vault backup: 2026-02-26 16:22:06

vault backup: 2026-02-23 16:57:23
vault backup: 2026-02-23 16:06:43
2026-02-26 16:22:06 +01:00 · 2026-02-23 16:57:23 +01:00 · 2026-02-23 16:06:43 +01:00 · 2026-02-20 14:51:42 +01:00 · 2026-02-20 14:05:20 +01:00 · 2026-02-19 14:56:07 +01:00
46 changed files with 105737 additions and 21 deletions
@@ -1,3 +1,4 @@
 {
-  "cssTheme": "Catppuccin"
+  "cssTheme": "Catppuccin",
+  "baseFontSize": 20
 }
@@ -1,4 +1,6 @@
 [
  "obsidian-git",
-  "obsidian-style-settings"
+  "obsidian-style-settings",
+  "obsidian-tikzjax",
+  "obsidian-completr"
 ]
@@ -0,0 +1,164 @@
+[
+  {
+    "displayName": "Note",
+    "replacement": "note",
+    "icon": "lucide-pencil",
+    "color": "#448aff"
+  },
+  {
+    "displayName": "Summary",
+    "replacement": "summary",
+    "icon": "lucide-clipboard-list",
+    "color": "#00b0ff"
+  },
+  {
+    "displayName": "Abstract",
+    "replacement": "abstract",
+    "icon": "lucide-clipboard-list",
+    "color": "#00b0ff"
+  },
+  {
+    "displayName": "TL;DR",
+    "replacement": "tldr",
+    "icon": "lucide-clipboard-list",
+    "color": "#00b0ff"
+  },
+  {
+    "displayName": "Info",
+    "replacement": "info",
+    "icon": "lucide-info",
+    "color": "#00b8d4"
+  },
+  {
+    "displayName": "To-Do",
+    "replacement": "todo",
+    "icon": "lucide-check-circle-2",
+    "color": "#00b8d4"
+  },
+  {
+    "displayName": "Tip",
+    "replacement": "tip",
+    "icon": "lucide-flame",
+    "color": "#00bfa6"
+  },
+  {
+    "displayName": "Hint",
+    "replacement": "hint",
+    "icon": "lucide-flame",
+    "color": "#00bfa6"
+  },
+  {
+    "displayName": "Important",
+    "replacement": "important",
+    "icon": "lucide-flame",
+    "color": "#00bfa6"
+  },
+  {
+    "displayName": "Success",
+    "replacement": "success",
+    "icon": "lucide-check",
+    "color": "#00c853"
+  },
+  {
+    "displayName": "Check",
+    "replacement": "check",
+    "icon": "lucide-check",
+    "color": "#00c853"
+  },
+  {
+    "displayName": "Done",
+    "replacement": "done",
+    "icon": "lucide-check",
+    "color": "#00c853"
+  },
+  {
+    "displayName": "Question",
+    "replacement": "question",
+    "icon": "lucide-help-circle",
+    "color": "#63dd17"
+  },
+  {
+    "displayName": "Help",
+    "replacement": "Help",
+    "icon": "lucide-help-circle",
+    "color": "#63dd17"
+  },
+  {
+    "displayName": "FAQ",
+    "replacement": "faq",
+    "icon": "lucide-help-circle",
+    "color": "#63dd17"
+  },
+  {
+    "displayName": "Warning",
+    "replacement": "warning",
+    "icon": "lucide-alert-triangle",
+    "color": "#ff9100"
+  },
+  {
+    "displayName": "Caution",
+    "replacement": "caution",
+    "icon": "lucide-alert-triangle",
+    "color": "#ff9100"
+  },
+  {
+    "displayName": "Attention",
+    "replacement": "attention",
+    "icon": "lucide-alert-triangle",
+    "color": "#ff9100"
+  },
+  {
+    "displayName": "Failure",
+    "replacement": "failure",
+    "icon": "lucide-x",
+    "color": "#ff5252"
+  },
+  {
+    "displayName": "Fail",
+    "replacement": "fail",
+    "icon": "lucide-x",
+    "color": "#ff5252"
+  },
+  {
+    "displayName": "Missing",
+    "replacement": "missing",
+    "icon": "lucide-x",
+    "color": "#ff5252"
+  },
+  {
+    "displayName": "Danger",
+    "replacement": "danger",
+    "icon": "lucide-zap",
+    "color": "#ff1744"
+  },
+  {
+    "displayName": "Error",
+    "replacement": "error",
+    "icon": "lucide-zap",
+    "color": "#ff1744"
+  },
+  {
+    "displayName": "Bug",
+    "replacement": "bug",
+    "icon": "lucide-bug",
+    "color": "#f50057"
+  },
+  {
+    "displayName": "Example",
+    "replacement": "example",
+    "icon": "lucide-list",
+    "color": "#7c4dff"
+  },
+  {
+    "displayName": "Quote",
+    "replacement": "quote",
+    "icon": "quote-glyph",
+    "color": "#9e9e9e"
+  },
+  {
+    "displayName": "Cite",
+    "replacement": "cite",
+    "icon": "quote-glyph",
+    "color": "#9e9e9e"
+  }
+]
@@ -0,0 +1,10 @@
+{
+    "id": "obsidian-completr",
+    "name": "Completr",
+    "version": "3.2.0",
+    "minAppVersion": "1.0.0",
+    "description": "This plugin provides advanced auto-completion functionality for LaTeX, Frontmatter and standard writing.",
+    "author": "tth05",
+    "authorUrl": "https://github.com/tth05",
+    "isDesktopOnly": true
+}
@@ -0,0 +1,895 @@
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+}
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+{
+	"id": "obsidian-tikzjax",
+	"name": "TikZJax",
+	"version": "0.5.2",
+	"minAppVersion": "0.12.0",
+	"description": "Render LaTeX and TikZ diagrams in your notes",
+	"author": "artisticat",
+	"authorUrl": "https://github.com/artisticat1",
+	"isDesktopOnly": false
+}
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+    "Def_graf1.png",
+    "MVT.png",
+    "d_ex_1.png",
+    "d1.png",
+    "conflict-files-obsidian-git.md",
+    "gv1.png",
+    "k2.png",
+    "k1.png",
    "Untitled.canvas"
  ]
 }
@@ -0,0 +1,10 @@
+- **Lokal maximum punkt**: *i $x=x_0$ om $\exists\;\;a,b\in\mathbb{R}$ så att $x_0\in\left(a,b\right),\left(a,b\right)\subseteq D_f$ och $f(x)\leq f(x_0)\forall x\in\left(a,b\right)$
+- **Global maximum punkt**: *i $x=x_0$ om $f(x)\leq f(x_ 0)\;\forall\;x\in{D_f}$*
+- **Global extrempumkt**
+	1. *Lokala extrampunkter*
+	2. *Värde på ändpunkten. (Eller gränsvärde)*
+	3. *Värde på punkter där derivata saknas(Kritiska punkter)*
+	4. *Jämför 1,2,3.*
+	- **Ex**: $$\begin{align}f(x)=1-\mid{x}\mid\\f'(0)\text{ Existerar inte}\end{align}$$![[Def_graf1.png]]
+- **ODE**/**Primärfunktioner**/**Integraler**
+	- $$\begin{align}F'(x)=f(x)\\F(x)=\int f(x)dx\end{align}$$
@@ -0,0 +1,51 @@
+- Derivata
+	- **Def**: *$f$ är deriverbar i punkten $a$ om $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$existerar.$$f'(x)=\frac{df}{dx}(a)=Df(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$är derivatan av $f$ i punkten $x=a$. Funktionen $f'$ är derivatan av $f$ och deinieras som $x\longmapsto f'(x)$ där det är definiead.* 
+	- **Defs**:
+		- $Df$: *Oendlig liten ändrig i $f$*
+		- $Dx$: *Oendlig liten ändrig i $x$*
+	- $\overset{\bullet}f=f'$
+![[d1.png]]
+- Egenskaper och regler
+	- $f$ deriverbar $\Rightarrow$ $f$ kontinuerlig. **Obs!** Inte alla kontinuerliga funktioner är deriverbara
+	- Derivering är linjär avbildning: $\left(\alpha f+\beta g\right)'=\alpha f'+\beta g'$
+	- **Produkt regel** (*Leibniz*): $\left(f\left(x\right)g\left(x\right)\right)'=f'\left(x\right)g\left(x\right)+f\left(x\right)g'\left(x\right)$
+	- **Sammansatt funktion**: $\left(f\circ g\right)'\left(x\right)=f'\circ g\left(x\right)g'\left(x\right)$
+	- **Kjedje regel**: $(f(g(x)))'=f'(f(x))g'(x)$
+	- **Division**: $\left(\frac{f\left(x\right)}{g\left(x\right)}\right)'=\frac{f'\left(x\right)g\left(x\right)-f\left(x\right)g'\left(x\right)}{g\left(x\right)^2}$
+	- **Ex**: ![[d_ex_1.png]]$$\begin{align}f(x)=\mid x\mid\\f\text{ är kontinuerlig på }\mathbb{R}.\\f\text{ är inte deriverbar i }0.\\\lim_{x\to0+}\frac{f\left(x\right)-f\left(0\right)}{x-0}=\lim_{x\to0+}\frac{\mid x\mid-0}x=\lim_{x\to0+}\frac xx=1\\\lim_{x\to0-}\frac{f\left(x\right)-f\left(0\right)}{x-0}=\lim_{x\to0-}\frac{\mid x\mid-0}x=\lim_{x\to0-}\frac{-x}x=-1\\\lim_{x\to0}\frac{f\left(x\right)-f\left(0\right)}{x-0}=f'(0)\text{ existerar inte-}\end{align}$$
+	- **Ex**: $$\begin{align}\text{Leibniz regel}\\\left(f\left(x\right)g\left(x\right)\right)'=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)g\left(x\right)}h\\=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}\\=\lim_{h\to0}\left(g(x+h)\frac{f(x+h)-f(x)}{h}+f(x)\frac{g(x+h)-g(x)}{h}\right)\\=g(x)f'(x)+f(x)g'(x)\end{align}$$
+	- **Ex**: $$\begin{align}h(x)=\frac1x\\h'(x)=-\frac1{x^2}\\h\circ g(x)=h(g(x))=\frac1{g(x)}\\(g\circ g)'(x)=\left(\frac1{g(x)}\right)^2\\=h'\circ g(x)g'(x)=h'(g(x))g'(x)=\frac{-1}{(g(x))^2}g'(x)\end{align}$$
+- Standerd derivarives
+	1. $f(x)=c\;\;\Rightarrow\;\;f'(x)=0$
+	2. $f(x)=n^n\;\;\Rightarrow\;\;f'(x)=nx^{n-1},\;n\in\mathbb{Z}$
+	3. $f(x)=x^\alpha\;\;\Rightarrow\;\;f'(x)=\alpha x^{\alpha-1},\;\alpha\in\mathbb{R},\;x>0$
+	4. $f(x)=e^x\;\;\Rightarrow\;\;f'(x)=e^x$
+	5. $f(x)=\ln\mid x\mid\;\;\Rightarrow\;\;f'(x)=x^{-1},\;x\neq0$
+	6. $f(x)=\sin x\;\;\Rightarrow\;\;f'(x)=\cos x$
+	7. $f(x)=\cos x\;\;\Rightarrow\;\;f'(x)=-\sin x$
+	8. $f(x)=\tan x\;\;\Rightarrow\;\;f'(x)=\sec^2x=1+\tan^2x$
+	9. $f(x)=a^x\;\;\Rightarrow\;\;f'(x)=a^x\ln a,\;a>0$
+	10. $f(x)=\log_a\mid x\mid\;\;\Rightarrow\;\;f'(x)=(x\ln a)^-1,\;a>0,\;x\neq0$
+- Implicit derviering
+	- **Ex**: Bestäm tangent & normal ekvation till kurvan $x^3y^2-x^2y^3=12$ i punkten $(2,-1)$ $$\begin{align}\text{Antag att }y=f(x)\text{ för någon funktion }f\text{ nära punkten }(2,-1)\\x^3(f(x))^2-x^2(f(x))^3=12\\\text{Derivera m.a.p. }x\\(x^3(f(x))^2)'-(x^2(f(x))^3)'=(12)'\\\Leftrightarrow(x^3)'(f(x))^2+x^3((f(x))^2)'\\-(x^2)'(f(x))^3-x^2((f(x))^3)'=0\\\text{(produkt regeln)}\\\Rightarrow3x^2(f(x))^2+x^3\times2f(x)f'(x)\\-2x(f(x))^3-x^2\times3(f(x))^2\times f'(x)=0\\\Leftrightarrow(2x^3f(x)-3x^2(f(x))^2)f'(x)\\=2x(f(x))^3-3x^2(f(x))^2\\\text{På punkten }(2,-1)\text{ har vi}\\y=f(2)=-1\\\text{sätt in }x=2\\\left(2\times2^3f(2)-3\times2^2\times(f(2))^2)f'(x)\right)=2\times2\times(f(2))^3-2\times2^2(f(2))^2\\\Leftrightarrow(-16-12)f'(2)=-4-12\Leftrightarrow f'(2)=\frac{-16}{-28}=\frac47\\\text{Tangent ekv: }y=f'(a)(x-a)+f(a)\\\Leftrightarrow y=\frac47(x-2)-1\Leftrightarrow4x-7y=15\\\text{Normal ekv: }y=-\frac1{f'(a)}(x-a)+f(a)\\\Leftrightarrow y=-\frac74(x-2)-1\Leftrightarrow7x+4y=10\end{align}$$
+	- **Kedje regeln**: $$\begin{align}\frac{df(y(x))}{dx}=\frac{df(y(x))}{dy}\times\frac{dy(x)}{dx}\\(f(y(x)))'=f'(y(x))y'(x)\end{align}$$
+- Invers
+	- **Theorem**: *Om $f$ är inverterbar och deriverbar i punkten $a$ så att $f'(a)\neq0$ då är inversen $f^{-1}$ deriverbar i punkten $b=f(a)$ med derivatan* $$\left(f^{-1}\right)'\left(b\right)=\frac1{f'(a)}$$
+	- Följdsats:
+		- **Theorem**: $$\begin{gather}\text{För }-1<x<1,\\>\;f(x)=\arcsin x\;\;\;\;\;\Rightarrow f'(x)=\frac1{\sqrt{1-x^2}}\\>\;f(x)=\arccos x\;\;\Rightarrow f'(x)=-\frac1{\sqrt{1-x^2}}\\>\;f(x)=\arctan x\;\;\;\;\;\;\;\;\Rightarrow f'(x)=\frac1{1+x^2}\\>\;f(x)=\arccos x\;\;\;\;\Rightarrow f'(x)=-\frac1{1+x^2}\end{gather}$$
+- Medelvärdessats
+	- **Theorem** *Om $f$ är kontinuerlig på slutet intervall $\left[a,b\right]$ och deriverbar på öppet intervall $\left(a,b\right)$, dår fins det minst en punkt $\xi\in\left(a,b\right)$ så att* $$f'(\xi)=\frac{f(b)-f(a)}{b-a}$$![[MVT.png]]
+- Egenskaper
+	- *Låt $f$ vara deriverbar i intevallet $\left(a,b\right). följande gäller$*
