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+#import "@preview/cetz:0.3.2";
+#import "@preview/cetz-plot:0.1.1": plot
+#import "@preview/physica:0.9.4": *
+#import "@preview/plotsy-3d:0.1.0": plot-3d-parametric-surface
+#import "@preview/fletcher:0.5.4" as fletcher: diagram, edge, node
+
+#set page(paper: "a4", margin: (x: 2.6cm, y: 2.8cm), numbering: "1 : 1")
+#set par(justify: true, leading: 0.52em)
+
+#let FONT_SIZE = 18pt;
+#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 heading.where(level: 2): it => [
+//   #set align(left)
+//   #set text(size: FONT_SIZE * 1.1, weight: "semibold")
+//   #(it.body)
+// ]
+
+#align(center)[
+  #text(size: FONT_SIZE * 2, weight: "bold")[#underline[exercise 1]]
+]
+
+these are my solutions to the first exercise set of TMA4135.
+
+i recommend using a PDF-reader with document rotation capabilities, like
+#link("https://wiki.archlinux.org/title/Zathura")[#text(blue.darken(5%))[zathura]].
+
+this document was created using
+#link("https://typst.app/")[#text(blue.darken(5%))[typst]].
+
+#v(42pt)
+
+#outline(title: none)
+
+
+= problem 1
+
+== a)
+
+#[#show math.equation: set text(size: (FONT_SIZE * 0.60))
+  #rotate(
+    -90deg,
+    reflow: true,
+    table(
+      $bold(u(x, y, t))$, $bold(u_y)$, $bold(u_t)$, $bold(u_(x x))$, $bold(u_(x y))$, $bold(u_(y x))$,
+
+      $t^4 - cos(x y)$,
+      $x sin(x y)$,
+      $4 t^3$,
+      $y^2 cos(x y)$,
+      $x y cos(x y)$,
+      $x
+      y cos(x y)$,
+
+      $-sin(t x y)$,
+      $- t x cos(t x y)$,
+      $- x y cos(t x y)$,
+      $t^2 y^2 sin(t x y)$,
+      $t^2 x y sin(t x y)$,
+      $t^2 x y cos(t x y)$,
+
+      $e^(-t) sin(x) ln(y)$,
+      $(e^(-t) sin(x)) slash y$,
+      $- e^(-t) sin(x) ln(y)$,
+      $-e^(-t) sin(x) ln(y)$,
+      $(e^(-t) cos(x)) slash y$,
+      $(e^(-t) cos(x)) slash y$,
+
+      $e^(-x) sqrt(x^3 + y^2)$,
+      $(2 y e^(-x)) slash sqrt(x^3 + y^2)$,
+      $0$,
+      $(dagger)$,
+      $(dagger.double)$,
+      $(dagger dagger)$,
+
+      $(t e^t) sin(x)$, $0$, $e^t (t + 1) sin(x)$, $- t e^t sin(x)$, $0$, $0$,
+
+      $sin(t) e^(-x) + cos(t) e^(-y)$, $- cos(t) e^(-y)$, $cos(t) e^(-x) - sin(t) e^(-y)$, $sin(t) e^(-x)$, $0$, $0$,
+
+      rows: 7,
+      columns: 6,
+    )
+      + [#set text(size: FONT_SIZE * 0.6, fill: gray.darken(35%))
+        #show math.equation: set text(size: FONT_SIZE * 0.5)
+        some calculations\
+        #table(
+          table.cell(
+            rowspan: 2,
+            $(dagger)$,
+          ),
+          table.cell(
+            rowspan: 2,
+            $
+              & quad pdv(, x, 2)e^(-x) sqrt(x^3 + y^2) \
+              & = pdv(, x) ((3 x^2 e^(-x) ) / (2 sqrt(x^3 + y^2))
+                  - e^(-x) sqrt(x^3 + y^2)) \
+              & = 3/2 dot ((2 x e^(-x) - x^2 e^(-x)) sqrt(x^3 + y^2)
+                - (3 x^4 e^(-x)) / (2 sqrt(x^3 + y^2))) / (x^3 + y^2)
+                - (3 x^2 e^(-x) ) / (2 sqrt(x^3 + y^2))
+                + e^(-x) sqrt(x^3 + y^2) \
+              & = (3 e^(-x) ((2 x - x^2) (x^3 + y^2) - 3 x^4))
+                / (4(x^3 + y^2)^(3/2))
+                - (6 x^2 e^(-x) (x^3 + y^2)) / (4 (x^3 + y^2)^(3/2))
