2

0, Let f () ={ 2,~T < I < 0 0 < I < TThen the Fourier series of f (x) isA) 1+2 2 1-(C1) cosnx n=1B) 1 + 4 2, 1 sin nxC) 1+?='64"sin nD) 1 +4 1...

Question

0, Let f () ={ 2,~T < I < 0 0 < I < TThen the Fourier series of f (x) isA) 1+2 2 1-(C1) cosnx n=1B) 1 + 4 2, 1 sin nxC) 1+?='64"sin nD) 1 +4 1cosnx n=1

0, Let f () ={ 2, ~T < I < 0 0 < I < T Then the Fourier series of f (x) is A) 1+2 2 1-(C1) cosnx n=1 B) 1 + 4 2, 1 sin nx C) 1+?='64"sin n D) 1 +4 1cosnx n=1



Answers

Fourier Series The following sum is a finite Fourier series.
$f(x)=\sum_{i=1}^{N} a_{i} \sin i x$ $\quad=a_{1} \sin x+a_{2} \sin 2 x+a_{3} \sin 3 x+\cdots+a_{N} \sin N x$
(a) Use Exercise 89 to show that the $n$ th coefficient $a_{n}$ is given by $a_{n}=(1 / \pi) \int_{-\pi}^{\pi} f(x) \sin n x d x$
(b) Let $f(x)=x$ . Find $a_{1}, a_{2},$ and $a_{3}$

Ok folks. So um today we're going to be looking at the following question. So we're looking at a finite furious series of a function. So that is we're expressing the function in this form and we're trying to find a one A two A M. So we're trying to find the coefficient. Okay, so the way we're going to do this is we're going to use these facts. These facts are actually proven in another question. So we're not going to prove them here because that's not really what this question is all about. But we're going to make use of them. So then all we need to do is we multiply both sides of this equation by sign of mx. So let's do that. So we have sign of mx multiplied by F of X. And then that will equal to saw a one sign X sign mx plus A to sign two X. Sign mx all the way up until a n sign nx sign mx. Then we integrate both sides with respect. Then we integrate both sides from pirate negative pyre. So the pie negative pi of sign mx F X D X is equal to integral of pirate a negative pi A one sign X sign mx D x plus yada yada yada plus integral of pied a negative pi of a N sign nx sign mx Okay dx. And we're gonna rip out all the coefficients. Yeah. Sign mx D X. And and we have pie to negative pi of sign nx sign em X. Dx. Okay. So now we're going to use our identity that we have before. So if M is not equal to end, then we're going to have the integral zero. Whereas if M is and then this integral is going to be pie. So what that means is all these integral in the some are going to be zero except the one where except the one corresponding to AM. So that means we're going to have a M pi negative pi sign mx sign mx D X. But we know this whole thing is equal to just pie. So then we have the AM is equal to one on pie pi negative pi of sign of mx F of X dx. And that's it. Okay. And that's how we can get all the coefficients. Okay, thank you very much.


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