+		1. *$f'(c)=0$ för något $c\in\left(a,b\right)\;\Rightarrow\;f$ har lokal extremvärde eller sadelpunkt i punkten $x=c$.*
+		2. *$f'(x)=0\;\forall{x}\in\left(a,b\right)\Rightarrow f(x)=C$, konstant funktion*
+		3. *$f'(x)=g'(x)\;\forall{x}\in\left(a,b\right)\Rightarrow f(x)=g(x)+C$, Där $C$ är någon konstant.*
+		4. *$f'(x)>0\;\forall{x}\in\left(a,b\right)\Rightarrow f(x)$ är strängt växande i $\left(a,b\right)$.*
+		5. *$f'(x)<0\;\forall{x}\in\left(a,b\right)\Rightarrow f(x)$ är strängt avtagande i $\left(a,b\right)$.*
+		6. *$f'(x)\geq0\;\forall{x}\in\left(a,b\right)$ med likhet i ändligt antal punkter $\Rightarrow f(x)$ är stängt växande i $\left(a,b\right)$.*
+		7. *$f'(x)\leq0\;\forall{x}\in\left(a,b\right)$ med likhet i ändligt antal punkter $\Rightarrow f(x)$ är stängt avtagande i $\left(a,b\right)$.*
+- Andra derivata
+	- **Betäkning**: $f''(x)$
+	- **Definition**: $\frac{d^2f}{dx^2}(x):=\frac{d}{dx}\left(\frac{df}{dx}(x)\right)$
+	- **Ex**: $f(x)=x^3\Rightarrow f'(x)=3x^2\Rightarrow f''(x)=6x$
+	- **Andra-derivatanstest**: $$\begin{align}\text{Låt }f\text{ vara deriverbar i punkten }x_0\;\&\;f'(x_0)=0\\1.\;\;f''(x_0)>0\Rightarrow x_0\text{ är en lokal minimum.}\\2.\;\;f''(x_0)<0\Rightarrow x_0\text{ är en lokal maximum.}\\3.\;\;f''(x_0)=0\Rightarrow\text{Vet ej.}\end{align}$$
@@ -0,0 +1,36 @@
+**DEF**: *En Determinant fins bara för kvadratiska matriser, t.ex: Låt $A$ vara en $m\times{n}$ matris. Denna matrisens determinant $\det(A)$ är det realla talet man får:  *$$\det(A)=\sum_{\sigma\in{S_n}}\operatorname{sgn}(\sigma)a_{1\sigma(1)}a_{2\sigma(2)}\dots{a_{n\sigma(n)}}$$
+- *Där $S_n$ mängden av alla penmutationer (samordningar) av talen $1,\;2,\;\dots,\;n$*
+- *$\sigma$ är en permutation av talen $1,\;2,\;\dots,\;n$*
+- *$\operatorname{sgn}(\sigma)$ är antigen $+1$ eller $-1$, beroende på antalet parytor som skiljer $\sigma$ från den vanliga ordningen*
+**EX** $$\begin{aligned}
+\text{Om vi har en $5\times5$ matris, då finns $5!=120$ sätt att omordna talen }1,\;2,\;3,\;4,\;5\\
+\text{Hur ser det termerna som motsvarar omordingen }\sigma=3-1-5-4-2. \text{iså fall är:}\\
+sgn(\sigma)a_{1\sigma(1)}a_{2\sigma(2)}a_{3\sigma(3)}a_{4\sigma(4)}a_{5\sigma(5)}=\underbracket{(-1)}\times a_{1\fbox{3}}a_{2\fbox{1}}a_{3\fbox{5}}a_{44}a_{5\fbox{2}}
+\end{aligned}$$
+*$\operatorname{sgn}$: Refererar till jämnt eller ojämt antal byten för att nå standerd ordning,*
+```cpp
+int sgn(int sigma_diff)
+{
+	return sigma_diff%2==0?1:-1;
+}
+```
+**EX**: $$\begin{aligned}\text{Vad är determinaten av $2\times2$ matrisen } A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}.\\\text{Det finns bata två sätt att omordna $1,\;2$: }1-2,\fbox{2}-\fbox{1}.\\\Rightarrow\text{determinatens summa har i det här fallet endast 2 termer}:\\\det(A)=\underbracket{+1}\times{a_{11}}\times{a_{22}}+\underbracket{-1}\times{a_{12}}\times{a_{21}}\\=a_{11}a_{22}-a_{12}a_{21}\end{aligned}$$
+**EX**: $$\begin{aligned}\text{Vad är determinaten av $3\times3$ matrisen }\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\\\text{Vad är dem 6 sätt att omordna?}\\\left.\begin{matrix}1-2-3&:&\operatorname{sgn}:&+1\\1-3-2&:&\operatorname{sgn}:&-1\\2-1-3&:&\operatorname{sgn}:&-1\\2-3-1&:&\operatorname{sgn}:&+1\\3-1-2&:&\operatorname{sgn}:&+1\\3-2-1&:&\operatorname{sgn}:&-1\\\end{matrix}\right\}\begin{aligned}\det(A)=\underbracket{+1}\times{a_{11}}a_{22}a_{33}+\\\underbracket{(-1)}\times{a_{11}}a_{23}a_{32}+\\\underbracket{(-1)}\times{a_{12}}a_{21}a_{33}+\\\underbracket{+1}\times{a_{12}}a_{23}a_{31}+\\\underbracket{+1}\times{a_{13}}a_{21}a_{32}+\\\underbracket{+1}\times{a_{13}}a_{22}a_{31}\end{aligned}\end{aligned}$$
+*Redan för $4\times4$ matriser skulle vi ha en summa med $24$ termer. Fins det något sätt att skriva deferminanten av $3\times3$ matrisen med hjälp av determinanten från $2\times2$ matrisen?*
+$$\begin{aligned}\det(A)&=a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}&-a_{12}a_{21}a_{33}+a_{12}a_{23}a_{31}&+a_{12}a_{21}a_{32}-a_{13}a_{22}a_{31}\\&=a_{11}\times\underbrace{\left(a_{22}a_{33}-a_{23}a_{32}\right)}_{\substack{\parallel\\\underline{\det\left(\begin{bmatrix}a_{22}&a_{23}\\a_{32}&a_{32}\end{bmatrix}\right)}\\A_{11}}}&-a_{12}\times\underbrace{\left(a_{21}a_{33}-a_{23}a_{31}\right)}_{\substack{\parallel\\\underline{\det\left(\begin{bmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{bmatrix}\right)}\\A_{12}}}&+a_{13}\times\underbrace{\left(a_{21}a_{32}-a_{22}a_{31}\right)}_{\substack{\parallel\\\underline{\det\left(\begin{bmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{bmatrix}\right)}\\A_{13}}}\end{aligned}$$
+**DEF**: *Låt $A$ vara en $m\times n$ matris. Imdermatrosem $A_{ij}$ är den $(m\times1)\times(n\times1)$ matrisen som fås genom att ta bort rad $i$ och kolumn $j$ från matrisen $A$.*
+**EX**: $$\begin{aligned}A=\begin{bmatrix}1&2&3&4\\5&6&7&8\\9&10&11&12\end{bmatrix}\Rightarrow\begin{aligned}A_{11}=\begin{bmatrix}6&7&8\\10&11&12\end{bmatrix}\\A_{23}=\begin{bmatrix}1&2&4\\9&10&12\end{bmatrix}\end{aligned}\end{aligned}$$
+**SATS**: *(RADUTVÄKLING): låt $A$ vara en $m\times{n}$ matris. För varje utvald index $i$ (mellan $1$ och $m$) gäller det att* $$\det(A)=\sum^{m}_{j=n}(-1)^{i+j}a_{ij}\det{A_{ij}}$$
+**Användiongs fall**
+*Vi vet att* $$\det(A)=\sum^{m}_{j=n}(-1)^{i+j}a_{ij}\det(A_ij)\left(\text{Radutväkling med avsende på rad $i$}\right)$$
+**SATS**: *Låt $A$ vara en $m±times{n}$ diaonal matris. Då gäller det att* $$\begin{aligned}\det(A)=\prod^{m}_{i=1}a_{ii}\\A=\begin{bmatrix}a_{11}&0&0&0\\0&a_{22}&0&0\\0&0&a_{33}&0\\0&0&0&a_{44}\end{bmatrix}\end{aligned}$$
+- **BEVIS**: $$\begin{aligned}\text{(tänk på 4x4 exemplet) Om vi radutvklar med avsende på rad $1$ ges:}\\\det(A)=\sum^{4}_{j=1}(-1)^{1+j}\underset{\text{Den enda termen som inte är $0$ är $a_{11}$}}{a_{1j}}\det(A_{1j})=a_{11}\times\det(A_{11})=\\a_{11}\times\det\left(\begin{bmatrix}a_{22}&0&0\\0&a_{33}&0\\0&0&a_{44}\end{bmatrix}\right)\Rightarrow\\\text{$m$ raduväklar igen, med avsende på rad $1$ i den nya mindre matrisen:}\\=a_{11}\times{a_{22}}\times\det\left(\begin{bmatrix}a_{33}&0\\0&a_44\end{bmatrix}\right)=a_{11}\times{a_{22}}\times{a_{33}}\times\det(\begin{bmatrix}a_{44}\end{bmatrix})\end{aligned}$$
+	- **OBS**: *Samma resultat gäller för både över- ohc under-triangul'ra matriser:* $$\begin{aligned}\det\left(\begin{bmatrix}a_{11}&0&0&0\\a_{21}&a_{22}&0&0\\a_{31}&a_{32}&a_{33}&0\\a_{41}&a_{42}&a_{43}&a_{44}\\\end{bmatrix}\right)=a_{11}\times{a_{22}}\times{a_{33}}\times{a_{44}}\\\det\left(\begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\0&a_{22}&a_{23}&a_{24}\\0&0&a_{33}&a_{34}\\0&0&0&a_{44}\\\end{bmatrix}\right)=a_{11}\times{a_{22}}\times{a_{33}}\times{a_{44}}\end{aligned}$$
+**SATS**: *Låt $A$ vara en $m\times{n}$ matris, och $\alpha\in\mathbb{R}$. Då gäller det att* $$\det(\alpha{A})=\underbracket{\alpha}\det(A)$$
+- **BEVIS**: $$\begin{aligned}\text{Kolla först $2\times2$ matriser: } A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\Rightarrow\alpha{A}=\begin{bmatrix}\alpha a_{11}&\alpha a_{12}\\\alpha a_{21}&\alpha a_{22}\end{bmatrix}\\\text{Då gäller det att: }\det\left(\begin{bmatrix}\alpha a_{11}&\alpha a_{12}\\\alpha a_{21}&\alpha a_{22}\end{bmatrix}\right)=(\alpha{a_{11}})\times(\alpha{a_{22}})-(\alpha{a_{12}})\times(\alpha{a_{21}})=\\a^2(a_{11}a_{22}-a_{12}a_{21})=a^2\det(A)\\\text{För störe matriser följer resultater ur radutväklingsformel}\end{aligned}$$
+**SATS** $$\begin{aligned}\text{Låt $A,B$ vara två $m\times{n}$ matriser. Då gäller det att}\\\det(AB)=\det(A)\times\det(B)\end{aligned}$$
+- **BEVIS** $$\begin{aligned}\text{Endast $2\times2$ matriser: }\\A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix},\;B=\begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix},\;AB=\begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\\\end{bmatrix}\\\Rightarrow\det(AB)=(a_{11}b_{11}+a_{12}b_{21})\times(a_{11}b_{12}+a_{12}b_{22})\\-(a_{21}b_{11}+a_{22}b_{21})\times(a_{21}b_{12}+a_{22}b_{22})\\=(\cancel{a_{11}b_{11}a_{21}b_{12}}+a_{11}b_{11}a_{22}b_{22}+a_{11}b_{11}a_{22}b_{22}+\cancel{a_{12}b_{21}a_{22}b_{22}})\\-(\cancel{a_{11}b_{11}a_{21}b_{12}}+a_{11}b_{12}a_{22}b_{22}+a_{12}b_{22}a_{21}b_{11}+\cancel{a_{12}b_{21}a_{22}b_{22}})\\=a_{11}b_{11}a_{22}b_{22}+a_{11}b_{11}a_{22}b_{22}-a_{11}b_{12}a_{22}b_{22}-a_{12}b_{22}a_{21}b_{11}\\=a_{11}a_{22}(b_{11}b_{22}-b_{12}b_{21})-a_{12}a_{21}(b_{11}b_{22}-b_{12}b_{21})\\=a_{11}a_{22}\times\det(B)-a_{12}a_{21}\times\det(B)=(a_{11}a_{22}-a_{12}a_{21})\det(B)=\det(A)\det(B)\end{aligned}$$
+**SATS**: *Låt $A$ vata en $m\times{n}$ matris. Då gäller: *$$\det(A)=\det(A^T)$$
+- **BEVIS**: $$\begin{aligned}\text{Endast $2\times2$: }\left.\begin{aligned}A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\Rightarrow\det(A)=a_{11}a_{22}-a_{12}a_{21}\\A^T=\begin{bmatrix}a_{11}&a_{21}\\a_{12}&a_{22}\end{bmatrix}\Rightarrow\det(A)=a_{11}a_{22}-a_{21}a_{12}\end{aligned}\right\}\text{Exakt samma}\end{aligned}$$
+**DEF**: *Låt $A$ vara $m\times{n}$ matris. Denna matrisen kofaktormatris är den $m\times{n}$ matrisen $\operatorname{cof}(A)$ vars element i rad $i$ och kolumn $j$ är *$$\begin{aligned}(-1)^{1+j}\det(A_{ij})\end{aligned}$$
+- **EX**: $$\begin{aligned}A=\begin{bmatrix}1&1&6\\-3&-4&-16\\3&5&13\end{bmatrix}\Rightarrow\operatorname{cof}(A)=\begin{bmatrix}+\begin{vmatrix}-4&-16\\5&13\end{vmatrix}&-\begin{vmatrix}-3&-16\\3&13\end{vmatrix}&+\begin{vmatrix}-3&-4\\3&5\end{vmatrix}\\-\begin{vmatrix}1&6\\5&13\end{vmatrix}&+\begin{vmatrix}1&6\\3&16\end{vmatrix}&-\begin{vmatrix}1&1\\3&5\end{vmatrix}\\+\begin{vmatrix}1&6\\-4&-16\end{vmatrix}&-\begin{vmatrix}1&6\\-3&-16\end{vmatrix}&+\begin{vmatrix}1&1\\-3&-4\end{vmatrix}\end{bmatrix}\\=\begin{bmatrix}28&-9&-3\\17&-5&-2\\8&-2&-1\end{bmatrix}\end{aligned}$$
@@ -0,0 +1,3 @@
+- **$dx$**: *oändlig liten förändring i $x$ värdet.*
+- **$df$**: *(motsvarande) oändligt liten förändring i $f$ värde.*
+- 
@@ -0,0 +1,84 @@
+**DEF**: *Låt $A$ vara $m\times{n}$ matris. Polynomet $$p_A(\lambda)=\det(A-\lambda I)$$. Kallas för matrisens kavaktieiska polynom. $\lambda\dots$ variabeln för detta polynom*
+**EX**: $$\begin{aligned}\text{Låt }A=\begin{bmatrix}2&-1\\3&-2\end{bmatrix}.\text{ Då är }A-\lambda{I}=\\\begin{bmatrix}2&-1\\3&-2\end{bmatrix}-\begin{bmatrix}\lambda&0\\0&\lambda\end{bmatrix}=\begin{bmatrix}2-\lambda&-1\\3&-2-\lambda\end{bmatrix}\\\Rightarrow\det(A-\lambda{I})=\begin{vmatrix}2-\lambda&-1\\3&-2-\lambda\end{vmatrix}=(2-\lambda)(-2-\lambda)-(-1)\times3\\=-4\cancel{-2\lambda}\cancel{+2\lambda}+\lambda^2+3=\underbrace{\lambda^2-4}\\\text{OBS: En $2\times2$ matris har en andragrads karaktieristisk polynom}\end{aligned}$$
