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Example 4.20. Investigate the convergence of the series
$$
1-\frac{3}{2}+\frac{5}{4}-\frac{7}{8}+\ldots
$$ | Solution. The general term of this series can be written as
$$
(-1)^{n+1} \frac{2 n-1}{2^{n-1}},(n=1,2, \ldots)
$$
Let $a_{n}:$
Let $a_{n}=\frac{2 n-1}{2^{n-1}}$, we will prove the monotonic tendency to zero:
$$
\begin{aligned}
& a_{n}-a_{n+1}=\frac{2 n-1}{2^{n-1}}-\frac{2 n+1}{2^{n}}=\frac{4 n-2-2 n-1}{2^{n}}= \\
... | S_{14} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,987 |
Example 4.21. Investigate the convergence of the series $\sum_{n=1}^{\infty} \frac{\sin n}{n}$. | Solution. In this case, $b_{n}=\frac{1}{n}$ monotonically tends to zero, and
$$
\begin{gathered}
S_{n}=\sum_{k=1}^{n} \sin k=\frac{1}{\sin \frac{1}{2}} \sum_{k=1}^{n} \sin \frac{1}{2} \sin k= \\
=\frac{1}{2 \sin \frac{1}{2}} \sum_{k=1}^{n}\left(\cos \left(\frac{1}{2}-k\right)-\cos \left(\frac{1}{2}+k\right)\right)= \\... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,988 |
Example 4.22. Investigate the convergence of the functional series
$$
\sum_{n=1}^{\infty}(-1)^{n} n^{-x}
$$ | Solution. For any fixed $x \in (0, +\infty) = D_{+}$, the series will be convergent by the Leibniz criterion (see Theorem 4.16). For $x \in (1, +\infty) = D_{1} \subset D_{+}$, the series will be absolutely convergent (see Example 4.20). If, however, $x \in (-\infty, 0] = D_{-}$, the series diverges because the necessa... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,989 |
Example 4.23. Investigate the convergence of the functional series
$$
\sum_{n=1}^{\infty}\left(\frac{x-2}{1-3 x}\right)^{n}
$$ | Solution. Apply Theorem 4.19:
$$
\lim _{n \rightarrow \infty}\left|\frac{x-2}{1-3 x}\right|^{n+1}:\left|\frac{x-2}{1-3 x}\right|^{n}=\left|\frac{x-2}{1-3 x}\right|
$$
Next, solve the inequality
$$
\begin{gathered}
\left|\frac{x-2}{1-3 x}\right|<1
\end{gathered}
$$
Thus, the domain of absolute convergence of the ser... | (-\infty,-\frac{1}{2})\cup(\frac{3}{4},+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,990 |
Example 4.24. Investigate the convergence of the functional series
$$
\sum_{n=1}^{\infty} \frac{\sin n x}{e^{n x}}
$$ | Solution. Apply the previous theorem:
$$
\varlimsup_{n \rightarrow \infty} \sqrt[n]{\frac{|\sin n x|}{e^{n x}}}=e^{-x} \varlimsup_{n \rightarrow \infty} \sqrt[n]{|\sin n x|}=e^{-x}
$$
From this, it is clear that for $x \in D=(0, +\infty)$, the series converges absolutely. For $x<0$, the series diverges. If $x=0$, the... | x\geqslant0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,991 |
Example 4.25. Investigate the uniform convergence of the series
$$
\sum_{n=1}^{\infty} \frac{(-1)^{n}}{x+n}, x \geqslant 0
$$ | Solution. In this case, the domain $D$ is all non-negative numbers. According to the Leibniz criterion (Theorem 4.16), the remainder of the series can be estimated as follows:
$$
\left|R_{n}\right|=\left|\sum_{n=1}^{\infty} \frac{(-1)^{n}}{x+n+k}\right|=\frac{1}{x+n+k} \leqslant \frac{1}{n+1}
$$
For any $\varepsilon>... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,992 |
Example 4.26. Investigate the convergence of the series $\sum_{n=1}^{\infty} \frac{(x-2)^{n}}{n^{2} 2^{n}}$. | Solution. In this case, $a_{n}=\frac{1}{n^{2} 2^{n}}$. Using the ratio test (Theorem 4.23), we find $R$:
$$
R=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2} 2^{n}}}{\frac{1}{(n+1)^{2} 2^{n+1}}}=\lim _{n \rightarrow \infty} \frac{2(n+1)^{2}}{n^{2}}=2
$$
The series under investigation will converge in the interval d... | x\in[0,4] | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,993 |
Example 4.27. Investigate the power series $\sum_{n=1}^{\infty} \frac{(x+2)^{n}}{n^{n}}$. | Solution. We apply the Cauchy formula (Theorem 4.23) to determine the radius of convergence $R$:
$$
R=\frac{1}{\varlimsup_{n \rightarrow \infty} \sqrt[n]{\frac{1}{n^{n}}}}=\frac{1}{\varlimsup_{n \rightarrow \infty} \frac{1}{n}}=+\infty
$$
This result means that the power series converges for all values of $x$. | +\infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,994 |
Example 4.28. Investigate the power series $\sum_{n=1}^{\infty} \frac{n^{2} x^{n}}{n!}$. | Solution. Here $a_{n}=\frac{n^{2}}{n!}$. By the D'Alembert's formula (Theorem 4.23):
$$
R=\lim _{n \rightarrow \infty} \frac{n^{2}}{n!}: \frac{(n+1)^{2}}{(n+1)!}=\lim _{n \rightarrow \infty} \frac{n^{2}(n+1)}{(n+1)^{2}}=+\infty
$$
Therefore, this series converges for all values of $x$. The convergence will be absolut... | +\infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,995 |
Example 4.30. Find the sum of the series:
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n}
$$ | Solution. As in Example 4.29, we will base our approach on the infinite geometric series $\sum_{n=1}^{\infty} x^{n}$. Let $|x|<1$, and apply Theorem 4.25\left(x_{0}=0\right)$ to this series:
$$
\int_{0}^{x}\left(\sum_{n=0}^{\infty} t^{n}\right) d t=\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}=\sum_{n=0}^{\infty} \frac{x^{n... | \ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,996 |
Example 4.31. Investigate the power series for the function $f(x)=a^{x}(a>0, a \neq 1)$ at the point $x_{0}=0$. | Solution. The function $f(x)=a^{x}=e^{x \ln a}$ is infinitely differentiable in the neighborhood of the point $x_{0}=0$. In this case,
$$
f^{(n)}(x)=\ln ^{n} a \cdot a^{x}, f^{(n)}(0)=\ln ^{n} a
$$
Thus,
$$
a^{x} \sim \sum_{n=0}^{\infty} \frac{\ln ^{n} a}{n!} x^{n}
$$
The remainder term of the Taylor formula, taken... | ^{x}=\sum_{n=0}^{\infty}\frac{\ln^{n}}{n!}x^{n} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,997 |
Example 4.32. Expand the function $f(x)=$ $=\ln \left(x^{2}+3 x+2\right)$ into a power series and determine the interval of convergence. | Solution. First of all, let's make some simple transformations:
$$
\begin{gathered}
f(x)=\ln \left(x^{2}+3 x+2\right)=\ln (x+1)+\ln (x+2)= \\
=\ln (x+1)+\ln 2+\ln \left(1+\frac{x}{2}\right)
\end{gathered}
$$
According to Remark 4.3 for $|x|<1$ and $|x|<2$ we have respectively
$$
\begin{aligned}
\ln (x+1) & =\sum_{n=... | \ln2+\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2^{n}+1}{2^{n}n}x^{n},\quad-1<x\leq1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,998 |
Example 4.33. Find the power series expansion for the function $f(x)=\ln \left(x+\sqrt{1+x^{2}}\right)$ and determine the interval of convergence of the obtained series. | Solution. We will use the results of Theorem 4.25 on term-by-term integration of power series. Apply the formula
$$
\ln \left(x+\sqrt{1+x^{2}}\right)=\int_{0}^{x} \frac{d t}{\sqrt{1+t^{2}}}
$$
According to Remark 4.3
$$
\begin{gathered}
\frac{1}{\sqrt{1+x}}=(1+x)^{-1 / 2}= \\
=1+\sum_{n=1}^{\infty} \frac{(-1 / 2)(-1... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,999 |
Example 4.34. Compute the integral $\int_{0}^{1} \frac{\sin t}{t} d t$ with accuracy $\varepsilon=10^{-4}$. | Solution. Note from the previous that
$$
\sin t=\sum_{n=0}^{\infty}(-1)^{n} \frac{t^{2 n+1}}{(2 n+1)!}, t \in \mathbb{R}
$$
From this it follows that
$$
\frac{\sin t}{t}=\sum_{n=0}^{\infty}(-1)^{n} \frac{t^{2 n}}{(2 n+1)!}
$$
Apply the theorem of term-by-term integration (Theorem 4.25) to the last series:
$$
\begi... | 0.9444 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,000 |
Example 4.36. Consider the trigonometric series $\sum_{k=1}^{\infty} a^{k} \sin k x(|a|<1)$. How many times can this series be differentiated term by term? | Solution. Consider the numerical series $\sum_{k=1}^{\infty} k^{s} a^{k} (s=1, 2, \ldots)$. We will investigate their absolute convergence using the D'Alembert's ratio test (Theorem 4.8):
$$
\lim _{k \rightarrow \infty} \frac{(k+1)^{s}|a|^{k+1}}{k^{s}|a|^{k}}=|a| \lim _{k \rightarrow \infty}\left(\frac{k+1}{k}\right)^... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,001 |
Example 4.37. How many times can the trigonometric series $\sum_{k=1}^{\infty} \frac{\cos k x}{k^{4}}$ be differentiated term by term? | Solution. In this example, $a_{k}=\frac{1}{k^{4}}$ and the series with the general term $k^{s} a_{k}$ will converge for $s=1$ and $s=2$ (for $s=3$ we get the harmonic series, and for $s>3$ the general term of the series will be an infinitely large quantity). Therefore, by Theorem 4.28, this series can be term-by-term d... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,002 |
Example 4.38. Write the Fourier series for the function $f(x)$ defined by the equations
$$
\begin{gathered}
f(x)=\left\{\begin{array}{l}
-\frac{\pi}{4},-\pi<x<0 \\
0, x=0, x=-\pi \\
\frac{\pi}{4}, 0<x<\pi
\end{array}\right. \\
f(x)=f(x+2 \pi) .
