#import "@preview/physica:0.9.6": *
#import "lib.typ": ccases

= problem 1

== a)

$
  f(x) := sin(3 x) + 5 cos(2 x)
$
is already in its fourier series form.

the coefficients can be read as
$
  a_0 & = 0, \
  a_n & = ccases(5, n = 2, 0) \
  b_n & = ccases(1, n = 3, 0)
$

== b)

let
$
  g(x) = x^2, quad -pi <= x <= pi
$

then we can compute the fourier series using the formulas for $2pi$-periodic
functions
$
  a_0 & = 1/pi integral_(-pi)^pi x^2 dd(x)
        = 2/pi integral_0^pi x^2 dd(x) \
      & = 2/(3 pi) [x^3]_0^pi = 2/3 pi^2
$

$
  a_n & = 1/pi integral_(-pi)^pi x^2 cos(n x) dd(x) \
      & = 1/pi [x^2/n sin(n x) + (2 x)/n^2 cos(n x)
          -2/n^3 sin(n x)]_(-pi)^pi \
      & = 2/(n pi) [x^2 sin(n x) + (2 x)/n cos(n x)
          - 2/n^2 sin(n x)]_0^pi \
      & = 2/(n pi) ((2 pi)/n cos(pi n)) \
      & = 4/n^2 (-1)^n
$

$
  b_n & = 1/pi integral_(-pi)^pi x^2 sin(n x) dd(x) \
      & = 1/pi [-x^2/n cos(n x) +(2 x)/n^2 sin(n x)
          + 2/n^3 cos(n x)]_(-pi)^pi \
      & = 1/pi [(cancel(-pi^2/n cos(pi n)) + (2 pi)/n^2 sin(pi n)
          + cancel(2/n^3 cos(pi n))) \
      & - (cancel(-pi^2/n cos(-pi n)) - (2 pi)/n^2 sin(-pi n)
          + cancel(2/n^3 cos(-pi n)))] \
      & = 2/n^2 (sin(pi n) + sin(-pi n)) \
      & = 2/n^2 (sin(pi n) - sin(pi n)) = 0
$

the fact that $b_n$ is zero is reassuring, since the oddness that it contributes
with would be undesirable for a function like $x^2$.

== c)

show
$
  h(x) & = f(x) * g(x) = integral_(-pi)^pi g(y) f(x - y) dd(y) \
       & = -(4 pi)/9 sin(3 x) + 5 pi cos(2 x)
$

#line(length: 100%, stroke: (dash: "dotted"))

$
  h(x) = integral_(-pi)^pi y^2 [sin(3 (x - y)) + 5 cos(2 (x - y))] dd(y)
$

this would be an awful lot of integration by parts, so let's be more clever
about this.

we can expand the trigonometric terms in $f(x - y)$ to obtain simpler terms that
cancel out under $integral_(-pi)^pi g(y) sin(n y) dd(y) = 0$

- $
    sin(3(x - y)) = sin(3x) cos(3y) - cos(3x) sin(3y)
  $
- $
    cos(2(x - y)) = cos(2x) cos(2y) + sin(2x) sin(2y)
  $

thus we obtain

$
  h(x) & = sin(3x) integral_(-pi)^pi y^2 cos(3y) dd(y) \
       & - cancel(cos(3 x) integral_(-pi)^pi y^2 sin(3 y) dd(y)) \
       & + 5 cos(2x) integral_(-pi)^pi y^2 cos(2y) dd(y) \
       & + cancel(5 sin(2x) integral_(-pi)^pi y^2 sin(2y) dd(y))
$

but these are simply the coefficients of $y^2$ from b), such that

$
  1/pi integral_(-pi)^pi y^2 cos(2y) dd(y) = [4/n^2 (-1)^n]_(n=2) = 1
$
$
  1/pi integral_(-pi)^pi y^2 cos(3y) dd(y) = [4/n^2 (-1)^n]_(n=3) = -4/9
$

which gives

$
  h(x) = - (4 pi)/9 sin(3x) + 5 pi cos(2x)
$

#h(1fr) #math.qed


== d)

$h(n)$ is also made up of basis vectors, such that the coefficients can be
written as
$
  a_0 & = 0 \
  a_n & = ccases(5pi, n = 2, 0) \
  b_n & = ccases(-(4pi)/9, n = 3, 0)
$

$
  c_n (f) & = a_n (f) + b_n (f) = ccases(5, n = 2, 1, n = 3, 0) \
  c_n (g) & = ccases(2/3 pi^2, n = 0, 4/n^2 (-1)^n,) \
  c_n (h) & = ccases(5pi, n = 2, -(4 pi)/9, n = 3, 0)
$

we can notice that $c_2 (h) = pi c_2 (f)$ and $c_3 (h) = pi c_3 (g)$.

if we look for a deeper pattern, we can also notice that $c_2 (g) = 1$ and
$c_3 (f) = 1$ such that we could say that
$
  c_2(h) = pi c_2(f) c_2(g) quad and quad c_3(h) = pi c_3(f) c_3(g)
$

this is a profound result -- well, the generalized statement is -- and is
related to the fourier transform. we are saying that convolving two functions is
the same as taking their point-wise product at certain frequencies.