+**DEF**: *Låt A vara en $m\times{n}$ matris. Nollställena till matrisens karakterisktiska polynom kalla för matrisens egenvärdarna.*$$P_A(\lambda)=0$$
+**OBS**:
+- *En $m\times{n}$ matris har alltid $m$ stycken egenvärden räknad med multiplicitet.* $$P_A(\lambda)=(\lambda-1)^3(\lambda-2)\Rightarrow\underbrace{4}.\text{ Lösninger: }\lambda=1,\lambda=1,\lambda=1,\lambda=2$$
+- *En matris med reella element behöver inte ha reella egenvärden* $$P_A(\lambda)=\lambda^2+1\Rightarrow\lambda^2+1=0\Rightarrow\lambda=+i,\lambda=-i$$
+**EX**: $$\begin{aligned}A=\begin{bmatrix}2&-1\\3&-2\end{bmatrix}\Rightarrow P_A(\lambda)=\lambda^2-1\Rightarrow\text{egenvärdena: }\lambda^2-1=0\Rightarrow\lambda=\pm1\end{aligned}$$
+**EX**: $$\begin{aligned}\text{Beräknaq egenvärdena av matrisen }A=\begin{bmatrix}13&4&8\\-6&-1&-4\\18&-6&-11\end{bmatrix}\\\text{Vi beräknar:}\\\det(A-\lambda I)=\begin{vmatrix}13-\lambda&4&8\\-6&-1-\lambda&-4\\-18&-6&-11-\lambda\end{vmatrix}=\\(13-\lambda)\begin{vmatrix}-1-\lambda&-4\\-6&-11-\lambda\end{vmatrix}-4\begin{vmatrix}-6&-4\\-18&-11\lambda\end{vmatrix}+8\begin{vmatrix}-6&-1-\lambda\\-18&-6\end{vmatrix}\\(13-\lambda)\left(11+\lambda+11\lambda+\lambda^2-24\right)-4(66+6\lambda-72)+8(36-18-18\lambda)\\=(13-\lambda)(\lambda^2+12\lambda-13)-4(64-6)+8(18-18\lambda)\\=13\lambda^2+12\times13\lambda-13^2-\lambda^3-12\lambda^2+13\lambda-24\lambda+24+144-144\lambda\\=-\lambda^3+\lambda^2+\lambda-1=-\lambda^2)(\lambda-1)+(\lambda-1)=(\lambda-1)(-\lambda^2-1)=\\(\lambda-1)\times(-1)\times(\lambda^2-1)=(\lambda^2-1)\times(-1)\times(\lambda-1)(\lambda+1)\\=-(\lambda-1)^2(\lambda+1)\end{aligned}$$
+**SATS**: *Låt $A$ vara en $m\times{n}$ matris, ich anta att $A$ antigen är diagonal eller triangulär (över eller under). Då sammanfaller engenvärdena med matrisens diagonala element*
+**BEVIS**: *Observera att matrisen $A-\lambda I$ är också diagonal eller tirangulär. Men för sådana matriser är determinanten lika med produkten av diagonala element (Föreläsning 12)*$$\begin{aligned}\Rightarrow P_A(\lambda)=\det(A-\lambda I)=\prod_{i=1}^{m}(a_{ii}-\lambda)\\\Rightarrow P_A(\lambda)=0\text{ precis för }\lambda=a_{11},\;\lambda=a_{22}\;\dots,\;\lambda=a_{mm}\end{aligned}$$
+**DEF**: *Låt $A$ vara en $m\times{n}$ matris och $\lambda$ vara ett av matrisens egenvärden. En $m\times1$ kolumnmatris $\overrightarrow{x}$ kallas för en egenvektor tillhörande $\lambda$ om $\overrightarrow{x}\neq\overrightarrow{0}$ och $A\overrightarrow{x}=\lambda\overrightarrow{x}$*
+**OBS**:
+- *Varje egenvärde har minst en egenskap*
+- *Om ett egenvärde upprepas, kan vi endast ha en linjärt oberoende egenvektor*
+- *Följande kan också hända: För ett egenvärde som upprepas $k$-gånger kan det finnas $k$ linjärt oberoende egenvektorer*
+- *Egenskaper räknas ut med hjälp av ett gauss shcema*
+**EX** $$\begin{aligned}
+A=\begin{bmatrix}
+2&-1\\
+3&-1
+\end{bmatrix},\text{ där vi redan har beröknat att }\lambda=\pm1\text{ egenvärdena}\\
+\text{Vad är de motsvarande egenvektorerna?}\\
+\begin{aligned}
+\text{Vilket schema?}\Rightarrow\begin{aligned}
+VL=A-\lambda I\\
+HL=\overrightarrow{o}
+\end{aligned}
+&&
+\begin{pmatrix}
+A\overrightarrow{x}&=\lambda\overrightarrow{x}\\
+A\overrightarrow{x}-\lambda\overrightarrow{x}&=\overrightarrow{0}\\
+\left(A-\lambda I\right)\overrightarrow{x}&\overrightarrow{0}
+\end{pmatrix}
+\end{aligned}\\
+\lambda=+1:\begin{pmatrix}
+1&-3&|&0\\
+3&-3&|&0
+\end{pmatrix}
+\begin{aligned}
+R_2-3R_1\rightarrow{R_2}\\
+\xrightarrow{}
+\end{aligned}
+\begin{pmatrix}
+1&-1&|&0\\
+0&0&|&0
+\end{pmatrix}\Rightarrow\overrightarrow{x}\\=\begin{bmatrix}
+x\\y
+\end{bmatrix}\text{ Där }\begin{aligned}
+y=t\text{ (fri variable)}\\
+x-y=0\Rightarrow x=t
+\end{aligned}\\\\
+\lambda=-1:\begin{pmatrix}
+3&-1&|&0\\
+3&-1&|&0
+\end{pmatrix}
+\begin{aligned}
+R_2-R_1\rightarrow{R_2}\\
+\xrightarrow{}
+\end{aligned}
+\begin{pmatrix}
+3&-1&|&0\\
+0&0&|&0
+\end{pmatrix}\\
+\begin{aligned}
+\frac13R_1\rightarrow{R_1}\\
+\xrightarrow{}
+\end{aligned}
+\begin{pmatrix}
+1&-\frac13&|&0\\
+0&0&|&0
+\end{pmatrix}\Rightarrow\begin{aligned}
+y=t\text{ (fri variable)}\\
+x-\frac13y=0\Rightarrow x=\frac13t\\\Rightarrow\begin{bmatrix}
+x\\y
+\end{bmatrix}
+=\begin{bmatrix}
+\frac13t\\
+t
+\end{bmatrix}=t\times\begin{bmatrix}
+\frac13\\
+1
+\end{bmatrix}
+\end{aligned}
+\end{aligned}$$
@@ -0,0 +1,73 @@
+**Def**: *Ett linjärt ekvationssystem med reella koefficienter är en samling av $m$ stycken ekvationer, där:*
+- *Varje ekvation innerhåller som m'st $m$-stycken variabler, och hat gemmesamma vatiabler för alla ekvationer*
+- *Varje vatiable förekommer om en första ordning moam $(x,\;4x,\;-3y,\cancel{x^2},\;\cancel{xy})$*
+- *En konstant term $(e,\;0,\;-5,\;\cancel{2+i})$*
+**Ex**: $$\begin{aligned}x_1-2x_2+3x_3=0\\x_1-x_4=-2\\\\\cancel{{x_1}^2}-2\cancel{x_2x_5}=0\\\cancel{\sin(x_1)}-x_4=\cancel{-2+}3i\end{aligned}$$
+*Ett allmänt linjär ekvationssystem med reella koefficienter herstamade av $m$ stycken ekvationer och $m$ stycken variablar ser ut så här: *$$\left.\begin{aligned}a_{11}x_1+a_{12}x_2+\dots+a_{1m}x_n=b_1\\a_{21}x_1+a_{22}x_2+\dots+a_{2n}x_n=b_2\\\vdots\\a_{m1}x_1+a_{m2}x_2+\dots+a_{mn}x_n=b_m\end{aligned}\right\}\begin{aligned}m\times{n}\text{ stycken koeffiencer }(a_{ij})\\m\text{ stycken koeffienter }(b_i)\end{aligned}$$
+**Ex**: $$\begin{aligned}x_1-2x_2-3x_3=0\\x_1-x_4=-2\\\Leftrightarrow x1_2-x_2-3x_3+0x_4=0\\x1+0x_2+0x_3-x_4=-2\\\end{aligned}$$
+**Def**: *En $m\times{n}$ matris med rella koeffienter är en samling av $m\times{n}$ stycken rella tal i en rektagulär schema med $m$ rader och $n$ koefiencer* $$A=\begin{bmatrix}a_{11}&a_{12}&\dots&a_{1n}\\a_{21}&a_{22}&\dots&a_{2n}\\\vdots&\vdots&\ddots&\vdots&\\a_{m1}&a_{m2}&\dots&a_{mn}\end{bmatrix}\leftarrow m\times{n}\text{ matris}$$
+*Variablar till häramde ett ekvationssystem samlas i en $n\times1$ matris $\overrightarrow{x}$ (också kallad för en kolomnvektor), och en koefficienterma $b_i$ som utgöt HL av en ekvationssystemet samlas i $m\times1$ matris $\overrightarrow{b}$(ett annat kolonnvektor)*$$\overrightarrow{x}=\left[\begin{aligned}x_1\\x_2\\x_3\\\vdots\;\\\vdots\;\\x_n\end{aligned}\right]\;\;\;\overrightarrow{b}=\left[\begin{aligned}b_1\\b_2\\\vdots\;\\\vdots\;\\b_m\end{aligned}\right]$$
+
+*Ex*: $$\begin{aligned}x_1-2x_2-3x_x=0\\x_1-x_4=-2\\\\A=\begin{bmatrix}1&-2&-3&0\\1&0&0&-1\end{bmatrix}\\\overrightarrow{x}=\left[\begin{aligned}x_1\\x_2\\x_3\\x_4\end{aligned}\right]\;\;\;\overrightarrow{b}=\left[\begin{aligned}0\\-2\end{aligned}\right]\end{aligned}$$
+- **Def**: *Ett gauss schema är en sammling av $A$, och $\overrightarrow{b}$ som tillhör ett ekvastions system:*$$\left(A\mid\overrightarrow{b}\right)=\begin{pmatrix}a_{11}&a_{12}&\dots&a_{1n}&|&b_1\\a_{21}&a_{22}&\dots&a_{2n}&|&b_2\\\vdots&\vdots&\ddots&\vdots&|&\vdots&\\a_{m1}&a_{m2}&\dots&a_{mn}&|&b_m\end{pmatrix}$$
+	- *Ett Jauss Schema (tillhörande ett ekvationssystem) har refuserats till sin trappform om följande gäller*
+		- *Varje rad börjar med en etta på $VL$(kallas för ledande etta), eller så är alla elementen på $VL$ lika med $0$*
+		- *Varje successivt rad börjar med en etta minst en kolumn senare, eller så är alla element på $VL$ lika med $0$*
+	- *Ur trappform kan vi läsa av ekvationssystemets egenskaper*
+		1. *Varje rad som har en ledande etta bestämmer en **pivåvariabel** - det är den variabeln som motsvarar kolumnen där ledande ettan befiner sig. Variabeln som inte har en motsvarande ledande etta är fria variabler*
+		2. *Ett ekvationssystem har:*
+			- *En entydlig lösning*: *Alla variablar är **privåvariabler***
+			- *Oändligt många lösningar*: *Mist en fri variable*
+			- *Saknar lösning*: *Om vi har en rad i trappformen där alla element på $VL$ är $0$, medans $HL$ är nollställen*
+	- *Ett ekvations system är antigen*
+		- *Exakt-bestämnd*: *Lika många ekvationer som variabler*
+		- *Över-bestämnd*: *Mera ekvationser än variabler*
+		- *Under-bestämnd*: *Mindre ekvationser än variablar*
+- **Ex**:
+	1. **Exakt bestämd system/Entydlig lösning**$$\begin{aligned}\begin{aligned}x-2y+z&=&3\\2x+y&=&1\\3y+2z&=&2\end{aligned}\Rightarrow\begin{pmatrix}1&-2&1&|&3\\2&-1&0&|&1\\0&3&2&|&2\end{pmatrix}\\\begin{aligned}\frac12R_2\rightarrow{R_2}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-2&1&|&3\\1&-\frac12&0&|&\frac12\\0&3&2&|&2\end{pmatrix}\begin{aligned}R_2-R_1\rightarrow{R_2}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-2&1&|&3\\0&\frac32&-1&|&-\frac52\\0&3&2&|&2\end{pmatrix}\\\begin{aligned}\frac32R_2\rightarrow{R_2}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-2&1&|&3\\0&1&-\frac23&|&-\frac53\\0&3&2&|&2\end{pmatrix}\begin{aligned}R_3-3R_2\rightarrow{R_3}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-2&1&|&3\\0&1&-\frac23&|&-\frac53\\0&0&4&|&7\end{pmatrix}\\\begin{aligned}\frac14R_3\rightarrow{R_3}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-2&1&|&3\\0&1&-\frac23&|&-\frac53\\0&0&1&|&\frac74\end{pmatrix}\end{aligned}$$
+		- *Vi hat nu bekräftat att ekvations systemet har en entydlig lösning*$$\begin{aligned}1\times{z}=\frac74\Longrightarrow{z=\frac74}\\1\times{y}-\frac23z=-\frac53\Rightarrow{y}=\frac23z-\frac53=\end{aligned}$$
+	2. **Exakt-bestämnd system/oändliga lösningar**$$\begin{aligned}\begin{aligned}x-2y+z&=&3\\2x-2y&=&1\\3x-4y+z&=&4\end{aligned}\Rightarrow\begin{pmatrix}1&-2&1&|&3\\2&-2&0&|&1\\3&-4&1&|&4\end{pmatrix}\begin{aligned}R_2-2R_1\rightarrow{R_2}\\R_3-3R_1\rightarrow{R_3}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-2&1&|&3\\0&2&-2&|&-5\\0&2&-2&|&-5\end{pmatrix}\\\begin{aligned}R_3-R_2\rightarrow{R_3}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-2&1&|&3\\0&2&-2&|&-5\\0&0&0&|&0\end{pmatrix}\begin{aligned}\frac12R_2\rightarrow{R_2}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-2&1&|&3\\0&1&-1&|&-\frac52\\0&0&0&|&0\end{pmatrix}\Rightarrow\\\begin{aligned}\text{Vi har två pivåvariabler $x$ och $y$, och en fri variabel, $z$. Vi har altså oändligt många lösningar.}\end{aligned}\end{aligned}$$
+		- *Eftersom $z$ är en fri variabler kan $z=t$, och $t\in\mathbb{R}$. sampt* $$\begin{aligned}y-z=-\frac52\Rightarrow{y}=z-\frac52=t-\frac52\\x-2y+z=3\Rightarrow{x}=2y-z+3=2\left(t-\frac52\right)-t+3=t-2\\\end{aligned}$$
+	3. **Exakt-bestämnd system/Saknar lösningar** $$\begin{aligned}\begin{aligned}x-3y+2z&=&3\\x-2y&=&2\\2x-5y+2z&=&4\end{aligned}\Rightarrow\begin{pmatrix}1&-3&2&|&3\\1&-2&2&|&2\\2&-5&2&|&-4\end{pmatrix}\begin{aligned}R_2-R_1\rightarrow{R_2}\\R_3-2R_1\rightarrow{R_3}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-3&2&|&3\\0&1&0&|&-1\\0&1&-2&|&-2\end{pmatrix}\\\begin{aligned}R_3-R_2\rightarrow{R_3}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-3&2&|&3\\0&1&2&|&-1\\0&0&0&|&-1\end{pmatrix}\end{aligned}$$