\end{gathered}
$$ | Solution. The considered function is odd and has discontinuities of the first kind at points $n \pi (n \in \mathbb{Z})$ with a jump of $\frac{\pi}{2}$. Further:
$$
\int_{-\pi}^{\pi} f^{2}(x) d x=2 \int_{0}^{\pi} f^{2}(x) d x=2 \cdot \frac{\pi^{2}}{16} \cdot \pi=\frac{\pi^{3}}{8}<+\infty
$$
We will calculate the Fouri... | f(x)\sim\sum_{n=1}^{\infty}\frac{\sin(2n-1)x}{2n-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,003 |
Example 4.39. Write the Fourier series for the function $f(x)$, where
$$
\begin{gathered}
f(x)=\left\{\begin{array}{l}
x,|x|<\pi \\
0, x=-\pi
\end{array}\right. \\
f(x)=f(x+2 \pi)
\end{gathered}
$$ | Solution. This function, like the one in Example 4.38, is odd and discontinuous (discontinuities of the first kind at $x=n \pi, n \in \mathbb{Z}$). The Fourier coefficients $a_{k}=0$, and the Fourier coefficients $b_{k}$ are calculated using formulas (4.11):
$$
\begin{gathered}
b_{k}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(... | f(x)\sim2\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}\sinkx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,004 |
Example 4.40. Write the Fourier series for the function $f(x)$ defined by the conditions
$$
\begin{gathered}
f(x)=f(x+2 \pi) \ (2 \pi \text{-periodicity }), \\
f(x)=f(-x) \text{ (evenness) } \\
f(x)=x, \ 0 \leqslant x \leqslant \pi .
\end{gathered}
$$ | Solution. Since the function in question is even, its coefficients $b_{k}$ will be zero, and the coefficients $a_{k}$ are calculated using formulas (4.11)
$$
a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) d x=\frac{2}{\pi} \int_{0}^{\pi} x d x=\left.\frac{2}{\pi} \frac{x^{2}}{2}\right|_{0} ^{\pi}=\pi
$$
For $k=1,2, \ldot... | f(x)\sim\frac{\pi}{2}-\frac{4}{\pi}\sum_{k=1}^{\infty}\frac{\cos(2k-1)x}{(2k-1)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,005 |
Example 4.41. Consider the function $f(x)$ defined by the conditions:
$$
\begin{gathered}
f(x)=f(x+2 \pi) \text{ (2$\pi$-periodicity) } \\
f(x)=-f(-x) \text{ (oddness) } \\
f(x)=\pi-x, 0<x \leqslant \pi, f(0)=0
\end{gathered}
$$ | Solution. This function has a piecewise continuous derivative, and therefore satisfies the Dirichlet condition by default. Let's write its Fourier series (due to the oddness of $f(x)$, the coefficients $a_{k}$ are 0). The coefficients $b_{k}$ are calculated using formulas (4.11):
$$
\begin{gathered}
b_{k}=\frac{1}{\pi... | f(x)=2\sum_{k=1}^{\infty}\frac{\sinkx}{k} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,006 |
Example 4.42. Prove the equalities:
\[
\begin{gathered}
\int_{-\pi}^{\pi} \cos k x \cos l x d x=\left\{\begin{array}{l}
0, k \neq l \\
\pi, k=l
\end{array}\right. \\
\int_{-\pi}^{\pi} \sin k x \sin l x d x=\left\{\begin{array}{l}
0, k \neq l \\
\pi, k=l
\end{array}\right. \\
\int_{-\pi}^{\pi} \cos k x \sin l x d x=0, ... | Solution. Let's check, for example, the validity of the formula $\int_{-\pi}^{\pi} \cos k x \sin l x d x=0$. Using known trigonometric formulas, we get
$$
\begin{gathered}
\int_{-\pi}^{\pi} \cos k x \sin l x d x=\frac{1}{2} \int_{-\pi}^{\pi}(\sin (l x+k x)+\sin (l x-k x)) d x= \\
=\frac{1}{2} \int_{-\pi}^{\pi}\left(\s... | proof | Calculus | proof | Yes | Yes | olympiads | false | 31,007 |
Example 4.43. Prove the validity of the equality
$$
\int_{-\pi}^{\pi}\left(f(x)-T_{n}(x)\right)^{2} d x=\int_{-\pi}^{\pi} f^{2}(x) d x-\frac{\pi a_{0}^{2}}{2}-\pi \sum_{k=1}^{n}\left(a_{k}^{2}+b_{k}^{2}\right)
$$
where \( T_{n}(x)=\frac{a_{0}}{2}+\sum_{k=1}^{n}\left(a_{k} \cos k x+b_{k} \sin k x\right) \),
$$
\begin... | Solution. Let's calculate the required integral directly, using the results of Example 4.42
$$
\begin{gathered}
\int_{-\pi}^{\pi}\left(f(x)-\frac{a_{0}}{2}+\sum_{k=1}^{n}\left(a_{k} \cos k x+b_{k} \sin k x\right)\right)^{2} d x= \\
=\int_{-\pi}^{\pi} f^{2}(x) d x-2 \frac{a_{0}}{2} \int_{-\pi}^{\pi} f(x) d x-
\end{gath... | proof | Calculus | proof | Yes | Yes | olympiads | false | 31,008 |
Example 2. Find the real solutions of the equation
$$
(4+2 i) x+(5-3 i) y=13+i
$$ | Solution. Let's separate the real and imaginary parts in the left side of the equation: $(4 x+5 y)+i(2 x-3 y)=13+i$. From here, according to the definition of equality of two complex numbers, we get
$$
\left\{\begin{array}{l}
4 x+5 y=13 \\
2 x-3 y=1
\end{array}\right.
$$
Solving this system, we find
$$
x=2, \quad y=... | 2,\quad1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,009 |
Example 3. Find the modulus and argument of the complex number
$$
z=-\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}
$$ | Solution. We have
$$
x=-\sin \frac{\pi}{8}<0, \quad y=-\cos \frac{\pi}{8}<0
$$
The principal value of the argument according to (1) will be
$$
\begin{aligned}
\arg z & =-\pi+\operatorname{arctg}\left(\operatorname{ctg} \frac{\pi}{8}\right)=-\pi+\operatorname{arctg}\left[\operatorname{tg}\left(\frac{\pi}{2}-\frac{\pi... | |z|=1,\operatorname{Arg}-\frac{5}{8}\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,010 |
Example 4. Write the complex number in trigonometric form
$$
z=-1-i \sqrt{3}
$$ | Solution. We have
$$
|z|=\sqrt{(-1)^{2}+(-\sqrt{3})^{2}}=2 ; \quad \operatorname{tg} \varphi=\frac{-\sqrt{3}}{-1}=\sqrt{3}, \quad \varphi=-\frac{2}{3} \pi
$$
Therefore,
$$
-1-i \sqrt{3}=2\left[\cos \left(-\frac{2}{3} \pi\right)+i \sin \left(-\frac{2}{3} \pi\right)\right]
$$ | -1-i\sqrt{3}=2[\cos(-\frac{2}{3}\pi)+i\sin(-\frac{2}{3}\pi)] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,011 |
Example 5. Find the real roots of the equation
$$
\cos x+i \sin x=\frac{1}{2}+\frac{3}{4} i
$$ | Solution. The given equation has no roots. Indeed, this equation is equivalent to the following: $\cos x=\frac{1}{2}, \sin x=\frac{3}{4}$. These equations are inconsistent, as $\cos ^{2} x+\sin ^{2} x=\frac{13}{16}$, which is impossible for any value of $x$.
