Question

Help Me.... for fourier series

Answer 1

What;s the question?

Answer 2

Find the fourier series for the periodic function

\(f \left( x \right) = x + \pi \) with interval \(-\pi < x < 0\)

\(f \left( x \right) = -x \) with interval \(0 < x < \pi\)

Answer 3

do you know what I mean by the terms:
a 0
a n
b n
...?

Answer 4

yess..., of course.,

Answer 5

so what is your value for a 0?

Answer 6

do you need a hand to find a 0?

Answer 7

@gerryliyana you still around dude?

Answer 8

yes i'm still here.., i need to find a0, a1...b1....and till get fourier series..,

Answer 9

Are you looking for the exponential Fourier Series or the Trigonometric? I see a0,a1,..,b1,... , so I suppose you are looking for the Trigonometric

Answer 10

yes..., i'm looking for that..,

Answer 11

You are going to use the formulas for the a's and b's and break each integral in two parts. One integral from -π to 0 plus one integral from 0 to π. Then you will replace f(x) with its value in each integral. In the first it would be x+π, in the second it will be -x.

Answer 12

yess.., i've done that..,

Answer 13

but i got wrong answer.., :(

Answer 14

Hi Please find attached .

Answer 15

This should help to get a picture .. Can u proceed furthur bcoz its a long time i had studied fourier series :)

Answer 16

\[f(x)=\begin{cases}x+\pi&\text{for }-\pi<x<0\\-x&\text{for }0<x<\pi\end{cases}\\
f(x) \sim a_0+\sum_{n=1}^\infty (a_n\cos(nx)+b_n\sin(nx))\]
\[\begin{align*}a_0&=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)~dx\\
&=\frac{1}{2\pi}\bigg[\int_{-\pi}^0(x+\pi)~dx+\int_0^\pi(-x)~dx\bigg]\\\\\\\\\\

a_n&=\frac{1}{\pi}\int_{-\pi}^\pi f(x)~\cos(nx)~dx\\
&=\frac{1}{\pi}\bigg[\int_{-\pi}^0(x+\pi)\cos(nx)~dx+\int_0^\pi(-x)\cos(nx)~dx\bigg]\\\\\\\\\\

b_n&=\frac{1}{\pi}\int_{-\pi}^\pi f(x)~\sin(nx)~dx\\
&=\frac{1}{\pi}\bigg[\int_{-\pi}^0(x+\pi)\sin(nx)~dx+\int_0^\pi(-x)\cos(nx)~dx\bigg]\\\\\\\\\\

\end{align*}\]

Answer 17

@Anu2401, are you saying that \(a_0=\dfrac{\pi}{2}\)?
\[\int_{-\pi}^0f(x)~dx=-\int_0^\pi f(x)~dx\]

Answer 18

@SithsAndGiggles :I guess there is some mistake in a0 . It should be 0. Thanks for pointing out
Actually in place of 0 it should be -pi becoz in general a0= Average value of the peak to peak at origin.
@gerryliyana Please have a look . & update .

Answer 19

You're welcome @Anu2401 :), but no, it's still 0: http://www.wolframalpha.com/input/?i=Average+value+of+Piecewise%5B%7B%7Bx%2Bpi%2C-pi%3Cx%3C0%7D%2C%7B-x%2C0%3Cx%3Cpi%7D%7D%5D+over+%5B-pi%2Cpi%5D

Answer 20

@SithsAndGiggles Yes i am also saying that a0=0
How : [pi(Peak from -π to 0)+(-pi) (Peak from 0 to π]/2

Answer 21

@SithsAndGiggles In 95% of cases the average value @origin is a0 . U can deduce a0 by just looking at the graph, no integration is required .

Answer 22

@Anu2401, oh I thought you were talking about \(a_0\) previously. By the way, I was under the impression that \(a_0\) \(is\) the average value.

Answer 23

I got

\[f(x) = \frac{ 1 }{ 2 } \pi + a_{1} \cos x + a_{2} \cos 2x + a_{3} \cos 3x .....+ b_{1} \sin x \]
\[= \frac{ \pi }{ 2 } + \frac{ 4 }{ \pi } \left( \cos x + \frac{ \cos 3x }{ 9 } + \frac{ \cos 5x }{ 25 } + \frac{ \cos 3x }{ 49 }\right)\]
\[+ 2 \left( \frac{ \sin 2x }{ 2 } + \frac{ \sin 4x }{ 4 } + \frac{ \sin 6x }{ 6 } + ...\right)\]
\[-4 \left( \sin x + \frac{ \sin 3x }{ 3 } + \frac{ \sin 5x }{ 25 } + \frac{ \sin 7x }{ 7 } + .....\right)\]

Answer 24

how about you guys ???

Answer 25

@oldrin.bataku

Answer 26

@gerryliyana, \(a_0\) should be 0.