+		- **OBS** *Rad $2$ och $3$ säger att det skall vara $-2$ medans de int har samma $VL$, detta går inte! samt säger det $0x+0y+0z=-1\Leftrightarrow{0=-1}$*
+	4. **Över-bestämd system/Entydlig Lösning** $$\begin{aligned}\begin{aligned}x-3y+2z&=&3\\x-2y&=&2\\x-y-z&=&2\\2x-5y+2z&=&5\end{aligned}\Rightarrow\begin{pmatrix}1&-3&2&|&3\\1&-2&0&|&2\\1&-1&-1&|&2\\2&-5&2&|&5\end{pmatrix}\begin{aligned}R_2-R_1\rightarrow{R_2}\\R_3-R_1\rightarrow{R_3}\\R_4-2R_1\rightarrow{R_4}\\\xrightarrow{}\end{aligned}\\\begin{pmatrix}1&-3&2&|&3\\0&1&-2&|&-1\\0&2&-3&|&-1\\0&1&-2&|&-1\end{pmatrix}\begin{aligned}R_3-2R_2\rightarrow{R_3}\\R_4-R_2\rightarrow{R_4}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-3&2&|&3\\0&1&-2&|&-1\\0&0&1&|&1\\0&0&0&|&0\end{pmatrix}\end{aligned}$$
+		- *Vi har fott en entydlig lösning med*$$\begin{aligned}z=1\\y-2z=-1\Rightarrow{}y=2z-1=1\\x-3y+2z=3\Rightarrow{}x=3y-2z+3=4\end{aligned}$$
+	5. **Över-bestämd system/oändliga lösningar** $$\begin{aligned}\begin{aligned}x-3y+2z=3\\x-2z=3\\-3y+4z=0\\3x-3y+2z=9\end{aligned}\Rightarrow\begin{pmatrix}1&-3&2&|&3\\1&0&-2&|&3\\0&-3&4&|&0\\3&-3&2&|&9\end{pmatrix}\begin{aligned}R_2-R_1\rightarrow{R_2}\\R_4-3R_1\rightarrow{R_4}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-3&2&|&3\\0&3&-4&|&0\\0&-3&4&|&0\\0&6&-8&|&0\end{pmatrix}\\\begin{aligned}R_3+R_2\rightarrow{R_3}\\R_4-2R_2\rightarrow{R_4}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-3&2&|&3\\0&3&-4&|&0\\0&0&0&|&0\\0&0&0&|&0\end{pmatrix}\begin{aligned}\frac13R_2\rightarrow{R_2}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-3&2&3\\0&1&-\frac34&|&0\\0&0&0&|&0\\0&0&0&|&0\end{pmatrix}\end{aligned}$$
+		- *Ty att vi har en fri variable i ekvations systemet* $$\begin{aligned}z=t,\;t\in\mathbb{R}\\y=-\frac43z=0\Rightarrow{}y=\frac43t\\x-3y+2z=3\Rightarrow x=3y-2x+3=2t+3\end{aligned}$$
+	6. **Över-bestämd system/Saknar lösning**$$\begin{aligned}\begin{aligned}x-4y+2z&=&2\\x-z&=&3\\4y-3z&=&1\\3x-4y&=&1\end{aligned}\Rightarrow\begin{pmatrix}1&-4&2&|&2\\1&0&-1&|&3\\0&4&-3&|&1\\3&-4&0&|&1\end{pmatrix}\begin{aligned}R_2-R_1\rightarrow{R_2}\\R_4-3R_1\rightarrow{R_4}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-4&2&|&2\\0&4&-3&|&1\\0&3&-3&|&1\\0&8&-6&|&-5\end{pmatrix}\\\begin{aligned}R_3-R_2\rightarrow{R_3}\\R_4-2R_2\rightarrow{R_4}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-4&2&|&2\\0&4&-3&|&1\\0&0&0&|&0\\0&0&0&|&-7\end{pmatrix}\end{aligned}$$
+		- *I sista raden ser vi att $0x+0y+0z=-7$, samt i näst sista som säger $0x+0y+0z=0$ dessa är motsägelse fulla, altså saknas det en lösning*
+	7. **Under-bestämd system/Entydlig lösning** *Falsk möjlighet! Ett under bestämt system har mindre antal ekvationer än antalet variabler. Men i så fall är det omöjligt att alal variabler vore pivåvariabler*
+	8. **Under-bestämd system/Oändliga lösningar**$$\begin{aligned}\begin{aligned}x-y-z&=&1\\x+z&=&2\end{aligned}\Rightarrow\begin{pmatrix}1&-1&-1&|&1\\1&0&1&|&2\end{pmatrix}\begin{aligned}R_2-R_1\rightarrow{R_2}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&-1&-1&|&1\\0&1&2&|&1\end{pmatrix}\end{aligned}$$
+		- *Ty att vi har en fri variablel så har ekvations systemet oändligt många lösningar*$$\begin{aligned}z=t,\;t\in\mathbb{R}\\y+2z=1\Rightarrow{}y=-2z+1=-2t+1\\x-y-z=1\Rightarrow{}x=y+z+1=-t+2\end{aligned}$$
+	9. **Under-bestämd system/Saknar lösningar**$$\begin{aligned}\begin{aligned}x-y-z&=&1\\x-y-z&=&2\end{aligned}\Rightarrow\begin{pmatrix}1&-1&-1&|&1\\1&-1&-1&|&2\end{pmatrix}\begin{aligned}R_2-R_1\rightarrow{R_2}\xrightarrow{}\end{aligned}\begin{pmatrix}1&-1&-1&|&1\\0&0&0&|&1\end{pmatrix}\end{aligned}$$
+		- *Sista ekvationer säger att $0=1\Rightarrow$ ekvationssystemet saknar lösning.*
+- **DEF**: *Ett ekvations system kallas homohen om hala $HL$ är noll* $$\text{EX: }\begin{aligned}
+x-y+z&=&0\\
+7x-3z&=&0
+\end{aligned}$$*För hohogena ekvations system gäller följande*
+	- *Exakt-bestämd + homogen $\Rightarrow$ Antigen: alla variablar är noll, eller oändligt många lösningar*
+	- *Över-bestämt system + homogen $\Rightarrow$ Antigen: alla variablar är noll($0,0,0,\dots,0$), eller oändligt många lösningar*
+	- *Under-bestämt system + homogen $\Rightarrow$ Oändligt många lösningar*
+- **För varje ekvations system med oändligt många lösningar kan lösningsmängden delas upp i två**: $$y''-2y'+y=x^2+1$$
+	- *Den homogena lösningen: den som löser samma ekvationer, fast med $HL$ lika med $0$*
+	- *Den partikulära lösningen: en lösning av ekvations systemet*
+	- **EX**$$\begin{aligned}\begin{aligned}x-3y+2z&=&3\\x-2z&=&3\\-3y+4z&=&0\\3x-3y-2z&=&9\end{aligned}\xRightarrow{\text{Lösning i EX 5}}\begin{aligned}z=t,\;t\in\mathbb{R}\\y=\frac43\\x=2t+3\end{aligned}\Leftrightarrow{}(x,y,z)=(2t+3,\frac43t,t)\\\underbracket{(3,0,0)}_{\text{Partikulära Lösningen}}+\underbracket{t\times(2,\frac43,1)}_{\text{Homogena Lösningen}}\end{aligned}$$
+- **Ex**: $$\begin{aligned}\begin{aligned}x_1-2x_2-3x_x&=&0\\x_1-x_4&=&-2\end{aligned}\\\\\Rightarrow\begin{pmatrix}1&-2&-3&0&|&0\\1&0&0&-1&|&-2\end{pmatrix}\end{aligned}$$
+- **Ex**: $$\left.\begin{aligned}x+2y-u+3v&=&2\\2x+3y+2z-2u+10v&=&0\\x+3y-2z-4u+2v&=&3\\\underbrace{-x-3y+2z+3u-v}_{\substack{\text{VL $4\times5$}\\\text{=20 platser i schemat}}}&=&\underbrace{-4}_{\substack{\text{HL $4$}\\\text{ platser}}}\\\end{aligned}\right.\Rightarrow\left(a\mid\overrightarrow{b}\right)=\begin{pmatrix}1&2&0&-1&3&|&2\\2&3&2&-2&10&|&0\\1&3&-2&-3&2&|&3\\-1&-3&2&3&1&|&-4\end{pmatrix}$$
+*Hur räknar man med ett gauss schema? Man räknar med hjälp av elemäntera radoperationer:*
+- **Radbyte**: *Vi byter plats på alla element i raderna $i$ och $j$ : $R_i\leftrightarrow{R_j}\;\;\left(R_1\leftrightarrow{R_3}\right)$*
+- **Radmultiplikation**: *Vi multiplicerar alla ellement i raden $i$ med en och samma nollstild tal $\lambda\in\mathbb{R}$: $\lambda\times{R_i}\rightarrow{R_i}\;\;\left(2R_1\leftarrow{R_1}\right)$*
+- **Radaddition**: *Vi adderar till varje element i raden $i$ en $\lambda$-mutipel av motsvarande element från raden $j$: $R_i+\lambda{R_j}\rightarrow{R_1}\;\;\left(R_1-3R_2\rightarrow{R_1}\right)$*
+**Ex**: $$\left(\begin{aligned}1\;-2\;3\;\;\;\;\;0:\;\;\;0\\1\;\;\;\;\;0\;0\;-1:-2\end{aligned}\right).\;\;R_2-R_1\rightarrow{R_2}\left(\begin{aligned}1\;-2\;-3\;\;\;\;\;0:\;\;\;0\\0\;\;\;\;\;2\;\;\;\;\;3\;-1:-2\end{aligned}\right).\;\;\frac12R_2\rightarrow{R_2}\left(\begin{aligned}1\;-2\;-3\;\;\;\;0\;\;:\;\;\;0\\0\;\;\;\;\;1\;\;\;\;\;\frac32\;\frac{-1}2:-1\end{aligned}\right)$$
+**Ex**: $$\begin{pmatrix}1&2&0&-1&3&|&2\\2&3&2&-2&10&|&0\\1&3&-2&-4&2&|&3\\-1&-3&2&2&-4&|&-4\end{pmatrix}\left.\begin{aligned}R_2-2R_1\rightarrow{R_2}\\R_3-R_1\rightarrow{R_3}\\\xrightarrow{}\\R_4+R_1\rightarrow{R_4}\end{aligned}\right.\begin{pmatrix}1&2&0&-1&3&|&2\\0&-1&2&0&4&|&-4\\0&1&-2&-3&-1&|&1\\0&-1&2&2&2&|&-2\end{pmatrix}$$
+
+**Avslutande av kapitle**: $$\begin{aligned}\begin{aligned}x+2y-u+3v&=&2\\2x+3y+2z-2u+10v&=&0\\x+3y-2z-4u+2v&=&3\\-x-3y+2z+3u-v&=&-4\end{aligned}\Rightarrow\begin{pmatrix}1&2&0&-1&3&|&2\\2&3&2&-2&10&|&0\\1&3&-2&-4&2&|&3\\-1&-3&2&3&-1&|&-4\end{pmatrix}\\\begin{aligned}R_2-2R_1\rightarrow{}R_2\\R_3-R_1\rightarrow{}R_3\\R_4+R_1\rightarrow{}R_4\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&2&0&-1&3&|&2\\0&-1&2&0&4&|&-4\\0&1&-2&-3&-1&|&1\\0&-1&2&2&2&|&-2\end{pmatrix}\begin{aligned}R_3+R_2\rightarrow{}R_3\\R_4-R_2\rightarrow{}R_4\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&2&0&-1&3&|&2\\0&-1&2&0&4&|&-4\\0&0&0&-3&3&|&-3\\0&0&0&2&-2&|&2\end{pmatrix}\\\begin{aligned}R_2\rightarrow{}R_2\\\frac13R_3\rightarrow{}R_3\\\frac12R_4\rightarrow{}R_4\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&2&0&-1&3&|&2\\0&1&-2&0&-4&|&4\\0&0&0&1&-1&|&1\\0&0&0&1&-1&|&1\end{pmatrix}\begin{aligned}R_4-R_3\rightarrow{}R_4\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&2&0&-1&3&|&2\\0&1&-2&0&-4&|&4\\0&0&0&1&-1&|&1\\0&0&0&0&0&|&0\end{pmatrix}\end{aligned}$$
+- *$z$ och $v$ är fria variablar i detta systemet*
+- $$\begin{aligned}
+n=s,\text{ där }s\in\mathbb{R}\text{ (frivariable)}\\
+u-n=1\Rightarrow{u=1-s}\\
+z=t,\text{ där }t\in\mathbb{R}\\
+y-2z-4v=4\Rightarrow{}y=2t+4s+4\\
+x+2y-u+3v=z\Rightarrow{}x=-2\left(2t+4s+4\right)+\left(1+s\right)-3s+2\\
+x=-4t-10s-5
+\end{aligned}$$
@@ -0,0 +1,54 @@
+- Begränsade funktioner
+	- Uppåt begränsad: $f(x)\leq{M}$, $\forall{x}\in{D_r}$
+		- Ex: $f(x)=-x^2-2x$
+	- Nedåt begränsad: $f(x)\geq{M}$, $\forall{x}\in{D_f}$
+		- Ex: $f(x)=x^2+2x+2$
+- Monoton funktion
+	- Växande: $x_1<x_2\Rightarrow{f(x_1)}\leq{f(x_2)}$
+	- Strängt växande: $x_1<x_2\Rightarrow{f(x_1)}<f(x_2)$
+	- Avtagande: $x_1<x_2\Rightarrow{f(x_1)}\geq{f(x_2)}$
+	- Avtagande: $x_1<x_2\Rightarrow{f(x_1)}>f(x_2)$
+	- *(Strängt) Monoton funktion är (Strängt) växande eller (Strängt) avtagande*
+- Jämna, Udda funktioner
+	- Jämna: $f(-x)=f(x)$
+		- Ex: $|x|,\;x^2,\;\cos{x}$
+		- $$\begin{align*}f\text{ är udda }, O\in{D_f}\\f(-x)=-f(x)\forall{x}\in{D_f}\\f(-o)=-f(o)\\\Leftrightarrow{f(o)=-f(o)}\Leftrightarrow{2f(o)=0}\\\Leftrightarrow f(o)=\frac{o}{2}=O\end{align*}$$
+	- Udda: $f(-x)=-f(x)$
+		- Ex: $x,\;x^3,\;\sin{x}$ 
+- Sammansatta funktion
+	- $g\circ{f(x)}=g(f(x))$ 
+	- **Egenskaper**:
+		- $V_{g\circ{f}}\subseteq{V_g}$
+		- $V_{f}\subseteq{D_g}$
+	- Ex: $$\begin{align*}f(x)=\sqrt{x}\text{ and }g(x)=(x+5)^2\\f\circ{g}(x)=f(g(x))=f((x+5)^2)=\sqrt{(x+5)^2}=|x+5|\\g\circ{f}(x)=g(f(x))=g(\sqrt{x})=(\sqrt{x}+5)^2\\\text{I allmänhet }f\circ{g(x)}\neq{g\circ{f(x)}}\end{align*}$$
+- Inverse
+	- **Def**: *En funktion $g$ är inverse till funktionen $f$ om $g\circ{f(x)}=x$ och $f\circ{g(x)}=x$ för varje $x\in{D_f}$*
+	- ![[f_inverse.png]]
+	- **OPS**: $f^{-1}(x)\neq{(f(x))^{-1}}$
+	- Betekning: $f^{-1}$ är inverse till $f$
+	- Graf till inversen $f^{-1}$ är spegling av grafen till f i linjen $y=x$
+	- Injektiv funktion: $\forall{x_1,x_2}\in{D_f},\;x_1\neq{f(x_2}\rightarrow{x_1}\neq{f(x_2)}=\frac{1}{f(x)}$$$\begin{align}f\\x_1\neq{x_2}\Rightarrow{f(x_1)}\neq{f(x_2)}\end{align}$$
+	- $f$ är stängt monoton $\Rightarrow\;x$ är injektiv (inverterbar) på $D_f$
+	- $f$ är inverterbar $\Rightarrow\;D_{f-1}=V_f$ och $V_{f-1}=D_f$
+	- Ex:
+		- $f(x)\left\{\begin{align}-x+5,\;0\leq{x}\leq2\\x-4,\;2\leq{x}<4\end{align}\right.$
+		- ![[g1.png]]
+		- $f(x)=x^2,\;x\in[0,1]$ $D_f=[0,1]$
+		- ![[g2.png]]
+		- $$\begin{align}f(x)=3x+5\\g(x)=\frac{x-5}{3}\end{align}$$
+- Exponential och logarithm
+	- Exponential: $f(x)=a^x$ för något $a>0$.
