diff --git a/exercise9/lib.typ b/exercise9/lib.typ
new file mode 100644
index 0000000..cc5f607
--- /dev/null
+++ b/exercise9/lib.typ
@@ -0,0 +1,60 @@
+#let title(cont, size: 18pt) = align(center)[
+  #text(size: size * 2, weight: "bold")[#underline[#cont]]
+]
+
+// writes a butcher tableau using the minimum required amount of numbers.
+#let butcher(s, nums) = {
+  let nums = nums.map(x => if type(x) == float or type(x) == int {
+    $#x$
+  } else { x })
+  table(
+    stroke: (x, y) => if x == 0 and y == s {
+      (right: 0.7pt + black, top: 0.7pt + black)
+    } else if x == 0 {
+      (right: 0.7pt + black)
+    } else if y == s {
+      (top: 0.7pt + black)
+    },
+    align: (x, y) => (
+      if x > 0 { center } else { left }
+    ),
+    columns: s + 1,
+    $0$, ..range(s).map(x => none), // first row
+    ..range(2, s + 1)
+      .map(i => {
+        let p(i) = calc.floor(i * i / 2 + i / 2 - 1)
+        (
+          nums.slice(p(i - 1), p(i - 1) + i),
+          range(i, s + 1).map(x => none),
+        ).flatten()
+      })
+      .flatten(),
+    none, ..nums.rev().slice(0, s).rev() // last row
+  )
+}
+
+// automatically adds aligned "if" and "otherwise" strings to a case block.
+//
+// example:
+// ```
+// $ccases(x, x > 0, -x, x <= 0)$
+// // is the same as
+// $cases(x & quad "if" space x > 0, -x & quad "if" space x <= 0$
+//
+// $ccases(x, x > 0, -x)$
+// // is the same as
+// $cases(x & quad "if" space x > 0, -x & quad "otherwise"$
+// ```
+#let ccases(..args) = {
+  let parsed = args.pos().map(x => if type(x) == content { x } else { $#x$ })
+  let result_array = parsed
+    .windows(2)
+    .enumerate()
+    .filter(x => calc.rem-euclid(x.first(), 2) == 0)
+    .map(array.last)
+    .map(x => x.intersperse($& quad "if" space$).join())
+  if calc.rem-euclid(parsed.len(), 2) != 0 {
+    result_array.push($#parsed.last() & quad "otherwise"$)
+  }
+  math.cases(..result_array)
+};
diff --git a/exercise9/main.pdf b/exercise9/main.pdf
new file mode 100644
index 0000000..cb9dabd
Binary files /dev/null and b/exercise9/main.pdf differ
diff --git a/exercise9/main.typ b/exercise9/main.typ
new file mode 100644
index 0000000..62e0d9c
--- /dev/null
+++ b/exercise9/main.typ
@@ -0,0 +1,43 @@
+#import "lib.typ"
+
+#let FONT_SIZE = 18pt;
+
+#set page(paper: "a4", margin: (x: 2.6cm, y: 2.8cm), numbering: "1 : 1")
+#set par(justify: true, leading: 0.52em)
+
+#set text(font: "FreeSerif", size: FONT_SIZE, lang: "us")
+#show math.equation: set text(
+  font: "Euler Math",
+  size: (FONT_SIZE * 1.0),
+  lang: "en",
+)
+
+#set heading(numbering: none)
+#show heading.where(level: 1): it => {
+  rect(inset: FONT_SIZE / 2)[#it]
+}
+
+#show ref: it => if it.element.func() != heading { it } else {
+  link(it.target, it.element.body)
+}
+
+#lib.title(size: FONT_SIZE)[exercise 9]
+
+these are my solutions to the ninth exercise set of TMA4135.
+
+this document was created using
+#link("https://typst.app/")[#text(blue.darken(5%))[typst]].
+
+\
+
+#outline(title: none)
+
+#pagebreak()
+
+#include "problem1.typ"
+
+#include "problem2.typ"
+
+#include "problem3.typ"
+
+#include "problem4.typ"
diff --git a/exercise9/problem1.typ b/exercise9/problem1.typ
new file mode 100644
index 0000000..a93f013
--- /dev/null
+++ b/exercise9/problem1.typ
@@ -0,0 +1,114 @@
+#import "@preview/physica:0.9.6": *
+
+= problem 1
+
+recall the definition of a periodic function $f$ for a $p > 0$
+$
+  f(x + p) = f(x) quad forall quad x in RR.
