmetric with respect to the origin, we conclude that the integral is zero; therefore, $M(X)=0$.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
120. Calculate $\left(\frac{9}{16}\right)^{-1 / 10}:\left(\frac{25}{36}\right)^{-3 / 2}-\left[\left(\frac{4}{3}\right)^{-1 / 2}\right]^{-2 / 5}\left(\frac{6}{5}\right)^{-3}$.
|
Solution. We will perform the actions sequentially:
1) $\left(\frac{9}{16}\right)^{-1 / 10}=\left[\left(\frac{3}{4}\right)^{2}\right]^{-1 / 10}=\left(\frac{3}{4}\right)^{-1 / 5}=\left(\frac{4}{3}\right)^{1 / 5}$;
2) $\left(\frac{25}{36}\right)^{-3 / 2}=\left[\left(\frac{5}{6}\right)^{2}\right]^{-3 / 2}=\left(\frac{5}{6}\right)^{-3}=\left(\frac{6}{5}\right)^{3}=\frac{216}{125}$;
3) $\left[\left(\frac{4}{3}\right)^{-1 / 2}\right]^{-2 / 5}=\left(\frac{4}{3}\right)^{1 / 5} ; \quad$ 4) $\left(\frac{6}{5}\right)^{-3}=\left(\frac{5}{6}\right)^{3}=\frac{125}{216}$.
Using the obtained results, we find
$$
\left(\frac{4}{3}\right)^{1 / 5} \cdot \frac{125}{216}-\left(\frac{4}{3}\right)^{1 / 5} \cdot \frac{125}{216}=0
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
137. $5^{x}=625$
|
Solution. Writing 625 as $5^{4}$, we get $5^{x}=5^{4}$, hence $x=4$. 138. $8^{x}=32$.
Solution. We have $32=2^{5} ; 8^{x}=\left(2^{3}\right)^{x}=2^{3 x}$. Therefore, $2^{3 x}=2^{5}$, hence $3 x=5$, i.e., $x=5 / 3$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
154. $(0,25)^{2-x}=\frac{256}{2^{x+3}}$.
154. $(0.25)^{2-x}=\frac{256}{2^{x+3}}$.
|
Solution. Let's convert all powers to base $2: 0.25=1/4=2^{-2}$; $256=2^{8}$. Therefore, $\left(2^{-2}\right)^{2-x}=\frac{2^{8}}{2^{x+3}}$. Applying the rule of dividing powers, we have
$$
\begin{gathered}
2^{-4+2 x}=2^{8-x-3} ; 2^{-4+2 x}=2^{5-x} ;-4+2 x=5-x ; 2 x+x=5+4 \\
3 x=9 ; x=3
\end{gathered}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
161. $2^{x}+2^{x-1}-2^{x-3}=44$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
161. $2^{x}+2^{x-1}-2^{x-3}=44$.
|
Solution. Since the smallest exponent is $x-3$, we factor out $2^{x-3}$:
$$
2^{x-3} \cdot\left(2^{3}+2^{2}-1\right)=44 ; 2^{x-3}(8+4-1)=44 ; 2^{x-3} \cdot 11=44
$$
Dividing both sides of the equation by 11, we get
$$
2^{x-3}=4 ; 2^{x-3}=2^{2} ; x-3=2 ; x=5
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
162. $7^{x}-3 \cdot 7^{x-1}+7^{x+1}=371$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
162. $7^{x}-3 \cdot 7^{x-1}+7^{x+1}=371$.
|
Solution. The least exponent is $x-1$; therefore, we factor out $7^{x-1}$:
$$
\begin{gathered}
7^{x-1} \cdot\left(7^{1}-3 \cdot 1+7^{12}\right)=371 ; 7^{x-1}(7-3+49)=371 \\
7^{x-1} \cdot 53=371 ; 7^{x-1}=7 ; x-1=1 ; x=2
\end{gathered}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
171. $7^{2 x}-48 \cdot 7^{x}=49$
171. $7^{2 x}-48 \cdot 7^{x}=49$
|
Solution. Let $7^{x}=y$, we get the quadratic equation $y^{2}-$ $-48 y-49=0$. Let's solve it. Here $a=1, b=-48, c=-49 ; \quad D=b^{2}-$ $-4 a c=(-48)^{2}-4 \cdot 1(-49)=2304+196=2500 ; \sqrt{D}=50$. Using the formula $y_{1,2}=\frac{-b \pm \sqrt{D}}{2 a}$, we find
$$
y_{1}=\frac{48-50}{2}=\frac{-2}{2}=-1 ; \quad y_{2}=\frac{48+50}{2}=\frac{98}{2}=49
$$
Since $7^{x}=y$, then $7^{x}=-1$ (this equality is impossible, as the exponential function can only take positive values); $7^{x}=49 ; 7^{x}=7^{2}$, i.e., $x=2$. Thus, we get the answer: $x=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
179. $5^{2}=25$.
|
Solution. Since the base of the power is 5, the exponent (logarithm) is 2, and the power is 25, then $\log _{5} 25=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
231. $\log _{4}(x+3)-\log _{4}(x-1)=2-3 \log _{4} 2$.
|
solution. Representing the number 2 as the logarithm of 16 to the base 4, we rewrite the given equation as
$$
\log _{4}(x+3)-\log _{4}(x-1)=\log _{4} 16-3 \log _{4} 2
$$
From this, we obtain
$$
\log _{4} \frac{x+3}{x-1}=\log _{4} \frac{16}{8}, \text { or } \frac{x+3}{x-1}=2
$$
We solve this equation:
$$
x+3=2(x-1) ; \quad x+3=2 x-2 ; \quad x=5
$$
To verify, substitute the value $x=5$ into the given equation:
$\log _{4}(5+3)-\log _{4}(5-1)=2-3 \log _{4} 2$
$\log _{4} 8-\log _{4} 4=\log _{4} 16-\log _{4} 8 ; \log _{4} \frac{8}{4}=\log \frac{16}{8} ; \log _{4} 2=\log _{4} 2$.
Thus, $x=5$.
Let's consider another type of logarithmic equations.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
320. $\sqrt{x-1} \cdot \sqrt{2 x+6}=x+3$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
320. $\sqrt{x-1} \cdot \sqrt{2 x+6}=x+3$.
|
Solution. First, we perform the multiplication of the roots:
$\sqrt{(x-1)(2 x+6)}=x+3 ; \sqrt{2 x^{2}-2 x+6 x-6}=x+3 ; \sqrt{2 x^{2}+4 x-6}=x+3$.
Now, we square both sides of the equation:
$\left(\sqrt{2 x^{2}+4 x-6}\right)^{2}=(x+3)^{2} ; 2 x^{2}+4 x-6=x^{2}+6 x+9 ; x^{2}-2 x-15=0$.
We solve the quadratic equation. Here $a=1, b=-2, c=-15, D=$ $=b^{2}-4 a c=4-4 \cdot 1 \cdot(-15)=64, \sqrt{D}=8$; therefore,
$$
x_{1,2}=\frac{-b \pm \sqrt{D}}{2 a} ; x_{1}=\frac{2-8}{2}=-3 ; x_{2}=\frac{2+8}{2}=5 ; x_{1}=-3, x_{2}=5
$$
Let's check both roots.
For $x=-3$ we have $\sqrt{(-3)-1} \cdot \sqrt{2(-3)+6}=-3+3$. We see that the first radical has no meaning in the set of real numbers; therefore, $x=-3$ is an extraneous root.
