Well that certainly explains it, says the guy who had to retake 9th grade algebra!

How about this:

Let's say we want to write out 1/3 as a decimal. We can think of decimals as a situation where the only primitives we have available are the powers of 10 (in this case negative powers of 10, so 1/10, 1/100, etc) and their multiples by a single digit (0-9). So our first approximation to 1/3 would be using just one of these primitives, 3 * (1/10), or 0.3. Our second approximation would be to use two of these primitives, 3 * (1/10) + 3 * (1/100) = 0.33. That's closer to 1/3 than our first approximation. The third approximation would be 0.333, closer still. As we add more and more terms to our finite sum, the answer will get closer and closer to 1/3. But it never equals 1/3 until we pass to the infinite sum.

With Fourier series we can do the same thing for a function f that is periodic with cyclic frequency ω. We can approximate it as an infinite sum of primitive functions, meaning that the longer and longer finite sums get closer and closer to our function f. And the primitives we use are the functions cos(k * ωt) and sin(k * ωt), where k is an integer at least 0. We get to multiply these function by an arbitrary real number constant, rather than just the digit 0-9.

So in post #12, we can do this for V(t). That one is easy, as it already a sine wave; and ω has been chosen to be 1 for simplicity. We only need one term, sin(t), and the multiple looks to be about 0.9.

But we can also do this for I(t). (I didn't check the expansion that is provided in the post but am just explicating it). The first term would be when k=0, in which case cos(0*t) = 1, so we just get a constant function. The coefficient for this term is 2/pi = 0.6366. So our first approximation for I(t) would be a constant current I(t) = 0.6366A. Not a great approximation, as the raw output of the full bridge rectifier has a lot of ripple.

So to better approximate I(t) we can use more terms. It turns out we don't need any sin terms, and all the cos terms are of the form cos(kt) where k is even. We say that I(t) has only even harmonics. The second harmonic gets a coefficient of 4/(-3pi) = -0.4244. So this approximation would be I(t) = 0.6366 - 0.4244 cos(2t). This approximation is graphed as the dashed line labeled "1 term" meaning we just used one cosine function. And it's graphed offset upwards, just so the functions don't pile up on each other.

We can keep doing this with more and more terms. The second approximation add a cos(4t) term or fourth harmonic; the coefficient is 4/(-15pi) = -0.08488. This approximation is I(t) = 0.6366 - 0.4244 cos(2t) - 0.08488 cos(4t) and is the dashed graph labeled "2 terms". It looks a little closer to the blue graph we are approximating.

Etc.

Cheers, Wayne