Question

Can someone help me understand the Fourier Series discussed at the end of Lecture 24? In particular, I'm trying to understand how, after taking the "dot product" of cosx with the series a0*(1)+a1cosx+b1sinx, etc., you're left with a1(cosx)^2? I understand how most of series ends up as zeroes when dotted with cosx (multiplied by cosx and the integrated), but I'm not clear how you end up with just cosx^2. Can anyone help?

Thanks Phi, but you still didn't answer how once you drop all the zeroes, you still end up with one integral of cosx^2. Which entry in the series after multiplying by cosx leaves the product cosx^2 rather than zero, which, according to Dr. Strang is the solution to the problem.

Answer 1

I think the idea is you have a vector cos(x) for many values of x (from x=0 to 2pi)
and ditto for the series (at least conceptually)
the dot product of the cos(x) vector with, for example, the b1*sin(x) vector
can be thought of as integrating
\[b_1 \int_0^{2\pi} \cos x \sin x \ dx \]
and it can be shown that this integral is zero.
cos x and sin x are orthogonal (i.e. the integral over 1 period is 0)
ditto for cos x and sin n x
and cos x and cos nx (except for n=1)

Answer 2

and to be complete,
the cos(x) vector dotted with a vector of 1's (i.e. the first term \(a_0 1 \) term)
gives the area of cos x, which sums to zero, because we treat the area under the x-axis as negative.

Answer 3

the 2nd term in the series is
\[a_1 \cos x\]
and the dot product of cos x with this term is equivalent to
\[ a_1\int_0^{2\pi} \cos^2x \ dx\]
that will be the only non-zero result.
all the other terms will integrate to zero

Answer 4

Let
\[ s(x)= a_0+a_1\cos x+b_1\sin x+... \\ ones(x)= [1\ 1\ 1 ...]\\ \cos(x) \cdot s(x)= a_0 \cos x \cdot ones(x)+a_1\cos x \cdot \cos x +b_1\cos x \cdot \sin x+...\] where x = 0 to 2pi in increments of dx i.e. finely sampled

each dot product is equivalent to numerically integrating from 0 to pi
\[ = a_0 \int_0^{2\pi} \cos x \ dx + a_1 \int_0^{2\pi} \cos^2 x \ dx +b_1 \int_0^{2\pi} \cos x \ \sin x\ dx + ... \]
only the 2nd term survives

Answer 5

Thanks so much, I get it now!

Answer 6

One last question. What happens to the first term a0*(1)? Or is a0 actually zero and not 1?

Answer 7

a0 is not 0.
as an integral it is
\[ a_0 \int_0^{2\pi} \cos x \ dx \]
if you don't know calculus, then intuitively, that calculates the area "under the curve"