141 lines
3.4 KiB
Typst
141 lines
3.4 KiB
Typst
#import "@preview/physica:0.9.6": *
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#import "lib.typ": ccases
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= problem 1
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== a)
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$
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f(x) := sin(3 x) + 5 cos(2 x)
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$
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is already in its fourier series form.
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the coefficients can be read as
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$
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a_0 & = 0, \
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a_n & = ccases(5, n = 2, 0) \
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b_n & = ccases(1, n = 3, 0)
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$
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== b)
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let
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$
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g(x) = x^2, quad -pi <= x <= pi
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$
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then we can compute the fourier series using the formulas for $2pi$-periodic
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functions
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$
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a_0 & = 1/pi integral_(-pi)^pi x^2 dd(x)
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= 2/pi integral_0^pi x^2 dd(x) \
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& = 2/(3 pi) [x^3]_0^pi = 2/3 pi^2
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$
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$
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a_n & = 1/pi integral_(-pi)^pi x^2 cos(n x) dd(x) \
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& = 1/pi [x^2/n sin(n x) + (2 x)/n^2 cos(n x)
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-2/n^3 sin(n x)]_(-pi)^pi \
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& = 2/(n pi) [x^2 sin(n x) + (2 x)/n cos(n x)
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- 2/n^2 sin(n x)]_0^pi \
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& = 2/(n pi) ((2 pi)/n cos(pi n)) \
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& = 4/n^2 (-1)^n
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$
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$
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b_n & = 1/pi integral_(-pi)^pi x^2 sin(n x) dd(x) \
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& = 1/pi [-x^2/n cos(n x) +(2 x)/n^2 sin(n x)
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+ 2/n^3 cos(n x)]_(-pi)^pi \
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& = 1/pi [(cancel(-pi^2/n cos(pi n)) + (2 pi)/n^2 sin(pi n)
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+ cancel(2/n^3 cos(pi n))) \
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& - (cancel(-pi^2/n cos(-pi n)) - (2 pi)/n^2 sin(-pi n)
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+ cancel(2/n^3 cos(-pi n)))] \
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& = 2/n^2 (sin(pi n) + sin(-pi n)) \
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& = 2/n^2 (sin(pi n) - sin(pi n)) = 0
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$
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the fact that $b_n$ is zero is reassuring, since the oddness that it contributes
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with would be undesirable for a function like $x^2$.
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== c)
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show
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$
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h(x) & = f(x) * g(x) = integral_(-pi)^pi g(y) f(x - y) dd(y) \
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& = -(4 pi)/9 sin(3 x) + 5 pi cos(2 x)
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$
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#line(length: 100%, stroke: (dash: "dotted"))
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$
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h(x) = integral_(-pi)^pi y^2 [sin(3 (x - y)) + 5 cos(2 (x - y))] dd(y)
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$
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this would be an awful lot of integration by parts, so let's be more clever
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about this.
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we can expand the trigonometric terms in $f(x - y)$ to obtain simpler terms that
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cancel out under $integral_(-pi)^pi g(y) sin(n y) dd(y) = 0$
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- $
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sin(3(x - y)) = sin(3x) cos(3y) - cos(3x) sin(3y)
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$
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- $
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cos(2(x - y)) = cos(2x) cos(2y) + sin(2x) sin(2y)
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$
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thus we obtain
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$
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h(x) & = sin(3x) integral_(-pi)^pi y^2 cos(3y) dd(y) \
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& - cancel(cos(3 x) integral_(-pi)^pi y^2 sin(3 y) dd(y)) \
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& + 5 cos(2x) integral_(-pi)^pi y^2 cos(2y) dd(y) \
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& + cancel(5 sin(2x) integral_(-pi)^pi y^2 sin(2y) dd(y))
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$
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but these are simply the coefficients of $y^2$ from b), such that
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$
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1/pi integral_(-pi)^pi y^2 cos(2y) dd(y) = [4/n^2 (-1)^n]_(n=2) = 1
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$
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$
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1/pi integral_(-pi)^pi y^2 cos(3y) dd(y) = [4/n^2 (-1)^n]_(n=3) = -4/9
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$
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which gives
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$
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h(x) = - (4 pi)/9 sin(3x) + 5 pi cos(2x)
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$
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#h(1fr) #math.qed
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== d)
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$h(n)$ is also made up of basis vectors, such that the coefficients can be
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written as
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$
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a_0 & = 0 \
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a_n & = ccases(5pi, n = 2, 0) \
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b_n & = ccases(-(4pi)/9, n = 3, 0)
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$
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$
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c_n (f) & = a_n (f) + b_n (f) = ccases(5, n = 2, 1, n = 3, 0) \
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c_n (g) & = ccases(2/3 pi^2, n = 0, 4/n^2 (-1)^n,) \
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c_n (h) & = ccases(5pi, n = 2, -(4 pi)/9, n = 3, 0)
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$
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we can notice that $c_2 (h) = pi c_2 (f)$ and $c_3 (h) = pi c_3 (g)$.
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if we look for a deeper pattern, we can also notice that $c_2 (g) = 1$ and
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$c_3 (f) = 1$ such that we could say that
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$
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c_2(h) = pi c_2(f) c_2(g) quad and quad c_3(h) = pi c_3(f) c_3(g)
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$
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this is a profound result -- well, the generalized statement is -- and is
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related to the fourier transform. we are saying that convolving two functions is
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the same as taking their point-wise product at certain frequencies.
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