+	- Logaritm: $g(x)=\log_a(x)$ för något $a>0$
+		- $f$ och $g$ inverse till varandra: $y=a^x\Leftrightarrow\log_a(y)=x$.
+		- $D_f=\mathbb{R}=V_g,\;\;V_f=(0,\infty)=D_g$.
+		- Om $a>1,\;f,\;g$ är strängt växande.
+		- $\log_a{(xy)}=\log_a(x)+\log_a(y),\;\log_a(x/y)=\log_a(x)-\log_a(y)$
+		- $\log_a(x^b)=b\log_a(x)$
+		- Basbyte: $\log_a(x)=\frac{\log_b(x)}{\log_b(a)}\Leftrightarrow\log_b(x)=\log_b(a)\log_a(x)$. $a^x=b^{x\log_b(a)}$
+		- Ex: $$\begin{align}\text{Räkna }D_f\text{ för }f(x)=\log_{10}(x^2+2x-3)\\f\text{ är definierad för }x^2+2x-3>0\\\Leftrightarrow(x+3)(x-1)>0\\\Leftrightarrow x\in(-\infty,-3)\cup(1,\infty)\\D_f=(-\infty,-3)\cup(1,\infty)\\\\2^{x+3}>4\\\Leftrightarrow\log_2(2^{x+3})>\log_24\\\Leftrightarrow x+3>2\\\Leftrightarrow x>-1\\\\\log_{10}36\\=\log_{10}(2^2\times3^2)\\=\log_{10}(2^2)+\log_{10}(3^2)\\=2\log_{10}2+2\log_{10}3\\\\2^x=e^{x\log_e2}=e^{x\ln2}\\\log_2x=(\log_2e)\ln{x}\\=\frac{\ln x}{\ln 2}\end{align}$$
+		- **Def**: $\log{x}=\log_{10}x$
+		- **Def**: $\ln{x}=\log_ex$
+		- **Def**: $a^x=e^{x\log_ea}=e^{x\ln a},\;a\in(0,\infty)$
+		- **Def**: $\log_a1=0$
+		- **Def**: $\log_aa=1$
+		- **Def**: $\log_ab=\frac{1}{\log_ba}$
@@ -1,4 +1,4 @@
- Talmängder:
+k- Talmängder:
 	- De natuliga talen: $\mathbb{N} = \{0, 1, 2, ...\}$
 	- De hela talen: $\mathbb{Z} = \{..., -2, -1, 0, 1, 2, ...\}$
 	- De rationella talen: $\mathbb{Q} = \{\frac{p}{q}: p, q \in \mathbb{Z}, q\neq0\}$
@@ -1,3 +1,42 @@
 - Graf
 	- Graf till funtion $f:\{(x,f(x)):x\in{D_f}$
-	- *Graf till $f$ med $y=V_f$ och $x=D_f$*
+	- *Graf till $f$ med $y=V_f$ och $x=D_f$*
+	- Ex: $$\begin{align*}f(x)=\left\{\begin{aligned}&2,\;0\leq{x}\leq{1}\\&x+3,\;1<x<2\\&-1,\;2\leq{x}<3\end{aligned}\right.\\D_f=[0,3)\\V_f=(-3,-2]\cup\{2\}\cup(4,5)\end{align*}$$
+- Variablebyte
+	- *Låt $f$ vara en funtion med $D_f=(x_1,x_2),\;V_f=(y_1,y_2)$*
+	- $g(x)=f(x-a)$, grafen flyttar $a$ enheter längst x-axeln. $$D_g=(x_1+a,x_2+a),\;V_g=(y_1,y_2)$$
+	- $g(x)=f(x)+b$, grafen flyttar $b$ enheter längt y-axeln $$D_g=(x_1,x_2),\;V_g=(y_1+b,y_2+b)$$
+	- $g(x)=f(cx),c\neq0$, "Scaling" längst x-axeln
+	- $g(x)=d\times{f(x)}$, "Scaling" längst y-axeln
+- Absolutbelopp
+	- **Def**: *Absolutbelopp funktion $|\dot{}|:\mathbb{R}\mapsto[0,\infty)$ definieras av $$|x|=\left\{\begin{aligned}x,\;\text{då }x\geq0,\\-x,\;\text{då }x<0.\end{aligned}\right.$$*
+	- Egenskapaer
+		- $|x|=\sqrt{x^2}\;\;\forall{x}\in\mathbb{R}$. (Alternativ definition av absolutbelopp)
+		- $|-x|=|x|\;\;\forall{x}\in\mathbb{R}$. (Jämn funktion)
+		- Multiplikation regle: $|x\times{y}|=|x|\times{|y|}\;\;\forall{x,y}\in\mathbb{R}$ 
+		- Triangel olikhet: $|x+y|\leq|x|+|y|$
+		- $|x-y|$ är avstånd mellan $x$ och $y$ på reell-linje. I synnerhet är $|x|$ avståndet mellan $x$ och $0$.
+	- Ex: Lös ekvationen $|x-3|=2$$$\begin{align*}|x-3|\Leftrightarrow\sqrt{(x-3)^2}=2\\\Leftrightarrow{}(x-3)^2=2^2\text{(kvadrering)}\\\Leftrightarrow{}(x-3)^2-2^2=0\Leftrightarrow(x-3+2)(x-3-2)=0\\\Leftrightarrow{}x_1=1,\;x_2=5\end{align*}$$
+	- Ex: Lös olikheten $|x-3|<2$$$\begin{align*}|x-3|=\left\{\begin{aligned}x-3,\;x-3\geq0\\3-x,\;x-3<0\end{aligned}\right.\\\text{Fall 1: }x-3\geq0\Leftrightarrow{x}\geq3\\|x-3|<2\Leftrightarrow{x}-3<2\\\Leftrightarrow{x}<2+4=5\\3\leq{x}<5\\\text{Fall 2: }x-3<0\Leftrightarrow{x}<3\\|x-3|<2\Leftrightarrow3-x<2\\\Leftrightarrow{x}>3-2=1\\1<x<3\\\\\text{Lösningmängd till }|x-3|<2\\(1,3)\cup{[3,5)}=(1,5)\end{align*}$$
+- Polynom
+	- **Def**: *En funtion i formen  $$p(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0=\sum^{n}_{k=0}{a_kx^k}$$är ett polynom. $a_k$ för $k=0,1,\dots,n$ är koefficienter. Om $a_n$ har polynomet grad $n$. Skrivs $grad(p)=n$*
+	- Nollställe/Rötter: Lösningar till $p(x)=0$
+	- Polynom av grad 0: $p(x)=c$, konstant function. Graf är parallel till x-axel.
+	- Polynom av grad 1 $p(x)=ax+b$, linjär function. Graf är en icke vertikal linje.
+- Andragradspolynom
+	- $p(x)=ax^2+bx+c,\;a\neq0$
+	- Faktorisering med kvadratkomplettering: $$\begin{align*}ax^2+bx+c=a\left(x^2+\frac{b}{a}\times\frac{c}{a}\right)\\=a\left(x^2+2\times{x}\times\frac{b}{2a}+\frac{b^2}{4a^2}-\frac{b^2}{3a^2}+\frac{c}{a}\right)\\=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right)\\=a\left(x+\frac{b}{2a}+\sqrt{\frac{b^2-4ac}{4a^2}}\right)\left(x+\frac{b}{2a}-\sqrt{\frac{b^2-4ac}{4a^2}}\right)\\=a\left(x+\frac{b+\sqrt{D}}{2a}\right)\left(x+\frac{b-\sqrt{D}}{2a}\right)\end{align*}$$Discriminant: $D=b²-4ac$
+	- Lösningar: $p(x)=ax^2+bx+c=0$ med $a\neq0$ har:
+		- Inga reella lösnngar om $D<0$. (Komplexa lösningar)
+		- En lösning (doubleroot) om $D=0$: $$x=-\frac{b}{2a}$$
+		- Två olika lösningar om $D>0$: $$x=\frac{-b\pm\sqrt{D}}{2a}$$
+		- Remark: Om $grad(p)=n,p(x)=0$ har max $n$ olika lösningar
+	- Ex Lös $x^2+2x-1=0$ $$\begin{align*}p(x)=x^2+2x-1=0\\=\end{align*}$$
+	- Ex: $$\begin{align*}p(x)=2x²+4x+4\\D=4^2-4\times2\times4<0\\p(x)=2x^2+4x+4\\=2(x^2+2x+2)\\=2(x^2+2x+1-1+2)\\=2\left((x+1)^2+1\right)\end{align*}$$
+	- Ex: $$\begin{align}p(x)=2x^2+2x+18\\D=12^2-4\times2\times18=0\\\text{en dubbel rot}\\p(x)=2x^2+12x+18\\=2(x^2+6x+18)\\=2(x+3)^2\end{align}$$
+	- Dubleroot vissar att det är två gånger samma factor i factorisering
+- Polynomdivision
+	- Rationell funktion: $f(x)=\frac{p(x)}{q(x)}$ där $p(x)$, $q(x)$ är polynom.
+	- **Def**: *$p(x)$ och $q(x)$ är polynom $\Rightarrow$ det fins polynom $k(x)$ (kvot) och $r(x)$ (rest) så att $$\begin{align}p(x)=q(x)k(x)+r(x)\\\frac{p(x)}{q(x)}=k(x)+\frac{r(x)}{q(x)}\end{align}$$, och $grad(r)<grad(q)$ om $grad(q)>0$*
+	- Remark: Om $r(x)=0$ för varje $x$ (nollpolynomet), divisionen får jämt ut. Vi har faktorisering $p(x)=q(x)k(x)$
+- 
@@ -0,0 +1,92 @@
+- Gränsvärden
+	- **Def**: *Om för varje $\epsilon>0$ existerar $\delta>0$ så att $$\mid{x-a}\mid<\delta\Rightarrow\mid{f(x)-L}\mid<\epsilon$$är talet $L$ gransvärde till $f(x)$ då $x$ får mot $a$. Betekning: $f(x)\longrightarrow{L}$ då $x\longrightarrow{a}$, eller $$\lim_{x\to{a}} f(x)=L$$*
+	- **Def**: *Om för varje $\epsilon>0$ existerar $M>0$ så att$$x>M\;\Rightarrow\;\mid{f(x)-L}\mid<\epsilon$$är talet $L$ gränsvärde till $f(x) då $x$ går mot oändlighit. Beteckning: $f(x)\longrightarrow{L}$ då $x\longrightarrow\infty$, eller $$\lim_{x\to\infty}f(x)=L$$*
+- Remarks
+	- *Om det inte fins sådant $L$ värde, saknar funktionen gränsvärde på punkten $a$,*
+		- **Ex**: $$\begin{align}f(x)=\sin x\\\lim_{x\to\infty}\sin x\\\text{Existerar inte}\end{align}$$
+		- **Ex**: $$\begin{align}f(x)=\sin\frac1x\\\lim_{x\to0}\sin\frac1x\\\text{Existerar inte}\end{align}$$
+	- *Punkten $a$ behöver inte vara i $D_f$.*
+	- *Beteende av funktionen kring "problempunkter" är intressant.*
+	- *Långsiktig beteende hos funktioner: $$\lim_{x\longrightarrow\infty}f(x)$$*
+	- *Derivator, integraler, asymptot etc definieras med hjälp av gränsvärde.*
+	- *Om $a$ int är "problempunkt" stoppar vi in $x=a$ i $f(x)$*
+	- **Def**: *"Problempunkt" t.ex $\lim_{x\to 0}\frac1x$ går inte att direkt lösa på grund av division med $0$*
+	- **Ex**: $$\begin{align}\lim_{x\to5}f(x)=\lim_{x\to5}\frac1x=\frac15\\\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac1x=0\\\lim_{x\to0}f(x)=\lim_{x\to0}\frac1x\text{ Existerar inte}\end{align}$$
+- One sided limits
+	- ![[gv1.png]]
+	- **Ex**: $$\begin{align}sgm(x)=\left\{\begin{aligned}1,\;x>0\\0,\;x=0\\-1,\;x<0\end{aligned}\right.\\D_{sgm}=\mathbb{R}\\\lim_{x\to0}sgm(x)\text{ Existerar inte}\\\lim_{x\to0^+}sgm(x)=\lim_{x\to0^+}1=1\\\lim_{x\to0^-}sgm(x)=\lim_{x\to0^-}(-1)=-1\end{align}$$
+	- **Ex**: $$\begin{align}\lim_{x\to a}f(x)\text{ existerar om}\\\lim_{x\to a+}f(x)\&\lim_{x\to a-}f(x)\\\text{ Existerarf och }\lim_{x\to a+}f(x)=\lim_{x\to a-}f(x)\\\\f(x)=\sqrt{x}, D_f=\left[0,\infty\right)\\\lim_{x\to0+}f(x)=\lim_{x\to0+}\sqrt{x}=0\\\\f(x)=\left\{\begin{aligned}x+1,\;x>0\\0,\;x=0\\2x+1,\ x<0\end{aligned}\right.\\D_f=\mathbb{R}\\\lim_{x\to0+}f(x)=\lim_{x\to0+}x+1\\=0+1=1\\\lim_{x\to0-}f(x)=\lim_{x\to0-}2x+1\\=2\times0+1=0\\\lim_{x\to0}f(x)=1\end{align}$$
+- Problem fall
+	- $\left[\frac00\right]$ form: **Ex**: $$\lim_{x\to1}\frac{x^2-3x+2}{x^2-1},\;\lim_{x\to0}\frac{e^x-1}x,\;\lim_{x\to\infty}\frac{\tan{x}}x$$
+	- $\left[\frac\infty\infty\right]$ form: **Ex**: $$\lim_{x\to\infty}\frac{x^2-3x+2}{x^2-1},\;\lim_{x\to\infty}\frac{x^3}{2^x}$$
+	- $\left[0\times\infty\right]$ form: **Ex**: $$\lim_{x\to\infty}x^2\ln\mid{x}\mid$$
+	- $\left[0^0\right]$ form: **Ex**: $$\lim_{x\to0+}x^x$$
+	- $\left[\infty^0\right]$ form **Ex**: $$\lim_{x\to\infty}x^{1/x}$$
+	- $\left[1^\infty\right]$ form: **Ex**: $$\lim_{x\to0}(1+x)^{1/x}$$
+	- $\left[\infty-\infty\right]$ form: $$\lim_{x\to\infty}\left(\sqrt{x^2+5x+1}-\sqrt{x^2+3x-5}\right)$$
+		- **Ex**: $$ \begin{align}\lim_{x\to\infty}\left(\sqrt{x^2+5x+1}-\sqrt{x^2+3x-5}\right)\\=\lim_{x\to\infty}\frac{\left(\sqrt{x^2+5x+1}-\sqrt{x^2+3x-5}\right)\left(\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}\right)}{\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}}\\=\lim_{x\to\infty}\frac{\left(\cancel{x^2}+5x+1\right)-\left(\cancel{x^2}+3x-5\right)}{\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}}\\=\lim_{x\to\infty}\frac{2x+6}{\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}}\\=\lim_{x\to\infty}\frac{x(2+\frac6x)}{\sqrt{x^2}\left(\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}\right)}\\=\lim_{x\to\infty}\frac{\cancel{x}(2+\frac6x)}{\cancel{x}\left(\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}\right)}\\=\lim_{x\to\infty}\frac{(2+\frac6x)}{\left(\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}\right)}\\=\frac{2+0}{\sqrt{1+0+0}+\sqrt{1+0-0}}=1\end{align}$$
+	- **Ex**: $$\begin{align}\lim_{x\to1}\frac{x^2-3x+2}{x^2-1}=\frac{0^2-3\times0+2}{0^2-1}=\frac{1+2}{1-1}=\frac{3}{0}\text{ Fins inget gränsvärde}\\\lim_{x\to1}\frac{x^2-3x+2}{x^2-1}\Longleftrightarrow\lim_{x\to1}\frac{(x-1)(x-2)}{(x-1)(x+1)}=\frac{x-2}{x+1}=\frac{1-2}{1+1}=-\frac12\end{align}$$
+	- **Ex**: $$\lim_{x\to\infty}\frac{x^2-3x+2}{x^2-1}=\lim_{x\to\infty}\frac{1-\frac3x+\frac2{x^2}}{1-\frac1{x^2}}=\frac{1-0+0}{1-0}=1$$
+	- **Ex**: $$$$
+- Räkneregler
+	- *Låt $f$ och $g$ vara funktioner så att $$\lim_{x\to a}f(x)=A,\;\lim_{x\to a}=B,\;\mid{A}\mid<\infty,\;\mid{B}\mid<\infty$$*
+	- $$\lim_{x\to a}\alpha(f(x)+\beta g(x))=\alpha A+\beta B$$
+	- $$\lim_{x\to a}f(x)\times g(x)=A\times B$$
+	- $$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{A}{B}\text{ om }B\neq0$$
+	- **Theorem**: *Instängningsregel $$\left.\begin{aligned}f(x)\leq g(x)\leq h(x),\;\forall x\\\lim_{x\to a}f(x)=L=\lim_{x\to a}h(x)\end{aligned}\right\}\Rightarrow\lim_{x\to a}g(x)=L$$*
+	- **Theorem**: $$f(X)\leq g(x),\;\forall x\Rightarrow\;\lim_{x\to a}f(x)\leq\lim_{x\to a}g(x)$$
+	- **Theorem**: *Sammansättningsregel $$\left.\begin{aligned}\lim_{x\to a}f(x)=b\\\lim_{x\to b}g(x)=L\end{aligned}\right\}\Rightarrow\lim_{x\to a}g\circ f(x)=L$$*
+	- **Variabelbyte**: $$\lim_{x\to a}g\circ f(x)=\lim_{t\to b}g(x)\text{ där }t=f(x)\longrightarrow b\text{ då }x\longrightarrow a$$
+	- **Ex**: $$
+\begin{align}\lim_{x\to0}x^2\sin\frac1x\\-1\leq\sin\frac1x\leq1,\; x\neq0\\\Rightarrow-x^2\leq x^2\sin\frac1x\leq x^2\\\lim_{x\to0}-x^2=0=\lim_{x\to0}x^2\\\text{Enlight instängningsregel, }\\\lim_{x\to0}x^2\sin\frac1x=0\\\end{align}
+$$
+	- **Ex**: $$\begin{align}\lim_{x\to0}\frac{\arcsin x}x\\\text{Låt }\arcsin x=y,x\in\left[-1,1\right]\\\Rightarrow x=\sin y,y\in\left[-\frac\pi2,\frac\pi2\right]\\t\to0\text{ då }x\to0\\\lim_{x\to0}\frac{\arcsin x}x=\lim_{y\to0}=\frac{y}{\sin y}\\=\lim_{y\to0}\frac1{\frac{\sin y}y}=\frac11=1\end{align}$$
+- Standerd gränsvärde
+	1. $\frac{x^\alpha}{a^x}\longrightarrow0$ då $x\longrightarrow\infty$ där $a>1$.