For \(a_n\) and \(b_n\), I get
\[\begin{align*}a_n&=\frac{1}{\pi}\left(\frac{1-\cos(n\pi)}{n^2}-\frac{n\pi \sin(n\pi)+\cos(n\pi)-1}{n^2}\right)\\
&=\frac{1}{\pi}\left(\frac{2-2\cos(n\pi)-n\pi \sin(n\pi)}{n^2}\right)\\\\\\\\
b_n&=\frac{1}{\pi}\left(\frac{\sin(n\pi)-n\pi}{n^2}-\frac{n\pi\cos(n\pi)-\sin(n\pi)}{n^2}\right)\\
&=\frac{1}{\pi}\left(\frac{2\sin(n\pi)-n\pi-n\pi\cos(n\pi)}{n^2}\right)
\end{align*}\]
So cosine and sine series that I get are, respectively,
\[\color{red}{\frac{1}{\pi}\sum_{n=1}^\infty\left(\frac{2-2\cos(n\pi)-n\pi \sin(n\pi)}{n^2}\right)\cos(nx)}\]
\[\color{blue}{\frac{1}{\pi}\sum_{n=1}^\infty \left(\frac{2\sin(n\pi)-n\pi-n\pi\cos(n\pi)}{n^2}\right)\sin(nx)}\]
Splitting up the sums for odd and even \(n\):
\[\color{red}{\frac{1}{\pi}\sum_{n=1}^\infty \left(\frac{2-2\cos(n\pi)-n\pi \sin(n\pi)}{n^2}\right)\cos(nx)}\\
=\frac{1}{\pi}\sum_{n=1}^\infty \left(\frac{2-2(-1)^n}{n^2}\right)\cos(nx)\\
=\frac{1}{\pi}\Bigg[\sum_{k=1}^\infty \left(\frac{2-2(-1)^{2k}}{(2k)^2}\right)\cos(2kx)+\sum_{k=0}^\infty \left(\frac{2-2(-1)^{2k+1}}{(2k+1)^2}\right)\cos((2k+1)x)\Bigg]\\
=\frac{1}{\pi}\Bigg[\sum_{k=1}^\infty \left(\frac{2-2}{(2k)^2}\right)\cos(2kx)+\sum_{k=0}^\infty \left(\frac{2-2(-1)}{(2k+1)^2}\right)\cos((2k+1)x)\Bigg]\\
=\frac{4}{\pi}\sum_{k=0}^\infty \frac{1}{(2k+1)^2}\cos((2k+1)x)\\
=\frac{4}{\pi}\left(\cos x+\frac{\cos(3x)}{9}+\frac{\cos(5x)}{25}\cdots\right)\]
\[\color{blue}{\frac{1}{\pi}\sum_{n=1}^\infty \left(\frac{2\sin(n\pi)-n\pi-n\pi\cos(n\pi)}{n^2}\right)\sin(nx)}\\
=\frac{1}{\pi}\sum_{n=1}^\infty \left(\frac{-n\pi-n\pi(-1)^n}{n^2}\right)\sin(nx)\\
=-\sum_{n=1}^\infty \left(\frac{1+(-1)^n}{n}\right)\sin(nx)\\
=-\Bigg[\sum_{k=1}^\infty \left(\frac{1+(-1)^{2k}}{2k}\right)\sin(2kx)+\sum_{k=0}^\infty \left(\frac{1+(-1)^{2k+1}}{2k+1}\right)\sin((2k+1)x)\Bigg]\\
=-\Bigg[\sum_{k=1}^\infty \left(\frac{1+1}{2k}\right)\sin(2kx)+\sum_{k=0}^\infty \left(\frac{1-1}{2k+1}\right)\sin((2k+1)x)\Bigg]\\
=-\sum_{k=1}^\infty \frac{1}{k}\sin(2kx)\\
=-\left(\sin(2x)+\frac{\sin(4x)}{2}+\frac{\sin(6x)}{3}+\cdots\right)\]
So,
\[f(x)\sim \frac{4}{\pi}\sum_{k=0}^\infty \frac{1}{(2k+1)^2}\cos((2k+1)x)-\sum_{k=1}^\infty \frac{1}{k}\sin(2kx)\]
Checking with WA, I've clearly made some mistake along the way... I think you're right about retaining the odd sine series, but for some reason mine disappears.

Here's your answer:
http://www.wolframalpha.com/input/?i=%28pi%2F2%29%2B%284%2Fpi%29*Sum%5BCos%5B%282k%2B1%29*x%5D%2F%282k%2B1%29%5E2%2C%7Bk%2C0%2C10%7D%5D+%2BSum%5BSin%5B2*k*x%5D%2F%284k%29%2C%7Bk%2C1%2C10%7D%5D-4*Sum%5BSin%5B%282k%2B1%29*x%5D%2F%282k%2B1%29%2C%7Bk%2C0%2C10%7D%5D

Here's your answer with \(a_0=0\):
http://www.wolframalpha.com/input/?i=%284%2Fpi%29*Sum%5BCos%5B%282k%2B1%29*x%5D%2F%282k%2B1%29%5E2%2C%7Bk%2C0%2C10%7D%5D+%2BSum%5BSin%5B2*k*x%5D%2F%284k%29%2C%7Bk%2C1%2C10%7D%5D-4*Sum%5BSin%5B%282k%2B1%29*x%5D%2F%282k%2B1%29%2C%7Bk%2C0%2C10%7D%5D

And here's my answer:
http://www.wolframalpha.com/input/?i=%284%2Fpi%29*Sum%5BCos%5B%282k%2B1%29*x%5D%2F%282k%2B1%29%5E2%2C%7Bk%2C0%2C10%7D%5D+-+Sum%5BSin%5B2*k*x%5D%2Fk%2C%7Bk%2C1%2C10%7D%5D

Answer 27

@SithsAndGiggles oh my bad..., thank you for correct me.., it's very clearly now.,