Any complex number $z \neq 0$ can be written in exponential ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,012 |
Example 6. Find all complex numbers $z \neq 0$, satisfying the condition $z^{n-1}=\bar{z}$. | Solution. Let $z=\rho e^{i \varphi}$. Then $\bar{z}=\rho e^{-i \varphi}$.
According to the condition
$$
\rho^{n-1} e^{i(n-1) \varphi}=\rho e^{-i \varphi} \text { or } \rho^{n-2} e^{i n \varphi}=1,
$$
from which $\rho^{n-2}=1$, i.e., $\rho=1$, and $i \varphi=2 k \pi i$, i.e., $\varphi=\frac{2 k \pi}{n}(k=0,1,2, \ldot... | z_{k}=e^{i2\pik/n}(k=0,1,2,\ldots,n-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,013 |
Example 7. Calculate $(-1+i \sqrt{3})^{60}$.
---
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution. Let's represent the number $z=-1+i \sqrt{3}$ in trigonometric form
$$
-1+i \sqrt{3}=2\left(\cos \frac{2}{3} \pi+i \sin \frac{2}{3} \pi\right)
$$
Applying the above formula for raising to a power, we get
$$
\begin{aligned}
(-1+i \sqrt{3})^{60} & =2^{60}\left[\cos \left(60 \cdot \frac{2}{3} \pi\right)+i \sin... | 2^{60} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,014 |
Example 8. Prove that the polynomial
$$
f(x)=(\cos \alpha+x \sin \alpha)^{n}-\cos n \alpha-x \sin n \alpha
$$
is divisible by $x^{2}+1$. | Solution. We have $x^{2}+1=(x+i)(x-i)$. By De Moivre's formula
$$
\begin{aligned}
f(i) & =(\cos \alpha+i \sin \alpha)^{n}-\cos n \alpha-i \sin n \alpha= \\
& =\cos n \alpha+i \sin n \alpha-\cos n \alpha-i \sin n \alpha=0 .
\end{aligned}
$$
Similarly, $f(-i)=0$. Therefore, $f(x)$ is divisible by $x^{2}+1$.
## Problem... | proof | Algebra | proof | Yes | Yes | olympiads | false | 31,015 |
Example 9. Find all values of $\sqrt[4]{1-i}$.
---
The provided text has been translated while preserving the original formatting and line breaks. | Solution. Convert the complex number $1-i$ to trigonometric form
$$
1-i=\sqrt{2}\left[\cos \left(-\frac{\pi}{4}\right)+i \sin \left(-\frac{\pi}{4}\right)\right]
$$
Therefore,
$$
\sqrt[4]{1-i}=\sqrt[8]{2}\left(\cos \frac{-\frac{\pi}{4}+2 k \pi}{4}+i \sin \frac{-\frac{\pi}{4}+2 k \pi}{4}\right)
$$
By setting $k=0,1,2... | \begin{aligned}&(k=0)\quad\sqrt[4]{1-i}=\sqrt[8]{2}(\cos\frac{\pi}{16}-i\sin\frac{\pi}{16})\\&(k=1)\quad\sqrt[4]{1-i}=\sqrt[8]{2}(\cos\frac{7\pi}{16}+i\sin\frac{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,016 |
Example 10. What set of points in the complex plane $z$ is defined by the condition
$$
\operatorname{Im} z^{2}>2 ?
$$ | Solution: Let $z=x+i y$. Then
$$
z^{2}=(x+i y)^{2}=\left(x^{2}-y^{2}\right)+i 2 x y
$$
Therefore, $\operatorname{Im} z^{2}=2 x y$.
By the condition $2 x y>2$, or $x y>1$. This inequality defines the set of points in the first and third quadrants, respectively above and below the hyperbola $x y=1$. in the first and t... | xy>1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,017 |
Example 11. What set of points in the complex plane is defined by the condition
$$
-\frac{\pi}{2} \leqslant \arg (z+1-i) \leqslant \frac{3}{4} \pi ?
$$ | Solution. The complex number
$$
z+1-i=z-(-1+i)
$$
is represented by a vector whose initial point is $-1+i$ and whose terminal point is $z$. The angle between this vector and the $O X$ axis is $\arg (z+1-i)$, and it varies from $-\frac{\pi}{2}$ to $\frac{3 \pi}{4}$. Therefore, the given inequality defines the angle be... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,018 |
Example 2. Into what curve is the unit circle $|z|=1$ mapped by the function $w=z^{2} ?$ | Solution. Since $|z|=1$ by condition, then
$$
|w|=|z|^{2}=1 \text {. }
$$
Thus, the image of the circle $|z|=1$ in the $z$-plane is the circle $|w|=1$ in the $w$-plane, traversed twice. This follows from the fact that since $\boldsymbol{w}=z^{2}$, then $\operatorname{Arg} w=2 \operatorname{Arg} z+2 k \pi$, so when th... | |w|=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,020 |
Example 3. Find the image of the circle
$$
z=R \cos t+i R \sin t \quad(0 \leqslant t<2 \pi)
$$
under the mapping $\boldsymbol{w}=\frac{\boldsymbol{z}}{\bar{z}}$. | Solution. Let $z=x+i y$. The given equation of the circle can be written as
$$
x=R \cos t, \quad y=R \sin t \quad(0 \leqslant t<2 \pi) .
$$
We separate the real and imaginary parts of the function $w=u+i v$. We have
$$
u+i v=\frac{z}{\bar{z}}=\frac{z^{2}}{z \bar{z}}=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}+i \frac{2 x y}{x^{... | u^2+v^2=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,021 |
Example 6. Write in algebraic form Arctg $(1+i)$.
---
Note: The translation maintains the original text's format and line breaks as requested. | Solution. Substituting $z=1+i$ into formula (9), we get
$$
\operatorname{Arctg}(1+i)=-\frac{i}{2} \operatorname{Ln} \frac{1+i(1+i)}{1-i(1+i)}=-\frac{i}{2} \operatorname{Ln} \frac{i}{2-i}=-\frac{i}{2} \operatorname{Ln}\left(-\frac{1}{5}+\frac{2}{5} i\right)
$$
Further,
$$
\operatorname{Ln}\left(-\frac{1}{5}+\frac{2}{... | \operatorname{Arctg}(1+i)=-\frac{1}{2}\operatorname{arctg}2+(2k+1)\frac{\pi}{2}+\frac{i}{2}\ln\sqrt{5}\quad(k=0,\1,\2,\ldots) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,023 |
Example 7. Solve the equation $\sin z=3$. | Solution. The problem reduces to finding the value of
$$
z=\operatorname{Arcsin} 3
$$
We will use formula (7):
$$
\operatorname{Arcsin} t=-i \operatorname{Ln}\left(t+\sqrt{1-t^{2}}\right)
$$
We will have
$$
z=\operatorname{Arcsin} 3=-i \operatorname{Ln}(3 i+\sqrt{-8})
$$
or, considering that
$$
\sqrt{-8}= \pm \s... | \frac{\pi}{2}+2k\pi-i\ln(3\\sqrt{8})\quad(k=0,\1,\2,\ldots) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,024 |
Example 1. Prove that the sequence
$$
z_{n}=\frac{n-i}{n+1}, \quad n=1,2 ; \ldots
$$
has the limit $a=1$. | Solution. Let an arbitrary number $\varepsilon>0$ be given. We will show that there exists a number $N$ such that $\left|z_{n}-\right| \mid\frac{\sqrt{2}}{\varepsilon}-1$. Therefore, we can take
$$
N=N(\varepsilon)=\left[\frac{\sqrt{2}}{\varepsilon}-1\right]+1
$$
Here the symbol $[x]$ denotes the integer part of the ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 31,025 |
Example 2. Let the sequence $\left\{z_{n}\right\}$ have a limit a. Prove that the sequence $\left\{\left|z_{n}\right|\right\}$ has a limit equal to $|a|$. | Solution. Indeed, since $\lim _{n \rightarrow \infty} z_{n}=a$, then
$$
\lim _{n \rightarrow \infty}\left|z_{n}-a\right|=0
$$
On the other hand, for any two complex numbers $z_{n}$ and $a$, the inequality holds (see p. 8)
$$
|| z_{n}|-| a|| \leqslant\left|z_{n}-a\right| .