For $x=5$ we have $\sqrt{5-1} \cdot \sqrt{5 \cdot 2+6}=5+3 ; \sqrt{4} \cdot \sqrt{16}=8 ; 2 \cdot 4=8$. Thus, the root of the given equation is $x=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
321. $\sqrt{2 x+5}+\sqrt{x-1}=8$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
321. $\sqrt{2 x+5}+\sqrt{x-1}=8$.
|
Solution. Isolate one of the radicals and square both sides of the equation:
$$
(\sqrt{2 x+5})^{2}=(8-\sqrt{x-1})^{2} ; 2 x+5=64-16 \sqrt{x-1}+(x-1)
$$
Move $16 \sqrt{x-1}$ to the left side, and all other terms to the right side:
$$
16 \sqrt{x-1}=64+x-1-2 x-5 ; 16 \sqrt{x-1}=58-x
$$
Square both sides of the equation again:
$(16 \sqrt{x-1})^{2}=(58-x)^{2} ; 256 x-256=3364-116 x+x^{2} ; x^{2}-372 x+3620=0$.
Solve the resulting quadratic equation. Since $a=1, b=-372$, $c=3620, \quad D=b^{2}-4 a c=(-372)^{2}-4 \cdot 1 \cdot 3620=138384-14480=123904$; $\sqrt{D}=352$, then
$$
x_{1}=\frac{-b-\sqrt{D}}{2 a}=\frac{372-352}{2}=10 ; \quad x_{2}=\frac{-b+\sqrt{D}}{2 a}=\frac{372+352}{2}=362
$$
Check both values $x_{1}=10 ; x_{2}=362$.
If $x=10$, then $\sqrt{2 \cdot 10+5}+\sqrt{10-1}=8 ; \sqrt{25}+\sqrt{9}=8 ; 5+3=8$; $8=8$
If $x=362$, then $\sqrt{2 \cdot 362+5}+\sqrt{362-1}=8 ; \sqrt{729}+\sqrt{361}=8 ; 27+$ $+19 \neq 8$; therefore, $x=362$ is an extraneous root.
The root of the given equation is $x=10$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
385. $(\tan \alpha+\cot \alpha)^{2}-(\tan \alpha-\cot \alpha)^{2}$.
|
Solution. We will use the formulas for the square of the sum and difference of two numbers:
$$
\begin{gathered}
(\operatorname{tg} \alpha+\operatorname{ctg} \alpha)^{2}-(\operatorname{tg} \alpha-\operatorname{ctg} \alpha)^{2}=\operatorname{tg}^{2} \alpha+2 \operatorname{tg} \alpha \operatorname{ctg} \alpha+\operatorname{ctg}^{2} \alpha-\operatorname{tg}^{2} \alpha+ \\
+2 \operatorname{tg} \alpha \operatorname{ctg} \alpha-\operatorname{ctg}^{2} \alpha=4 \operatorname{tg} \alpha \operatorname{ctg} \alpha=4
\end{gathered}
$$
(after combining like terms, we applied identity IV).
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
26. Compute the second-order determinants:
a) $\left|\begin{array}{rr}2 & 5 \\ -3 & -4\end{array}\right|$;
b) $\left|\begin{array}{ll}a^{2} & a b \\ a b & b^{2}\end{array}\right|$.
|
Solution. a) $\left|\begin{array}{rr}2 & 5 \\ -3 & -4\end{array}\right|=2(-4)-5(-3)=-8+15=7$;
b) $\left|\begin{array}{ll}a^{2} & a b \\ a b & b^{2}\end{array}\right|=a^{2} \cdot b^{2}-a b \cdot a b=a^{2} b^{2}-a^{2} b^{2}=0$.
27-32. Calculate the determinants:
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
33. Calculate the determinants of the third order:
a) $\left|\begin{array}{lll}3 & 2 & 1 \\ 2 & 5 & 3 \\ 3 & 4 & 3\end{array}\right|$
b) $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$
|
Solution.
a) $\left|\begin{array}{lll}3 & 2 & 1 \\ 2 & 5 & 3 \\ 3 & 4 & 3\end{array}\right|=3 \cdot 5 \cdot 3+2 \cdot 3 \cdot 3+2 \cdot 4 \cdot 1-1 \cdot 5 \cdot 3-2 \cdot 2 \cdot 3-$ $-3 \cdot 3 \cdot 4=45+18+8-15-12-36=71-63=8$;
b) $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=a c b+b a c+c b a-c \cdot c \cdot c-b \cdot b \cdot b-a \cdot a \cdot a=$ $=3 a b c-a^{3}-b^{3}-c^{3}$.
34-39. Calculate the determinants:
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
92. Show that as $t \rightarrow \infty$ the limit of the variable
$x=\frac{6 t^{3}-9 t+1}{2 t^{3}-3 t}$ is 3.
|
Solution. We find the difference between the variable $x$ and the number 3:
$$
\begin{gathered}
x-3=\frac{6 t^{3}+9 t+1}{2 t^{3}+3 t}-3=\frac{6 t^{3}+9 t+1-3\left(2 t^{3}+3 t\right)}{2 t^{3}+3 t}=\frac{6 t^{3}+9 t+1-6 t^{3}-9 t}{2 t^{3}+3 t}= \\
=\frac{1}{2 t^{3}+3 t}
\end{gathered}
$$
If $t \rightarrow \infty$, then $\frac{1}{2 t^{3}+3 t} \rightarrow 0$. Therefore, the condition $|x-3|<\varepsilon$ is satisfied, and consequently, $\lim _{t \rightarrow \infty} x=3$.
|
3
|
Calculus
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
94. Find $\lim _{x \rightarrow 2}\left(3 x^{2}-2 x\right)$.
|
Solution. Using properties 1, 3, and 5 of limits sequentially, we get
$$
\begin{gathered}
\lim _{x \rightarrow 2}\left(3 x^{2}-2 x\right)=\lim _{x \rightarrow 2}\left(3 x^{2}\right)-\lim _{x \rightarrow 2}(2 x)=3 \lim _{x \rightarrow 2} x^{2}-2 \lim _{x \rightarrow 2} x= \\
=3\left(\lim _{x \rightarrow 2} x\right)^{2}-2 \lim _{x \rightarrow 2} x=3 \cdot 2^{2}-2 \cdot 2=8 .
\end{gathered}
$$
|
8
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
95. Find $\lim _{x \rightarrow 4} \frac{x^{2}-2 x}{x-3}$.
|
Solution. To apply the property of the limit of a quotient, we need to check whether the limit of the denominator is not zero when $x=4$. Since $\lim _{x \rightarrow 4}(x-3)=$[^3]$=\lim _{x \rightarrow 4} x-\lim 3=4-3=1 \neq 0$, in this case, we can use property 4:
$$
\lim _{x \rightarrow 4} \frac{x^{2}-2 x}{x-3}=\frac{\lim _{x \rightarrow 1}\left(x^{2}-2 x\right)}{\lim _{x \rightarrow 4}(x-3)}
$$
But $\lim _{x \rightarrow 4}\left(x^{2}-2 x\right)=\left(\lim _{x \rightarrow 4} x\right)^{2}-2 \lim _{x \rightarrow 4} x$ and, therefore,
$$
\lim _{x \rightarrow 4} \frac{x^{2}-2 x}{x-3}=\frac{\left(\lim _{x \rightarrow 4} x\right)^{2}-2 \lim _{x \rightarrow 4} x}{\lim _{x \rightarrow 4} x-\lim 3}=\frac{4^{2}-2 \cdot 4}{4-3}=8
$$
96-98. Find the limits:
|
8
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
115. Find $\lim _{x \rightarrow 3} \frac{x^{2}-9}{3-x}$.
|
Solution. A direct transition to the limit is impossible here, since the limit of the denominator is zero: $\lim _{x \rightarrow 3}(3-x)=3-3=0$. The limit of the dividend is also zero: $\lim _{x \rightarrow 3}\left(x^{2}-9\right)=9-9=0$. Thus, we have an indeterminate form of $0 / 0$. However, this does not mean that the given function does not have a limit; to find it, we need to preliminarily transform the function by dividing the numerator and the denominator by the expression $x-3$:
$$
\frac{x^{2}-9}{3-x}=\frac{(x-3)(x+3)}{3-x}=-\frac{(x-3)(x+3)}{x-3}=-(x+3)
$$
Fig. 95
$$
\begin{aligned}
& \text { For the expression }-(x+3) \text {, the limit as } \\
& x \rightarrow 3 \text { is easily found: } \lim _{x \rightarrow 3}(-(x+3))= \\
&=-\lim _{x \rightarrow 3}(x+3)=-6 .