+	2. $\frac{\ln x}{x^\alpha}\longrightarrow0$ då $x\longrightarrow\infty$ där $\alpha>0$.
+	3. $x^\alpha\ln x\longrightarrow0$ då $x\longrightarrow0+$ där $\alpha>0$.
+	4. $\frac{\sin x}x\longrightarrow1$ då $x\longrightarrow0$
+	5. $\left(1+x\right)^{1/x}\longrightarrow e$ då $x\longrightarrow0$
+	6. $\frac{\ln\left(1+x\right)}x\longrightarrow1$ då $x\longrightarrow0$
+	7. $\frac{e^x-1}x\longrightarrow0$ då $x\longrightarrow0$
+	8. $\left(1+\frac1n\right)^n\longrightarrow e$ då $n\longrightarrow\infty$
+	9. $\frac{a^n}{n!}\longrightarrow0$ då $n\longrightarrow\infty$
+	10. $\sqrt[n]{n!}\longrightarrow\infty$ då $n\longrightarrow\infty$
+	- **Ex**: $$$$
+- Definitions
+	- **Def**: *Funktionen $f$ är kontinuerling i punkten $a$ om*
+		1. $a\in D_f$ *och*
+		2. $$\lim_{x\to a}f(x)=f(a)$$
+	- *På samma sätt, kontinuitet från höger och vänster med en-sidig gränsvärde.*
+	- **Def**: *Funktion $f$ är en kontinuerlig funktion på intervallet $I$ om $f$ är kontinuerlig i varje punkten $a\in I$*
+- Remarks
+	- Eöementära funktioner är kontinuerliga på sina definitionsmöngder.
+	- **Ex**: $x^n;\;\;x^\alpha,\;x>0;\;\;a^x,\;a>0;\;\;\log_ax,\;a>0;\;\;\sin x;\;\;\arcsin x,\;x\in\left[-1,1\right]\;\;\text{etc.}$
+	- $f,\;g$ kontinuerlig då är följande kontinuerlig: $f+g,f\times g,\text{ och }f\circ g$
+	- $\frac{f}g$ kontinuerlig på definitionsmängden av $\frac{f}g$
+	- $f$ är strängt monoton kontinuerlig funktion $\Longrightarrow f^{-1}$ är kontinuerlig.
+	- **Ex**: 
+		1. **Språng**: $f(x)=\left\{\begin{aligned}x+2,\;x\geq1\\x+1,\;x<1\end{aligned}\right.$
+			- <graf 1>
+		2. **Hävbar**: $f(x)=\left\{\begin{aligned}x^2+1,\; x\neq0\\-1,\;x=0\end{aligned}\right.$
+			- <graf 1>
+		3. **Lodrät asymptot**: $f(x)=\left\{\begin{aligned}\frac1{x+1},\;x\neq1\\0,\;x=1\end{aligned}\right.$
+			- <graf 1>
+			- <graf 2>
+			- <graf 3>
+		- $f(x)=\frac1x, x\in\left(0,\infty\right)$ $f$ är kontinuerlig på $\left(0,\infty\right)$. $f$ saknar *störta*/*minsta* värde
+	- Egenskaper:
+		- Satsen om mellanliggandevärden:
+		- **Theorem**: *Funktionen $f$ kontinuerlig i $\left[a,b\right]\Rightarrow f$ tar alla värde mellan $f(x)$ och $f(b)$ minst en gång*
+		- **Ex**: $f$ kontinuerlig funktion så att $f(-5)=3$ och $f(x)=-2$. Enlight satsen har $f$ minst ett nollställe.
+		- Extreamvärde:
+		- **Theorem**: *Funktionen $f$ är kontinuerlig på $\left[a,b\right]\Rightarrow f$ har ett största och ett minsta värde på $\left[a,b\right]$*
+	- Asymptoter
+		- Sned asymptot:
+		- **Def**: *En rät linje $y=kx+m$ är en (**sned**) asymptot till kurvan $y=f(x)$ då $x\longrightarrow\infty$ om $$\lim_{x\to\infty}(f(x)-(ax+b))=0$$. Formel: $$a=\lim_{x\to\infty}\frac{f(x)}x$$ och $$b=\lim_{x\to\infty}(f(x)-ax$$*
+		- På samma sätt för $x\longrightarrow-\infty$.
+		- Lodrät asymptot:
+		- **Def**: *En rät linje $x=a$ är en lodrät asymptot till kurvan $y=f(x)$ om $$\lim_{x\to a+}f(x)=\pm\infty$$ eller $$\lim_{x\to a-}f(x)=\pm\infty$$*
+- Vanliga tenta frågor
+	- $$\begin{align}f(x)=\left\{\begin{aligned}\frac{\sin\alpha x}{x^3},\;x>0\\\beta,\;x=0\\\frac{\sqrt{1+2x^2}-1}{x^2},\;x<0\end{aligned}\right.\\\text{Bestäm }\alpha,\;\beta\text{ så att }f\text{ är kontinuerlig på }\mathbb{R}\\\text{Lös: }f(x)\text{ är kontinuerlig på }\left(0,\infty\right)\text{ eftersom }\sin\alpha x,x^3+x\\\text{är kontinuerlig \& däsmed }\frac{\sin\alpha x}{x^3}\text{ är kontinuerlig på }\left(0,\infty\right)\\f(x)\text{ ---||--- }\left(-\infty,0\right)\\\text{---}\sqrt{1+2x^2}-1,x^2\text{---}\\\text{---}\frac{\sqrt{1+2x^2}-1}{x^2}\text{---}\left/-\infty,0\right).\\f\text{ är kontinuerlig på }\mathbb{R}\text{ om det är kontinuerlig i x=0}\end{align}$$
+	- 
@@ -0,0 +1,3 @@
+- **Insättning**
+	- **Theorem**: *$F$ är en primitiv funktion till funktionen $f$. Bestämd integralen ges av*$$\int_a^bf(y)dy=F(b)-F(a)$$![[Int1.png]]
+	- 
@@ -0,0 +1,30 @@
+- Komplexa tal
+	- **Def**: $x^2+1=0$ saknar reell lösning. Vi antar talet $i\notin\mathbb{R}$ löser ekvationen, d.v.s $i^2=-1$
+	- Mängd av komplexa talen: $\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}$
+	- Om $z=a+bi,a=Re(z)$ och $b=Im(z)$
+	- **Konjugat**: $z=a+bi\Rightarrow\bar{z}=a-bi$
+	- **Regler**:
+		- $\bar{\bar{z}}=z$
+		- $\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}$
+		- $\overline{z_1\times{z_2}}=\overline{z_1}\times{z_2}$
+	- **Absolut belopp**: $$\mid{z}\mid=\mid\overline{z}\mid=\sqrt{z\overline{z}}=\sqrt{a^2+b^2}\text{ om }z=a+bi$$
+	- **Triangelsformeln**: $\mid{z_1+z_2}\mid\leq\mid{z_1}\mid+\mid{z_2}\mid$
+	- **Ex**: $$\begin{align}z_1=2+3i\\z_2=2-i\\\\z_1+z_2=(2+3i)+(2-1)\\=4+2i\\\overline{z_1+z_2}=4-2i\\\overline{z_1}=2-3i,\;\overline{z_2}=2+i\\\overline{z_1}+\overline{z_2}=2-3i+2+i\\=3-2i\\\\z_1\times{z_2}=(2+3i)(2-i)\\=4-2i+6i-3i^2\\=4+4i+3\\=7+4i\\\overline{z_1\times{z_2}}=7-4i\\\overline{z_1}=2-3i,\;\overline{z_2}=2+i\\\overline{z_1}\times\overline{z_2}=(2-3i(2+i)\\=4+2i-6i-3i^2\\=4-2i+3\\=7-4i\end{align}$$
+	- **Ex 2**: $$\begin{align}z=a+bi\\\overline{z}=a-bi\\z\times\overline{z}=(a+bi)(a-bi)\\=a^2-\left(bi\right)^2\\=a^2-b^2i^2\\=a^2+b^2\end{align}$$
+	- **Ex 3**: $$\begin{align}\mid{z_1+z_2}\mid=\mid4+2i\mid\\=\sqrt{4^2+2^2}\\=\sqrt{16+4}=2\sqrt{5}\\\mid{z_1}\mid=\mid2+3i=\sqrt{2^2+3^2}\\=\sqrt{13}\\\mid{z_2}\mid=\mid2-i\mid=\sqrt{2^2+(-i)^2}=\sqrt{5}\end{align}$$
+	- **Ex 4**: $$\begin{align}\frac{z_1}{z_2}=\frac{2+3i}{2-i}\\=\frac{2+3i}{2-i}\times\frac{2+i}{2+i}\\=\frac{4+2i+6i+3i^2}{2^2-i^2}\\=\frac{1+8i}{5}\\=\frac{1}{5}+\frac{8}{5}i\end{align}$$
+- Grafer
+	- ![[k1.png]]
+	- ![[k2.png]]
+- Polär form
+	- **Eulers formel**: $e^{i\theta}=\cos\theta+i\sin\theta$
+	- Varje komplex tal $z=x+yi$ kan skrivas på pol'r form som $$z=re^{i\theta}$$ där $$r=\sqrt{x^2+y^2}$$ och $arg(z)=\theta$ är så att $$\cos\theta=\frac{x}{\sqrt{x^2+y^2}}\text{ och }\sin\theta=\frac{y}{x^2+y^2}$$
+	- **de Moivre**: $z=re^{i\theta}\Rightarrow z^n=r^ne^{in\theta}=r^n(\cos(n\theta)+i\sin(n\theta))$
+	- **Ex**: Lös $z^3=1+i\sqrt3$ $$\begin{align*}1+i\sqrt3=n_\circ e^{i\theta}, \theta\in\left[0,2\pi\right)\\n_\circ=\sqrt{1^2+\left(\sqrt3\right)^2}=2\\\theta\in\left[0,2\pi\right)\text{ uppfyller}\\\cos\theta=\frac12,\sin\theta=\frac{\sqrt3}2\\\Rightarrow\theta=\frac\pi3\\z^3=1+i\sqrt3=2e^{i\frac\pi3}\\\text{Låt }z=n_1e^{i\phi}\\\text{Då är }z^3=n_1^3e^{i3\phi}\\\Leftrightarrow\left\{\begin{aligned}n_1^3=2,n\in\mathbb{R}\\e\phi=\frac\pi3+2\pi{k},k\in\mathbb{Z},\phi\in\left[0,2\pi\right)\end{aligned}\right.\\\Leftrightarrow\left\{\begin{aligned}n_1=\sqrt[3]{2}\\\phi=\frac\pi9+\frac{2\pi k}{3},k=0,1,2\end{aligned}\right.\\k=0:\;\phi=\frac\pi9+\frac{2\pi}3\times0=\frac\pi9\\k=1:\;\phi=\frac\pi9+\frac{2\pi}3=\frac{7\pi}9\\k=2:\;\phi=\frac\pi9+\frac{2\pi}4\times2=\frac{13\pi}9\end{align*}$$
+	- **Ex 2**: $$\begin{align}z=-\frac{\sqrt3}{2}+\frac{1}{2}i\\z=ne^{i\theta}\\n=\sqrt{\left(-\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2}=1\\\theta\text{ är så att }\cos\theta=\frac{\sqrt{3}}{2},\sin\theta=\frac{1}{2}\\\text{En lösning}:\theta=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\\\text{Alla lösningar}:\theta=\frac{5\pi}{6}+2\pi{n},n\in\mathbb{Z}\\z=e^{i\left(\frac{5\pi}{6}+2\pi{n}\right)},n\in\mathbb{Z}\\\text{Svar: }z=e^{i\frac{5\pi}{6}}\end{align}$$
+- Polynom
+	- **Theorem**: *Algebrans huvudsats: Polynomet$$p(z)=c_nz^n+c_{n-1}z^{n-1}\dots+c_0,\;c_k\in\mathbb{C}$$har en rot i $\mathbb{C}$. D.v.s det finns en $z_1\in\mathbb{C}$ så att $p(z_1)=0$.*
+	- **Faktorsats**: $p(z)=(z-z_1)q(z)$
+	- **Theorem**: *Polynomet ovan kan skrivas som $p(z)=c_n(z-z_1)\dots(z-z_n)$. Alla polynom har $n$ komplexa rötter (och faktorer).*
+	- **Theorem**: *Polynom med reella koefficienter:$p(x)=a_nz^n+a_{n-1}z^{n-1}\dots+a_0,\;a_k\in\mathbb{R}$. Om $z_0$ är en rot så är $\overline{z_0}$*
+	- **Ex**: $$\begin{align}p(z)=3z^3-7z^2+17z-5\\p(1+2i)=0\\\text{Polynomet har reella koefficienten. även konjugatet 1-2i är en rot.}\\\text{Enlight faktorsatsen}\\p(z)=(z-1-2i)(z-1+2i)q(z)\\\text{för något polynom }q(z)\\p(z)=\left(\left(z-1\right)^2-\left(2i\right)^2\right)q(z)\\=\left(z^2-2z+1+4\right)q(z)\\=\left(z^2-2z+5\right)q(z)\\\text{Polynomdivision: }\\\frac{3z-1}{z^2-2z+5}\\p(z)=\left(z-1-2i\right)\left(z-1+2i\right)\left(3z-1\right)\\\text{Rötter: }1+2i,1-2i,\frac13\end{align}$$
@@ -0,0 +1,13 @@
+# TODO: Fyll i info från bilder
+
+- *I rummer $R^2$ kan en linje $l$ anges på flera sätt*
+	- $y=kx+m$: Funkar inte för vertikala linjer
+	- $ax=ky+c=0,\text{ där minst en av }a,k\text{ mellanskild kallas för nirmalform av en linje}$
+	- *Paramaterformen som ges av en punkt $P$ och en vektor $\overrightarrow{v}$*
+- **DEF**: *Låt $l$ vara en linje i $\mathbb{R}^2$ som ges av $P$ och $\overrightarrow{v}$. Denna linjens normalvektor $\overrightarrow{m}$ definieras som $\overrightarrow{m}=\left(-v_2,-v_1\right)$ där $\overrightarrow{v}=\left(v_1,v_2\right)$*
+	- **OBS**: *Det gäller att $<\overrightarrow{m},\overrightarrow{v}>=-v_2\times v_1+v_1\times v_2=0$*
+	- **OBS**: *Hur kan man beskriva tangentelinjen till grafen av fuktionen $f$ med hjälp av parameterformen*
+		- *För att beskriva en linje behöver vi $P$ och $\overrightarrow{v}$. Vad kam vi välja som $P$ och $\overrightarrow{v}$ i ett sådant fall fall* $$\begin{align}P=\left(a,f\left(a\right)\right)\\\overrightarrow{v}=\left(1,f'\left(a\right)\right)\end{align}$$
+- **Area**
+	- **Sats**: *Den sigmerade volum (dvs. volum med tetraheden $+/-$)  av tetrahdeden som spänns upp av tre linjärt vektorere $\overrightarrow{u},\;\overrightarrow{v},\;\overrightarrow{w}$ är lika med: $$\frac16<\overrightarrow{u}\times\overrightarrow{v},\;\overrightarrow{w}>=\frac16\mid\mid\overrightarrow{u}\times\overrightarrow{v}\mid\mid\mid\mid\overrightarrow{w}\mid\mid\cos(\alpha)$$*