$$
From (1) and (2), we obtain that $\lim _... | proof | Algebra | proof | Yes | Yes | olympiads | false | 31,026 |
Example 5. Let
$$
x_{n}=1+\rho \cos \alpha+\rho^{2} \cos 2 \alpha+\ldots+\rho^{n} \cos n \alpha
$$
where $0<\rho<1, n=1,2, \ldots$. Find $\lim _{n \rightarrow \infty} x_{n}$. | Solution. Let
$$
y_{n}=\rho \sin \alpha+\rho^{2} \sin 2 \alpha+\ldots+\rho^{n} \sin n \alpha
$$
and consider the limit of the sequence of complex numbers
$$
\begin{gathered}
z_{n}=x_{n}+i y_{n}=1+\rho(\cos \alpha+i \sin \alpha)+\rho^{2}(\cos 2 \alpha+i \sin 2 \alpha)+\ldots \\
\ldots+\rho^{n}(\cos n \alpha+i \sin n ... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,027 | |
Example 7. Show that the function $w=z^{2}$ is continuous for any value of $z$. | Solution. Let us take an arbitrary point $z_{0}$ and an arbitrary number $\varepsilon>0$. Since the value of the function $f(z)=z^{2}$ at the point $z_{0}$ is $f\left(z_{0}\right)=z_{0}^{2}$, we will show that there exists a number $\delta(\varepsilon)>0$ such that $\left|z^{2}-z_{0}^{2}\right|<\varepsilon$ whenever $|... | proof | Calculus | proof | Yes | Yes | olympiads | false | 31,028 |
Example 1. Show that the function $w=e^{z}$ is analytic throughout the entire complex plane. | Solution. We have $e^{z}=e^{z}(\cos y+i \sin y)$, so
$$
u(x, y)=e^{z} \cos y, \quad v(x, y)=e^{x} \sin y
$$
The functions $u(x, y)$ and $v(x, y)$, as functions of real variables $x$ and $y$, are differentiable at any point $(x, y)$ (they have continuous partial derivatives of any order) and satisfy conditions (2).
T... | proof | Calculus | proof | Yes | Yes | olympiads | false | 31,029 |
Example 2. Is the function $w=z \bar{z}$ analytic at least at one point | Solution. We have $z \bar{z}=x^{2}+y^{2}$, so
$$
u(x, y)=x^{2}+y^{2}, \quad v(x, y) \equiv 0
$$
The Cauchy-Riemann conditions in this case are
$$
\left\{\begin{array}{l}
2 x=0 \\
2 y=0
\end{array}\right.
$$
and are satisfied only at the point $(0,0)$.
Therefore, the function $w=z \bar{z}$ is differentiable only at... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,030 |
Example 3. Is the function $w=\bar{z}=x-i y$ analytic? | Solution. Here $u=(x, y)=x, v(x, y)=-y$ - are differentiable functions of variables $x$ and $y$. Further,
$$
\frac{\partial u}{\partial x}=1, \quad \frac{\partial u}{\partial y}=0, \quad \frac{\partial v}{\partial x}=0, \quad \frac{\partial v}{\partial y}=-1
$$
So $\frac{\partial u}{\partial x} \neq \frac{\partial v}... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,031 |
Example 4. Find the analytic function $w=f(z)$, given its real part $u(x, y)=2 e^{x} \cos y$ and the additional condition $f(0)=2$. | Solution. First method. We have $\frac{\partial u}{\partial x}=2 e^{x} \cos y$. According to the first of the Cauchy-Riemann conditions, it should be $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$, so $\frac{\partial v}{\partial y}=2 e^{x} \cos y$. From this,
$$
v(x, y)=\int 2 e^{x} \cos y d y=2 e^{x} \... | f(z)=2e^{z} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,032 |
Example 5. Find the analytic function $w=f(z)$ given its imaginary part $v(x, y)=3 x+2 x y$ under the condition that $f(-i)=2$. | Solution. We will use formula (6). In our example, $v(x, y)=3 x+2 x y, z_{0}=-i, C_{0}=2$, so
$$
f(z)=2 i\left(3 \frac{z+i}{2}+2 \frac{z+i}{2} \cdot \frac{z-i}{2 i}\right)+2=3 i z+z^{2}
$$
## Problems for Independent Solution
Reconstruct the function $f(z)$, analytic in a neighborhood of the point $z_{0}$, given the... | 3iz+z^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,033 |
Example 7. Find the coefficient of stretching and the angle of rotation for the mapping $w=z^{2}$ at the point $z_{0}=\sqrt{2}+i \sqrt{2}$. | Solution. We have $w^{\prime}(z)=2 z$, so $\left.w^{\prime}\right|_{z=\sqrt{2}+i \sqrt{2}}=2 \sqrt{2}+i 2 \sqrt{2}$.
Transitioning from the algebraic form of the complex number $2 \sqrt{2}+i 2 \sqrt{2}$ to the trigonometric form, we get
$$
2 \sqrt{2}+i 2 \sqrt{2}=4\left(\frac{\sqrt{2}}{2}+i \frac{\sqrt{2}}{2}\right)=... | r=4,\varphi=\frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,034 |
Example 8. The point $z=x+i y$ describes the segment
$$
x=1, \quad-1 \leqslant y \leqslant 1
$$
What is the length of the line obtained by mapping this segment using the function $w=z^{2}$? | Solution. First method. We have $w=z^{2}$, or
$$
u+i v=x^{2}-y^{2}+i 2 x y
$$
i.e.
$$
\left\{\begin{array}{l}
u=x^{2}-y^{2} \\
v=2 x y
\end{array}\right.
$$
Obviously, on the line (10) we will have
$$
\left\{\begin{array}{l}
u=1-y^{2} \\
v=2 y
\end{array}\right.
$$
where, as $y$ changes from -1 to +1, $v$ will ch... | 2\sqrt{2}+\ln(3+2\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,035 |
Example 1. Compute the integral
$$
\int_{C}(1+i-2 \bar{z}) d z
$$
along the lines connecting the points $z_{1}=0$ and $z_{2}=1+i$,
1) along a straight line;
2) along the parabola $y=x^{2}$;
3) along the broken line $z_{\mathrm{j}} z_{3} z_{2}$, where $z_{3}=1$. | Solution. Rewrite the integrand function as
$$
1+i-2 \bar{z}=(1-2 x)+i(1+2 y)
$$
Here $u=1-2 x, v=1+2 y$.
Applying formula (1), we get
$$
\int_{C}(1+i-2 \bar{z}) d z=\int_{C}(1-2 x) d x-(1+2 y) d y+i \int_{C}(1+2 y) d x+(1-2 x) d y .
$$
1) The equation of the line passing through the points $z=0$ and $z_{2}=1+i$ i... | 2(i-1),-2+\frac{4}{3}i,-2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,036 |
Example 2. Compute the integral
$$
\int_{C}\left(z^{2}+z \bar{z}\right) d z
$$
where $C-$ is the arc of the circle $\{z \mid=1(0 \leqslant \arg z \leqslant \pi)$. | Solution. Let $z=e^{i \varphi}$, then $d z=i e^{i \varphi} d \varphi$ and
$$
\begin{aligned}
\int_{C}\left(z^{2}+z \bar{z}\right) d z & =\int_{0}^{\pi} i e^{i \varphi}\left(e^{i 2 \varphi}+1\right) d \varphi= \\
& =i \int_{0}^{\pi}\left(e^{i 3 \varphi}+e^{i \varphi}\right) d \varphi=\left.\left(\frac{1}{3} e^{i 3 \var... | -\frac{8}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,037 |
Example 3. Compute the integral $\int_{C} e^{\bar{z}} d z$, where $C$ is the line segment $y=-x$, connecting the points $z_{1}=0$ and $z_{2}=\pi-i \pi$. | Solution. The parametric equations of the curve $C$ are
$$
x=t, \quad y=-t
$$
or in complex form
$$
z=t-i t
$$
where the real variable $t$ varies from 0 to $\pi$.
