Szeregi Fouriera

Wskazówka

Binder <- interaktywna sesja notebooka

from random import choice

from IPython.display import display, Markdown, Latex

import generator_zadan.generatory as gz

print(gz.__version__)

0.2.11

ile_zadan_przykladowych = 10

zadanie = gz.szereg_Fouriera(
    typ_l=choice([0, 1, 2, 3, 4]),
    typ_p=choice([0, 1, 2, 3, 4]),
    bez_wykresu=True,
    tylko_koncowy=True)
Hide code cell source 
print("\033[34m** Zadanie **" + '*' * 81 + '\033[0m')
print(zadanie[0])
print("\033[34m*\033[0m" * 95)
print("\033[32m** Rozwiązanie **" + '*' * 78 + '\033[0m')
print(zadanie[1])
print("\033[32m*\033[0m" * 95)

** Zadanie ***********************************************************************************
Rozwinąć w szereg Fouriera funkcję 
	\[
		f(x)=\left\{\begin{matrix}
			- x - 1 & \textnormal{ dla } & x\in\left(- \frac{\pi}{2},0\right]\\
			- \cos{\left(x \right)} & \textnormal{ dla } & x\in\left(0,\frac{\pi}{2}\right]
		\end{matrix}\right.
	\]
o okresie zasadniczym $2T=\pi.$ Naszkicować wykres funkcji, do której zbieżny jest uzyskany szereg.
***********************************************************************************************
** Rozwiązanie ********************************************************************************
$a_0=\frac{2 \left(- \frac{\pi}{2} - 1 + \frac{\pi^{2}}{8}\right)}{\pi},\quad a_n=\frac{2 \left(8 \left(-1\right)^{n} n^{2} - \left(-1\right)^{n} - 4 n^{2} + 1\right)}{\pi \left(16 n^{4} - 4 n^{2}\right)},\quad b_n=\frac{- 8 n^{2} + \left(4 n^{2} - 1\right) \left(- 2 \left(-1\right)^{n} + \left(-1\right)^{n} \pi + 2\right)}{2 \pi n \left(4 n^{2} - 1\right)},$\\
	$S(x) = \frac{\left(4 - 3 \pi\right) \sin{\left(2 x \right)}}{6 \pi} + \frac{\left(-32 + 15 \pi\right) \sin{\left(4 x \right)}}{60 \pi} + \frac{\left(68 - 35 \pi\right) \sin{\left(6 x \right)}}{210 \pi} - \frac{5 \cos{\left(2 x \right)}}{3 \pi} + \frac{2 \cos{\left(4 x \right)}}{15 \pi} - \frac{53 \cos{\left(6 x \right)}}{315 \pi} + \frac{- \frac{\pi}{2} - 1 + \frac{\pi^{2}}{8}}{\pi} + \dots $\\
	\includegraphics[width = 224pt]{../pics/szereg_Fouriera_1_funkcja}
	\includegraphics[width = 224pt]{../pics/szereg_Fouriera_1_inf}

***********************************************************************************************
Hide code cell source 
for i in range(1, ile_zadan_przykladowych + 1):
    zadanie = gz.szereg_Fouriera(
        typ_l=choice([0, 1, 2, 3, 4]),
        typ_p=choice([0, 1, 2, 3, 4]),
        bez_wykresu=True,
        tylko_koncowy=True,
        nr_zadania=i)
    print(f"\033[34m** Zadanie {i} **" + '*' * 80 + '\033[0m')
    display(Markdown(zadanie[0].split('\n')[0]))
    display(Latex('$$' + zadanie[0].split('\[')[1].split('\]')[0] + '$$'))
    display(Markdown(zadanie[0].split('\n')[7].replace('$', '$$')))
    print("\033[34m*\033[0m" * 95)
    print("\033[32m** Rozwiązanie **" + '*' * 78 + '\033[0m')
    display(Latex('$' + zadanie[1].split('$')[1] + '$'))
    display(Latex('$' + zadanie[1].split('$')[3] + '$'))
    display(Markdown(f'![](./pics//szereg_Fouriera_{i}_funkcja.png)'))
    display(Markdown(f'![](./pics//szereg_Fouriera_{i}_inf.png)'))
    print("\033[32m*\033[0m" * 95)

** Zadanie 1 **********************************************************************************

Rozwinąć w szereg Fouriera funkcję

\[\begin{split} f(x)=\left\{\begin{matrix} - \cos{\left(x \right)} & \textnormal{ dla } & x\in\left[- \pi,0\right)\\ - \cos{\left(x \right)} & \textnormal{ dla } & x\in\left[0,\pi\right) \end{matrix}\right. \end{split}\]

o okresie zasadniczym $$2T=2 \pi.$$ Naszkicować wykres funkcji, do której zbieżny jest uzyskany szereg.

