#import "@preview/physica:0.9.6": *
#import "lib.typ": ccases
#import "@preview/lilaq:0.5.0" as lq

= problem 4

== a)

to find the fourier series of
$
  f(x) = ccases(
    0, -pi < x < 0 "or" pi/2 < x <= pi,
    x, 0 <= x <= pi/2,
  )
$
we can use the formulae for the coefficients
$
  a_0 & = 1/pi integral_(-pi)^pi f(x) dd(x) \
      & = 1/pi integral_0^(pi slash 2) x dd(x) \
      & = 1/(2 pi) [x^2]_0^(pi slash 2) \
      & = pi/8
$
$
  a_n & = 1/pi integral_(-pi)^pi f(x) cos(n x) dd(x) \
      & = 1/pi integral_0^(pi slash 2) x cos(n x) dd(x) \
      & = 1/pi [x/n sin(n x) + 1/n^2 cos(n x)]_0^(pi slash 2) \
      & = 1/(2 n) sin(pi/2 n) + 1/(pi n^2) (cos (pi/2 n) - 1)
$
then for the different cases of $n mod 4$
$
  a_n = ccases(
    0, n equiv 0,
    1/(2 n) - 1/(pi n^2), n equiv 1,
    -2/(pi n^2), n equiv 2,
    -1/(2 n) - 1/(pi n^2), n equiv 3,
  )
$
then for the $sin$ terms
$
  b_n & = 1/pi integral_(-pi)^pi f(x) sin(n x) dd(x) \
      & = 1/pi integral_0^(pi slash 2) x sin(n x) dd(x) \
      & = 1/pi [1/n^2 sin(n x) - x/n cos(n x)]_0^(pi slash 2) \
      & = 1/(pi n^2) sin(pi/2 n) - 1/(2 n) cos(pi/2 n)
$
then similar $n mod 4$ computations
$
  b_n = ccases(
    -1/(2 n), n equiv 0,
    1/(pi n^2), n equiv 1,
    1/(2 n), n equiv 2,
    -1/(pi n^2), n equiv 3
  )
$
then putting it all together
$
  f(x) = a_0/2 + sum_(n=1)^oo a_n cos(n x) + sum_(n=1)^oo b_n sin (n x)
$
which is our final fourier series for $f(x)$.


#{
  let a_0 = calc.pi / 8
  let a(n) = {
    if calc.rem-euclid(n, 4) == 0 {
      0
    } else if calc.rem-euclid(n, 4) == 1 {
      1 / (2 * n) - 1 / (calc.pi * n * n)
    } else if calc.rem-euclid(n, 4) == 2 {
      -2 / (calc.pi * n * n)
    } else if calc.rem-euclid(n, 4) == 3 {
      -1 / (2 * n) - 1 / (calc.pi * n * n)
    }
  }
  let b(n) = {
    if calc.rem-euclid(n, 4) == 0 {
      -1 / (2 * n)
    } else if calc.rem-euclid(n, 4) == 1 {
      1 / (calc.pi * n * n)
    } else if calc.rem-euclid(n, 4) == 2 {
      1 / (2 * n)
    } else if calc.rem-euclid(n, 4) == 3 {
      -1 / (calc.pi * n * n)
    }
  }
  let fourier(x, n: 50) = (
    a_0 / 2
      + range(1, n).map(n => a(n) * calc.cos(n * x)).sum()
      + range(1, n).map(n => b(n) * calc.sin(n * x)).sum()
  )
  let f(x) = {
    let t = calc.rem-euclid(x, 2 * calc.pi)
    if 0 <= t and t <= calc.pi / 2 { t } else { 0 }
  }
  let xs = lq.linspace(-3 * calc.pi, 3 * calc.pi, num: 100)
  let configs = (
    (n: 5),
    (n: 20),
    (n: 100),
  )
  for config in configs {
    lq.diagram(
      title: [the fourier series of $f(x)$ with $N = #config.n$],
      xlabel: $x$,
      ylabel: $y$,
      width: 100%,
      height: 32%,
      xlim: (-3 * calc.pi, 3 * calc.pi),
      xaxis: (
        locate-ticks: lq.locate-ticks-linear.with(unit: calc.pi),
        format-ticks: lq.format-ticks-linear.with(suffix: $pi$),
      ),
      lq.plot(xs, x => fourier(x, n: config.n), mark: none),
      lq.plot(xs, f, mark: none),
    )
  }
}

