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Fourier coefficients for sine terms

Video transcript

many videos ago we first looked at the idea of representing a periodic function as a set of weighted cosines and sines as a sum as the infinite sum of weighted cosines and sines and then we did some work in order to get some basics in terms of some of these integrals which we then started to use to derive formulas for the various coefficients and we are almost there we have figured out a formula for a sub-zero we've figured out a coefficient in general for any of the coefficients on the cosines and now let's figure out a way to figure out the coefficients on the sines so how can we do that what we're going to use a very similar technique that we did to figure out the formula for a sub n what we're going to do is multiply both sides of this equation right over here by sine of n T so let's do that so sine of NT and so if we multiply the right-hand side by sine of NT that means we're going to multiply every one of the infinite terms by sine of NT so sine of n T sine of n T I think you well this is always a little time-consuming but I'll get there sine of n T sine of n T sine of n T sine of n T sine of NT and of course there's these dot dots so we're just showing in general we're taking each of the each of these terms and multiplying by sine of NT and we'll do it for all of the terms we're going to keep on doing this on and on forever for an infinite number of terms well now let's take the integral of both sides of this equation the definite integral from 0 to 2pi and once again I'm picking that interval because 2pi is the period of the periodic function that we care about and everything else that I've been doing has been over that interval of 2pi so I'll just keep focusing it on that so from 0 to 2pi DT and I could say the integral of this whole thing but we know that the integral of a sum of a bunch is a is also the same things the sum of the integrals and we also know that we could take these Conte's constant multipliers of expressions outside of integrals and so this is going to be equal to if I take the integral of the right hand side the definite integral is going to be from 0 to 2 pi DT 0 to 2 pi DT 0 to 2 pi DT 0 to 2 pi DT we're almost there is 0 to 2 pi DT 0 to 2 pi DT and 0 to 2 pi DT and I would be doing this for every term in this Fourier in this Fourier expansion now this is where some of that integration work is going to be valuable we've already shown that sign of the the definite integral from 0 to 2 pi of sine of NT DT is going to be equal to 0 for n being any integer so we saw that we saw that right over there so that tells us that that term is going to be 0 so that's going to be equal to 0 and we also know that when we take a cosine of something times T times the sine of something times T where these are integers and you take that definite integral we know that's going to be 0 so all of those are going to be 0 and we also know that when you take the sine of two different non 2 different coefficients right over or sine of some integer coefficient times T times sine of some other integer coefficient times T that these are also going to be equal to 0 we saw that we saw that right over here that is this property and so really all you're left with the only one that does not become 0 is this term right over here where it's sine of NT times sine of NT and what is that going to be equal to well this inside of the integral that's the same thing let me do this in a different color since I'm using orange to cross that everything that ends up becoming zero so that is the same thing as the definitely go from 0 to 2 pi of sine squared of NT DT and I'll do the or the DT is still there so I'm just replacing the blue stuff with the sine squared of n T and we know what that's going to be we know that when the coefficient on the T is a nonzero integer that this is going to result in pi so we know that all of this stuff right over here let me do that in a different color all of we know that all of this stuff is going to evaluate actually me interested all of this stuff is going to become pi so we know that B sub n times pi is going to be equal to this definite integral or we could write let me write it this way B sub n times pi is equal to because everything else ends up becoming zero is going to be equal to zero the definite integral from zero to 2pi I'll put my DT out here of F of T F of T times sine of NT sine of n T DT and so we could then divide both sides by divide both sides by PI and we get a little bit of a drumroll we get B sub N and actually I could write that I'll write that in that same blue color we get B sub n I'll do it actually right here B sub n is going to be equal to 1 over pi times the definite integral from 0 to 2 pi 0 to 2 pi over DT here you have your f of T here and then you have your sine sine and T so once again that was pretty straightforward now that we knew these these these properties of integrals we knew how to evaluate these integrals that are definite integrals where we're taking of cosine or sine or products of cosines and sines and using these three formulas we can now attempt to to find the Fourier expansion the Fourier series find the coefficients for our square wave