Applying formula (2), we get
$$
\int_{C} \mathrm{e}^{\bar{z}} d z=\int_{0}^{\pi} e^{t+i t}(1-i) d t=(1-i) \int_{0}^{\pi} e^{(1+i) t} d t=\left.\frac{1-... | (e^{\pi}+1)i | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,038 |
Example 4. Calculate the integral
$$
\int_{1-i}^{2+i}\left(3 z^{2}+2 z\right) d z
$$ | Solution. Since the integrand $f(z)=3 z^{2}+2 z$ is analytic everywhere, we can apply the Newton-Leibniz formula to find
$$
\int_{1-i}^{2+i}\left(3 z^{2}+2 z\right) d z=\left.\left(z^{3}+z^{2}\right)\right|_{1-i} ^{2+i}=(2+i)^{3}+(2+i)^{2}-(1-i)^{3}-(1-i)^{2}=7+19 i
$$ | 7+19i | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,039 |
Example 5. Calculate the integral
$$
\int_{0}^{i} z \cos z d z
$$ | Solution. The functions $f(z)=z$ and $\varphi(z)=\cos z$ are analytic everywhere. Applying the formula for integration by parts, we get
$$
\begin{aligned}
\int_{0}^{i} z \cos z d z & =\int_{0}^{i} z(\sin z)^{\prime} d z=\left.(z \sin z)\right|_{0} ^{i}-\int_{0}^{i} \sin z d z= \\
& =i \sin i+\left.\cos z\right|_{0} ^{... | \frac{1-e}{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,040 |
Example 7. Calculate the integral
$$
I=\int_{1}^{i} \frac{\ln ^{3} z}{z} d z
$$
along the arc of the circle $|z|=1$ (where $\ln z$ is the principal value of the logarithm, $\ln 1=0$). | Solution. First method. Applying the Newton-Leibniz formula, we get
$$
I=\int_{1}^{i} \frac{\ln ^{3} z}{z} d z=\int_{1}^{i} \ln ^{3} z d(\ln z)=\left.\frac{\ln ^{4} z}{4}\right|_{1} ^{1}=\frac{\ln ^{4} i-\ln ^{4} 1}{4}=\frac{\ln ^{4} i}{4}=\frac{1}{4}\left(\frac{\pi i}{2}\right)^{4}=\frac{\pi^{4}}{64}
$$
Second metho... | \frac{\pi^{4}}{64} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,041 |
Example 1. Compute the integral
$$
\int_{|z|=2} \frac{\operatorname{ch} i z}{z^{2}+4 z+3} d z
$$ | Solution. Inside the circle $|z|=2$, the denominator of the fraction becomes zero at the point $z_{0}=-1$. To apply formula (1), we rewrite the integral as follows:
$$
\int_{|z|=2} \frac{\operatorname{ch} i z}{z^{2}+4 z+3} d z=\int_{|z|=2} \frac{\operatorname{ch} i z}{(z+1)(z+3)} d z=\int_{|z|=2} \frac{\frac{\operator... | \pii\cos1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,042 |
Example 1. Investigate the convergence of the series
$$
\sum_{n=1}^{\infty} \frac{e^{i n}}{n^{2}}
$$ | Solution. We have $e^{i n}=\cos n+i \sin n$. Thus, the question of the convergence of the given series reduces to the question of the convergence of the series with real terms:
$$
\sum_{n=1}^{\infty} \frac{\cos n}{n^{2}} \text { and } \sum_{n=1}^{\infty} \frac{\sin n}{n^{2}} .
$$
Each of these series converges absolu... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,044 |
Example 2. Investigate the convergence of the series
$$
\sum_{n=1}^{\infty} \frac{e^{i \pi / n}}{n}
$$ | Solution. We have
$$
e^{i \pi / n}=\cos \frac{\pi}{n}+i \sin \frac{\pi}{n}
$$
The series $\sum_{n=1}^{\infty} \frac{\cos \frac{\pi}{n}}{n}$ diverges, while the series $\sum_{n=1}^{\infty} \frac{\sin \frac{\pi}{n}}{n}$ converges. Therefore, the given series diverges.
## Problems for Independent Solution
Investigate ... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,045 |
Example 3. Determine the radius of convergence of the power series
$$
\sum_{n=0}^{\infty} \cos i n \cdot z^{n}
$$ | Solution. We have
$$
c_{n}=\cos i n=\frac{e^{-n}+e^{n}}{2}=\operatorname{ch} n
$$
To find the radius of convergence $R$, we apply formula (6):
$$
\begin{aligned}
R=\lim _{n \rightarrow \infty} \frac{|\operatorname{ch} n|}{|\operatorname{ch}(n+1)|} & =\lim _{n \rightarrow \infty} \frac{\operatorname{ch} n}{\operatorn... | e^{-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,046 |
Example 4. Find the radius of convergence of the power series
$$
\sum_{n=0}^{\infty}(1+i)^{n} z^{n}
$$ | Solution. We find the modulus of the coefficient $c_{n}=(1+i)^{n}$:
$$
\left|c_{n}\right|=\left|(1+i)^{n}\right|=|1+i|^{n}=(\sqrt{2})^{n}=2^{n / 2}
$$
Applying formula (7), we find the radius of convergence of the given power series:
$$
R=\lim _{n \rightarrow \infty} \frac{1}{\sqrt[n]{2^{n / 2}}}=\frac{1}{\sqrt{2}}
... | \frac{1}{\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,047 |
Example 5. Expand the function
$$
f(z)=\frac{z}{z^{2}-2 z-3}
$$
into a Taylor series in the neighborhood of the point $z_{0}=0$ using expansion (12), and find the radius of convergence of the series. | Solution. Let's decompose the given function into partial fractions:
$$
\frac{z}{z^{2}-2 z-3}=\frac{1}{4} \frac{1}{z+1}-\frac{3}{4} \frac{1}{z-3}
$$
Transform the right-hand side as follows:
$$
f(z)=\frac{1}{4} \frac{1}{1+z}-\frac{1}{4} \frac{1}{1-\frac{2}{3}}
$$
Using the expansion (12) of the function $\frac{1}{1... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,048 |
Example 6. Expand the function in powers of the difference $z-3$
$$
f(z)=\frac{1}{3-2 z}
$$ | Solution. Let's transform the given function as follows:
$$
\frac{1}{3-2 z}=\frac{1}{3-2(z-3+3)}=\frac{1}{-3-2(z-3)}=-\frac{1}{3} \frac{1}{1+\frac{2}{3}(z-3)}
$$
By substituting $z$ with $\frac{2}{3}(z-3)$ in the expansion (12), we get
$$
\begin{aligned}
\frac{1}{3-2 z} & =-\frac{1}{3}\left[1-\frac{2}{3}(z-3)+\frac{... | -\frac{1}{3}+\frac{2}{3^{2}}(z-3)-\frac{2^{2}}{3^{3}}(z-3)^{2}+\frac{2^{3}}{3^{4}}(z-3)^{3}-\cdots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,049 |
Example 7. Find the first few terms of the expansion in a power series of the function $f(z)=\operatorname{tg} z$ and find the radius of convergence of the series. | Solution. Let the desired series have the form
$$
f(z)=c_{0}+c_{1} z+c_{2} z^{2}+c_{3} z^{3}+\ldots
$$
where
$$
c_{n}=\frac{f^{(n)}(0)}{n!} \quad(n=0,1,2, \ldots), \quad f^{(0)}(0)=f(0)=0
$$
To find the values of the derivatives $f^{(n)}(z)$ at the point $z=0$, we differentiate the given function. We have
$$
\begi... | \operatorname{tg}z+\frac{2}{3!}z^{3}+\frac{16}{5!}z^{5}+\ldots | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,050 |
Example 8. Find the region of convergence of the series
$$
\sum_{n=1}^{\infty} \frac{(1+i)^{n+1}}{z^{n}}
$$ | Solution. Here $c_{-n}=(1+i)^{n+1}, c_{-n-1}=(1+i)^{n+2}, z_{0}=0$. Therefore,
$$
r=\lim _{n \rightarrow \infty} \frac{\left|(1+i)^{n+2}\right|}{\left|(1+i)^{n+1}\right|}=\lim _{n \rightarrow \infty}|1+i|=\sqrt{2}
$$
The series converges in the region $|z|>\sqrt{2}$. | |z|>\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,051 |
Example 9. Find the region of convergence of the series
$$
\sum_{n=1}^{\infty} \frac{\sin i n}{(z+i)^{n}}
$$ | Solution. We have
$$
c_{-n}=\sin i n=i \operatorname{sh} n, \quad c_{-n-1}=i \operatorname{sh}(n+1).
