From f6912cf338500409b3bb71ddd6d27dbf05069d84 Mon Sep 17 00:00:00 2001
From: fredrikr79 <fredrikrobertsen7@gmail.com>
Date: Tue, 28 Oct 2025 15:03:09 +0100
Subject: [PATCH] ex10: problem 1

---
 exercise10/main.typ     |   2 +-
 exercise10/problem1.typ | 136 ++++++++++++++++++++++++++++++++++++++++
 2 files changed, 137 insertions(+), 1 deletion(-)

diff --git a/exercise10/main.typ b/exercise10/main.typ
index 28cadb8..60ac8c5 100644
--- a/exercise10/main.typ
+++ b/exercise10/main.typ
@@ -21,7 +21,7 @@
   link(it.target, it.element.body)
 }
 
-#lib.title(size: FONT_SIZE)[exercise 9]
+#lib.title(size: FONT_SIZE)[exercise 10]
 
 these are my solutions to the tenth exercise set of TMA4135.
 
diff --git a/exercise10/problem1.typ b/exercise10/problem1.typ
index 617b32c..969499f 100644
--- a/exercise10/problem1.typ
+++ b/exercise10/problem1.typ
@@ -1,4 +1,140 @@
 #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.