***********************************************************************************************
** Rozwiązanie ********************************************************************************
\[\begin{split}a_0=0,\quad a_n=\frac{\begin{cases} 0 & \text{for}\: n > 1 \\- \pi & \text{otherwise} \end{cases}}{\pi},\quad b_n=0,\end{split}\]
\[S(x) = - \cos{\left(x \right)} + \dots \]

***********************************************************************************************

** Zadanie 2 **********************************************************************************

Rozwinąć w szereg Fouriera funkcję

\[\begin{split} f(x)=\left\{\begin{matrix} x^{2} & \textnormal{ dla } & x\in\left(- \frac{3 \pi}{2},0\right]\\ 0 & \textnormal{ dla } & x\in\left(0,\frac{3 \pi}{2}\right] \end{matrix}\right. \end{split}\]

o okresie zasadniczym $$2T=3 \pi.$$ Naszkicować wykres funkcji, do której zbieżny jest uzyskany szereg.

***********************************************************************************************
** Rozwiązanie ********************************************************************************
\[a_0=\frac{3 \pi^{2}}{4},\quad a_n=\frac{9 \left(-1\right)^{n}}{2 n^{2}},\quad b_n=\frac{9 \left(\left(-1\right)^{n} \pi^{2} n^{2} - 2 \left(-1\right)^{n} + 2\right)}{4 \pi n^{3}},\]
\[S(x) = \frac{9 \left(4 - \pi^{2}\right) \sin{\left(\frac{2 x}{3} \right)}}{4 \pi} + \frac{9 \pi \sin{\left(\frac{4 x}{3} \right)}}{8} + \frac{\left(4 - 9 \pi^{2}\right) \sin{\left(2 x \right)}}{12 \pi} - \frac{9 \cos{\left(\frac{2 x}{3} \right)}}{2} + \frac{9 \cos{\left(\frac{4 x}{3} \right)}}{8} - \frac{\cos{\left(2 x \right)}}{2} + \frac{3 \pi^{2}}{8} + \dots \]

***********************************************************************************************

** Zadanie 3 **********************************************************************************

Rozwinąć w szereg Fouriera funkcję

\[\begin{split} f(x)=\left\{\begin{matrix} x & \textnormal{ dla } & x\in\left[-1,0\right)\\ 0 & \textnormal{ dla } & x\in\left[0,1\right) \end{matrix}\right. \end{split}\]

o okresie zasadniczym $$2T=2.$$ Naszkicować wykres funkcji, do której zbieżny jest uzyskany szereg.

***********************************************************************************************
** Rozwiązanie ********************************************************************************
\[a_0=- \frac{1}{2},\quad a_n=\frac{1 - \left(-1\right)^{n}}{\pi^{2} n^{2}},\quad b_n=\frac{\left(-1\right)^{n + 1}}{\pi n},\]
\[S(x) = \frac{\sin{\left(\pi x \right)}}{\pi} - \frac{\sin{\left(2 \pi x \right)}}{2 \pi} + \frac{\sin{\left(3 \pi x \right)}}{3 \pi} + \frac{2 \cos{\left(\pi x \right)}}{\pi^{2}} + \frac{2 \cos{\left(3 \pi x \right)}}{9 \pi^{2}} - \frac{1}{4} + \dots \]

***********************************************************************************************

** Zadanie 4 **********************************************************************************

Rozwinąć w szereg Fouriera funkcję

\[\begin{split} f(x)=\left\{\begin{matrix} - x^{2} & \textnormal{ dla } & x\in\left(- \frac{\pi}{2},0\right]\\ 2 & \textnormal{ dla } & x\in\left(0,\frac{\pi}{2}\right] \end{matrix}\right. \end{split}\]

o okresie zasadniczym $$2T=\pi.$$ Naszkicować wykres funkcji, do której zbieżny jest uzyskany szereg.