$$
Therefore,
$$
r=\lim _{n \rightarrow \infty} \frac{|i \operatorname{sh}(n+1)|}{|i \operatorname{sh} n|}=\lim _{n \rightarrow \infty} \frac{\operatorname{sh}(n+1)}{\operatorname{sh} n}=\lim _{n \rightarrow \infty} ... | |z+i|>e | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,052 |
Example 10. Determine the region of convergence of the series
$$
\sum_{n=1}^{\infty} \frac{e^{i n}}{(z+1)^{n}}+\sum_{n=0}^{\infty} \frac{(z+1)^{n}}{e^{i n+1 / 2}}
$$ | Solution. For the series $\sum_{n=1}^{\infty} \frac{e^{\text {in }}}{(z+1)^{n}}$ we have
$$
c_{-n}=e^{i n}, \quad c_{-n-1}=e^{i(n+1)}
$$
Therefore,
$$
r=\lim _{n \rightarrow \infty} \frac{\left|e^{i(n+1)}\right|}{\left|e^{i n}\right|}=1
$$
so the first series converges in the region $\{z+1 \mid>1\}$.
$$
\begin{ali... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,053 |
Example 11. Determine the region of convergence of the series
$$
\sum_{n=1}^{\infty} \frac{(3+4 i)^{n}}{(z+2 i)^{n}}+\sum_{n=0}^{\infty}\left(\frac{z+2 i}{6}\right)^{n}
$$ | Solution. For the series $\sum_{n=1}^{\infty} \frac{(3+4 i)^{n}}{(z+2 i)^{n}}$ we have
$$
c_{-n}=(3+4 i)^{n}, \quad c_{-n-1}=(3+4 i)^{n+1}
$$
Therefore,
$$
r=\lim _{n \rightarrow \infty} \frac{\left|(3+4 i)^{n+1}\right|}{\left|(3+4 i)^{n}\right|}=\lim _{n \rightarrow \infty}|3+4 i|=5
$$
The first series converges i... | 5<|z+2i|<6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,054 |
Example 12. Expand into a Laurent series in the annulus $0<|z-1|<2$ the function
$$
f(z)=\frac{1}{\left(z^{2}-1\right)^{2}}
$$ | Solution: The first method. The function $f(z)=\frac{1}{\left(z^{2}-1\right)^{2}}$ is analytic in the annulus $0<|z-1|<2$, i.e., $n>-3$. Applying formula (2) from § 6 for the derivative of any order of an analytic function, we get
$$
\begin{aligned}
& c_{n}=\frac{1}{2 \pi i} \int_{\Gamma} \frac{\frac{1}{(z+1)^{2}}}{(z... | \frac{1}{(z^{2}-1)^{2}}=\frac{1}{4}\frac{1}{(z-1)^{2}}-\frac{1}{4}\frac{1}{z-1}+\frac{3}{16}-\frac{1}{8}(z-1)+\frac{5}{64}(z-1)^{2}-\frac{3}{64}(z | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,055 |
Example 13. Expand the function
$$
f(z)=z^{2} \cos \frac{1}{z}
$$
into a Laurent series in the neighborhood of the point $z_{0}=0$. | Solution. For any complex $\zeta$ we have
$$
\cos \zeta=-\frac{\zeta^{2}}{2!}+\frac{\zeta^{4}}{4!}-\frac{\zeta^{6}}{6!}+\cdots
$$
Setting $\zeta=\frac{1}{z}$, we get
$$
z^{2} \cos \frac{1}{z}=z^{2}\left(1-\frac{1}{2!z^{2}}+\frac{1}{4 \cdot z^{4}} \frac{1}{6!z^{6}}+\ldots\right)=z^{2}-\frac{1}{2!}+\frac{1}{4 \cdot z^... | z^{2}\cos\frac{1}{z}=-\frac{1}{2}+z^{2}+\frac{1}{4!z^{2}}-\frac{1}{6!z^{4}}+\ldots | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,056 |
Example 14. Consider various Laurent series expansions of the function
$$
f(z)=\frac{2 z+1}{z^{2}+z-2}
$$
taking $z_{0}=0$. | Solution. The function $f(z)$ has two singular points: $z_{1}=-2$ and $z_{2}=1$. Therefore, there are three "rings" centered at the point $z_{0}=0$, in each of which $f(z)$ is analytic:
a) $\mathrm{kpyr}|z|1$. Therefore, we transform $f(z)$ as follows:
$$
f(z)=\frac{1}{2} \frac{1}{1+\frac{z}{2}}+\frac{1}{z} \frac{1}{1... | \frac{2z+1}{z^{2}+z-2}=\sum_{n=1}^{\infty}\frac{1}{z^{n}}+\frac{1}{2}\sum_{n=0}^{\infty}\frac{z^{n}}{2^{n}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,057 |
Example 15. Expand the function
$$
f(z)=\frac{2 z-3}{z^{2}-3 z+2}
$$
into a Laurent series in the neighborhood of its singular points. | Solution. The special points of the function $f(z): z_{1}=1, z_{2}=2$.
1) Expansion of $f(z)$ in the neighborhood of the point $z_{1}=1$, i.e., in the annulus $0<|z-1|<1$. Represent the function $f(z)$ as a sum of elementary fractions
$$
\frac{2 z-3}{z^{2}-3 z+2}=\frac{1}{z-1}+\frac{1}{z-2}
$$
Transform the right-ha... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,058 |
Example 2. Show that the infinite product
$$
\prod_{k=1}^{\infty}\left(1+\frac{1-k}{k}\right)
$$
diverges. | Solution. Indeed, since $1+\frac{1-k}{k}=\frac{1}{k}$, then
$$
p_{n}=\prod_{k=1}^{n}\left(1+\frac{1-k}{k}\right)=1 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \ldots \cdot \frac{1}{n}=\frac{1}{n!}, \quad p=\lim _{n \rightarrow \infty} p_{n}=\lim _{n \rightarrow \infty} \frac{1}{n!}=0
$$
Since $p$ is zero, by definition... | proof | Algebra | proof | Yes | Yes | olympiads | false | 31,059 |
Example 3. Show that the infinite product
$$
\prod_{k=1}^{\infty}\left[1+(-1)^{k+1} \frac{1}{k}\right]=(1+1)\left(1-\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \ldots
$$
converges conditionally. | Solution. It is not difficult to verify that
$$
p_{n}=\left\{\begin{array}{cl}
\frac{n+1}{n}, & \text { if } n \text { is odd, } \\
1, & \text { if } n \text { is even }
\end{array}\right.
$$
from which it follows that $\lim _{n \rightarrow \infty} p_{n}=1$, i.e., the product (10) converges.
On the other hand, the s... | proof | Algebra | proof | Yes | Yes | olympiads | false | 31,060 |
Example 5. Investigate the convergence of the infinite product
$$
\prod_{k=1}^{\infty}\left(1-\frac{1}{k+1}\right)=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \ldots\left(1-\frac{1}{k+1}\right) \ldots
$$ | Solution. Here all $u_{k}=-\frac{1}{k+1}$ are negative and the series (14)
$$
\sum_{k=1}^{\infty} u_{k}=-\sum_{k=1}^{\infty} \frac{1}{k+1}=-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots\right)
$$
obviously diverges.
Then, by Theorem 4, the infinite product (15) diverges.
