Question

Развить в ряд Фурье 2π-периодическую функцию на интервале (-π;π]

Answer 1

Пошаговое объяснение:

\begin{gathered}f(x)=6x-2\; ,\; \; (\pi ,\pi )a_0=\frac{1}{\pi }\int\limits^{\pi }_{-\pi }f(x)\, dx =\frac{1}{\pi } \int\limits^{\pi }_{-\pi }(6x-2)\, dx=\frac{(6x-2)^2}{\pi \cdot 6\cdot 2}\Big |_{-\pi }^{\pi }==\frac{1}{12\pi }\cdot ((6\pi -2)^2-(-6\pi -2)^2)=-\frac{4\pi }{\pi }=-4a_{n}= \frac{1}{\pi }\int\limits^{\pi }_{-\pi }f(x)\cdot cosnx\, dx=\frac{1}{\pi } \int\limits^{\pi }_{-\pi }(6x-2)cosxnx\, dx==[u=6x-2,\; du=6dx,\; dv=cosnx,\; v=\frac{1}{n}sinnx]=\end{gathered}

f(x)=6x−2,(π,π)

a

0

=

π

1

−π

∫

π

f(x)dx=

π

1

−π

∫

π

(6x−2)dx=

π⋅6⋅2

(6x−2)

2

∣

∣

∣

∣

−π

π

=

=

12π

1

⋅((6π−2)

2

−(−6π−2)

2

)=−

π

4π

=−4

a

n

=

π

1

−π

∫

π

f(x)⋅cosnxdx=

π

1

−π

∫

π

(6x−2)cosxnxdx=

=[u=6x−2,du=6dx,dv=cosnx,v=

n

1

sinnx]=

\begin{gathered}=\frac{1}{\pi }\Big (\frac{6x-2}{n}\cdot sin\, nx\Big |_{-\pi }^{\pi }-\frac{6}{n}\int\limits^{\pi }_{-\pi }sin\, nx\, dx\Big )==\frac{1}{\pi }\cdot \Big (0+\frac{6}{n^2}\cdot cos\, nx\Big |_{-\pi }^{\pi }\Big )=\frac{1}{\pi }\cdot \frac{6}{n^2}\cdot \Big (cos\pi n-cos(-\pi n)\Big )=0 \end{gathered}

=

π

1

(

n

6x−2

⋅sinnx

∣

∣

∣

∣

−π

π

−

n

6

−π

∫

π

sinnxdx)=

=

π

1

⋅(0+

n

2

6

⋅cosnx

∣

∣

∣

∣

−π

π

)=

π

1

⋅

n

2

6

⋅(cosπn−cos(−πn))=0

\begin{gathered}b_{n}=\frac{1}{\pi }\int\limits^{\pi }_{-\pi }f(x)sin\, nx\, dx=\frac{1}{\pi }\int\limits^{\pi }_{-\pi }(6x-2)\cdot sin\, nx\, dx==[u=6x-2,\; du=6dx,\; dv=sin\, nxdx,\; v=-\frac{1}{n}cos\, nx]==\frac{1}{\pi }\cdot \Big (-\frac{6x-2}{n}cos\, nx\Big |_{-\pi }^{\pi }+\frac{6}{n}\int\limits^{\pi }_{-\pi }cos\, nx\, dx\Big )==-\frac{1}{\pi n}\cdot \Big ((6\pi -2)\cdot cos\, \pi n-(-6\pi -2)\cdot cos(-\pi n)\Big )++\frac{6}{\pi n^2}\cdot sin\, nx\Big |_{-\pi }^{\pi }==-\frac{1}{\pi n}\cdot \Big ((6\pi -2)\cdot (-1)^{n}+(6\pi +2)\cdot (-1)^{n}\Big )+0=\end{gathered}

b

n

=

π

1

−π

∫

π

f(x)sinnxdx=

π

1

−π

∫

π

(6x−2)⋅sinnxdx=

=[u=6x−2,du=6dx,dv=sinnxdx,v=−

n

1

cosnx]=

=

π

1

⋅(−

n

6x−2

cosnx

∣

∣

∣

∣

−π

π

+

n

6

−π

∫

π

cosnxdx)=

=−

πn

1

⋅((6π−2)⋅cosπn−(−6π−2)⋅cos(−πn))+

+

πn

2

6

⋅sinnx

∣

∣

∣

∣

−π

π

=

=−

πn

1

⋅((6π−2)⋅(−1)

n

+(6π+2)⋅(−1)

n

)+0=

\begin{gathered}= \frac{(-1)^{n}}{\pi n}\cdot (6\pi -2+6\pi +2)=\frac{(-1)^{n}\cdot 12\pi }{\pi n}=\frac{(-1)^{n}\cdot 12}{n}f(x)\sim \frac{a_0}{2}+\sum\limits _{n=1}^{\infty }\Big (a_{n}\cdot cos\, nx+b_{n}\cdot sin\, nx\Big )f(x)=-4+\sum \limits _{n=1}^{\infty }\Big (\frac{(-1)^{n}\cdot 12}{n}\cdot sin\, nx\Big )= -4+12\sum \limits _{n=1}^{\infty }\frac{(-1)^{n}sin\, nx}{n}\end{gathered}

=

πn

(−1)

n

⋅(6π−2+6π+2)=

πn

(−1)

n

⋅12π

=

n

(−1)

n

⋅12

f(x)∼

2

a

0

+

n=1

∑

∞

(a

n

⋅cosnx+b

n

⋅sinnx)

f(x)=−4+

n=1

∑

∞

(

n

(−1)

n

⋅12

⋅sinnx)=−4+12

n=1

∑

∞

n

(−1)

n

sinnx