3. Suppose (f) and (g) are two functions on ([-pi, pi]) with...

3. Suppose (f) and (g) are two functions on ([-pi, pi]) with (f(pi) = f(-pi)) and (g(pi) = g(-pi)). Let us write the complex Fourier coefficients of (f) and (g) as (hat{f}(n) = frac{1}{2pi} int_{-pi}^{pi} f(x)e^{-inx}dx), (hat{g}(n) = frac{1}{2pi} int_{-pi}^{pi} g(x)e^{-inx}dx). Define the convolution of (f) and (g) to be the function (h = f ast g) defined by (h(x) = frac{1}{2pi} int_{-pi}^{pi} f(x-y)g(y)dy). Suppose (h(x) = sum_{n=-infty}^{infty} hat{h}(n)e^{inx}), (hat{h}(n) = frac{1}{2pi} int_{-pi}^{pi} h(x)e^{-inx}dx) Show that (hat{h}(n) = hat{f}(n)hat{g}(n)). (So the (n)-th Fourier coefficient of the convolution of two functions is the product of their (n)-th Fourier coefficients).

Transcript

00:01 Okay, in this question we are given two functions.

00:03 F and g, we'll assume zero some functions that are synch -synch -bo with f -pi.

00:13 Those two functions satisfy f -pi is equal to f -native pi and g -py equal to g -n negative -py.

00:24 And we define the fourier coefficient.

00:28 The end -fureure coefficient is defined to be to 1 over 2 pi.

00:34 Times the integral from negative pi to pi, fx leaves the power negative i nx d x, the same for g.

00:49 Then we define the convolution of f and g.

00:57 Let's define h x to be equal to 1 over 2 pi times the integral from negative pi to pi to pi f x minus y, g y, d y, we want to show the fourier coefficient of h is just equal to the product of the coefficient for f and g.

01:24 The fourier transformation just transforms the convolution to the pointwise product.

01:32 Here, to avoid some confusion, we want to first consider this convolution, because here's the definition in this question, this this this xx is, will bring us some confusion because it is not well defined for x next to pi.

02:00 Why? because our f, notice our f is only defined from next to pi to pi.

02:06 Here if we take, for example if we take xb pi over 2 you can see this function is not always well define.

02:17 Sometimes we don't have that because for some x we don't have a definition for our f that means this integral is meaningless.

02:26 Why? because you can see our y goes from negative pi to pi.

02:30 So when y is equal to negative pi, then our f will add pi over 2 plus pi which is equal to f3 pi over 2.

02:40 But as our f is defined on this interval, that means this guy does not have any definitions.

02:53 So we can find a value for our h.

02:57 That means the definition for h in this question is not so right.

03:02 So to define the composition, we need to use some extension of f.

03:09 I mean, let's define f to be equal to fx.

03:16 One x is between negative pi to pi, and we define to be equal to x.

03:21 To 0 for some other x and we'll extend, we just extend our definition for f to the whole real line.

03:30 The same for g, we define g1 to equal to g x x x x x and z and sit and divide to equal to 0 else.

03:43 So and our h is defined to be the convolution for h and g1 and the integral from negative.

03:57 To positive infinity to positive infinity.

04:00 Then our definition is right.

04:01 I mean, for, to use this definition, we don't have change.

04:09 We don't change the principle of our convolution...

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