+                + (4 e^(-x) (x^3 + y^2)^2) / (4 (x^3 + y^2)^(3/2)) \
+              & = (3 e^(-x) (2 x y^2 - x^5 - x^2 y^2 - x^4)
+                - 6 x^2 e^(-x) (x^3 + y^2)
+                + 4 e^(-x) (x^6 + 2 x^3 y^2 + y^4))
+                / (4 (x^3 + y^2)^(3/2)) \
+              & = (6 x y^2 e^(-x) - 3 x^5 e^(-x) - 3 x^2 y^2 e^(-x) - 3 x^4 e^(-x)
+                - 6 x^5 e^(-x) + 6 y^2 e^(-x)
+                + 4 x^6 e^(-x) + 8 x^3 y^2 e^(-x) + 8 y^4 e^(-x))
+                / (4 (x^3 + y^2)^(3/2)) \
+              & = (6 x y^2 e^(-x) - 9 x^5 e^(-x) - 3 x^2 y^2 e^(-x) - 3 x^4 e^(-x)
+                + 6 y^2 e^(-x)
+                + 4 x^6 e^(-x) + 8 x^3 y^2 e^(-x) + 8 y^4 e^(-x))
+                / (4 (x^3 + y^2)^(3/2)) \
+              & = e^(-x) dot (6 x y^2 - 9 x^5 - 3 x^2 y^2 - 3 x^4
+                + 6 y^2
+                + 4 x^6 + 8 x^3 y^2 + 8 y^4)
+                / (4 (x^3 + y^2)^(3/2)) \
+              & #[a few errors somewhere, but close enough...]
+            $,
+          ),
+
+          $(dagger.double)$,
+          $
+            & quad pdv(, y, x) e^(-x) sqrt(x^3 + y^2) \
+            & = pdv(, y) ((3 x^2 e^(-x) ) / (2 sqrt(x^3 + y^2))
+                - e^(-x) sqrt(x^3 + y^2)) \
+            & = (-3 x^2 y e^(-x))/2 dot (x^3 + y^2)^(-3/2)
+              - (2 y e^(-x))/(2 sqrt(x^3 + y^2))
+          $,
+
+          $(dagger dagger)$,
+          $
+            & quad pdv(, x, y) e^(-x) sqrt(x^3 + y^2) \
+            & = pdv(, x) space (-y e^(-x))/(sqrt(x^3 + y^2)) \
+            & = (y e^(-x) sqrt(x^3 + y^2) + y e^(-x) dot 1 slash 2 dot (x^3
+                + y^2)^(-1/2)) / (x^3 + y^2) \
+            & = (y e^(-x) (sqrt(x^3 + y^2) + 1 slash 2 dot (x^3
+                  + y^2)^(-1/2))) / (x^3 + y^2)
+          $,
+
+          columns: 4,
+          stroke: none,
+        )
+      ],
+  )
+]
+
+
+== b)
+
+
+define $ f^i_k := (partial f^i) / (partial y^k)
+quad quad f^i_(k l) := (partial f^i) / (partial y^k partial y^l) $
+
+and let the jacobian matrix $ frak(J) = [ (f_y)_(i j) = f^i_j ] $
+
+where $f^2_y$ is the matrix product and $f_(y y)$ has entries $(f^i_(k l))$.
+
+consider $ f : RR^m -> RR^m,
+space y |-> (f^1 (y^1, ..., y^m), ..., f^m (y^1, ..., y^m))^T $
+
+show that $ (f_y f)_y f = f^T f_(y y) f + f^2_y f $
+
+#align(center)[#line(length: 75%)]
+
+$
+  y |-> #text(red)[$f(y)$] |-> pdv(, y) (pdv(f, y)(f(#text(red)[$x$])))
+$ $
+  => (f_y f)_y f & = pdv(, y) (pdv(f, y) (f(f(y)))) \
+                 & = pdv(, y) (pdv(f, y) (f^2(y)))
+$
+
+$
+  y |-> #text(red)[$f(y)$] |-> f^T (pdv(f, y, 2)(#text(red)[$x$])) + pdv(f, y)
+  (pdv(f, y) (#text(red)[$x$]))
+$ $
+  => f^T f_(y y) f + f^2_y f & = (f(pdv(f, x, 2)(f(y))))^T + pdv(f, x) (pdv(f, x) (f(y)))
+$
+
+we can see through $eta$-reduction that we only need to show $ (f_y f)_y = f^T f_(y y) + f^2_y $
+
+and further that $ f_y f = f^T f_y + f_y f $
+
+thus we need to prove that $ f^T f_y = 0 $
+
+but $ f^T f_y = (f(pdv(f, y)))^T = (f^1(f^1_y, ..., f^m_y), ..., f^m (f^1_y, ..., f^m_y)) $
+
+
+
+= problem 2
+
+== a)
+
+let $f(x) = x^4 + 3 x^3 - 2 x + 5$; find all taylor polynomials around
+$x_0 = -2$.