Remark. Calculating the $n$-th partial pr... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,061 |
Example 6. Determine the convergence of the infinite product
$$
\prod_{k=1}^{\infty}\left(1+\frac{1}{k^{\alpha}}\right)=(1+1)\left(1+\frac{1}{2^{\alpha}}\right)\left(1+\frac{1}{3^{\alpha}}\right) \ldots\left(1+\frac{1}{k^{\alpha}}\right) \ldots
$$
for $\alpha>1$. | Solution. Series
$$
\sum_{k=1}^{\infty} u_{k}=\sum_{k=1}^{\infty} \frac{1}{k^{\alpha}}
$$
converges when $\alpha>1$. Then, according to Theorem 4, the infinite product (16) also converges.
## Problems for Independent Solution
Prove the convergence and find the values of the following infinite products:
276. $\prod... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,062 |
Example 1. Find the zeros of the function $f(z)=1+\cos z$ and determine their orders. | Solution. Setting $f(z)$ to zero, we get $\cos z=-1$, from which $z_{n}=$ $(2 n+1) \pi(n=0, \pm 1, \pm 2, \ldots)$ - these are the zeros of the given function. Next,
$$
\begin{aligned}
& f^{\prime}[(2 n+1) \pi]=-\sin (2 n+1) \pi=0 \\
& f^{\prime \prime}[(2 n+1) \pi]=-\cos (2 n+1) \pi=1 \neq 0
\end{aligned}
$$
Therefo... | z_{n}=(2n+1)\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,063 |
Example 2. Find the zeros of the function $f(z)=1-e^{z}$ and determine their orders. | Solution. By setting $f(z)$ to zero, we find the zeros $z_{n}=2 n \pi i (n=0, \pm 1, \ldots)$ of the function $f(z)$. Next,
$$
f^{\prime}(2 n \pi i)=-e^{2 n \pi i}=-1 \neq 0 .
$$
Thus, $f(2 n \pi i)=0, f^{\prime}(2 n \pi i) \neq 0$, therefore, the points $z_{n}=2 n \pi i (n=0, \pm 1, \pm 2, \ldots)$ are simple zeros ... | z_{n}=2n\pii(n=0,\1,\2,\ldots) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,064 |
Example 3. Find the order of the zero $z_{0}=0$ for the function
$$
f(z)=\frac{z^{8}}{z-\sin z}
$$ | Solution. Using the Taylor series expansion of the function $\sin z$ in the neighborhood of the point $z_{0}=0$, we obtain
$$
\begin{aligned}
f(z) & =\frac{z^{8}}{z-\sin z}=\frac{z^{8}}{z-\left(z-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-\ldots\right)}= \\
& =\frac{z^{8}}{\frac{z^{3}}{3!}-\frac{z^{5}}{5!}+\ldots}=\frac{z^{5}}... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,065 |
Example 4. Find the zeros of the function $f(z)=\left(z^{2}+1\right)^{3} \sin z$ and determine their orders. | Solution. Assuming $f(z)=0$; we get $\left(z^{2}+1\right)^{3} \operatorname{sh} z=0$, from which $z^{2}+1=0$ or $\operatorname{sh} z=0$. Solving these equations, we find the zeros of the function $f(z)$:
$$
z=-i, \quad z=i, \quad z=k \pi i \quad(k=0, \pm 1, \pm 2, \ldots)
$$
Let $z=-i$, then $f(z)$ can be represented... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,066 |
Example 8. Determine the nature of the singular point $z=0$ of the function
$$
f(z)=e^{1 / z^{2}}
$$ | Solution. Consider the behavior of this function on the real and imaginary axes. On the real axis $z=x$ and $f(x)=e^{1 / z^{2}} \rightarrow \infty$ as $x \rightarrow 0$. On the imaginary axis $z=i y$ and $f(i y)=e^{-1 / y^{2}} \rightarrow 0$ as $y \rightarrow 0$. Therefore, the limit of $f(z)$ at the point $z=0$ does n... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,067 |
Example 9. Determine the nature of the singular point $z=0$ of the function
$$
f(z)=\frac{1}{2+z^{2}-2 \operatorname{ch} z}
$$ | Solution. The point $z=0$ is a pole of the function $f(z)$, since it is a zero of the denominator. Consider the function $\varphi(z)=\frac{1}{f(z)}=2+z^{2}-2 \operatorname{ch} z$. For this function, $\varphi(0)=0$. Let's find the order of the zero $z=0$ of this function. We have
$$
\begin{array}{ll}
\varphi^{\prime}(z... | 0ispoleofthefourthorderforf(z) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,068 |
Example 10. Determine the nature of the singular point $z=1$ of the function
$$
f(z)=\frac{\sin \pi z}{2 e^{z-1}-z^{2}-1}
$$ | Solution. Consider the function
$$
\varphi(z)=\frac{1}{f(z)}=\frac{2 e^{z-1}-z^{2}-1}{\sin \pi z}
$$
The point $z=1$ is a zero of the third order for the numerator
$$
\psi(z)=2 e^{z-1}-z^{2}-1
$$
since
$$
\begin{gathered}
\psi(1)=0 ; \quad \psi^{\prime}(1)=\left.\left(2 e^{z-1}-2 z\right)\right|_{z=1}=0 ; \\
\psi^... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,069 |
Example 11. Determine the nature of the singular point $z_{0}$ of the function
$$
f(z)=\frac{1-e^{-z}}{z}
$$ | Solution. Using the Taylor series expansion for the function $e^{-z}$ in the neighborhood of the point $z_{0}=0$, we obtain the Laurent series expansion of the function $f(z)$ in the neighborhood of zero
$$
\begin{aligned}
f(z)=\frac{1}{z}\left(1-e^{-z}\right) & =\frac{1}{z}\left[1-\left(1-z+\frac{z^{2}}{2!}-\frac{z^{... | z_{0}=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,070 |
Example 12. Determine the nature of the singular point $z_{0}=0$ of the function
$$
f(z)=\frac{1-\cos z}{z^{7}}
$$ | Solution. Expanding the function $\cos z$ into a Taylor series in powers of $z$, we obtain the Laurent series expansion of the function $f(z)$ in the neighborhood of zero:
$$
\begin{aligned}
f(z) & =\frac{1}{z^{7}}\left(\frac{z^{2}}{2!}-\frac{z^{4}}{4!}+\frac{z^{6}}{6!}-\frac{z^{8}}{8!}+\frac{z^{10}}{10!}-\cdots\right... | z_{0}=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,071 |
Example 13. Determine the nature of the singular point $z=1$ of the function
$$
f(z)=(z-1) e^{1 /(z-1)}
$$ | Solution. Using the expansion
$$
e^{u}=1+u+\frac{u^{2}}{2!}+\frac{u^{3}}{3!}+\ldots
$$
and setting $u=\frac{1}{z-1}$, we obtain the Laurent series expansion of the function $f(z)$ in the neighborhood of the point $z_{0}=1$:
$$
\begin{aligned}
f(z) & =(z-1)\left[1+\frac{1}{z-1}+\frac{1}{2!(z-1)^{2}}+\frac{1}{3!(z-1)^... | z_{0}=1isanessentialsingularityofthefunctionf(z) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,072 |
Example 3. Find the residues of the function $f(z)=\frac{1}{z^{4}+1}$ at its singular points. | Solution. The singular points of $f(z)$ are the zeros of the denominator, i.e., the roots of the equation $z^{4}+1=0$. We have
$$
z_{1}=e^{i \pi / 4}, \quad z_{2}=e^{i 3 \pi / 4}, \quad z_{3}=e^{-i 3 \pi / 4}, \quad z_{4}=e^{-i \pi / 4}
$$
Using formula (5), we get
$$
\begin{aligned}
& \operatorname{res} f\left(z_{1... | \begin{aligned}&\operatorname{res}f(z_{1})=\frac{1}{4}e^{-i3\pi/4}\\&\operatorname{res}f(z_{2})=\frac{1}{4}e^{-i9\pi/4}\\&\operatorname{res}f(z_{3})=\frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,073 |
Example 4. Find the residue of the function
$$
f(z)=z^{3} \cdot \sin \frac{1}{z^{2}}
$$
at its singular point. | Solution. A singular point of the function $f(z)$ is the point $z=0$. It is an essential singular point of the function $f(z)$. Indeed, the Laurent series expansion of the function in the neighborhood of the point $z=0$ is
$$
f(z)=z^{3}\left(\frac{1}{z^{2}}-\frac{1}{3!z^{6}}+\frac{1}{5!z^{10}}-\cdots\right)=z-\frac{1}... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,074 |
Example 5. Find the residue at the point $z=0$ of the function
$$
f(z)=\frac{\sin 3z - 3 \sin z}{(\sin z - z) \sin z}