\end{aligned}
$$
Remark. By canceling the fraction by $x-3$, we assume that $x \rightarrow 3$, but $x \neq 3$.
|
-6
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
116. Find $\lim _{x \rightarrow \infty} \frac{3}{x+5}$.
|
Solution. When $x \rightarrow \infty$, the denominator $x+5$ also tends to infinity, and its reciprocal $\frac{1}{x+5} \rightarrow 0$. Therefore, the product $\frac{1}{x+5} \cdot 3=\frac{3}{x+5}$ tends to zero if $x \rightarrow \infty$. Thus, $\lim _{x \rightarrow \infty} \frac{3}{x+5}=0$.
Identical transformations under the limit sign are applicable not only when the argument tends to a finite limit, but also when $x \rightarrow \infty$.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
117. Find $\lim _{x \rightarrow \infty} \frac{2 x^{3}+x}{x^{3}-1}$.
|
Solution. Here, the numerator and denominator do not have a limit, as both increase indefinitely. In this case, it is said that there is an indeterminate form of $\infty / \infty$. We will divide the numerator and denominator term by term by $x^{3}$ (the highest power of $x$ in this fraction):
$$
\lim _{x \rightarrow \infty} \frac{2 x^{3}+x}{x^{3}-1}=\lim _{x \rightarrow \infty} \frac{2+\frac{1}{x^{2}}}{1-\frac{1}{x^{3}}}=\frac{\lim _{x \rightarrow \infty}\left(2+\frac{1}{x^{2}}\right)}{\lim _{x \rightarrow \infty}\left(1-\frac{1}{x^{3}}\right)}=2
$$
since $1 / x^{2}$ and $1 / x^{3}$ tend to zero as $x \rightarrow \infty$.
Remark. The method of dividing the numerator and denominator of a fraction by the highest power of the variable $x$, used to resolve indeterminate forms of $\infty / \infty$, cannot be used to find the limits of functions that do not lead to the indeterminate form of the specified type.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
118. Find $\lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{\alpha}$.
|
Solution. We will transform this limit into the form (1). For this, we multiply the numerator and the denominator of the fraction by 2, and take the constant factor 2 outside the limit sign. We have
$$
\lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{\alpha}=\lim _{\alpha \rightarrow 0} \frac{2 \sin 2 \alpha}{2 \alpha}=2 \lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{2 \alpha}
$$
Considering that if $\alpha \rightarrow 0$, then $2 \alpha \rightarrow 0$, we get
$$
\lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{\alpha}=2 \lim _{2 \alpha \rightarrow 0} \frac{\sin 2 \alpha}{2 \alpha}=2 \lim _{x \rightarrow 0} \frac{\sin x}{x}=2 \cdot 1=2
$$
If $x \rightarrow \infty$, then the following relation holds
$$
\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e \approx 2.7182 \ldots
$$
which is known as the second remarkable limit. The proof of its validity is provided in detailed courses of mathematical analysis.
The number $e$ is of great importance in mathematics. Logarithms with base $e$ are called natural, and for them, the notation $\ln$ is used. Thus, $\ln x=\log _{e} x$. For example, $\ln 2 \approx 0.6931$.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
120. Find $\lim _{x \rightarrow 3}\left(x^{2}-7 x+4\right)$.
|
Solution. To find the limit of the given function, we will replace the argument $x$ with its limiting value:
$$
\lim _{x \rightarrow 3}\left(x^{2}-7 x+4\right)=3^{2}-7 \cdot 3+4=-8
$$
|
-8
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
135. Find $\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+4}}{x}$.
|
Solution. As the argument $x$ tends to infinity, we have an indeterminate form of $\infty / \infty$. To resolve it, we divide the numerator and the denominator of the fraction by $x$. Then we get
$$
\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+4}}{x}=\lim _{x \rightarrow \infty} \frac{\sqrt{\frac{x^{2}+4}{x^{2}}}}{1}=\lim _{x \rightarrow \infty} \sqrt{1+\frac{4}{x^{2}}}=1
$$
since $4 / x^{2} \rightarrow 0$ as $x \rightarrow \infty$.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
136. Find $\lim _{x \rightarrow \infty} \frac{3 x}{\sqrt{x^{2}-2 x+3}}$.
|
Solution. The limit transition as $x \rightarrow \infty$ can always be replaced by the limit transition as $\alpha \rightarrow 0$, if we set $\alpha=1 / x$ (the method of variable substitution).
Thus, setting $x=1 / \alpha$ in this case, we find that $\alpha \rightarrow 0$ as $x \rightarrow \infty$. Therefore,
$$
\begin{aligned}
\lim _{x \rightarrow \infty} \frac{3 x}{\sqrt{x^{2}-2 x+3}}= & \lim _{\alpha \rightarrow 0} \frac{3 \cdot \frac{1}{\alpha}}{\sqrt{\frac{1}{\alpha^{2}}-2 \cdot \frac{1}{\alpha}+3}}=\lim _{\alpha \rightarrow 0} \frac{3 \cdot \frac{1}{\alpha}}{\sqrt{\frac{1-2 \alpha+3 \alpha^{2}}{\alpha^{2}}}}= \\
& =\lim _{\alpha \rightarrow 0} \frac{3}{\sqrt{1-2 \alpha+3 \alpha^{2}}}=\frac{3}{1}=3
\end{aligned}
$$
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
137. Find $\lim _{x \rightarrow \infty} \frac{3 x^{2}+5 x+1}{x^{2}-2}$.
|
Solution. Method I. Dividing the numerator and the denominator by $x^{2}$, we find
$$
\lim _{x \rightarrow \infty} \frac{3 x^{2}+5 x+1}{x^{2}-2}=\lim _{x \rightarrow \infty} \frac{3+\frac{5}{x}+\frac{1}{x^{2}}}{1-\frac{2}{x^{2}}}=\frac{3}{1}=3
$$
Method II. Let $x=1 / a$; then $a \rightarrow 0$ as $x \rightarrow \infty$. Therefore,
$$
\lim _{x \rightarrow \infty} \frac{3 x^{2}+5 x+1}{x^{2}-2}=\lim _{a \rightarrow 0} \frac{3 \cdot \frac{1}{a^{2}}+5 \cdot \frac{1}{a}+1}{\frac{1}{a^{2}}-2}=\lim _{a \rightarrow 0} \frac{3+5 a+a^{2}}{1-2 a^{2}}=3
$$
138-141. Find the limits:
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
142. Find $\lim _{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})$.
|
Solution. Here, it is required to find the limit of the difference of two quantities tending to infinity (indeterminate form of $\infty-\infty$). By multiplying and dividing the given expression by its conjugate, we get
$$
\sqrt{x+1}-\sqrt{x}=\frac{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})}{\sqrt{x+1}+\sqrt{x}}=\frac{1}{\sqrt{x+1}+\sqrt{x}}
$$
Therefore,
$$
\lim _{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})=\lim _{x \rightarrow \infty} \frac{1}{\sqrt{x+1}+\sqrt{x}}=0
$$
Let's consider examples where remarkable limits are used.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
168. Find the derivative of the function $y=5x$.
|
Solution. $1^{0} . y_{\mathrm{H}}=5(x+\Delta x)=5 x+5 \Delta x$.
