upload diverse filer

01001 - Matematik 1a/Mat1a gennemgang/Mat1a gennemgang.typ

@@ -668,7 +668,7 @@ Er et polynomium $Z^n=w$
 $Z,w in CC, n in ZZ_(>=0)$
 
 $
-    Z= root(n, |w|) dot e^(i ((a r g(w))/n) + p (2 pi)/n), p = {0,dots,n-1}
+    Z= root(n, |w|) dot e^(i ((a r g(w))/n + p (2 pi)/n)), p = {0,dots,n-1}
 $
 
 #example()[
@@ -699,7 +699,7 @@ Et n'te grads polynomie har n komplekse rødder regnet med multiplicitet.
 Man kan omskrive alle polynomier som et produkt af førstegradspolynomier.\
 Hvis $p(z)$ har rødder $lambda_1, dots, lambda_n$, kan det omskrives til:
 $
-P(z)=a_n(z-lambda_1) dot dots dot (z-lambda_n)
+P(z)=a_n (z-lambda_1) dot dots dot (z-lambda_n)
 $
 eller med multiplicitet:
 $
@@ -840,7 +840,7 @@ sum^(n-1)_(k=1)k=((n-1)(n-1+1))/2=((n-1)dot n)/2
 $
 
 $
-sum^(n)_(k=1)k=underbrace(1+2+3+dots+(n-2)+(n-1)+, sum^(n-1)_(k=1)k) n
+sum^(n)_(k=1)k=underbrace(1+2+3+dots+(n-2)+(n-1), sum^(n-1)_(k=1)k) + n
 $
 $
 sum^(n)_(k=1)k&=sum^(n-1)_(k=1)k + n\
@@ -1022,7 +1022,7 @@ $u = a_1+b_1Z\ v=a_2+b_2Z$
 
 Betragt
 $
-u+ alpha dot v &= (a_1+b_1Z)+alpha (a_2 + b_2 Z)
+u+ alpha dot v &= (a_1+b_1Z)+alpha (a_2 + b_2 Z)\
 &= a_1 + b_1Z + alpha a_2+ alpha b_2Z = underbrace((a_1 + alpha a_2), a in CC) + underbrace((b + alpha b_2), b in CC)Z
 $
 ]
@@ -1164,7 +1164,7 @@ $f$ er injektiv hvis og kun hvis $ker(f)={0}$ (og $f$ er lineær)
 
 Derfor, find nulrummet. $"rref"(amat(f, gamma, beta))=mat(1,0), x_1 = 0$ dvs.
 
-$amat(f, gamma, beta)) underbrace(vec(0,t), = t vec(0,1))=0$ for *alle* $t in RR$
+$amat(f, gamma, beta) underbrace(vec(0,t), = t vec(0,1))=0$ for *alle* $t in RR$
 
 $f(vec(0,1))=0$. Tilbage til normale koordinater.
 

