![]() So it's gonna be cosine mt, cosine mt, evaluated at two pi and Of this business right over here, we know is cosine mt. But this is useful because now we can say this is equal to negative one over m and now the antiderivative If we take the product they're gonna cancel out, we're gonna get our original expression. I could also multiply by negative one over m. That would change the value of the expression, but So what if I put a negative m there, but I can't just do that, Negative m sine of mt, I just wanna have a negative m here. Of mt, or we could write this is gonna be equal to negative m sine of mt, and I could put a The derivative of mt with respect to t times the derivative ofĬosine mt with respect to mt. Integral from zero to two pi of sine mt, dt, now we know we won't take the antiderivative of sine of mt, so we know that the derivative with respect to t ofĬosine mt, is equal to, what is this, this is So let's first do this top one, so let me just re-write the integral. It, 'cause it's actually a good review of some Or you've already proven it to yourself, and if so, youĬould actually skip this video. Take this for granted or you feel good about it, Of mt, dt, is equal to zero for any non-zero integer m. That the integral from zero to two pi of cosine For non-zero, non-zero integer, integer m, and I also wanna establish That that is equal to zero for any non-zero integer m. The definite integral from zero to two pi of sine of mx, dx, actually let me stay in t, since our original function is in terms of t. So the first I wanna establish, I wanna establish that So let's just establish some things about definite integrals of trig functions. Intervals of that period, but I'm focusing on two piīecause it makes the math a little bit cleaner,Ī little bit simpler, and then we can generalize in the future. Pi, completes one cycle from zero to two pi, weĬould've done it over other intervals of length two pi, and if this period was other than two pi, we would've done it over The intervals zero to pi over this video and the next few videos because the function we're approximating has a period of two Wanna do, the first thing I'm gonna do is establish ![]() Straightforward for us to find these coefficients that ![]() And what I'm gonna startĭoing in this video, is starting to establish Video we introduced the idea that we could represent anyĪrbitrary, periodic function by a series of weighted cosines and sines. ![]()
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