$2^{0} . \Delta y=y_{\mathrm{k}}-y=(5 x+5 \Delta x)-5 x=5 \Delta x$.
$3^{0} . \frac{\Delta y}{\Delta x}=\frac{5 \Delta x}{\Delta x}=5$.
|
5
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
329. Find the value of the arbitrary function $f(x)=\sin ^{4} x-$ $-\cos ^{4} x$ at $x=\pi / 12$.
|
Solution. First, we transform the given function:
$$
f(x)=\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)=-\cos 2 x
$$
(since $\sin ^{2} \alpha+\cos ^{2} \alpha=1, \cos ^{2} \alpha-\sin ^{2} \alpha=\cos 2 \alpha$). Using the rule for differentiating a composite function, we get
$$
f^{\prime}(x)=(-\cos 2 x)^{\prime}=-(\cos 2 x)^{\prime}=-(-\sin 2 x) \cdot(2 x)^{\prime}=2 \sin 2 x
$$
At $x=\frac{\pi}{12}$, we have $f^{\prime}\left(\frac{\pi}{12}\right)=2 \sin \frac{\pi}{6}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
374. Find the slope of the tangent line drawn to the curve $y=x^{3}$ at the point $C(-2;-8)$.
|
Solution. Let's find the derivative of the function $y=x^{3}$ at the point $x=-2$:
$$
y^{\prime}=\left(x^{3}\right)^{\prime}=3 x^{2} ; \quad y_{x=-2}^{\prime}=3(-2)^{2}=12
$$
Thus, the slope of the tangent line to the curve $y=x^{3}$ at the point $C(-2 ;-8)$ is 12 (Fig. 104).
Fig. 104
|
12
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
461. The law of temperature change $T$ of a body depending on time $t$ is given by the equation $T=0.2 t^{2}$. At what rate is this body heating at the moment of time $10 \mathrm{c}$?
|
Solution. The heating rate of a body is the derivative of temperature $T$ with respect to time $t$:
$$
\frac{d T}{d t}=\left(0.2 t^{2}\right)^{\prime}=0.4 t
$$
Determine the heating rate of the body at $t=10$:
$$
\left(\frac{d T}{d t}\right)_{t=10}=0.4 \cdot 10=4(\text{ deg } / \mathrm{s})
$$
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
564. $y=x^{2}+2$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
564. $y=x^{2}+2$.
|
Solution. $1^{0}$. Find the derivative: $y^{\prime}=\left(x^{2}+2\right)^{\prime}=2 x$.
$2^{0}$. Set it to zero; $2 x=0$, from which $x=0$ - the critical point.
$3^{\circ}$. Determine the sign of the derivative at the value $x0$, for example at $x=1$: $y_{x=1}^{\prime}=2 \cdot 1=2$. Since the derivative changes sign from negative to positive as we pass through $x=0$, the function has a minimum at $x=0$.
$4^{0}$. Find the minimum value of the function, i.e., $f(0)=0^{2}+2=2$.
Now we can sketch the curve near the point $A(0 ; 2)$ (Fig. 121).
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
582. $y=4x-x^{2}$.
|
Solution. $1^{0}$. Find the first derivative: $y^{\prime}=4-2 x$. $2^{0}$. Solving the equation $4-2 x=0$, we get the critical point $x=2$. $3^{0}$. Calculate the second derivative: $y^{\prime \prime}=-2$.
$4^{0}$. Since $y^{\prime \prime}$ is negative at any point, then $y^{\prime \prime}(2)=-2<0$. This means that the function has a maximum at the point $x=2$. Let's find its value: $y(2)=4 \cdot 2-2^{2}=4$, i.e., $y_{\max }=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
585. $y=x^{6}$.
|
Solution. $1^{0} . y^{\prime}=6 x^{5}$.
$2^{0} .6 x^{5}=0$, i.e. $x=0$.
$3^{0} \cdot y^{\prime \prime}=30 x^{4}$.
$4^{0}$. Since $y^{\prime \prime}(0)=0$, the method of investigation using the second derivative is not applicable.
We will investigate the function for an extremum using the first derivative. If $x<0$, then $y^{\prime}<0$; if $x>0$, then $y^{\prime}>0$. Therefore, at the point $x=0$ the function has a minimum $y=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
614. There is a square sheet of tin, the side of which $a=$ $=60 \mathrm{~cm}$. By cutting out equal squares from all its corners and folding up the remaining part, a box (without a lid) needs to be made. What should be the dimensions of the squares to be cut out so that the box has the maximum volume?
|
Solution. According to the condition, the side of the square $a=60$. Let the side of the squares cut out from the corners be denoted by $x$. The bottom of the box is a square with side $a-2 x$, and the height of the box is equal to the side $x$ of the cut-out square. Therefore, the volume of the box can be expressed by the function
$$
V=(a-2 x)^{2} x
$$
To find the value of $x$ at which the function will take the maximum value, we first transform the function and then investigate it for an extremum:
$$
\begin{gathered}
V=\left(a^{2}-4 a x+4 x^{2}\right) x ; \quad V=4 x^{3}-4 a x^{2}+a^{2} x \\
V^{\prime}=12 x^{2}-8 a x+a^{2} ; \quad 12 x^{2}-8 a x+a^{2}=0 \\
x=\frac{4 a \pm \sqrt{16 a^{2}-12 a^{2}}}{12}=\frac{4 a \pm \sqrt{4 a^{2}}}{12}=\frac{4 a \pm 2 a}{12} ; \quad x_{1}=\frac{a}{2}, \quad x_{2}=\frac{a}{6}
\end{gathered}
$$
Obviously, the value $x=a / 2$ does not meet the condition, as in this case the square would be cut into four equal parts and no box would be formed. Therefore, we investigate the function for an extremum at the critical point $x_{2}=a / 6$ :
$$
V^{\prime \prime}=24 x-8 a ; V^{\prime \prime}\left(\frac{a}{6}\right)=24 \frac{a}{6}-8 a=4 a-8 a=-4 a<0
$$
i.e., the maximum is achieved at $x=a / 6$. Thus, the side of the cut-out square should be equal to $x=a / 6$. In this specific case, when $a=60$ cm, we get $x=10$ cm.
|
10
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
228. Given the function $f(x)=2 x+4$. Find the increment of any of its antiderivatives when $x$ changes from -2 to 0.
|
Solution. Let's find the antiderivative of the given function:
$$
F=\int(2 x+4) d x=x^{2}+4 x+C
$$
Consider, for example, the antiderivatives $F_{1}=x^{2}+4 x, F_{2}=x^{2}+4 x+2$, $F_{3}=x^{2}+4 x-1$, and compute the increment of each of them on the interval $[-2,0]: \quad F_{1}(0)-F_{1}(-2)=0-(-4)=4 ; \quad F_{2}(0)-F_{2}(-2)=2-(-2)=4 ;$ $F_{3}(0)-F_{3}(-2)=1-(-5)=4$. Thus, the increments of these antiderivatives are equal to each other.