+
+#align(center)[#line(length: 75%)]
+
+recall that each term is given by $ P_k (x) = (f^((k)) (x_0)) / k! (x - x_0)^k $
+for $k in [0, deg(f)] inter ZZ$.
+
+first compute
+$
+       f'(x) & = 4x^3 + 9x^2 - 2 \
+      f''(x) & = 12x^2 + 18x \
+  f^((3))(x) & = 24 x + 18 \
+  f^((4))(x) & = 24
+$
+
+then
+$
+  P_0 & = f(-2) = 16 - 24 + 4 + 5 = 1 \
+  P_1 & = f'(-2) dot (x + 2) = 2 (x+2) = 2x + 4 \
+  P_2 & = (f''(-2))/2 dot (x + 2)^2 = 6 x^2 + 24x + 24 \
+  P_3 & = (f^((3)) (-2))/6 dot (x + 2)^3 = -5 (x + 2)^3 \
+      & = -5x^3 - 30 x^2 - 60 x - 40 \
+  P_4 & = (x + 2)^4 = x^4 + 8 x^3 + 24 x^2 + 32 x + 16
+$
+
+then the $k$-th taylor polynomial can be expressed as
+$
+  T_k = sum_(i=0)^k P_i
+$
+
+or alternatively
+$
+  T_k = T_(k-1) + P_k
+$
+
+for $k in NN^+$.
+
+thus the taylor polynomials for $f$ are
+$
+  T_0 & = P_0 = 1 \
+  T_1 & = P_0 + P_1 = 2x + 5 \
+  T_2 & = 6x^2 + 26x + 28 \
+  T_3 & = -5x^3 - 24x^2 + 34x - 12 \
+  T_4 & = x^4 + 3x^3 + 66x + 4
+$
+
+== b)
+
+let $g(x) = ln(1 + x)$; calculate its maclaurin series.
+
+#align(center)[#line(length: 75%)]
+
+first we differentiate
+$
+        g'(x) & = 1/(1+x) \
+       g''(x) & = -1/(1+x)^2 \
+              & dots.v \
+  g^((k)) (x) & = (-1)^(k-1) dot (k-1)! dot (1 + x)^(-k)
+$
+
+for $k in NN^+$.
+
+recall that the maclaurin series of a non-analytic function $f$ is
+$
+  f(x) = sum_(i=0)^k (f^((i)) (0))/(i!) x^i + O(x^(k+1))
+$
+
+now lucky us, since
+$
+  g^((k)) (0) = (-1)^(k-1) dot (k - 1)!
+$
+
+so
+$
+  g(x) & = sum_(i=1)^k (g^((i)) (0)) / (i!) x^i + O(x^(k+1)) \
+       & = sum_(i=1)^k (-1)^(i-1) (i-1)! / (i!) x^i + O(x^(k+1)) \
+       & = sum_(i=1)^k ((-1)^(i-1) x^i) / i + O(x^(k+1)) \
+       & = x - x^2 / 2 + x^3 / 3 + dots.c + O(x^(k+1))
+$
+
+we may choose $O(x^(k+1)) = 0$.
+
+= problem 3
+
+== a)
+
+are $1+x$, $1-x$ and $x-x^2$ linearly independent in $P_2$?
+
+#align(center)[#line(length: 75%)]
+
+let us denote these in vector notation as
+$
+  vec(0, 1, 1), vec(0, -1, 1) #[and] vec(-1, 1, 0)
+$
+respectively. thus we can determine their dependency via gauss-jordan
+elimination
+$
+  mat(0, 0, -1; 1, -1, 1; 1, 1, 0) ~
+  mat(1, 0, 0; 0, 1, 0; 0, 0, 1) = I_3
+$
+
+the three vectors are thus linearly independent. then what is their span?
+since they are linearly independent in $P_2$, they form a basis for the vector
+space and thus span out $P_2$. of note: $P_2$ is itself isomorphic to $RR^3$.
+
+== b)
+
+our affine space has two conditions, $p(1) = 1$ and $p(2) = 2$.