== b)

let
$
  f(x) = ccases(
    0, -pi < x < 0,
    x, 0 < x < pi/2,
    pi - x, pi/2 < x <= pi
  )
$

then we can proceed as usual to calculate the coefficients of the fourier
series, utilising symmetries of $x$ and $pi - x$ on the given intervals.
$
  a_0 & = 1/pi integral_(-pi)^pi f(x) dd(x) \
      & = 1/pi (integral_0^(pi slash 2) x dd(x)
          + integral_(pi slash 2)^pi (pi - x) dd(x)) \
      & = 2/pi integral_0^(pi slash 2) x dd(x) \
      & = 1/pi [x^2]_0^(pi slash 2) \
      & = pi / 4
$

$
  a_n & = 1/pi integral_(-pi)^pi f(x) cos(n x) dd(x) \
      & = 1/pi (integral_0^(pi slash 2) x cos(n x) dd(x)
          + integral_(pi slash 2)^pi (pi - x) cos(n x) dd(x)) \
      & = 1/pi ([x/n sin(n x) + 1/n^2 cos(n x)]_0^(pi slash 2) \
      & + [(pi-x)/n sin(n x) - 1/n^2 cos(n x)]_(pi slash 2)^pi) \
      & = 1/pi ([cancel(pi/(2 n) sin(pi/2 n)) + 1/n^2 cos(pi/2 n) - 1/n^2] \
      & + [1/n^2 (-1)^(n+1) - cancel(pi/(2 n) sin(pi/2 n)) + 1/n^2 cos(pi/2 n)]) \
      & = 1/(pi n^2) (2 cos(pi/2 n) - 1 + (-1)^(n + 1))
$
notice that for odd $n$, $a_n = 0$, since the cosine is zero and the alternating
sign is fixed to 1, canceling with the constant term. thus we can shorten it
further with a definition for $n = 2 k$
$
  a_n & = 1/(4 pi k^2) (2 cos(pi k) - 1 + (-1)^(2k + 1)) \
      & = 1/(2 pi k^2) ((-1)^k - 1) \
      & = 2/(pi n^2) ((-1)^(n slash 2) - 1)
$

similarly $b_n$, we can recognize the symmetry earlier in our calculations
$
  b_n & = 1/pi integral_(-pi)^pi f(x) sin(n x) dd(x) \
      & = 1/pi (integral_0^(pi slash 2) x sin(n x) dd(x)
          + integral_(pi slash 2)^pi (pi - x) sin(n x) dd(x)) \
      & = 1/pi (1 - (-1)^n) integral_0^(pi slash 2) x sin(n x) dd(x)
$
notice that for even $n$, $b_n = 0$, let $n = 2k + 1$
$
  b_n & = 1/pi (1 - (-1)^(2k + 1))
        integral_0^(pi slash 2) x sin ((2k + 1)x) dd(x) \
      & = 1/pi (1 + (-1)^(2k)) [1/n^2 sin(n x)
          - x/n cos(n x)]_0^(pi slash 2) \
      & = 2/(pi n^2) sin(pi/2 n) \
      & = 2/(pi n^2) (-1)^((n-1)/2)
$

#{
  let a_0 = calc.pi / 4
  let a(n) = {
    if calc.rem-euclid(n, 2) == 0 {
      2 / (calc.pi * n * n) * (calc.pow(-1, n / 2) - 1)
    } else { 0 }
  }
  let b(n) = {
    if calc.rem-euclid(n, 2) == 1 {
      2 / (calc.pi * n * n) * calc.pow(-1, (n - 1) / 2)
    } else { 0 }
  }
  let fourier(x, n: 50) = (
    a_0 / 2
      + range(1, n).map(n => a(n) * calc.cos(n * x)).sum()
      + range(1, n).map(n => b(n) * calc.sin(n * x)).sum()
  )
  let f(x) = {
    let t = calc.rem-euclid(x, 2 * calc.pi)
    if 0 < t and t < calc.pi / 2 {
      t
    } else if calc.pi / 2 < t and t <= calc.pi {
      calc.pi - t
    } else {
      0
    }
  }
  let xs = lq.linspace(-3 * calc.pi, 3 * calc.pi, num: 100)
  let configs = (
    (n: 5),
    (n: 20),
    (n: 100),
  )
  for config in configs {
    lq.diagram(
      title: [the fourier series of $f(x)$ with $N = #config.n$],
      xlabel: $x$,
      ylabel: $y$,
      width: 100%,
      height: 32%,
      xlim: (-3 * calc.pi, 3 * calc.pi),
      xaxis: (
        locate-ticks: lq.locate-ticks-linear.with(unit: calc.pi),
        format-ticks: lq.format-ticks-linear.with(suffix: $pi$),
      ),
      lq.plot(xs, x => fourier(x, n: config.n), mark: none),
      lq.plot(xs, f, mark: none),
    )
  }
}

