A finite Fourier series is given by the sum f(x) = ∑n =1^N an sinnx f(x) = a1 sinx + a2 sin2x +

A finite Fourier series is given by the sum f(x) = ∑n =1^N...

A finite Fourier series is given by the sum
$$ f(x) = \sum_{n =1}^{N} a_n \sin nx $$
$$ f(x) = a_1 \sin x + a_2 \sin 2x + \cdots + a_N \sin Nx $$

Show that the $ m $ th coefficient am is given by the formula

$$ a_m = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin mx dx $$

Transcript

00:01 A finite 4ea series is given by the sum of a little n of sine of nx, where the ends range from 1 all the way up to a large n.

00:14 So an expanded form, here's the finite 4a series for some function f.

00:21 And we like to show that the m coefficient, am, is given by the following formula.

00:32 So the key fact for this problem is comes from problem 68 that the integral negative pi to pi, sine nx, sign mx is 0 or pi depending on whether or not m or equal.

01:09 So it's pi when m equals n and it's zero otherwise.

01:19 So before we proceed, let's just make a few remarks about this identity and where it's coming from.

01:24 So we broke this into two cases in problem 68, so an m was not equal to n.

01:37 We used the formula cosine a minus b minus cosine of a plus b.

01:55 So we used this formula to rewrite the intergrand, and then we evaluated each term in the intigrant.

02:04 And in the other case, in which m and n were equal, we wrote the intergram is sine squared mx, and then using a pythagron identity, we can write this as 1 minus cosine 2mx over 2, and that makes it easier to integrate.

02:27 So this is where this identity comes from, and this is a key tool for this following problem.

02:38 Okay, so going on to the next page, we have 1 over pi times the integral, negative pi to pi, f of x, sine mx, dx, now using the formula that was given for f, the finite 4iase sum, we can go ahead and replace f with its sum...

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