Now consider the increment of any antiderivative $F=x^{2}+4 x+$ $+C$ on the interval $[-2,0]$. We have
$$
\left(x^{2}+4 x+C\right)_{x=0}-\left(x^{2}+4 x+C\right)_{x=-2}=4
$$
Therefore, the increment of any antiderivative of the function $f(x)=$ $=2 x+4$ when $x$ changes from -2 to 0 is 4, i.e., the increments of all antiderivatives of this function are equal to each other and do not depend on $C$.
Let's find the increment of any antiderivative of the function $f(x)$ when the argument $x$ changes from $x=a$ to $x=b$:
$$
(F(x)+C)_{x=b}-(F(x)+C)_{x=0}=F(b)-F(a)
$$
The obtained result means that when the argument $x$ changes from $x=a$ to $x=b$, all antiderivatives of the given function have the same increment, equal to $F(b)-F(a)$.
This increment is called the definite integral.
Definition 2. If $F(x)+C$ is an antiderivative of $f(x)$, then the increment $F(b)-F(a)$ of the antiderivatives when the argument $x$ changes from $x=a$ to $x=b$ is called the definite integral and is denoted by the symbol
$$
\begin{aligned}
& \text { } \int_{a}^{b} f(x) d x, \text { i.e., } \\
& \qquad \int_{a}^{b} f(x) d x=F(b)-F(a)
\end{aligned}
$$
where $a$ is the lower limit, and $b$ is the upper limit of the definite integral.
The symbol $\int_{a}^{b} f(x) d x$ is read as: "the definite integral from $a$ to $b$ of $f$ of $x$ $d x$."
The function $f(x)$ is assumed to be continuous in the interval of the argument $x$ from $a$ to $b$.
To compute the definite integral $\int_{a}^{b} f(x) d x$, one finds:
1) the indefinite integral $\int f(x) d x=F(x)+C$;
2) the value of the integral $F(x)+C$ at $x=b, C=0$, i.e., compute $F(b)$
3) the value of the integral $F(x)+C$ at $x=a, C=0$, i.e., compute $F(a)$;
4) the difference $F(b)-F(a)$.
The process of computation is evident from the formula
$$
\int_{a}^{b} f(x) d x=\left.F(x)\right|_{a} ^{b}=F(b)-F(a)
$$
Equality (2) is called the Newton-Leibniz formula.
Remarks. 1. Under $F(x)$ in formula (2), the simplest of the antiderivatives is meant, for which $C=0$.
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
245. Find $\int_{\pi / 2}^{\pi} \frac{2 \sin x d x}{(1-\cos x)^{2}}$.
|
Solution. Let's use the substitution $u=1-\cos x$, from which $d u=$ $=\sin x d x$. Then we will find the new limits of integration; substituting into the equation $u=1-\cos x$ the values $x_{1}=\pi / 2$ and $x_{2}=\pi$, we will respectively obtain $u_{1}=1-\cos (\pi / 2)=1$ and $u_{2}=1-\cos \pi=2$. The solution is written as follows:
$$
\begin{aligned}
& \int_{\pi / 2}^{\pi} \frac{2 \sin x d x}{(1-\cos x)^{2}}=\left|\begin{array}{l}
u=1-\cos x, u_{1}=1-\cos (\pi / 2)=1, \\
d u=\sin x d x ; u_{2}=1-\cos \pi=2
\end{array}\right|=2 \int_{u}^{u_{2}} \frac{d u}{u^{2}}=2 \int_{1}^{2} u^{-2} d u= \\
& =-\left.2 u^{-1}\right|_{1} ^{2}=-1+2=1
\end{aligned}
$$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
294. $\int_{1}^{e} \frac{3 \ln ^{2} x d x}{x}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
294. $\int_{1}^{e} \frac{3 \ln ^{2} x \, dx}{x}$.
|
Solution. $\int_{1}^{e} \frac{3 \ln ^{2} x d x}{x}=\left|\begin{array}{l}\ln x=t, t_{1}=\ln 1=0, \\ \frac{d x}{x}=d t ; t_{2}=\ln e=1\end{array}\right|=3 \int_{0}^{1} t^{2} d t=\left.t^{3}\right|_{0} ^{1}=1$.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
380. Two bodies simultaneously started linear motion from a certain point in the same direction with speeds $v_{1}=$ $=\left(6 t^{2}+4 t\right) \mathrm{m} /$ s and $v_{2}=4 t \mathrm{m} / \mathrm{s}$. After how many seconds will the distance between them be 250 m?
|
Solution. Let $t_{1}$ be the moment of the meeting. Then
$$
s_{1}=\int_{0}^{t_{1}}\left(6 t^{2}+4 t\right) d t=\left.\left(2 t^{3}+2 t^{2}\right)\right|_{0} ^{t_{1}}=2 t_{1}^{3}+2 t_{1}^{2} ; \quad s_{2}=\int_{0}^{t_{1}} 4 t d t=2 t_{1}^{2}
$$
Since $s_{1}-s_{2}=250$, we get the equation $2 t_{1}^{3}+2 t_{1}^{2}-2 t_{1}^{2}=250$ or $2 t_{1}^{3}=250$, from which $t_{1}^{3}=125$, i.e., $t=5$ (s).
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
195. In a motorboat with an initial speed of $v_{0}=$ $=5 \mathrm{m} / \mathrm{s}$, the engine was turned off. During movement, the boat experiences water resistance, the force of which is proportional to the square of the boat's speed, with a proportionality coefficient of $m / 50$, where $m$ is the mass of the boat. After what time interval will the boat's speed decrease by half, and what distance will the boat travel during this time?
|
Solution. $1^{0}$. According to the condition, we can take the function as the path $s$, and the argument as time $t$. Using Newton's second law $F=m a$, we obtain the equation
$$
m s_{(t)}^{\prime \prime}=-\frac{m}{50}\left(s_{(t)}^{\prime}\right)^{2}, \text { or } s_{(t)}^{\prime \prime} \doteq-\frac{1}{50}\left(s_{(t)}^{\prime}\right)^{2}
$$
where the minus sign indicates that the resistance force is directed opposite to the motion.
$2^{0}$. This is a second-order differential equation. We reduce its order. Since $s_{(t)}^{\prime}=v_{(t)}$ and $s^{\prime \prime}(t)=v^{\prime}(t)$, we arrive at the first-order equation
$$
\frac{d v}{d t}=-\frac{1}{50} v^{2}
$$
We separate the variables and integrate:
$$
\frac{d v}{v^{2}}=-\frac{1}{50} d t ; \frac{1}{v}=\frac{t}{50}+C_{1}
$$
$3^{0}$. Using the condition $v_{0}=5$ m $/ \mathrm{c}$, we find $C_{1}=\frac{1}{5}$, from which $v=$ $=\frac{50}{t+10} \cdot$ Then we get
$$
\frac{d s}{d t}=\frac{50}{t+10} ; d s=\frac{50 d t}{t+10} ; s=50 \ln (t+10)+C_{2}
$$
From the condition $s=0$ at $t=0$ we find $C_{2}=-50 \ln 10$, from which
$$
s=50 \ln \frac{t+10}{10}
$$
We need to determine the time interval during which the speed will decrease by half. Substituting into the formula for $v$ the value $\frac{1}{2} v_{0}=2.5$, we find $2.5=\frac{50}{t+10} ; t=10$ s. During this time, the boat will travel the distance $s=50 \ln 2 \approx 34.5$ (m).
|
10
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
150. Two shooters shoot at a target. The probability of a miss with one shot for the first shooter is 0.2, and for the second shooter it is 0.4. Find the most probable number of volleys in which there will be no hits on the target, if the shooters will make 25 volleys.
|
Solution. Misses by the shooters are independent events, so the multiplication theorem of probabilities of independent events applies. The probability that both shooters will miss in one volley, $p=0.2 \cdot 0.4=0.08$.