+	- **Proff**: *Volymen av en tetrahden som en geometriska figur ges av en formel: $$\frac13\times\text{Area av bas ytan}\times\text{Höjden}\Rightarrow\frac13\times\frac12\mid\mid\overrightarrow{u}\times\overrightarrow{v}\mid\mid\times\mid\mid\overrightarrow{w}\mid\mid\times\cos(\alpha)\Rightarrow\text{Klar!}$$*
@@ -0,0 +1,13 @@
+- **Def**: *Om $f$ har kontinuerliga derivator till och med orning $n+1$ i en omgivning $\left(-\epsilon,\epsilon\right)$, då gäller för $x\in\left(-\epsilon,\epsilon\right), och $0\leq\theta\leq1$* $$f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\dots+\frac{f^{(n)}(0)}{n!}x^n+\frac{f^{(n+1)}(0)}{(n+1)!}x^{n+1}$$
+- **Def**: *Maclaurinpolynom av ordning $n$ för $f$:* $$P_{f,n}(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\dots+\frac{f^{(n)}(0)}{n!}x^n$$
+- **Ordo form**: $$f(x)=\sum^n_{i=0}\frac{f^{(i)}(0)}{i!}x^i+\frac{O(x^{n+1})}{ordo}$$
+  Ordo form: $O(x^{n+1})=x^{n+1}B_{n+1}(x)$ där $B_{n+1}$ är begränsad
+- **Ex**: $$\begin{align}f(x)=\sin2x\\\text{Bestäm Maclarin Polynom till }f\text{ av ordning 4}\\\underline{\text{Lösn}}:\;f(x)=\sin2x\Rightarrow f(0)=0\\f'(x)=2\cos2x\Rightarrow f'(0)=2\\f''(x)=-4\sin2x\Rightarrow f''(0)=0\\f'''(x)=-8\cos2x\Rightarrow f'''(0)=-8\\f''''(x)=-8\cos2x\Rightarrow f''''(0)=0\\\text{Maclarin Polynomet ordning 4:}\\P_{f,4}(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\frac{f''''(0)}{4!}x^4\\=0+\frac21x+\frac02x^2-\frac86x^3+\frac0{24}x^4\\=2x-\frac43x^3\\\text{Felet: }R_5(x)=\frac{f'''''(\theta x)}{5!}x^5=\frac{32\cos2(\theta x)}{120}x^5\\\text{sum}: \mid{R_5(x)}\mid=\frac{32\mid\cos{2\theta x}\mid}{120}\mid{x}\mid^5\leqslant\frac4{15}\mid{x}\mid^5\\\text{Till ex, }\mid{R_5\left(10^1\right)}\mid\leqslant\frac4{15}\times10^{-5}\end{align}$$
+- **
+- **Taylors forml**
+	- **Def**: *Om $f$ har kontonuerliga derivator till och med ordning $n+1$ i en omgiving $\left(a-\theta,a+\theta\right)$, då gäller för $x\in\left(a-\theta,a+\theta\right)$ och $\xi$ mellan $a$ och $x$* $$f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\dots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{x+1}$$
+	- **Def**: *Taylorpolynom av ording $n$ för $f$:*$$P_{f,n}(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\dots+\frac{f^{(n)}(a)}{n!}(x-a)^n$$
+	- **Rest**: *$R_{n+1}=\frac{f^{n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}=(x-1)^{n+1}B_{n+1}(x)$ där $B_{n+1}$ är en begränsad funktion nära $a$*
+- **L'Hôpitals regel**
+	- *Om $f(a)=0=g(a)$, $f,g$ är deriverbara på $a,g'(a)\neq0$* $$\begin{align}\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f(x)+f'(a)(x-a))+(x-a)^2B_2(x)}{g(a)+g'(a)(x-a)+(x-a)^2\bar{B}_2(x)}\\=\lim_{x\to a}\frac{\cancel{(x-a)}(f'(a)+(x+a)^2B_2(x))}{\cancel{(x-a)}(g'(a)+(x-a)\bar{B}_2(x))}\\=\frac{f'(a)}{g'(a)}\left(B_2,\bar{B}_2\text{ är begränsade funktioner}\right)\end{align}$$
+	- **Regel**: $$\begin{align}\left[\begin{aligned}\frac00\end{aligned}\right]\text{ eller }\left[\begin{aligned}\frac\infty\infty\end{aligned}\right]\text{ form}\\\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}\end{align}$$
@@ -0,0 +1,82 @@
+**DEF**: *En matris med reella koefficienter är en samling av $m\times{n}$ reella tal, uppdelade i $m$ rader och $n$ kolumner*$$\begin{aligned}A=\begin{bmatrix}a_{11}&a_{12}&\dots&a_{1n}\\a_{21}&a_{22}&\dots&a_{2n}\\\vdots&\vdots&\ddots&\vdots\\a_{m1}&a_{m2}&\dots&a_{mn}\end{bmatrix}\end{aligned}$$*Antalet rader och kolumner utgör matrisens dimension: $m\times{n}=$"$m$ gånger $n$"*
+
+**Räknavis**
+- **DEF**: *För två (eller flera) matriser vara samma dimension defimiras addition och skalär multiplikation positionsvis*
+- **EX**: $$\begin{aligned}A=\begin{bmatrix}1&-3&4\\0&3&5\end{bmatrix},B=\begin{bmatrix}-3&-3&2\\1&0&1\end{bmatrix},\lambda=3\\Rightarrow{}A+B=\begin{bmatrix}-2&-6&8\\1&3&6\end{bmatrix},3\times{A}=\begin{bmatrix}3&-9&12\\0&9&15\end{bmatrix}\end{aligned}$$
+- $$\begin{bmatrix}1&3\\2&4\end{bmatrix}+\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}=\text{Går ej att addera matriser i olika dimensioner}$$
+**Vanliga räkne regler gäller**
+- $A+B=B+A$
+- $(A+B)+C=A+(B+C)$
+- $\lambda\times(\mu{A)}=(\mu\lambda)*A$
+- $\lambda(A+B)=\lambda{A}+\lambda{B}$
+- $(\lambda+\mu)=\lambda{A}+\mu{A}$
+
+
+**DEF**: *Låt $A$ vara en $m\times{n}$ matris och $B$ vara en $n\times{p}$ matris. I så fall definieras matrisprodukten $AB$ som *$$(AB)_{ij}=\sum^n_{k=1}(A)_{1k}\times{(B)_{k1}}$$*Resultatet $AB$ är en $m\times{p} matris$*
+
+**EX**: $$\begin{aligned}\left.\begin{aligned}A=\begin{bmatrix}1&-3&4\\0&3&5\end{bmatrix}\text{ En $2\times3$ matris}\\B=\begin{bmatrix}-3&-3&1&4\\1&0&1&-2\\2&-1&6&1\end{bmatrix}\text{ En $3\times4$ matris}\end{aligned}\right\}AB=\begin{bmatrix}1&-7&22&14\\14&-5&33&-1\end{bmatrix}\end{aligned}$$**Transponering**: 
+- **DEF**: *Låt $A$ vara en $m\times{n}$ matris. Denna matrisen transponat $A^T$ är den $m\times{n}$ matrisen som fås genom att använda alla rader från matrisen $A$ till kolumner.*
+-  **EX**: *om* $$\begin{aligned}A=\begin{bmatrix}1&2&3\\1&2&4\end{bmatrix},\text{ Då är}\\A^T=\begin{bmatrix}1&1\\2&2\\3&4\end{bmatrix}\end{aligned}$$
+- **Vilka räkneregler gäller?**$$\begin{aligned}-&&\left(A^T\right)^T&=A\\-&&\left(A+B\right)^T&=A^T+B^T\\-&&\left(\alpha\times{A}\right)^T&=\alpha\times{A^T}\\-&&\left(AB\right)^T&=B^TA^T!!\end{aligned}$$
+- **DEF**: *En kvadratisk matris $A$ kallas för symmetrisk om $A^T=A$*
+	- **EX**: $$\left.\begin{aligned}A=\begin{bmatrix}\underline{1}&2&3\\2&\underline{5}&6\\3&6&\underline{9}\end{bmatrix},&\;\;B=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}\\A^T=\begin{bmatrix}\underline{1}&2&3\\2&\underline{5}&6\\3&6&\underline{9}\end{bmatrix},&\;\;B^T=\begin{bmatrix}1&4\\2&5\\3&6\end{bmatrix}\end{aligned}\right\}\begin{aligned}A^T=A\\B^T\neq{B}\end{aligned}$$
+	- **DEF**: *I en kvadratisk matris $A$ kallas:*
+		- *Element $a_{ij}$ med $i=j\Leftrightarrow$ diagonala element*
+		- *Element $a_{ij}$ med $i<j\Leftrightarrow$ över-diagonala element*
+		- *Element $a_{ij}$ med $i>j\Leftrightarrow$ under-diagonala element*
+		- **EX**: $$A=\begin{bmatrix}a_{11}&\overline{a_{12}}&\overline{a_{12}}&\overline{a_{13}}\\\underline{a_{21}}&a_{22}&\underline{a_{22}}&\overline{a_{23}}\\\underline{a_{31}}&\underline{a_{12}}&a_{12}&\overline{a_{33}}\\\underline{a_{41}}&\underline{a_{42}}&\underline{a_{43}}&a_{44}\\\end{bmatrix},\;\begin{aligned}\text{OBS: en kvadratisk matris}\\\text{ $A$ är symetrisk om}\\\underline{\underline{a_{ij}=a_{ji},\text{ för }i\neq{j}}}\end{aligned}$$
+	- **DEF**: *En kvadratisk matris $A$ kallas för*
+		- *Diagonal matris $\Leftrightarrow$ alla över- och under-diagonala element är $0$*
+		- *Över-triangulär matris $\Leftrightarrow$ alla under-diagonala element är $0$*
+		- *Under-triangulär matris $\Leftrightarrow$ alla över-diagonala element är $0$*
+		- **EX**: $$\begin{aligned}A=\begin{bmatrix}1&\overline{0}&\overline{0}\\\underline{0}&5&\overline{0}\\\underline{0}&\underline{0}&9\end{bmatrix},\;\;B=\begin{bmatrix}1&\overline{2}&\overline{3}\\\underline{0}&5&\overline{6}\\\underline{0}&\underline{0}&9\end{bmatrix},\;\;C=\begin{bmatrix}1&\overline{0}&\overline{0}\\\underline{4}&5&\overline{0}\\\underline{7}&\underline{8}&9\end{bmatrix}\end{aligned}$$
+		- **OBS**:
+			- *Transponanten av en diagonal matris är en diagonal matris, samt alla diagonala matriser är symetriska*
+			- *Transponanten av en över-triangulär matris är en under-triangulär matris*
+			- *Transponanten av en under-triangulär matris är en över-triangulär matris*
+			- **EX**: $$\begin{aligned}A=\begin{bmatrix}1&\overline{0}&\overline{0}\\\underline{0}&5&\overline{0}\\\underline{0}&\underline{0}&9\end{bmatrix},\;\;B=\begin{bmatrix}1&\overline{0}&\overline{0}\\\underline{2}&5&\overline{0}\\\underline{3}&\underline{6}&9\end{bmatrix},\;\;C=\begin{bmatrix}1&\overline{4}&\overline{7}\\\underline{0}&5&\overline{8}\\\underline{0}&\underline{0}&9\end{bmatrix}\end{aligned}$$
+	- **DEF**: *Den diagonala matrisen vars alla diagonala element är $1$ kallas för identitetsmatrisen och betänkas $I$.* 
+		- **EX**: $$
+\begin{aligned}
+I=\begin{bmatrix}
+1&0\\0&1
+\end{bmatrix}\\
+\shortparallel\;\;\;\;\;\\
+I_2\;\;\;\:
+\end{aligned},\;\;
+\begin{aligned}
+I=\begin{bmatrix}
+1&0&0\\
+0&1&0\\
+0&0&1
+\end{bmatrix}\\
+\shortparallel\;\;\;\;\;\;\;\;\\
+I_2\;\;\;\;\;\;\:
+\end{aligned}$$
+		- **OBS**: *Om $X$ är en $m\times{n}$ matris och $I$ identitersmatrisen av samma dimension, då gäller:* $$\begin{aligned}IX=XI=X&&\left(\underbracket{1}\times{x}=x\times\underbracket{1}=x\right)\end{aligned}$$
+		- **DEF**: *låt $A$ vara en $m\times{n}$ matris. Denna matrisen invers matris $A^{-1}$ är den matrisen som uppfyller $AA^{-1}=A^{-1}A=I$ (om en sådan matris $A^{-1}$ fins)* 
+			- **EX**: $$\begin{aligned}\text{Har matrisen }A=\begin{bmatrix}0&2\\0&0\end{bmatrix}\text{ en invers?}\\\text{Om den har en invers }A^{-1}=\begin{bmatrix}x&y\\z&w\end{bmatrix},\text{ då ska }\\AA^{-1}=A^{-1}A=I\\\text{Vad är }AA^{-1}\begin{bmatrix}0&2\\0&0\end{bmatrix}\begin{bmatrix}x&y\\z&w\end{bmatrix}=\begin{bmatrix}2z&2w\\0&0\end{bmatrix}\overset{?}{\text{=}}\begin{bmatrix}1&0\\0&1\end{bmatrix}\Rightarrow0=1\\\text{Detta går inte eftersom $0\neq1$}\\A\text{ har ingen invers}\end{aligned}$$
+		- **Räkneregler**: *(låt $A,B$ vara $m\times{x}$ matriser som har inverser*
+			- $\left(A^{-1}\right)^{-1}=A$
+			- $\left(A^{-1}\right)^{T}=\left(A^T\right)^{-1}$
+			- $\left(AB\right)^{-1}=B^{-1}A^{-1}$!!