+$
+the smallest such $p$ is called the fundamental period of $f$.
+
+== a)
+
+"every periodic function has a fundamental period" is a false statement.
+
+examine $f(x) = 1,$ which is periodic since given a $p = 1$, then $f(x + 1)
+= f(x) = 1$ for all $x in RR$. it is trivially periodic. however, there is no
+smallest $p$ for which this holds:
+
+given a $delta > 0$ there is always a $hat(delta) > 0$ for which $hat(delta)
+< delta$ holds. this is a property of the real number line, thus there is no
+fundamental period for this periodic function.
+
+== b)
+
++ "$"Per"_p := {f: RR -> RR | f #[is $p$-periodic]}$ is a vector space" is
+  a true statement since all periodic functions that are added together or
+  scaled by some scalar are still periodic. this comes from the linear property
+  of periodic functions.\
+  we can prove this by taking two periodic functions $f$ and $g$ and seeing if
+  their linear combination is an element of the space
+  $
+    h(t) & := a dot f(t) + b dot g(t) \
+         & = a dot f(t + p_f) + b dot g(t + p_g) \
+         & = a dot f(t + p_f dot p_g) + b dot g(t + p_g dot p_f) \
+         & = h(t + p_f dot p_g) = h(t + p_h)
+  $
+  thus the linear combination $h(t)$ must be periodic itself and therefore an
+  element of $"Per"_p$. this proves scalar multiplication and vector addition
+  axioms, rest is trivial.
+
++ "let $phi.alt : RR -> RR => phi.alt compose f$ is $p$-periodic" is true, since
+  it 'captures' the input of the function $phi.alt$ such that it becomes
+  periodic itself, always reiterating over the same values. \
+  $
+    (phi.alt compose f)(t) & = phi.alt(f(t)) \
+                           & = phi.alt(f(t + p)) \
+                           & = (phi compose f)(t + p)
+  $
+
++ "let $n in NN, a in RR => f(x + a), f(n x), f(x slash n)$ are $p$-periodic" is
+  false, since scaling the input parameter changes the period. \
+  $
+    (1) &quad f(x + a) = f((x + p) + a) = f((x + a) + p)) #underline[ok] \
+    (2) &quad f(n x) = f(n (x + p)) = f(n x + n p) = f(n x + p) #underline[ok] \
+    (3) &quad f(x slash n) = f((x + p) slash n) = f(x slash n + p slash n) #underline[not ok]
+  $
+  (2) works because $n in NN$, such that the period repeats $n$ times. (3)
+  doesn't work because dividing by a natural number causes the period to
+  contract and thus isn't $p$-periodic anymore.
+
++ "the absolute difference between two periods $p$ and $p'$ is also a period of
+  $f$" is true, since it just means that the periods occur periodically. \
+  $
+    f(x) & = f(x + p') = f(x - p') \
+         & = f((x + p) - p') = f(x + (p - p'))
+  $
+  given that $p > p'$.
+
++ "let $a, b in RR => integral_a^(a + p) f(x) dd(x) = integral_b^(b + p) f(x) dd(x)$"
+  is true, since it says that integrating over the period is the same regardless
+  of where you start integrating from. \
+
+  we can choose a $c in RR$ such that
+  $
+    integral_a^(a + p) f(x) dd(x)
+    & = integral_a^c f(x) dd(x) + integral_c^(a + p) f(x) dd(x) \
+    & = integral_c^(a + p) f(x) dd(x) + integral_a^c f(x) dd(x)
+    = integral_b^(b + p) f(x) dd(x)
+  $
+  because $c$ is a midpoint chosen to displace the integral sum such that the
+  startpoints of the integration are the same in terms of the period of the
+  function $f$.
+
++ "if $f$ is differentiable, $f'$ is also $p$-periodic" is true \
+  $
+    f(x) = f(x + p) ==> f'(x) = f'(x + p)
+  $
+
+== c)
+
+- $f(x) = cos(2x + 3)$ has a fundamental period $p = pi$ since the $+ 3$ doesn't
+  affect the period and then the usual period of $2 pi$ is halved by the
+  coefficient in-front of $x$.