***********************************************************************************************
** Rozwiązanie ********************************************************************************
\[a_0=\frac{2 \left(\pi - \frac{\pi^{3}}{24}\right)}{\pi},\quad a_n=- \frac{\left(-1\right)^{n}}{2 n^{2}},\quad b_n=\frac{2 \left(-1\right)^{n} + n^{2} \left(- \left(-1\right)^{n} \pi^{2} - 8 \left(-1\right)^{n} + 8\right) - 2}{4 \pi n^{3}},\]
\[S(x) = \frac{\left(\pi^{2} + 12\right) \sin{\left(2 x \right)}}{4 \pi} - \frac{\pi \sin{\left(4 x \right)}}{8} + \frac{\left(9 \pi^{2} + 140\right) \sin{\left(6 x \right)}}{108 \pi} + \frac{\cos{\left(2 x \right)}}{2} - \frac{\cos{\left(4 x \right)}}{8} + \frac{\cos{\left(6 x \right)}}{18} + \frac{\pi - \frac{\pi^{3}}{24}}{\pi} + \dots \]

***********************************************************************************************

** Zadanie 5 **********************************************************************************

Rozwinąć w szereg Fouriera funkcję

\[\begin{split} f(x)=\left\{\begin{matrix} x^{2} & \textnormal{ dla } & x\in\left[- \pi,0\right)\\ - \sin{\left(x \right)} & \textnormal{ dla } & x\in\left[0,\pi\right) \end{matrix}\right. \end{split}\]

o okresie zasadniczym $$2T=2 \pi.$$ Naszkicować wykres funkcji, do której zbieżny jest uzyskany szereg.

***********************************************************************************************
** Rozwiązanie ********************************************************************************
\[\begin{split}a_0=\frac{-2 + \frac{\pi^{3}}{3}}{\pi},\quad a_n=\frac{\begin{cases} \frac{2 \left(-1\right)^{n} \pi \left(n^{2} - 1\right) + n^{2} \left(\left(-1\right)^{n} + 1\right)}{n^{2} \left(n^{2} - 1\right)} & \text{for}\: n > 1 \\\frac{2 \left(-1\right)^{n} \pi}{n^{2}} & \text{otherwise} \end{cases}}{\pi},\quad b_n=\frac{\begin{cases} \frac{\left(-1\right)^{n} \pi^{2} n^{2} - 2 \left(-1\right)^{n} + 2}{n^{3}} & \text{for}\: n > 1 \\\frac{\left(-1\right)^{n} \pi^{2}}{n} - \frac{2 \left(-1\right)^{n}}{n^{3}} - \frac{\pi}{2} + \frac{2}{n^{3}} & \text{otherwise} \end{cases}}{\pi},\end{split}\]
\[S(x) = \frac{\left(- \pi^{2} - \frac{\pi}{2} + 4\right) \sin{\left(x \right)}}{\pi} + \frac{\pi \sin{\left(2 x \right)}}{2} + \frac{\left(\frac{4}{27} - \frac{\pi^{2}}{3}\right) \sin{\left(3 x \right)}}{\pi} - 2 \cos{\left(x \right)} + \frac{\left(\frac{2}{3} + \frac{\pi}{2}\right) \cos{\left(2 x \right)}}{\pi} - \frac{2 \cos{\left(3 x \right)}}{9} + \frac{-2 + \frac{\pi^{3}}{3}}{2 \pi} + \dots \]

***********************************************************************************************

** Zadanie 6 **********************************************************************************

Rozwinąć w szereg Fouriera funkcję

\[\begin{split} f(x)=\left\{\begin{matrix} \cos{\left(x \right)} & \textnormal{ dla } & x\in\left(- \frac{3 \pi}{2},0\right]\\ - x^{2} & \textnormal{ dla } & x\in\left(0,\frac{3 \pi}{2}\right] \end{matrix}\right. \end{split}\]

o okresie zasadniczym $$2T=3 \pi.$$ Naszkicować wykres funkcji, do której zbieżny jest uzyskany szereg.

