Question

\(f(x)\) as a sine series

Answer 1

\[
\qquad\text{\(f(x)\) as a sine series}
\begin{equation*} % f(x)
f(x)=1,\qquad0<x<\pi
\end{equation*}\]

Answer 2

\[\begin{equation*} % g(x)
g(x)=
\begin{cases}
1,&0<x<\pi\\
-1,&-\pi<x<0
\end{cases}\qquad\text{odd extension of \(f(x)\)}
\end{equation*}\]

\begin{equation*} % a_0, a_n
a_0=a_n=0\qquad\text{odd function}
\end{equation*}

\begin{align*} % b_n
b_n&=\frac1\pi\int\limits_{-\pi}^{\pi} g(x)\sin(n x)\,\text dx\\
&=\frac2\pi\int\limits_0^\pi \sin(n x)\,\text dx\qquad\text{even integrand}\\
&=\frac2\pi\left(\frac{-\cos(n x)}{n}\right)\Big|_0^\pi\\
&=-\frac2\pi\left(\frac{\cos(0)-\cos(n\pi )}{n}\right)\\
&=-\frac2{\pi}\left(\frac{1-(-1)^n}{n}\right)\\
\end{align*}

\begin{align*}
S(x)&=-\frac2\pi\sum\limits_{n=1}^\infty\left(\frac{1-(-1)^n}{n}\right)\sin(nx)\\
&=-\frac2\pi\sum\limits_{n=1,3,5,\dots}^\infty\left(\frac{2}{n}\right)\sin(nx)\\
&=-\frac4\pi\sum\limits_{r=1}^\infty\frac{\sin\big((2r-1)x\big)}{2r-1}
\end{align*}

Answer 3

just change L to pi
http://mathworld.wolfram.com/FourierSeriesSquareWave.html

Answer 4

you got the opposite one .. shift the period by +pi

Answer 5

where did i go wrong?

Answer 6

don't know .. don't have much time right now!! but i often use that link as reference!!

Answer 7

sin(nx) is not even .. get rid of that minus

Answer 8

i'll see later!!

Answer 9

why does (4) on that link have a sin^2 term?

Answer 10

ah i found my mistake

Answer 11

\[\begin{align*} % b_n
b_n&=\frac1\pi\int\limits_{-\pi}^{\pi} g(x)\sin(n x)\,\text dx\\
&=\frac2\pi\int\limits_0^\pi \sin(n x)\,\text dx\qquad\text{even integrand}\\
&=\frac2\pi\left(\frac{-\cos(n x)}{n}\right)\Big|_0^\pi\\
&=\frac2\pi\left(\frac{\cos(0)-\cos(n\pi )}{n}\right)\\
&=\frac2{\pi}\left(\frac{1-(-1)^n}{n}\right)\\
\end{align*}\]

Answer 12

\begin{align*}
S(x)&=\frac2\pi\sum\limits_{n=1}^\infty\left(\frac{1-(-1)^n}{n}\right)\sin(nx)\\
&=\frac4\pi\sum\limits_{n=1,3,5,\dots}^\infty\frac{\sin(nx)}{n}\\
&=\frac4\pi\sum\limits_{r=1}^\infty\frac{\sin\big((2r-1)x\big)}{2r-1}
\end{align*}

Answer 13

just a minus sign

Answer 14

guy please help me with this. This is impossible tough for me

Answer 15

i havent studies solving PDEs yet sorry @Echdip