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Numerade Educator



Problem 70 Hard Difficulty

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

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


=\frac{1}{\pi} \sum_{n=1}^{N} a_{n} \int_{-\pi}^{\pi} \sin n x \sin m x=\frac{1}{\pi} \times a_{m} \pi=a_{m}


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Video Transcript

a finite for yea. Siri's is given by the sun of a little end of a sign of an ex, where the ends range from one all the way up to a large and so an expanded form. Here's the find a four year Siri's for some function F, and we like to show that the empty coefficient AM is given by the following formula. So the key fact for this problem is comes from Problems sixty eight that the integral negative pion a pie sign annex signed M IX zero or pie, depending on whether or not memory can enter equal. So it's pie when m equals in and in general otherwise. So before we proceed, let's just make a few remarks about this identity and where it's coming from. So we had we broke this into two cases in problem sixty eight, so name was not not equal to him. We use the Formula Co sign a minus B minus cho sun of a plus B. So we use this formula to rewrite the end of rand and then we evaluated each term and answer grand. And in the other case in which eminent were equal, we wrote the anagram is sine squared mx and then using our pathetic and identity weaken right, this is one minus co sign to MX over to, and that makes it easier to integrate. So this is where this identity comes from and this is a key tool for this following problem. Okay, so going on to the next page, we have one over pi times the integral negative part of pie s events. Sina mix the X now using the formula that was given for us the finite for years. Some we could go on to replace f with it some Let's do that one over pi. Now if becomes there's some from once a capital in of a little inn Sign Little Anna Vicks. So this is your f f fx and we multiply this bye sign MX What? So for the next step, let's go ahead and pull out the summation time outside of the integral. That's one of the properties of the integral. We know that the integral of some is that some of the intervals. So here this is a finite sum, so weaken rate is a sum of intervals and in the end they're just constant so we could pull those outside of the integral sign. So here all we've done so far is just use some properties of it definition of the integral to rewrite this in a roll. And then now, from the previous problem, we know that this integral here, this is equal to So this is equal to pi when m equals in and zero otherwise. So when we take the sum of these on ly, one of these in a girls is going to be not zero and the only time as we let little and go from one so big, there's only one valued and that's going to make this integral zero. And that's precisely when little and happens to be m because this Emma's fixed. Where is this? M depends on the index. So we have times I am times pi because once again, for any little and other than m, we have a m. They're sorry am time zero So those all go to zero but in the one instance in which and equals m, we have a m times pi from our previous problem and this is exactly a M. And this proves the identity that we wanted to show We wanted to show that I am was one over pi integral negative pika pi of f of x times sine m x So this completes the proof.