***********************************************************************************************
** Rozwiązanie ********************************************************************************
\[a_0=\frac{2 \left(- \frac{9 \pi^{3}}{8} - 1\right)}{3 \pi},\quad a_n=\frac{2 \left(-1\right)^{n} \left(9 n^{2} - \frac{27 \pi \left(4 n^{2} - 9\right)}{4}\right)}{3 \pi n^{2} \left(4 n^{2} - 9\right)},\quad b_n=\frac{9 \left(-1\right)^{n} \pi^{2} n^{2} \left(4 n^{2} - 9\right) - 16 n^{4} + 18 \left(1 - \left(-1\right)^{n}\right) \left(4 n^{2} - 9\right)}{4 \pi n^{3} \left(4 n^{2} - 9\right)},\]
\[S(x) = - \frac{\left(-196 + 45 \pi^{2}\right) \sin{\left(\frac{2 x}{3} \right)}}{20 \pi} + \frac{\left(-256 + 252 \pi^{2}\right) \sin{\left(\frac{4 x}{3} \right)}}{224 \pi} + \frac{\left(- 2187 \pi^{2} - 324\right) \sin{\left(2 x \right)}}{2916 \pi} + \frac{2 \left(9 + \frac{135 \pi}{4}\right) \cos{\left(\frac{2 x}{3} \right)}}{15 \pi} + \frac{\left(36 - \frac{189 \pi}{4}\right) \cos{\left(\frac{4 x}{3} \right)}}{42 \pi} - \frac{2 \left(81 - \frac{729 \pi}{4}\right) \cos{\left(2 x \right)}}{729 \pi} + \frac{- \frac{9 \pi^{3}}{8} - 1}{3 \pi} + \dots \]

***********************************************************************************************

** Zadanie 7 **********************************************************************************

Rozwinąć w szereg Fouriera funkcję

\[\begin{split} f(x)=\left\{\begin{matrix} \sin{\left(x \right)} & \textnormal{ dla } & x\in\left(-1,0\right]\\ x^{2} & \textnormal{ dla } & x\in\left(0,1\right] \end{matrix}\right. \end{split}\]

o okresie zasadniczym $$2T=2.$$ Naszkicować wykres funkcji, do której zbieżny jest uzyskany szereg.

***********************************************************************************************
** Rozwiązanie ********************************************************************************
\[a_0=- \frac{2}{3} + \cos{\left(1 \right)},\quad a_n=\frac{2 \left(-1\right)^{n} \left(\pi^{2} n^{2} - 1\right) + \pi^{2} n^{2} \left(\left(-1\right)^{n + 1} \cos{\left(1 \right)} + 1\right)}{\pi^{2} n^{2} \left(\pi^{2} n^{2} - 1\right)},\quad b_n=\frac{\left(-1\right)^{n + 1} \pi^{4} n^{4} \sin{\left(1 \right)} + \left(-1\right)^{n + 1} \pi^{2} n^{2} \left(\pi^{2} n^{2} - 1\right) + 2 \left(\left(-1\right)^{n} - 1\right) \left(\pi^{2} n^{2} - 1\right)}{\pi^{3} n^{3} \left(\pi^{2} n^{2} - 1\right)},\]
\[S(x) = \frac{\left(- 4 \pi^{2} + 4 + \pi^{4} \sin{\left(1 \right)} + \pi^{2} \left(-1 + \pi^{2}\right)\right) \sin{\left(\pi x \right)}}{\pi^{3} \left(-1 + \pi^{2}\right)} + \frac{\left(- 4 \pi^{2} \left(-1 + 4 \pi^{2}\right) - 16 \pi^{4} \sin{\left(1 \right)}\right) \sin{\left(2 \pi x \right)}}{8 \pi^{3} \left(-1 + 4 \pi^{2}\right)} + \frac{\left(- 36 \pi^{2} + 4 + 81 \pi^{4} \sin{\left(1 \right)} + 9 \pi^{2} \left(-1 + 9 \pi^{2}\right)\right) \sin{\left(3 \pi x \right)}}{27 \pi^{3} \left(-1 + 9 \pi^{2}\right)} + \frac{\left(- 2 \pi^{2} + 2 + \pi^{2} \left(\cos{\left(1 \right)} + 1\right)\right) \cos{\left(\pi x \right)}}{\pi^{2} \left(-1 + \pi^{2}\right)} + \frac{\left(-2 + 4 \pi^{2} \left(1 - \cos{\left(1 \right)}\right) + 8 \pi^{2}\right) \cos{\left(2 \pi x \right)}}{4 \pi^{2} \left(-1 + 4 \pi^{2}\right)} + \frac{\left(- 18 \pi^{2} + 2 + 9 \pi^{2} \left(\cos{\left(1 \right)} + 1\right)\right) \cos{\left(3 \pi x \right)}}{9 \pi^{2} \left(-1 + 9 \pi^{2}\right)} - \frac{1}{3} + \frac{\cos{\left(1 \right)}}{2} + \dots \]

***********************************************************************************************

** Zadanie 8 **********************************************************************************

Rozwinąć w szereg Fouriera funkcję

\[\begin{split} f(x)=\left\{\begin{matrix} - x^{2} & \textnormal{ dla } & x\in\left[-2,0\right)\\ 2 & \textnormal{ dla } & x\in\left[0,2\right) \end{matrix}\right. \end{split}\]

o okresie zasadniczym $$2T=4.$$ Naszkicować wykres funkcji, do której zbieżny jest uzyskany szereg.