01001 - Matematik 1a/Noter/Andenordens differentialligninger/python-mobius.ipynb

@@ -0,0 +1,142 @@
+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "8d87a703",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import sympy as sp\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "dffe7f14",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/latex": [
+       "$\\displaystyle \\left[\\begin{matrix}52 & -8 & -8\\\\-4 & 50 & 2\\\\-4 & 2 & 50\\end{matrix}\\right]$"
+      ],
+      "text/plain": [
+       "Matrix([\n",
+       "[52, -8, -8],\n",
+       "[-4, 50,  2],\n",
+       "[-4,  2, 50]])"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "{60: 1, 48: 1, 44: 1}\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Opgave 1\n",
+    "A = sp.Matrix([[52, -8, -8], [-4, 50, 2], [-4, 2, 50]])\n",
+    "display(A)\n",
+    "eigenvalsA = A.eigenvals()\n",
+    "print(eigenvalsA)\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 36,
+   "id": "debdde8b",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/latex": [
+       "$\\displaystyle \\left[\\begin{matrix}9 & -2 & 1\\\\-1 & 10 & -1\\\\1 & -2 & 9\\end{matrix}\\right]$"
+      ],
+      "text/plain": [
+       "Matrix([\n",
+       "[ 9, -2,  1],\n",
+       "[-1, 10, -1],\n",
+       "[ 1, -2,  9]])"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[(8, 2, [Matrix([\n",
+      "[2],\n",
+      "[1],\n",
+      "[0]]), Matrix([\n",
+      "[-1],\n",
+      "[ 0],\n",
+      "[ 1]])]), (12, 1, [Matrix([\n",
+      "[ 1],\n",
+      "[-1],\n",
+      "[ 1]])])]\n",
+      "hej\n",
+      "Matrix([[2], [1], [0]])\n",
+      "Matrix([[-1], [0], [1]])\n",
+      "Matrix([[1], [-1], [1]])\n",
+      "Matrix([[12], [20], [-20]])\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Opgave 2\n",
+    "\n",
+    "B = sp.Matrix([[9, -2, 1], [-1, 10, -1], [1, -2, 9]])\n",
+    "display(B)\n",
+    "eigenvectsB = B.eigenvects()\n",
+    "\n",
+    "print(eigenvectsB)\n",
+    "print(\"hej\")\n",
+    "print(eigenvectsB[0][2][0])\n",
+    "print(eigenvectsB[0][2][1])\n",
+    "print(eigenvectsB[1][2][0])\n",
+    "\n",
+    "print(sp.Matrix([[9, -2, 1], [-1, 10, -1], [1,-2,9]]) * sp.Matrix([2, 2, -2]))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "fa460b84",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Opgave 3\n",
+    "\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.13.7"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}

01001 - Matematik 1a/Noter/Matrixalgebra og determinanter/Matrixalgebra og determinanter.typ

@@ -1,7 +1,7 @@
 #import "@local/dtu-template:0.5.1":*
 
 
-#show: dtu-math-assignment.with(
+#show: dtu-note.with(
   course: "01001",
   course-name: "Mathematics 1a (Polytechnical Foundation)",
   title: "Matrixalgebra og determinanter",
@@ -94,3 +94,49 @@ $
 ]
 
 *Gange matricer sammen:*
+Bygger videre på før.
+$
+bold(A) in FF^(m times n), quad bold(B) in FF^(n times L)
+$
+Antal søjler i $bold(B)$ skal være lig med antal rækker i $bold(A)$.
+
+$bold(B) = mat(bold(b_1),dots,bold(b_L))$
+
+$
+bold(A) dot bold(B) = mat(bold(A) dot bold(b_1), bold(A) dot bold(B_2), dots, bold(A) dot bold(b_L))
+$
+Bemærk $bold(A) dot bold(B) in FF^(m times L)$
+
+#note-box()[
+Bemærk: $bold(A) dot bold(B) eq.not bold(B) dot bold(A)$
+]
+
+= Lineære ligningssystem
+$
+cases(a_(11) x_1 + dots + a_(1 n) x_n = b_1, dots.v, a_(m 1) x_1 + dots + a_(m n) x_n = b_n)
+$
+Er det samme som:
+$
+mat(a_(11), dots, a_(1 n);dots.v;a_(m 1), dots, a_(m n)) dot vec(x_1, dots.v, x_n) = vec(b_1, dots.v, b_n)
+$
+
+#definition(title: "Identitetsmatrix")[
+$
+bold(I)_n = mat(1,0,dots,0;0,,,dots.v;dots.v,,,0;0,dots,0,1) in FF^(n times n)
+$
+Kaldes den $n times n$ identitetsmatrix
+]
+
+#definition(title: "7.3.1: Invers matrix")[
+Givet $bold(A) in FF^(n times n)$ hvis der findes en matrix $bold(B) in FF^(n times n)$ således at $bold(A) dot bold(B) = bold(I)_n$ og $bold(B) dot bold(A) = bold(I)_n$, så siges at $bold(A)$ er invers og $bold(B)$ kaldes $bold(A)$'s inverse matrix:
+$
+bold(B) = bold(A)^(-1)
+$
+
+Man finder den ved:
+$
+mat(bold(A),bold(I)_n; augment: #(-1)) arrow.long dots arrow.long mat(bold(I)_n, bold(A)^(-1);augment: #(-1))
+$
+]
+
+