2

1 Find the Fourier series of flx) = e-x cos X,-T<x Stt its periodic with period ZTt: With help of this formula: ax eax cOS bx dx (a cos bx + b sin bx 1 a2 + b2 a...

Question

1 Find the Fourier series of flx) = e-x cos X,-T<x Stt its periodic with period ZTt: With help of this formula: ax eax cOS bx dx (a cos bx + b sin bx 1 a2 + b2 ax eax sin bx dx (a sin bx b cos bx ) a2 + b2cos X cos nx(cos(1 +n) x + cos(1 _ n)x) 2cos x sin nx(sin(1 + n) x - sin(1 _ n) x) 2

1 Find the Fourier series of flx) = e-x cos X,-T<x Stt its periodic with period ZTt: With help of this formula: ax eax cOS bx dx (a cos bx + b sin bx 1 a2 + b2 ax eax sin bx dx (a sin bx b cos bx ) a2 + b2 cos X cos nx (cos(1 +n) x + cos(1 _ n)x) 2 cos x sin nx (sin(1 + n) x - sin(1 _ n) x) 2



Answers

A finite Fourier series is given by the sum
$$ f(x) = \sum_{n =1}^{N} a_n \sin nx $$
$$ f(x) = a_1 \sin x + a_2 \sin 2x + \cdots + a_N \sin Nx $$








Show that the $ m $ th coefficient am is given by the formula

$$ a_m = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin mx dx $$

Okay. What we want to do is we want um to show um multiple things and we're gonna start with what is called a finite four year series um is given by F of X. Is equal to the summation of S. A. V. Sign of I. Of X is equal to a one sign X plus A to sign of two X plus a three sign of three X plus. And it goes on to a seven is equal to sign in fx. And the first thing we want to do is to show that the inthe coefficient is going to be given by one of the pie um times the integral from pi to negative pi of F of X. Sine of X Dx. So that in um coefficient is given by that integration. Okay, so what we want to start with is that we know that F of X is equal to this summation, right? Um And so what we want to start with is that F of X times Sine of Innovex is going to be equal to this summation. And I guess this should be an end here, upper case of this eighth of I sign of I've X. And then this is going to be times a sign of Innovex which is going to be um a one sine of X. Times sine of Innovex plus a two sign of two vax times Sine of Innovex plus dot dot dot. We have um a seven sign of Innovex times another sign of innovex. Okay. And we also know so this then will be um now will be actually equal to the integral. And so we're gonna kind of do the integral. So this will be if we do the integral from negative pi two pi of fx sign of inner backs D. X. And so this is going to be um the integral from negative pi to pie of this summation times. This sign Innovex dx. Which we then will have these individual and the first one would be this um whoops. Um this first one would be kind of this ace of one. Um Sine of X times sine of interfax dx plus. The integral from negative pi, the pie of sf two sign of two X sign of the backs dx plus. And then we can keep going and then we have plus and then of course this is going to be a sine squared. The integral from negative pika pi of ace event sine squared of innovex DX. Um and we know we know previously um we know that the integral from negative pi to pie of sign of M. X. Times sine of in the next D. X. Is equal to zero for em not equal to end. So all of these, all of these, we'll go to zero. The only one that is not going to go to zero is this last one right here. So what we have is the integral from negative pi two pi of F of X times sine in of X Dx is actually going to be equal to the integral from negative hi two pi of aces in which I actually can pull him out. So let's go ahead and pull him out of the integral of sine squared of X. Dx. Okay, so now all I have to do is find that integral. That's the only thing I have to now do. And so we're actually gonna be using a trig identity. I think it's gonna be a double angle formula I think is what we're gonna do is that sign squared of innovex is actually going to be equal to one half and then it's going to be a minus co sign of two innovex over two. So we're going to be actually using this identity to help us out. And so this will actually be equal to um ace of end times. Um And let me do over two because I'm gonna pull pull that two out of there. Um And then it's going to be um the integral from negative pi to pie of D. X minus. The integral from negative pie to negative two pi of co sign of two innovex D. X. And so this will be equal to a 7/2 times X. And we're gonna evaluate that at pie and negative pi. And then of course this guy um actually if I do use of on and let you be too in of X. Then this becomes um sign of to end of X divided by two in. And we're gonna evaluate at upper limit of pie and the lower limit of negative pi. And so this will actually become a sieve end of pie. And so that tells me now that if I divide both sides by pi a C van is one over pi times the integral from negative pi to pie of F of X. Sign of innovex T. X. And so I've gone ahead and prove that out and now what we're going to do is um we're actually gonna find some of the coefficients and we are actually going to let um now we're gonna let ethel Becks equal X. Can. Um And then we're gonna go ahead and find a one A two and a three. So so for a one we have one over pie integral from negative pi two pi F of X. Now is X and n is one. So this would be sign of eggs D. X. Um And of course we are going to do integration by parts on this. So we're gonna let U equal X. Do you use equal to dx T. V. Is equal to sign of X. D. X. Which means V. Is going to be equal to negative coastline of X. Okay. Um And so now we have this is going to be equal to one over pie. And you ve so this will be negative X. Co sign of X then plus it's actually a minus V. D. You but it's gonna be a plus. Um And we're gonna evaluate at pi and negative pi on this. The integral from negative pi day pie of co sign of xcX. And the integral of co sign of course is sine of X. Which sign of eggs That will all be zero. And so this will actually be equal to just a two. Okay. Um And then of course we do the same process for um A to this would be one over pie. Um The integral from negative pi to pie of X. Sign of two X. D. X. And we're gonna go through and do the exact same process. Um We're gonna go ahead and do immigration bar parts which kind of works out the same way. Um And then accept you just have a two in here. So you would have a two X. And then of course you would have a one half out here and this will be equal to negative one if you go through the same process and then we're going to do the exact same thing for a three. So this be one of the pie integral from negative pika pi of X. And now we're doing sign of three X. And you would do the exact same process. You would do the um you would actually do integration by parts using the same format and now you're just gonna have a one third stuck out there someplace. And so this will actually be equal to a two thirds.


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