Since the product $n p=25 \cdot 0.08=2$ is an integer, the most probable number of volleys in which there will be no hits, $k_{0}=n p=2$.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
172. After the student answers the questions in the examination ticket, the examiner asks the student additional questions. The teacher stops asking additional questions as soon as the student shows ignorance of the given question. The probability that the student will answer any additional question is 0.9. Required: a) construct the distribution law of the random discrete variable $X$ - the number of additional questions the teacher will ask the student; b) find the most probable number $k_{0}$ of additional questions asked to the student.
|
Solution. a) The discrete random variable $X$ - the number of additional questions asked - has the following possible values: $x_{1}=1, x_{2}=2, x_{3}=3, \ldots, x_{k}=k, \ldots$ Let's find the probabilities of these possible values.
The variable $X$ will take the possible value $x_{i}=1$ (the examiner will ask only one question) if the student does not answer the first question. The probability of this possible value is $1-0.9=0.1$. Thus, $P(X=1)=0.1$.
The variable $X$ will take the possible value $x_{2}=2$ (the examiner will ask only two questions) if the student answers the first question (the probability of this event is 0.9) and does not answer the second (the probability of this event is 0.1). Thus, $P(X=2)=0.9 \cdot 0.1=0.09$.
Similarly, we find
$$
P(X=3) \leftrightharpoons 0.9^{2} \cdot 0.1=0.081, \ldots, P(X=k)=0.9^{k-1} \cdot 0.1, \ldots
$$
Let's write the required distribution law:
| $X$ | 1 | 2 | 3 | $\ldots$ | $k$ | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $p$ | 0.1 | 0.09 | 0.081 | $\ldots$ | $0.9^{k-1} \cdot 0.1$ | $\ldots$ |
b) The most probable number $k_{0}$ of questions asked (the most probable possible value of $X$), i.e., the number of questions asked by the teacher, which has the highest probability, as follows from the distribution law, is one.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
189. Find the mathematical expectation of the random variable $Z$, if the mathematical expectations of $X$ and $Y$ are known:
a) $Z=X+2 Y, M(X)=5, M(Y)=3 ;$ b) $Z=3 X+4 Y$, $M(X)=2, M(Y)=6$
|
Solution. a) Using the properties of mathematical expectation (the expectation of a sum is equal to the sum of the expectations of the terms; a constant factor can be factored out of the expectation), we get
$$
\begin{aligned}
M(Z)=M(X+2 Y)= & M(X)+M(2 Y)=M(X)+2 M(Y)= \\
& =5+2 \cdot 3=11 .
\end{aligned}
$$
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
277. The random variable $X$ in the interval (-c, c) is given by the probability density function $f(x)=1 /\left(\pi \sqrt{c^{2}-x^{2}}\right)$; outside this interval, $f(x)=0$. Find the mathematical expectation of the variable $X$.
|
Solution. We use the formula $M(X)=\int_{a} x f(x) \mathrm{d} x$. Substituting $a=-c, b=c, f(x)=1 /\left(\pi \sqrt{c^{2}-x^{2}}\right)$, we get
$$
M(X)=\frac{1}{\pi} \int_{-c}^{c} \frac{x d x}{\sqrt{c^{2}-x^{2}}}
$$
Considering that the integrand is an odd function and the limits of integration are symmetric with respect to the origin, we conclude that the integral is zero. Therefore, $M(X)=0$.
This result can be obtained immediately if we take into account that the distribution curve is symmetric about the line $\boldsymbol{x}=\mathbf{0}$.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
280. Find the mathematical expectation of the random variable $X$, given by the distribution function
$$
F(x)=\left\{\begin{array}{ccr}
0 & \text { for } & x \leqslant 0 \\
x / 4 & \text { for } & 0 < x \leqslant 4 \\
1 & \text { for } & x > 4
\end{array}\right.
$$
|
Solution. Find the density function of the random variable $\boldsymbol{X}$:
$$
f(x)=F^{\prime}(x)=\left\{\begin{array}{ccc}
0 & \text { for } & x4 .
\end{array}\right.
$$
Find the required expected value:
$$
M(X)=\int_{0}^{4} x f \cdot(x) \mathrm{d} x=\int_{0}^{4} x \cdot(1 / 4) \mathrm{d} x=2
$$
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
647. Why, when checking the hypothesis of exponential distribution of the population by the Pearson criterion, is the number of degrees of freedom determined by the equation $k=s-2$, where $s$ is the number of sample intervals?
|
Solution. When using the Pearson criterion, the number of degrees of freedom is $k=s-1-r$, where $r$ is the number of parameters estimated from the sample. The exponential distribution is defined by one parameter $\lambda$. Since this parameter is estimated from the sample, $r=1$, and therefore, the number of degrees of freedom is $k=s-1-1=s-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
657. Why, when testing the hypothesis of a uniform distribution of the population $X$ using the Pearson's criterion, is the number of degrees of freedom determined by the equation $k=s-3$, where $s$ is the number of intervals in the sample?
|
Solution. When using the Pearson criterion, the number of degrees of freedom $k=s-1-r$, where $r$ is the number of parameters estimated from the sample. The uniform distribution is defined by two parameters $a$ and $b$. Since these two parameters are estimated from the sample, $r=2$, and therefore, the number of degrees of freedom $k=s-1-2=s-3$.
|
-3
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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)), and the limit of a constant is the constant itself (formula (1.48)), we sequentially obtain
$$
\begin{gathered}
\lim _{x \rightarrow 2}\left(4 x^{2}-6 x+3\right)=\lim _{x \rightarrow 2} 4 x^{2}-\lim _{x \rightarrow 2} 6 x+\lim _{x \rightarrow 2} 3= \\
=4 \lim _{x \rightarrow 2} x^{2}-6 \lim _{x \rightarrow 2} x+\lim _{x \rightarrow 2} 3=4\left(\lim _{x \rightarrow 2} x\right)^{2}-6 \lim _{x \rightarrow 2} x+3= \\
=4 \cdot 2^{2}-6 \cdot 2+3=16-12+3=7 .
\end{gathered}
$$
Remark 1. The calculation of the limit of a second-degree polynomial has been reduced to calculating its value at the limit of the argument.
Remark 2. To calculate the limit of an n-th degree polynomial
$$
P_{n}(x)=b_{0}+b_{1} x+b_{2} x^{2}+\ldots+b_{n} x^{n} \text { as } x \rightarrow b
$$
it is sufficient to find \( P_{n}(b) \), i.e., its value at \( x=b \). For example,
$$
\lim _{x \rightarrow 1}\left(2 x^{5}-4 x^{4}+3 x^{3}-6 x^{2}+8 x+2\right)=2 \cdot 1^{5}-4 \cdot 1^{4}+3 \cdot 1^{3}-6 \cdot 1^{2}+8 \cdot 1+2=5
$$
Example 2. Find \(\lim _{x \rightarrow 1} \frac{3 x^{2}-4 x+7}{2 x^{2}-5 x+6}\).
