anonymous
  • anonymous
Help Me.... for fourier series
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
What;s the question?
anonymous
  • anonymous
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 \(0 < x < \pi\)
Jack1
  • Jack1
do you know what I mean by the terms: a 0 a n b n ...?

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anonymous
  • anonymous
yess..., of course.,
Jack1
  • Jack1
so what is your value for a 0?
Jack1
  • Jack1
do you need a hand to find a 0?
Jack1
  • Jack1
@gerryliyana you still around dude?
anonymous
  • anonymous
yes i'm still here.., i need to find a0, a1...b1....and till get fourier series..,
anonymous
  • anonymous
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
anonymous
  • anonymous
yes..., i'm looking for that..,
anonymous
  • anonymous
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.
anonymous
  • anonymous
yess.., i've done that..,
anonymous
  • anonymous
but i got wrong answer.., :(
anonymous
  • anonymous
Hi Please find attached .
1 Attachment
anonymous
  • anonymous
This should help to get a picture .. Can u proceed furthur bcoz its a long time i had studied fourier series :)
anonymous
  • anonymous
\[f(x)=\begin{cases}x+\pi&\text{for }-\pi
anonymous
  • anonymous
@Anu2401, are you saying that \(a_0=\dfrac{\pi}{2}\)? \[\int_{-\pi}^0f(x)~dx=-\int_0^\pi f(x)~dx\]
anonymous
  • anonymous
@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 .
anonymous
  • anonymous
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
anonymous
  • anonymous
@SithsAndGiggles Yes i am also saying that a0=0 How : [pi(Peak from -π to 0)+(-pi) (Peak from 0 to π]/2
anonymous
  • anonymous
@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 .
anonymous
  • anonymous
@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.
anonymous
  • anonymous
I got \[f(x) = \frac{ 1 }{ 2 } \pi + a_{1} \cos x + a_{2} \cos 2x + a_{3} \cos 3x .....+ b_{1} \sin x \] \[= \frac{ \pi }{ 2 } + \frac{ 4 }{ \pi } \left( \cos x + \frac{ \cos 3x }{ 9 } + \frac{ \cos 5x }{ 25 } + \frac{ \cos 3x }{ 49 }\right)\] \[+ 2 \left( \frac{ \sin 2x }{ 2 } + \frac{ \sin 4x }{ 4 } + \frac{ \sin 6x }{ 6 } + ...\right)\] \[-4 \left( \sin x + \frac{ \sin 3x }{ 3 } + \frac{ \sin 5x }{ 25 } + \frac{ \sin 7x }{ 7 } + .....\right)\]
anonymous
  • anonymous
how about you guys ???
anonymous
  • anonymous
@oldrin.bataku
anonymous
  • anonymous
@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
anonymous
  • anonymous
@SithsAndGiggles oh my bad..., thank you for correct me.., it's very clearly now.,

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