+
+let
+$
+  p(x) = a x^3 + b x^2 + c x + d
+$
+such that
+$
+  p(1) = a + b + c + d = 1
+$
+and
+$
+  p(2) = 8 a + 4 b + 2 c + d = 2
+$
+
+we can represent this system with a matrix
+$
+  mat(1, 1, 1, 1 | 1; 8, 4, 2, 1 | 2) tilde
+  mat(1, 1, 1, 1 | 1; 7, 3, 1, 0 | 1)
+$
+
+thus we can see that there are two linearly independent basis vectors that may
+form a two-dimensional linear space.
+
+== c)
+
+we can create a basis out of the three vectors
+$
+  x-1, x^2 - 1 #[and] x^3 - 1
+$
+to form a three-dimensional space, since they are linearly independent.
+
+from the results of the last task we can intuitively guess that the three
+conditions will lead to a system of equations with three unknowns. thus the
+system is solvable and we may indeed choose arbitrary values for our
+conditions.
+
+so let
+$
+  p(x) = alpha (x - 1) + beta (x^2 - 1) + gamma (x^3 - 1)
+$
+such that
+$
+  p(0) & = - alpha - beta - gamma = y_0, \
+  p(1) & = 0 = y_1, \
+  p(2) & = alpha + 3 beta + 7 gamma = y_2
+$
+
+this is different from my original expectation. we can tell that $y_1$ must be
+0, thus cannot be chosen arbitrarily. then we effectively only have two
+equations, thus ending up in the same situation as the last subtask, meaning we
+will not be able to choose the remaining values arbitrarily, since the system of
+equations will be underdetermined.
+
+
+= problem 4
+
+== a)
+
+#let inner(f, g) = $angle.l #f, #g angle.r$
+
+prove that ${sin(t), cos(t), 1}$ is orthogonal in the space $C[0, 2 pi]$ with
+inner product
+$
+  inner(f, g) = integral_0^(2 pi) f(s) g(s) dd(s)
+$
+i.e. that it is an orthogonal basis for $C[0, 2 pi]$.
+
+#align(center)[#line(length: 75%)]
+
+we must compute the pair-wise inner product of each base vector
+$
+  inner(sin, cos) & = integral_0^(2 pi) sin(t) cos(t) dd(t) = \
+    inner(sin, 1) & = integral_0^(2 pi) sin(t) dd(t) = \
+    inner(1, cos) & = integral_0^(2 pi) cos(t) dd(t) = 0
+$
+because $sin(t) cos(t)$, $sin(t)$ and $cos(t)$ all have a period of $2 pi$.
+
+thus they are all orthogonal and they form a basis under this definition of
+inner product. to make it an orthonormal basis, we can scale each base component
+by its length, such that
+$
+  frak(O) := {sin(t)/a, cos(t)/b, 1/c}
+$
+forms an orthonormal basis, where
+$
+  a & = sqrt(inner(sin, sin)) = (integral_0^(2 pi) sin^2 t dd(t))^(1 slash 2) = sqrt(pi) \
+  b & = sqrt(inner(cos, cos)) = (integral_0^(2 pi) cos^2 t dd(t))^(1 slash 2) = sqrt(pi) \
+  c & = sqrt(inner(1, 1)) = (integral_0^(2 pi) dd(t)) = sqrt(2 pi)
+$
+
+== b)
+
+to form an orthonormal basis for the monomials ${1, x, x^2}$, we use the
+gram-schmidt method with $vectorbold(v_1) = 1$.
+
+then
+$
+  vectorbold(v_2) & = x - "proj"_(vectorbold(v_1)) (x) \
+                  & = x - (inner(vectorbold(v_1), x) / inner(
+                        vectorbold(v_1),
+                        vectorbold(v_1)
+                      )) vectorbold(v_1) \
+                  & = x - inner(1, x) / inner(1, 1) \
+                  & = x - 1/2 integral_(-1)^1 x dd(x) \
+                  & = x
+$
+
+then similarly for the last vector
+$
+  vectorbold(v_3) & = x^2
+                    - "proj"_(vectorbold(v_1))(x^2)
+                    - "proj"_(vectorbold(v_2))(x^2) \
+                  & = x^2
+                    - 1/2 integral_(-1)^1 x^2 dd(x)
+                    - (integral_(-1)^1 x^2 dd(x))^(-1)
+                    dot integral_(-1)^1 x^3 dd(x) \
+                  & = x^2 - 1/3
+$
+
+thus we have found an orthonormal base using gram-schmidt
+$
+  frak(O) := {1/a, x/b, (x^2 - 1/3)/c}
+$
+
+where $a & := sqrt(2), b & := sqrt(2/3)$ and $c & := sqrt(2/5)$.
+