Solution. Since the limit of a quotient is equal to the quotient of the limits, based on formula (1.46), and using formulas (1.42), (1.44), (1.45), and (1.48), we sequentially obtain
$$
\lim _{x \rightarrow 1} \frac{3 x^{2}-4 x+7}{2 x^{2}-5 x+6}=\frac{\lim _{x \rightarrow 1}\left(3 x^{2}-4 x+7\right)}{\lim _{x \rightarrow 1}\left(2 x^{2}-5 x+6\right)}=\frac{\lim _{x \rightarrow 1} 3 x^{2}-\lim _{x \rightarrow 1} 4 x+\lim _{x \rightarrow 1} 7}{\lim _{x \rightarrow 1} 2 x^{2}-\lim _{x \rightarrow 1} 5 x+\lim _{x \rightarrow 1} 6}=
$$
$$
=\frac{3 \lim _{x \rightarrow 1} x^{2}-4 \lim _{x \rightarrow 1} x+\lim _{x \rightarrow 1} 7}{2 \lim _{x \rightarrow 1} x^{2}-5 \lim _{x \rightarrow 1} x+\lim _{x \rightarrow 1} 6}=\frac{3 \cdot 1^{2}-4 \cdot 1+7}{2 \cdot 1^{2}-5 \cdot 1+6}=\frac{6}{3}=2
$$
Remark 1. The calculation of the limit of a rational function (the ratio of two polynomials) has been reduced to calculating the value of this function at the limit of the argument.
Remark 2. To calculate the limit of a rational function
$$
R(x)=\frac{b_{0}+b_{1} x+b_{2} x^{2}+\ldots+b_{n} x^{n}}{c_{0}+c_{1} x+c_{2} x^{2}+\ldots+c_{m} x^{m}}
$$
as \( x \rightarrow a \), it is sufficient to find its value at \( x=a \).
For example,
$$
\lim _{x \rightarrow 2} \frac{x^{5}-2 x^{4}+3 x^{2}-6 x+9}{3 x^{4}-12 x^{2}+7 x-11}=\frac{2^{5}-2 \cdot 2^{4}+3 \cdot 2^{2}-6 \cdot 2+9}{3 \cdot 2^{4}-12 \cdot 2^{2}+7 \cdot 2-11}=\frac{9}{3}=3
$$
Remark 3. The limit of an elementary function \( f(x) \) as \( x \rightarrow a \), where \( a \) belongs to its domain, is equal to the value of the function at \( x=a \), i.e.,
$$
\lim _{x \rightarrow a} f(x)=f(a)
$$
For example,
$$
\text { 1) } \begin{gathered}
\lim _{x \rightarrow 4} \frac{\sqrt{x^{2}-7}+\sqrt[3]{x+60}}{2 x^{2}-5 x-8}=\frac{\sqrt{4^{2}-7}+\sqrt[3]{4+60}}{2 \cdot 4^{2}-5 \cdot 4-8}= \\
=\frac{\sqrt{9}+\sqrt[3]{64}}{32-20-8}=\frac{3+4}{4}=\frac{7}{4}
\end{gathered}
$$
2). \(\lim \left[\lg \left(t+\sqrt{t^{2}+80}\right)\right]+t \sqrt{t^{2}+8}=\lg (1+\sqrt{81})+1 \sqrt{1+8}=1+3=4\). \(t \rightarrow 1\)
|
7
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
$$
x^{2}+p x+q=\left(x-x_{1}\right)\left(x-x_{2}\right)
$$
where $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}+p x+q=0$.
Since the equation $x^{2}-6 x+8=0$ has roots $x_{1}=2, x_{2}=4$, we have $x^{2}-6 x+8=(x-2)(x-4)$. Substituting this expression into the given function and canceling the common factor $(x-4) \neq 0$, we get
$$
\lim _{x \rightarrow 4} \frac{x^{2}-6 x+8}{x-4}=\lim _{x \rightarrow 4} \frac{(x-2)(x-4)}{x-4}=\lim _{x \rightarrow 4}(x-2)=4-2=2
$$
Remark 1. When canceling the common factor, we assumed that $x-4 \neq 0$. This is indeed the case. According to the definition of the limit of a function, the argument $x$ approaches its limiting value $a$ without ever coinciding with it, i.e., $x \neq a$ and $x-a \neq 0$. (See formula (1.26).)
Remark 2. The limit of a function does not depend on whether the function is defined at the limiting point or not.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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, we get
$$
\begin{gathered}
\lim _{x \rightarrow+\infty} \frac{6 x^{2}+5 x+4}{3 x^{2}+7 x-2}=\lim _{x \rightarrow+\infty} \frac{6+\frac{5}{x}+\frac{4}{x^{2}}}{3+\frac{7}{x}-\frac{2}{x^{2}}}=\frac{\lim _{x \rightarrow+\infty}\left(6+\frac{5}{x}+\frac{4}{x^{2}}\right)}{\lim _{x \rightarrow+\infty}\left(3+\frac{7}{x}-\frac{2}{x^{2}}\right)}= \\
=\frac{6+0+0}{3+0-0}=2
\end{gathered}
$$
Here it is taken into account that
$$
\lim _{x \rightarrow+\infty} \frac{c}{x}=0, \lim _{x \rightarrow+\infty} \frac{c}{x^{2}}=0(c=\text { const })
$$
(See formula (1.54)).
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
$$
\begin{aligned}
\lim _{x \rightarrow+\infty} \frac{7 x^{2}+6 x-3}{9 x^{3}+8 x^{2}-2}= & \lim _{x \rightarrow+\infty} \frac{\frac{7}{x}+\frac{6}{x^{2}}-\frac{3}{x^{3}}}{9+\frac{8}{x}-\frac{2}{x^{3}}}=\frac{\lim _{x \rightarrow+\infty}\left(\frac{7}{x}+\frac{6}{x^{2}}-\frac{3}{x^{3}}\right)}{\lim _{x \rightarrow+\infty}\left(9+\frac{8}{x}-\frac{2}{x^{3}}\right)}= \\
& =\frac{0+0-0}{9+0-0}=\frac{0}{9}=0
\end{aligned}
$$
Remark. To resolve the indeterminacy of the form $\frac{\infty}{\infty}$, given by the ratio of two polynomials, one must divide the numerator and the denominator by the highest power of $x$ present in them, and then take the limit.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
\[
\begin{gathered}
\lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}=\lim _{n \rightarrow \infty} \frac{\frac{1+(2 n-1)}{2} \cdot n}{\frac{1+n}{2} \cdot n}= \\
=\lim _{n \rightarrow \infty} \frac{1+(2 n-1)}{1+n}=\lim _{n \rightarrow \infty} \frac{2 n}{n+1}=2 \lim _{n \rightarrow \infty} \frac{n}{n+1}=2 \lim _{n \rightarrow \infty} \frac{1}{1+\frac{1}{n}}=2 .
\end{gathered}
\]
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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:
$$
\lim _{n \rightarrow \infty} \frac{12 n+5}{\sqrt[3]{27 n^{3}+6 n^{2}+8}}=\lim _{n \rightarrow \infty} \frac{12+\frac{5}{n}}{\sqrt[3]{27+\frac{6}{n}+\frac{8}{n^{3}}}}=\frac{12}{\sqrt[3]{27}}=\frac{12}{3}=4
$$
(Here it is taken into account that $\lim _{n \rightarrow \infty} \frac{5}{n}=0, \lim _{n \rightarrow \infty} \frac{6}{n}=0, \lim _{n \rightarrow \infty} \frac{8}{n^{3}}=0$.)
## Problems
Find the limits:
1. $\lim _{x \rightarrow 3}\left(2 x^{2}-7 x+6\right)$.
2. $\lim _{x \rightarrow 1}\left(3 x^{4}-5 x^{3}+6 x^{2}-4 x+7\right)$.
