Formula for first term in Fourier Series

Several videos ago, we introduced the idea of a Fourier series. I could take a periodic function, we started with the example of this square wave, and I could represent it as the sum of weighted sine and cosine functions. Then we took a little bit of an interlude, building up some of our mathematical foundations, just establishing a bunch of properties of taking the definite integral over the period of that periodic function of sine and cosine. We established all of these properties, and now we're going to get the benefit from establishing all those because we're going to start actually finding at least formulas for the Fourier coefficients. Then we can apply it to our original square wave to see that, hey, this could actually be a pretty straightforward thing.

So, right over here, I have rewritten a Fourier series expression, or I've rewritten a Fourier series for a periodic function f of T. Let's say its period is 2π, and I'm going to use this in some of the properties that we have established to start solving for these actual coefficients. What I'm going to do in this video is first try to solve for a₀, and then in the next video, we're going to solve for an arbitrary aₙ. In either that one or the next one, we'll also solve for an arbitrary bₙ.

To solve for a₀, what we're going to do is take the definite integral of both sides from 0 to 2π—so from 0 to 2π dt of f of T. Well, that's going to be the same thing as going from 0 to 2π of all of this stuff. Remember, this is an infinite series right over here. We have an infinite number of terms, and then we would have a dt out there. But we know from our integration properties that taking the definite integral of a sum, even an infinite sum, is the same thing as the sum of the definite integrals. So that's going to be the same thing as taking this integral dt plus this integral.

I could take the scale out; actually, let me not just do that. Let me just write it like this: 0 to 2π dt, 0 to 2π dt, 0 to 2π dt. This is getting a little monotonous, but it'll be worth it—0 to 2π dt, 0 to 2π dt, 0 to 2π dt. We'll do it for every single one of the terms. Now what's nice is we can look at our integration properties. This right over here, we could take those coefficients out. We could take this a₁, put it in front of the integral sign. The a₂ put it in front of the integral sign. The b₁ put it in front of the integral sign. Then all you're left with is the integral from 0 to 2π of cosine of some integer multiple of T dt.

Well, we established a couple of videos ago that that's always going to be equal to zero. The integral from 0 to 2π of cosine of some nonzero integer multiple of T dt is equal to zero. The same thing is true for sine of mt, so this is going to be fun. This is going to be zero based on what we just saw. If you just take that factor out of that integral, take that a₁ out of the integral, it's going to be a₁ * 0. This is going to be a₂ * 0. That's going to be zero, that's going to be zero, that's going to be zero, that's going to be zero. Every term is going to be zero except for this one involving a₀.

So, what is this going to be equal to? Well, let me write it this way—let's take the definite integral. Let me see where I have some space. So we're going to take the definite integral from 0 to 2π of a₀ dt. Well, that's the same thing. Once again, we could take the coefficient out of a₀, and I could just put the dt like this. So, that's going to be equal to a₀ * T evaluated at 2π and 0, which is going to be equal to a₀ * (2π - 0) = 2π a₀.

So I could just write this as a₀ * 2π. This expression right here is a₀ * 2π. So let me scroll down a little bit so I can rewrite this thing up here: the integral from 0 to 2π of f of T dt, which is equal to this integral, but we've just figured out that the integral from 0 to 2π of a₀ dt is the same thing as a₀ * 2π, is equal to a₀ * 2π * 2π. So now it's actually pretty straightforward to solve for a₀.

a₀ is going to be equal to a₀ = 1/(2π) * the definite integral from 0 to 2π of f of T dt. This is pretty cool because think about what this is. This over here is the average value of our function. This is the average value of f over the interval from 0 to 2π. Hopefully, that actually makes intuitive sense because, if you just think of it from an engineering point of view, you're just trying to engineer this, trying to just play around with these numbers.

All of these cosines and sines, they oscillate between positive one and negative one. So in order to actually represent this function, you're going to have to shift that oscillation. Some of a bunch of oscillations still going to be an oscillation that's going to vary between a positive one and our negative one. In order to shift it, well, that's what our a₀ is going to do. It makes sense that you would want to shift the oscillation so it oscillates around the average value of the function, or you could say the average value of the function over an interval that's representative of a period of that function.

So that is what a₀ is doing; a₀ is just going to be that average value of the function.