+
+- $f(x) = pi sin(3/2 pi x)$ has a fundamental period $p = 4 slash 3$ since the
+  $pi$ in-front of the $sin$-expression only affects the amplitude and $(2 pi)
+  / ((3 slash 2) pi) = 4 slash 3$.
+
+- $f(x) = cos(pi/(m+1) x) + sin(pi/(n-1) x)$ for $m in ZZ \\ {-1}, n in ZZ \\ {1}$
+  can be broken into two functions that can be analyzed separately first.
+  - $g(x) := cos(pi/(m + 1) x)$ has a fundamental period $p_g = 2(m + 1)$.
+  - $h(x) := sin(pi/(n + 1) x)$ has a fundamental period $f_h = 2(n + 1)$.
+  - we can combine these two to obtain $f(x) = g(x) + h(x)$.
+  to find the fundamental period, we can draw inspiration from number theory to
+  see that the combined period must be
+  $
+    gcd(p_g, f_H) & = gcd(2(m + 1), 2(n + 1)) \
+                  & = gcd(m + 1, n + 1) \
+                  & <= (m + 1)(n + 1) \
+                  & = m n + m + n + 1
+  $
+  but we cannot shorten this further using $p$-period algebra, so this must be
+  the fundamental period for this wave.
+
diff --git a/exercise9/problem2.typ b/exercise9/problem2.typ
new file mode 100644
index 0000000..6fbd0e3
--- /dev/null
+++ b/exercise9/problem2.typ
@@ -0,0 +1,62 @@
+#import "@preview/physica:0.9.6": *
+
+= problem 2
+
+== a)
+
+$h(x) := f(x) g(x)$ is odd for odd $f$ and even $g$, since
+$
+  h(x) = -f(-x) g(-x) = - (f(-x) g(-x)) = -h(-x)
+$
+
+== b)
+
+if $f$ and $g$ are both odd or even, then $h(x) := f(x) g(x)$ is even
+
+1. for the first case: $f$ and $g$ are both even
+  $
+    h(x) = f(x) g(x) = f(-x) g(-x) = h(-x)
+  $
+
+2. for the second case: $f$ and $g$ are both odd
+  $
+    h(x) & = f(x) g(x) \
+         & = (-f(-x)) (-g(-x)) = f(-x) g(-x) = h(-x)
+  $
+
+== c)
+
+if $f$ is odd and $g$ is even then both $f compose g$ and $g compose f$ are
+even, since
+1. $f(g(x)) = f(g(-x))$, since g is even
+2. $g(f(x)) = g(-f(-x)) = g(f(-x))$
+
+== d)
+
+if $f$ is odd and $L > 0$,
+$
+  integral_(-L)^L f(x) dd(x) & = integral_(-L)^0 f(x) dd(x)
+                               + integral_0^L f(x) dd(x) \
+                             & = integral_0^L f(-x) (-1) dd(x)
+                               + integral_0^L f(x) dd(x) \
+                             & = -integral_0^L f(-x) dd(x)
+                               + integral_0^L f(x) dd(x) \
+                             & = -integral_0^L f(x) dd(x)
+                               + integral_0^L f(x) dd(x) \
+                             & = 0
+$
+
+== e)
+
+if $f$ is even and $L > 0$,
+$
+  integral_(-L)^L f(x) dd(x) & = integral_(-L)^0 f(x) dd(x)
+                               + integral_0^L f(x) dd(x) \
+                             & = integral_0^L f(-x) dd(x)
+                               + integral_0^L f(x) dd(x) \
+                             & = integral_0^L f(x) dd(x)
+                               + integral_0^L f(x) dd(x) \
+                             & = 2 integral_0^L f(x) dd(x)
+$
+
+
diff --git a/exercise9/problem3.typ b/exercise9/problem3.typ
new file mode 100644
index 0000000..ef32449
--- /dev/null
+++ b/exercise9/problem3.typ
@@ -0,0 +1,91 @@
+#import "@preview/physica:0.9.6": *
+
+= problem 3
+
+$
+  a_0(f') = 0, quad a_n (f') = n b_n (f), quad b_n(f') = -n a_n(f)
+$
+
+if $f(-pi) != f(pi)$ then the formulas don't work anymore, since there will be