3. $\lim _{x \rightarrow 2} \frac{4 x^{2}-5 x+2}{3 x^{2}-6 x+4}$.
4. $\lim _{x \rightarrow 4} \frac{x^{2}-5 x+4}{x^{2}-7 x+6}$.
5. $\lim _{x \rightarrow 5} \frac{x^{2}-7 x+12}{x^{2}-6 x+5}$.
6. $\lim _{t \rightarrow 3}\left[2 t \sqrt{t^{2}-8}+\lg \left(3 t+\sqrt{t^{2}-8}\right)-\frac{\sqrt[3]{t^{2}-1}+\sqrt[4]{t^{3}-2 t^{2}+7}}{t^{3}-2 t^{2}+t-11}\right]$.
7. $\lim _{x \rightarrow 6} \frac{x^{2}-8 x+12}{x^{2}-7 x+6}$.
8. $\lim _{x \rightarrow 2} \frac{3 x^{2}-7 x+2}{4 x^{2}-5 x-6}$.
9. $\lim _{x \rightarrow 1} \frac{x^{3}-3 x+2}{x^{4}-4 x+3}$.
10. $\lim _{x \rightarrow \infty} \frac{10 x^{3}-6 x^{2}+7 x+5}{8-4 x+3 x^{2}-2 x^{3}}$.
11. $\lim _{x \rightarrow \infty} \frac{2 x^{4}-5 x^{3}+7 x^{2}+8 x-9}{3 x^{5}-6 x^{3}+4 x^{2}-2 x+11}$.
12. $\lim _{x \rightarrow \infty} \frac{x^{7}+8 x^{6}+5 x^{4}-3 x^{2}-12}{10 x^{6}+7 x^{5}-6 x^{3}-4 x+17}$.
13. $\lim _{x \rightarrow 5} \frac{x^{2}-25}{2-\sqrt{x-1}}$.
14. $\lim _{n \rightarrow \infty} \frac{\sqrt{4 n^{2}-3}}{2 n+1}$.
15. $\lim _{x \rightarrow \infty} \frac{6 x-5}{1+\sqrt{x^{2}+3}}$.
16. $\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^{2}-1}$.
17. $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin 2 x-\cos 2 x-1}{\cos x-\sin x}$.
18. $\lim _{x \rightarrow \pi} \frac{\sqrt{1-\operatorname{tg} x}-\sqrt{1+\operatorname{tg} x}}{\sin 2 x}$.
19. $\lim _{x \rightarrow \infty}\left(\frac{2 x^{2}}{1-x^{2}}+3^{\frac{1}{x}}\right)$.
## Answers
1.3. 2. 7. 3. 2. 4.0. 5. $\infty$. 6.4. 7. $\frac{4}{5}$. 8. $\frac{15}{44}$. 9. $\frac{1}{2}$ 10. -5 . 11. 0 . 12. $\infty$.
13. -40 . 14. $\frac{3}{4}$. Hint. Let $x=t^{12}$. 15. 1. 16.6. 17. $\frac{1}{2}$.
14. $-\sqrt{2} \cdot 19 \cdot-\frac{1}{2} \cdot 20 \cdot-1$.
§ 1.5. The number e, $\lim _{\alpha \rightarrow 0} \frac{\sin \alpha}{\alpha}=1$
1. The number e is the limit
$$
\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e
$$
or
$$
\lim _{\alpha \rightarrow 0}(1+\alpha)^{\frac{1}{\alpha}}=e
$$
The number $e$ is useful for resolving indeterminate forms of the type $1^{\infty}$.
If the base of logarithms is the number $e$, then the logarithms are called natural:
$$
\ln x=\log _{e} x
$$
2. If the angle $\alpha$ is expressed in radians, then
$$
\lim _{\alpha \rightarrow 0} \frac{\sin \alpha}{\alpha}=1
$$
3. When finding limits of the form
$$
\lim _{x \rightarrow a}[\varphi(x)]^{\psi(x)}=C
$$
the following should be kept in mind:
1) if there exist finite limits
$$
\lim _{x \rightarrow a} \varphi(x)=A \text { and } \lim _{x \rightarrow a} \psi(x)=B
$$
then
$$
C=A^{B}
$$
2) if
$$
\lim _{x \rightarrow a} \varphi(x)=A \neq 1 \text { and } \lim _{x \rightarrow a} \psi(x)=\infty \text {, }
$$
then the limit is found using formulas (1.55) and (1.56);
3) if
$$
\lim _{x \rightarrow a} \varphi(x)=1, \lim _{x \rightarrow a} \psi(x)=\infty
$$
then we set $\varphi(x)=1+\alpha(x)$, where $\alpha(x) \rightarrow 0$ as $x \rightarrow a$ and, consequently,
$$
\begin{gathered}
C=\lim _{x \rightarrow a}\left\{[1+\alpha(x)]^{\frac{1}{\alpha(x)}}\right\}^{\alpha(x) \psi(x)}=e^{\lim _{x \rightarrow \alpha} \alpha(x) \psi(x)}= \\
=e^{\lim _{x \rightarrow a}[\varphi(x)-1] \psi(x)}
\end{gathered}
$$
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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,
$$
\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=1
$$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
$$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\operatorname{tg} x}{x} & =\lim _{x \rightarrow 0}\left(\frac{\sin x}{\cos x} \cdot \frac{1}{x}\right)=\lim _{x \rightarrow 0}\left(\frac{1}{\cos x} \cdot \frac{\sin x}{x}\right)= \\
& =\lim _{x \rightarrow 0} \frac{1}{\cos x} \cdot \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \cdot 1=1
\end{aligned}
$$
Thus,
$$
\lim _{x \rightarrow 0} \frac{\operatorname{tg} x}{x}=1
$$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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)
$$
\begin{gathered}
\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+9}-3}=\lim _{x \rightarrow 0} \frac{\sin x(\sqrt{x+9}+3)}{(\sqrt{x+9}-3)(\sqrt{x+9}+3)}= \\
=\lim _{x \rightarrow 0} \frac{\sin x(\sqrt{x+9}+3)}{(x+9)-9}=\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \lim _{x \rightarrow 0}(\sqrt{x+9}+3)=1(3+3)=6
\end{gathered}
$$
|
6
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
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 (1.55)
$$
\lim _{x \rightarrow \infty}\left(\frac{2 x-1}{3 x+4}\right)^{x^{2}}=0
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
$$
\begin{aligned}
& \lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}=\lim _{x \rightarrow 0}[1-(1-\cos x)]^{\frac{1}{x}}=\lim _{x \rightarrow 0}\left(1-2 \sin ^{2} \frac{x}{2}\right)^{\frac{1}{x}}= \\
& =\lim _{x \rightarrow 0}\left[\left(1-2 \sin ^{2} \frac{x}{2}\right)^{-\frac{1}{2 \sin ^{2} \frac{x}{2}}}\right]^{-\frac{2 \sin ^{2} \frac{x}{2}}{x}}=e^{\lim _{x \rightarrow 0}\left(-\frac{2 \sin ^{2} \frac{x}{2}}{x}\right)}
\end{aligned}
$$
Since
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(-\frac{2 \sin ^{2} \frac{x}{2}}{x}\right)=-\lim _{x \rightarrow 0} \frac{\sin ^{2} \frac{x}{2}}{\frac{x}{2}}= \\
= & -\lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}} \cdot \lim _{x \rightarrow 0} \sin \frac{x}{2}=-1 \cdot 0=0
\end{aligned}
$$
$$
\lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}=e^{0}=1
$$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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