$$ | Solution. The point $z=0$ is a zero of both the numerator $\varphi(z)=\sin 3 z-3 \sin z$ and the denominator $\psi(z)=(\sin z-z) \sin z$. Let's determine the orders of these zeros using the Taylor series expansion of $\sin z$ around the point $z=0$:
$$
\sin z=z-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-\ldots
$$
We have
$$
... | 24 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,075 |
Example 6. Find the residues of the function
$$
f(z)=\frac{e^{1 / z}}{1-z}
$$
at its singular points. | Solution. The singular points of the given function are $z=1$ and $z=0$. The point $z=1-$ is a simple pole, therefore
$$
\operatorname{res}_{z=1} f(z)=\left.\frac{e^{1 / z}}{-1}\right|_{z=1}=-e
$$
To determine the nature of the singular point $z=0$, we expand the function into a Laurent series in the neighborhood of ... | -ee-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,076 |
Example 7. Find the residue of the function
$$
f(z)=\cos z \sin \frac{1}{z}
$$
at its singular point $z=0$. | Solution. To determine the nature of the singular point, we expand the given function into a Laurent series in the neighborhood of the point $z=0$. We have
$$
\begin{aligned}
& \cos z=1-\frac{z^{2}}{2!}+\frac{z^{4}}{4!}-\ldots \\
& \sin \frac{1}{z}=\frac{1}{z}-\frac{1}{3!z^{3}}+\frac{1}{5!z^{5}}-\ldots
\end{aligned}
$... | \sum_{n=0}^{\infty}\frac{1}{(2n)!(2n+1)!} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,077 |
Example 8. Find the residue of the function
$$
w=z^{2} \sin \frac{1}{z+1}
$$
at its singular point. | Solution. A singular point of the given function is the point $z=-1$. To determine the nature of this point, we expand the function into a Laurent series in the neighborhood of the point $z=-1$. For this, we express $z^{2}$ in terms of powers of the difference $z-(-1)=z+1$. We have
$$
z^{2}=[(z+1)-1]^{2}=(z+1)^{2}-2(z... | \frac{5}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,078 |
Example 9. Find the residue of the function
$$
f(z)=e^{1 / z^{2}} \cos z
$$
at the point $z=0$. | Solution. Since the residue at the point $z=0$ is equal to the coefficient of $z^{-1}$, we immediately obtain that in this case the residue is zero, since the function $f(z)$ is even and its expansion in the neighborhood of the point $z=0$ cannot contain odd powers of $z$.
## Problems for Independent Solution
Find th... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,079 |
Example 1. Compute the integral
$$
\int_{|z|=4} \frac{e^{z}-1}{z^{2}+z} d z
$$ | Solution. In the region $|z|<4$, the function $f(z)=\frac{e^{z}-1}{z^{2}+z}$ is analytic everywhere except at $z=0$ and $z=-1$.
By the residue theorem of Cauchy,
$$
\int_{|z|=4} \frac{e^{z}-1}{z^{2}+z} \mathrm{~d} z=2 \pi i(\operatorname{res} f(0)+\operatorname{res} f(-1))
$$
The point $z=0$ is a removable singulari... | 2\pii(1-e^{-1}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,080 |
Example 2. Compute the integral
$$
\int_{|z|=2} \tan z \, dz
$$ | Solution. In the region $D:|z|<2$, the function $f(z)=\operatorname{tg} z$ is analytic everywhere except at the points $z=\frac{\pi}{2}$ and $z=-\frac{\pi}{2}$, which are simple poles. All other singular points $z_{k}=\frac{\pi}{2}+k \pi$ of the function $f(z)=\operatorname{tg} z$ lie outside the region $D$ and are the... | -4\pii | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,081 |
Example 4. Compute the integral
$$
\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
$$
To determine the nature... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,083 |
Example 6. Compute the integral
$$
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... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,084 |
Example 7. Compute the integral
$$
I=\int_{|z|=3} \frac{z^{17}}{\left(z^{2}+2\right)^{3}\left(z^{3}+3\right)^{4}} d z
$$ | Solution. The integrand function
$$
f(z)=\frac{z^{17}}{\left(z^{2}+2\right)^{3}\left(z^{3}+3\right)^{4}}
$$
inside the contour $|z|=3$ has five singular points, which are multiple poles. Using the main theorem of residues leads to extensive calculations. For the calculation of this integral, it is more convenient to ... | 2\pii | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,085 |
Example 8. Calculate the integral
$$
I=\int_{0}^{\infty} \frac{x^{2} d x}{\left(x^{2}+a^{2}\right)^{2}} \quad(a>0)
$$ | Solution. Since the integrand $f(x)=\frac{x^{2}}{\left(x^{2}+a^{2}\right)^{2}}$ is an even function,
$$
I=\frac{1}{2} \int_{-\infty}^{+\infty} \frac{x^{2} d x}{\left(x^{2}+a^{2}\right)^{2}}
$$
Introduce the function $f(z)=\frac{z^{2}}{\left(z^{2}+a^{2}\right)^{2}}$; which on the real axis, i.e., when $z=x$, coincides... | \frac{\pi}{4a} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,086 |
Example 9. Calculate the integral
$$
I=\int_{0}^{\infty} \frac{x \sin a x}{x^{2}+k^{2}} d x \quad(a>0, k>0)
$$ | Solution. Introduce the auxiliary function
$$
f(z)=\frac{z e^{i a z}}{z^{2}+k^{2}}
$$
It is easy to see that if $z=x$, then $\operatorname{Im} f(x)$ coincides with the integrand $\varphi(x)=\frac{x \sin a x}{x^{2}+k^{2}}$. Consider the contour shown in Fig. 7. For sufficiently large $R$, on the contour $C_{R}$, the f... | \frac{\pi}{2}e^{-} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,087 |
Example 10. Find the integral representation of the unit function (Heaviside function)
$$
f(t)=\left\{\begin{array}{lll}
0 & \text { for } \quad t<0 \\
1 & \text { for } \quad t>0
\end{array}\right.
$$ | Solution. Consider the function
$$
f(t)=\frac{1}{2 \pi i} \int_{C} \frac{e^{-i z t}}{z} d z
$$
where the contour $C$ is shown in Fig. 8.
By closing the contour with a semicircle $C_{R}$ in the upper half-plane, we notice that for $t > 0$, by Jordan's lemma, the integrals
$$
\int_{c_{:}^{\prime}} \frac{e^{-i z t}}{z... | f()=\frac{1}{2\pii}\int_{C}\frac{e^{-iz}}{z}=\begin{cases}0&\text{for}<0\\1&\text{for}>0\end{cases} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,088 |
Example 11. Calculate the integral
\[
\begin{gathered}
I=\int_{0}^{\infty} \frac{\sin a x}{x\left(x^{2}+b^{2}\right)} d x \\
(a>0, b>0)
\end{gathered}
\] | Solution. Let us introduce the function
Fig. 9
$$
f(z)=\frac{e^{i a z}}{z\left(z^{2}+b^{2}\right)}
$$
such that for $z=x$, $\operatorname{Im} f(z)$ coincides with the integrand in (6). Th... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,089 |
Example 12. Calculate the Fresnel integrals
$$
\begin{aligned}
& I_{1}=\int_{0}^{\infty} \cos x^{2} d x \\
& I_{2}=\int_{0}^{\infty} \sin x^{2} d x
\end{aligned}
$$
Fig. 10
$$
\int_{0}^{\... | Solution. Consider the auxiliary function $f(z)=e^{i z^{2}}$ and the contour shown in Fig. 10 (a circular sector $O B A O$, where $O A=O B=R$ and $\angle B O A=\frac{\pi}{4}$). Inside this contour, $f(z)$ is analytic, and by Cauchy's theorem,
$$
\int_{O B A O} e^{i z^{2}} d z=\int_{0}^{R} e^{i z^{2}} d x+\int_{C_{R}} ... | \int_{0}^{\infty}\cosx^{2}=\frac{1}{2}\sqrt{\frac{\pi}{2}},\quad\int_{0}^{\infty}\sinx^{2}=\frac{1}{2}\sqrt{\frac{\pi}{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,090 |
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