+a discreet jump every time the period is restarted. to fix the formulas we need
+to account for this. let $[f] = f(pi) - f(-pi)$
+$
+  cases(
+    a_0 (f') & = 1/(2 pi) integral_(-pi)^pi f'(x) dd(x)
+               = 1/(2 pi) [f(x)]_(-pi)^pi = [f]/(2 pi), ,
+    a_n (f') & = 1/pi integral_(-pi)^pi f'(x) cos(n x) dd(x) \
+             & = 1/pi ([f(x) cos(n x)]_(-pi)^pi
+                 + n integral_(-pi)^pi f(x) sin(n x) dd(x)) \
+             & = 1/pi ((-1)^n [f] + n pi b_n (f)) \
+             & = n b_n (f) + 1/pi (-1)^n [f], ,
+    b_n (f') & = - n a_n (f)
+  )
+$
+
+recall the fourier series for a $p$-periodic function $f$
+$
+  f(x) approx f_N (x) = a_0 + sum_(n=1)^N
+  (a_n cos((2 pi)/p x) + b_n sin((2 pi)/p x))
+$
+where
+$
+  cases(
+    a_0 = 1/p integral_p f(x) dd(x),
+    a_n = 2/p integral_p f(x) cos((2 pi)/p n x) dd(x),
+    b_n = 2/p integral_p f(x) sin((2 pi)/p n x) dd(x)
+  )
+$
+
+let $f(x) = sin^2(x) + 3 x^2 - 4 x + 5$ be periodically continued based on
+$[-pi, pi]$.
+
+notice that this series has a jump $[f] > 0$ so we need to use the modified
+formulas we found above.
+
+also remark that $sin^2(x) = (1 - cos(2x))/2$, which is already its fourier
+series.
+
+since fourier series are linear, we can compute the fourier series of each
+term and we can use the provided properties to compute the coefficients using
+the derivative of $f$.
+$
+  f'(x) = dots.c + 6 x - 4
+$
+we then need to compute the fourier series for $x^2$, $x$ and $1$ using
+$
+    "fourier"(1) & = cases(
+                     a_0 (1) = 1/(2 pi) integral_(-pi)^pi 1 dd(x) = 1,
+                     a_n (1) = 1/pi integral_(-pi)^pi cos(x) dd(x) = 0,
+                     b_n (1) = 0
+                   ) \
+    "fourier"(x) & = cases(
+                     a_0 (x) = 1/(2 pi) integral_(-pi)^pi x dd(x) = 0,
+                     a_n (x) = 1/n (b_n (1)) = 0,
+                     b_n (x) = 1/n (a_n (1) - 1/pi (-1)^n [x])
+                     = 2/n (-1)^(n + 1)
+                   ) \
+  "fourier"(x^2) & = cases(
+                     a_0 (x^2) = 1/(2 pi) integral_(-pi)^pi x^2 dd(x)
+                     = 1/(6 pi) [x^3]_(-pi)^pi = 1/3 pi^2,
+                     a_n (x^2) = -1/n b_n (2 x)
+                     = -2/n b_n (x) = 4/n^2 (-1)^n,
+                     b_n (x^2) = 1/n (a_n (2 x) - 1/pi (-1)^n [x^2]) = 0
+                   )
+$
+
+combining these we get
+$
+  "fourier"(f(x)) & = "fourier"(sin^2(x)) \
+                  & + "fourier"(3 x^2) - "fourier"(4 x) + "fourier"(5) \
+                  & = 1/2 (1 - cos(2 x)) + 3 "fourier"(x^2) \
+                  & - 4 "fourier"(x) + 5 "fourier"(1) \
+                  & = 1/2 (1 - cos(2 x)) x + cases(
+                      a_0 (f(x)) & = 3 a_0 (x^2) - 4 a_0 (x) + 5 a_0 (1) \
+                                 & = pi^2 + 5,
+                      a_n (f(x)) & = 12/n^2 (-1)^n,
+                      b_n (f(x)) & = -8/n (-1)^(n+1) = 8/n (-1)^n
+                    ) \
+                  & = pi^2 + 11/2 - 1/2 cos(2 x) \
+                  & + sum_(n=1)^oo (12/n cos(n x)
+                      + 8 sin(n x))(-1)^n / n
+$
diff --git a/exercise9/problem4.typ b/exercise9/problem4.typ
new file mode 100644
index 0000000..903aef6
--- /dev/null
+++ b/exercise9/problem4.typ
@@ -0,0 +1,324 @@
+#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.