***********************************************************************************************
** Rozwiązanie ********************************************************************************
\[a_0=\frac{2}{3},\quad a_n=- \frac{8 \left(-1\right)^{n}}{\pi^{2} n^{2}},\quad b_n=\frac{2 \left(4 \left(-1\right)^{n} + \pi^{2} n^{2} \left(1 - 3 \left(-1\right)^{n}\right) - 4\right)}{\pi^{3} n^{3}},\]
\[S(x) = \frac{2 \left(-8 + 4 \pi^{2}\right) \sin{\left(\frac{\pi x}{2} \right)}}{\pi^{3}} - \frac{2 \sin{\left(\pi x \right)}}{\pi} + \frac{2 \left(-8 + 36 \pi^{2}\right) \sin{\left(\frac{3 \pi x}{2} \right)}}{27 \pi^{3}} + \frac{8 \cos{\left(\frac{\pi x}{2} \right)}}{\pi^{2}} - \frac{2 \cos{\left(\pi x \right)}}{\pi^{2}} + \frac{8 \cos{\left(\frac{3 \pi x}{2} \right)}}{9 \pi^{2}} + \frac{1}{3} + \dots \]

***********************************************************************************************

** Zadanie 9 **********************************************************************************

Rozwinąć w szereg Fouriera funkcję

\[\begin{split} f(x)=\left\{\begin{matrix} - x^{2} & \textnormal{ dla } & x\in\left(- \pi,0\right]\\ x^{2} & \textnormal{ dla } & x\in\left(0,\pi\right] \end{matrix}\right. \end{split}\]

o okresie zasadniczym $$2T=2 \pi.$$ Naszkicować wykres funkcji, do której zbieżny jest uzyskany szereg.

***********************************************************************************************
** Rozwiązanie ********************************************************************************
\[a_0=0,\quad a_n=0,\quad b_n=\frac{2 \left(- \left(-1\right)^{n} \pi^{2} n^{2} + 2 \left(-1\right)^{n} - 2\right)}{\pi n^{3}},\]
\[S(x) = \frac{2 \left(-4 + \pi^{2}\right) \sin{\left(x \right)}}{\pi} - \pi \sin{\left(2 x \right)} + \frac{2 \left(-4 + 9 \pi^{2}\right) \sin{\left(3 x \right)}}{27 \pi} - \frac{\pi \sin{\left(4 x \right)}}{2} + \frac{2 \left(-4 + 25 \pi^{2}\right) \sin{\left(5 x \right)}}{125 \pi} + \dots \]

***********************************************************************************************

** Zadanie 10 **********************************************************************************

Rozwinąć w szereg Fouriera funkcję

\[\begin{split} f(x)=\left\{\begin{matrix} 0 & \textnormal{ dla } & x\in\left(- \frac{\pi}{2},0\right)\\ - x^{2} & \textnormal{ dla } & x\in\left[0,\frac{\pi}{2}\right] \end{matrix}\right. \end{split}\]

o okresie zasadniczym $$2T=\pi.$$ Naszkicować wykres funkcji, do której zbieżny jest uzyskany szereg.

***********************************************************************************************
** Rozwiązanie ********************************************************************************
\[a_0=- \frac{\pi^{2}}{12},\quad a_n=- \frac{\left(-1\right)^{n}}{2 n^{2}},\quad b_n=\frac{\left(-1\right)^{n} \pi^{2} n^{2} - 2 \left(-1\right)^{n} + 2}{4 \pi n^{3}},\]
\[S(x) = \frac{\left(4 - \pi^{2}\right) \sin{\left(2 x \right)}}{4 \pi} + \frac{\pi \sin{\left(4 x \right)}}{8} + \frac{\left(4 - 9 \pi^{2}\right) \sin{\left(6 x \right)}}{108 \pi} + \frac{\cos{\left(2 x \right)}}{2} - \frac{\cos{\left(4 x \right)}}{8} + \frac{\cos{\left(6 x \right)}}{18} - \frac{\pi^{2}}{24} + \dots \]

***********************************************************************************************