== c)

let
$
  f(x) = ccases(
    -pi - x, -pi < x < -pi/2,
    x, -pi/2 < x < pi/2,
    pi - x, pi/2 < x <= pi,
  )
$

then
$
  a_0 & = 1/pi integral_(-pi)^pi f(x) dd(x) \
      & = 1/pi (integral_(-pi)^(-pi slash 2) (-pi - x) dd(x)
          + cancel(integral_(-pi slash 2)^(pi slash 2) x dd(x))
          + integral_(pi slash 2)^pi (pi - x) dd(x)) \
      & = 2/pi integral_(pi slash 2)^pi (pi - x) dd(x) \
      & = 2/pi ([pi^2 - pi^2/2] - 1/2 [pi^2 - pi^2/4]) \
      & = 2 pi - pi - pi + pi/2 \
      & = pi/2
$

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

$
  b_n & = 1/pi integral_(-pi)^pi f(x) sin(n x) dd(x) \
      & = 2/pi (integral_(pi slash 2)^pi (pi - x) sin(n x) dd(x)
          + integral_0^(pi slash 2) x sin(n x) dd(x)) \
      & = 2/pi ([(x - pi)/n cos(n x) - 1/n^2 sin(n x)]_(pi slash 2)^pi \
      & + [1/n^2 sin(n x) - x/n cos(n x)]_0^(pi slash 2)) \
      & = 2/pi (- pi/(2 n) cos(pi/2 n) - cancel(1/n^2 sin(pi/2 n)) \
      & + cancel(1/n^2 sin(pi/2 n)) - pi/(2 n) cos(pi/2 n)) \
      & = 2/n cos(-pi/2 n)
$

we could figure out what these expressions equal in each case ($mod 4$), but for
the sake of brevity, i'll leave it at that.

#{
  let a_0 = calc.pi / 2
  let a(n) = (
    2
      / calc.pi
      * (
        1 / (n * n) * calc.pow(-1, n)
          - 1 / (n * n) * calc.cos(calc.pi / 2 * n)
          - calc.pi / (2 * n) * calc.sin(calc.pi / 2 * n)
      )
  )
  let b(n) = (
    2 / n * calc.cos(-calc.pi / 2 * n)
  )
  let fourier(x, n: 50) = (
    a_0 / 2
      + range(1, n).map(n => a(n) * calc.cos(n * x)).sum()
      + range(1, n).map(n => b(n) * calc.sin(n * x)).sum()
  )
  let f(x) = {
    let t = calc.rem-euclid(x, 2 * calc.pi)
    if -calc.pi < t and t < -calc.pi / 2 {
      -calc.pi - t
    } else if -calc.pi / 2 < t and t < calc.pi / 2 {
      t
    } else {
      calc.pi - t
    }
  }
  let xs = lq.linspace(-3 * calc.pi, 3 * calc.pi, num: 100)
  let configs = (
    (n: 5),
    (n: 20),
    (n: 100),
  )
  for config in configs {
    lq.diagram(
      title: [the fourier series of $f(x)$ with $N = #config.n$],
      xlabel: $x$,
      ylabel: $y$,
      width: 100%,
      height: 32%,
      xlim: (-3 * calc.pi, 3 * calc.pi),
      xaxis: (
        locate-ticks: lq.locate-ticks-linear.with(unit: calc.pi),
        format-ticks: lq.format-ticks-linear.with(suffix: $pi$),
      ),
      lq.plot(xs, x => fourier(x, n: config.n), mark: none),
      lq.plot(xs, f, mark: none),
    )
  }
}

as we can see, i've made a mistake in my calculations.