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Current time:0:00Total duration:6:25

in the last video we introduced the idea that we could represent any arbitrary periodic function by a series of weighted cosines and sines and what I'm going to start doing in this video is starting to establish our mathematical foundation so it'll be pretty straightforward for us to find these coefficients that give us that function so the first thing I want to do the first thing I'm going to do is establish some truths using you or some truths with the definite integrals I'm going to focus over the interval 0 to 2pi over this video in the next few videos because the function we're approximating has a period of 2pi it completes one cycle from 0 to 2pi we could have done it over other intervals of length 2 pi and if this period was other than 2pi we would have done it over intervals of that period but I'm focusing on 2pi because it makes the math a little bit cleaner and a little bit simpler and then we can generalize in the future so let's just establish some things about integrated integrals of trig functions so the first thing I want to establish I want to establish that the definite integral from 0 to 2pi of sine of M X DX actually let me stay in T since we're since our original function is in terms of T sine of M T DT I want to establish that that is equal to 0 for any non-zero integer M for non non zero nonzero integer integer M and I also want to establish that the integral from 0 to 2pi of cosine of M T DT is equal to 0 for any nonzero integer M and you might already take this for granted or you feel good about it or you've already proven it to yourself and if so you could actually skip this video but let's work through it because it's it's actually a good review of some integral calculus here so let's first do this this top one so let me just rewrite the integral we're gonna take the integral from 0 to 2pi of sine M T DT now we know we want to take the antiderivative sine of M T so we know that the we know that the derivative with respect to T of cosine M T is equal to what is this this is going to be equal to M the derivative of M T with respect to T times the derivative of cosine M T with respect to M T so times negative sine of M T or we could write this is going to be equal to negative M sine of M T I could put a parenthesis there if I like and so I almost have negative M sine of empty I just don't have a negative M here so what if I put a negative M there but I can't just do that that would change the value of the expression but I could I could also multiply by negative 1 over m now these two would if we take the product they're going to cancel out it we're going to get our original expression but this is useful because now we can say this is equal to negative 1 over m and now the antiderivative of this business right over here we know is cosine M T so it's going to be cosine M T cosine M T evaluated at 2 pi and 0 2 pi and 0 and so this is going to be equal to this is equal to negative 1 over m times cosine of M times 2 pi let me write this way cosine cosine of M times 2 pi minus cosine of well it's going to be M times 0 which is just going to be we could just write that as 0 and so let's see cosine of any multiple of 2 pi well that's just going to be 1 and cosine of 0 well that's also 1 so you have 1 minus 1 is 0 times negative 1 over M well this is all going to evaluate to this is the result we wanted this is all going to evaluate to 0 so we have just proven that first statement so now let's prove the second one it's going to be a very similar argument so let's rewrite it we're going to get the integral from 0 to 2pi of cosine of M T DT and now let me engineer this a little bit we know that the derivative of sine of M T is M cosine M T so let me multiply and divide by M multiplied by M and divide by M not changing the actual value and so this is going to be equal to 1 over m and then the antiderivative of we find a nice color here the antiderivative of that right over there let me see I should say the antiderivative of this right over here is sine of M T so sine of M T notice the derivative of sine of M T is M cosine empty and we're going to evaluate that at from 0 to 2pi so this is going to be equal to we still have our 1 over m out front 1 over m and so this is going to be sine of sine of M times 2 pi or we could say 2 pi M 2 pi M minus sine of M times 0 so sine I'm going to write that as sine of 0 and what's the sine of any multiple of 2 pi remember M is a nonzero integer so any multiple it's going to be a multiple of 2 pi here well that's just going to be 0 and sine of 0 is just going to be 0 so this whole thing this whole thing is just going to be 0 and so we have established our second statement there so this is this is going to be a nice base to to build from and now we're going to we're going to do slightly more complex integrals in the next few videos so that it's going to be hopefully pretty straightforward to find our Fourier coefficients with a little bit of calculus and algebraic manipulation