+- **EX**: *Hur löser vi ekvationen $AX=B$, där $A,B$ kända $m\times{n}$ matriser, $X$ är okänd $m\times{n}$ matris?* $$
+\begin{aligned}\begin{aligned}
+AX=B\Leftrightarrow\left(\begin{aligned}
+X=BA^{-1}?\\
+X=A^{-1}B?
+\end{aligned}\right)\end{aligned}\\\begin{aligned}
+AX=B&\Rightarrow\underbracket{A^{-1}}AX=A^{-1}B\Rightarrow{IX=A^{-1}B}\Rightarrow{X=A^{-1}B}\\
+&\Rightarrow{AX\underbracket{A^{-1}}}=B\underbracket{A^{-1}}\Rightarrow???
+\end{aligned}\end{aligned}$$
+- **FAKTA**: *Om $A$ är em $m\times{n}$ matris och anta att $A$ har en invers. Då beräknas $A^{-1}$ genom: *$$\left(A\mid{I}\right)\longrightarrow\left(I\mid{A}\right),$$*dvs. Vi skriver $A$ som $VL$ och $I$ som $HL$ i ett gauss-chema, och sen genom radoperationer säkerställer att $I$ find på $VL$ till slut, och då är $A^{-1}$ kvar i $HL$.* 
+	- **EX**: $$\begin{aligned}\text{Låt }A=\begin{bmatrix}1&2\\2&7\end{bmatrix}\text{. beräkna }A^{-1}\\\left(A\mid{I}\right)=\begin{pmatrix}1&2&|&1&0\\2&7&|&0&1\end{pmatrix}\begin{aligned}R_2-2R_1\rightarrow{R_2}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&3&|&1&0\\0&1&|&-2&1\end{pmatrix}\\\begin{aligned}R_1-3R_2\rightarrow{R_1}\\\xrightarrow{}\end{aligned}\begin{pmatrix}1&0&|&7&-3\\0&1&|&-2&1\end{pmatrix}\Rightarrow{A^{-1}}=\begin{bmatrix}7&-3\\-2&1\end{bmatrix}\end{aligned}$$ $$\begin{aligned}
+A=\begin{bmatrix}
+1&2\\
+3&4
+\end{bmatrix}\Rightarrow?\\
+A^{-1}=\begin{bmatrix}
+4&-2\\
+-3&1
+\end{bmatrix}?
+\end{aligned}$$
@@ -0,0 +1,8 @@
+**OBS**: *En $m\times{n}$ matris kan tänkas bestå av $n$ stycken $m\times1$ kolumner*$$A=\begin{bmatrix}a_{11}&1_{12}&\dots&a_{1n}\\a_{21}&a_{22}&\dots&a_{2n}\\\vdots&\vdots&\ddots&\vdots\\a_{m1}&a_{m2}&\dots&a_{mn}\end{bmatrix}\Rightarrow A=\begin{bmatrix}|&|&\dots&|\\\overrightarrow{a_1}&\overrightarrow{a_2}&\dots&\overrightarrow{a_m}\\|&|&\dots&|\end{bmatrix}$$
+**EX**: $$A=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}\Rightarrow\overrightarrow{a_1}=\begin{bmatrix}1\\5\end{bmatrix},\;\overrightarrow{a_2}=\begin{bmatrix}2\\5\end{bmatrix},\;\overrightarrow{a_3}=\begin{bmatrix}3\\6\end{bmatrix}$$
+**OBS (fortsätning)**: *Transponaten av en matris lyfter rader mot kolumner och kolumner mot rader*$$A^T=\begin{bmatrix}\textemdash&\overrightarrow{a_1}^T&\textemdash\\\textemdash&\overrightarrow{a_2}^T&\textemdash\\&\vdots\\\textemdash&\overrightarrow{a_m}^T&\textemdash\end{bmatrix}\;\;\begin{aligned}\text{EX: }A=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}\Rightarrow A^T=\begin{bmatrix}1&4\\2&5\\3&6\end{bmatrix}\\\Rightarrow \overrightarrow{a_1}^T=\begin{bmatrix}1&4\end{bmatrix},\;\overrightarrow{a_2}^T=\begin{bmatrix}2&5\end{bmatrix},\;\overrightarrow{a_3}^T=\begin{bmatrix}3&6\end{bmatrix}\end{aligned}$$
+**OBS**: *Vad händer om vi har tvp $3\times1$ kolumnmatriser* $$\overrightarrow{a}=\begin{bmatrix}
+1\\2\\3
+\end{bmatrix},\overrightarrow{l}=\begin{bmatrix}
+7\\8\\9
+\end{bmatrix}$$
@@ -0,0 +1,46 @@
+**ODE** $\Longleftrightarrow$ **Ordinära differentialekvation**
+**PDE** $\Longleftrightarrow$ **Partiell differentialekvation**
+- Separabel ODE
+- Linjär ODE av ordning 1
+- Linjär ODE av ordning 2 med konstant koefficienter
+
+
+- **Ex**: *Newtons lag* $m\frac{d^2}{df^2}\stackrel{\rightarrow}{s}(t)=\stackrel{\rightarrow}{F}(t)$
+- **Ex**: **PDE** *Maxwellsekvation, Schrödingerekvation* $$\begin{align}\text{Okänd funktion }y(x)\\\text{ODE: }F\left(x,y(x),y'(x),\dots,y^{(n)}(x)\right)=0\\\text{Ording: }n\end{align}$$
+- **Ex**: *ODE av ordning 3*: $xy'''(x)+x^{1/4}y'(x)+\left(y(x)\right)^2=7x+3$
+- **Linjär ODE**
+	- $$\begin{align}a_n(x)y^{(x)}(x)+\dots+a_1(x)y'(x)+a_0(x)y(x)=h(x)\\a_k\text{ är funktionen av }x\end{align}$$
+	- **Ex**: $$\begin{align}\sqrt{x}y''+\frac1xy'+\pi{y}=e^x\_\_(\star)\\\text{Om }y_{_1}\&y_{_2}\text{ uppfylen}\\\sqrt{x}y''+\frac1xy'+\pi{y}=0\_\_(1)\\\text{så är }\alpha{y_1}+\beta{y_2},\;\alpha,\beta\in\mathbb{R}\text{ också lösning till }(1)\end{align}$$
+	- $(\star)$ är en *linjär ODE*
+		- **Ex**: $yy'=x+2$: *Icke-linjär*
+		- **Ex**: $\left.\begin{aligned}y'+\underline{\sqrt{y}}=x+2\\\underline{\sqrt{y'}}+y=2x+3\end{aligned}\right\}$: *Icke-linjär*
+		- **Ex**: $\underline{e^y}+\underline{\sin y}+y'=0$: *Icke-linjär*
+		- **Ex**: $(\sin x)y'+\sqrt{x}=\pi$: *Linjär*
+	- **Ex**: $$\begin{align}5y''=x+\sin x\\\Leftrightarrow y''=\frac15(x+\sin x)\\\text{Integrera m.a.p. }x\\y'=\frac15\int(x+\sin x)dx=\frac15\left(\frac{x^2}2-\cos x\right)+C\\y=\int\left(\frac15\left(\frac{x^2}2-\cos x\right)+C_1\right)dx\\=\frac15\left(\frac{x^3}6-\sin x\right)+C_1x+C_2\\\text{där }C_1,C_2\text{ är konstanter}\end{align}$$
+- $$\begin{align}\text{ODE: }g(y)y'=h(x)\\\text{Lösning: }g(y)y'=h(x)\\g(y)y'dx=h(x)dy\\\int g(x)dy=\int h(x)dx\\G(y)=H(x)+C\end{align}$$Där $G$ är primitiv till $g$ och $H$ är primitiv till $h$
+- **Ex**: $$\begin{align}y^2y'=x\sqrt{y}\;\;\left.\begin{aligned}\text{Icke-kin.}\\\text{ODE av}\\\text{ordning 1}\end{aligned}\right.\\\text{för }y\not\equiv0\\y^2y'=2x\sqrt{y}\\\Leftrightarrow\frac{y^2}{\sqrt{y}}y'=2x\Leftrightarrow y^{3/2}y'=2x\\\text{Integrera m.a.p. x}\\\int y^{3/2}y'dx=\int 2xdx\\\Leftrightarrow\int y^{3/2}dy=\cancel{2}\frac{x^{1+1}}{\cancel{1+1}}+C\\\Leftrightarrow y^{5/2}=\frac52\left(x^2+C\right)\\\Rightarrow y=\left[\frac52\left(x^2+C\right)\right]^{}2/5,C\in\mathbb{R}\\\text{Om }y(x)=0\;\forall{x}\in\mathbb{R},\text{ så är }y'(x)=0\\\left.\begin{aligned}\text{VL: }y^2y'?0^2\times0=0\\\text{HL: }2x\sqrt{y}=2x\times+=0\end{aligned}\right\}\;\;y(x)=0\text{ är en lösning}\\\underline{\text{Svar}}:y(x)=\left[\frac52\left(x^2+C\right)\right]^{2/5},\;x\in\mathbb{R}\\\text{eller }y(x)=0\end{align}$$
+	- **Initialvärdersproblem**
+		- **Ex**: *Lös* **IVP** $$\begin{align}y^2y'=2x\sqrt{y},\;\;y(1)=1\\\underline{\text{Lösn}}:y=\left[\frac52\left(x^2+C\right)\right]^{2/5}\text{ eller }y=0\\y=0\text{ uppfyller inte vilkor }y(1)=1\\y(1)=1\\\Leftrightarrow\left[\frac52\left(1^2+C\right)\right]^{2/5}=1\\\Leftrightarrow\left[\frac52(10C)\right]^2=1^5=1\\\Leftrightarrow\frac52(1+C)=\pm1\\\Leftrightarrow1+C=\pm\frac25\Leftrightarrow\left\{\begin{aligned}-1+\frac25\\-1-\frac25\end{aligned}\right.\\\Leftrightarrow C=\frac{-3}5\text{ eller }\frac{-7}5\\\underline{\text{svar}}:y=\left[\frac52\left(x^2-\frac35\right)\right]^{2/5}\text{ eller}\\y=\left[\frac52\left(x^2-\frac75\right)\right]^{2/5}\end{align}$$
+		- **Kontrol** $$\begin{align}y=\left[\frac52\left(x^2+C\right)\right]^{2/5}\\\Rightarrow y'=\frac52\times\left(\frac52\left(x^2\right)\right)^{2/5-1}\times\frac{\cancel{}}\\y^2y'=\end{align}$$
+- **Separabel**
+	- $y'+y=\sin x$: *Inte Separabel*.
+	- $yy'=\sin x$: *Separabel*.
+	- $y'+y=2$: *Separabel*.
+		- **Lösn**: $$\begin{align}y'+y=2\\\Leftrightarrow y'=2-y\\\text{För }y(x)\neq2\\y'=2-y\Leftrightarrow\frac{y'}{2-y}=1\\\text{Integrera m.a.p. }x\\\int\frac1{2-y}dy=\int1dx\Leftrightarrow\ln\mid2-y\mid=x+D\\\Leftrightarrow\mid2-y\mid=e^{x-D}=Ce^x,\text{ där }C=e^D>0\\\Leftrightarrow2-y=Ce^x,C>0,y\leq2\\\text{eller }y-2=Ce^x,C>0,y\geq2\\\Leftrightarrow y=2-Ce^x,C>0\\\text{eller }y=2+Ce^x,C>0.\\\text{Om }y(x)=2\forall x\in\mathbb{R},\text{ blir }y'(x)=VL_1=y'ý=0+2=2=HL_1\\y(x)=2\forall x\in\mathbb{R}\text{ är också en lönsning}\\\underline{\text{Svar}}: y(x)=2+C_0e^x,x\in\mathbb{R}\\\text{där }C_\in\mathbb{R}\text{är en bestämning}\end{align}$$
+- **ODE av ordning 2 med konstant koeffienter**
+	- **ODE**: $y''+ay'+by=h(x)$
+	- **Homogen ODE**: $y''+ay+by=0$
+	- *Eftersom **ODE** är linjär, superpositionsprincip ger att* $\left.\begin{aligned}y_h\;\;\text{homohen lösning}\\y_p\;\;\text{Partikulär lösning}\end{aligned}\right\}\Longrightarrow y_h+y_p\;\;\text{också en lösning.}$
+	- **Karakteristiska polynomet**: $p(r)=r^2+ar+b$
+	- **Karakteristiska ekvationen**: $p(r)=0$
+	- **Homogena lösningar**: *Fall 1: Karakteristiska polynomet har reella rötter $r_1$ och $r_2$, $r_1\neq r_2$. Alla hommogena lösningar ges av*$$C_1e^{r_1x}+C_2e^{r_2x}$$*Fall 2: Karakteristiska polynomet har reel dubbelrot $r_0$. Alla homogena lösningar ges av*$$\left(C_1x+C_2\right)e^{r_0x}$$*Fall 3: Karakteristiska polynomet har komplexa rötter $k+i\omega$.*$$\left(A\sin\omega{x}+B\cos\omega{x}\right)e^{kx}$$
+	- **Ex Homohena**$$\begin{align}\text{Fall 1: }y''-3y'+2y=0\\\text{Karakteristiska polynomet}\\P(n)=n^2-3n+2\\P(n)=0\Leftrightarrow\left(n-2\right)\left(n-1\right)\\\Leftrightarrow{n}=1\text{ eller }2\\y_h=C_1e^xĆ_2e^{2x}\\\\\text{Fall 2: }y''-4y'+4y=0\\P(n)=n^2-4n-4=\left(b-2\right)^2\\P(n)=0\Leftrightarrow\left(n-2\right)^2=0\Leftrightarrow n=2\\y_h=\left(C_1c+C_0\right)e^{2x}\\\\\text{Fall 3: }y''-4y+5y=0\\P(n)=n^2+4n+5=\left(n-2\right)^2+1\\P(n)=0\Leftrightarrow\left(n-2\right)^2+1=0\\\Leftrightarrow n=2\pm i\end{align}$$
+	- **Ansatser**
+		- $h(x)=P(x)\Rightarrow y_p(x)=x^mA(x),\;grad(A)=grad(1).$
+		  **Ex**: $h(x)=x^2\Rightarrow y_p(x)=x^m\left(a_2x^2+a_1x+a_0\right)$ $$\begin{align}y''-3y'+2y=x^2+1\\y_p=ax²+bx+c\\\Rightarrow y_p'=2ax+b\\\Rightarrow y_p''=2a\\\text{Sätt in i ODE}\\3a-3\left(2ax+b\right)+2\left(ax^2+bx+c\right)=x^2+1\\\Leftrightarrow 2ax^2+\left(2b-6a\right)x+2a-3b+2c=x^2+1\\\text{Jämför koeffieinten:}\\x^2:\;\;2a=1\Leftrightarrow a=\frac12\\x^2:\;\;2b-6a=0\Leftrightarrow b=3a=\frac32\\x^0:\;\;2a-3b+2x=1\Leftrightarrow2x=1-2a+3b=1-1+\frac92\\\Leftrightarrow c=\frac94\\\underline{\text{sum}}:\;y_p==\frac12x^2+\frac32x+\frac94\\\text{Almän lösning till ODE:}\\y=t_h+y_p=C_1e^x+C_2e^{2x}+\frac12x^2+\frac32x+\frac94\end{align}$$$$\begin{align}y''=x+1\\y_h=C_x+C_0\\y_p=x^2\left(ax+b\right)=ax^3+bx^2\\\Rightarrow y'_p=3ax^2+2bx\\\Rightarrow y''_p=6ax+2b\\\text{Sätt in }y_p\text{ i ODE: }y''_p=x+1\\\Leftrightarrow6ax+2b=x+1\\\Leftrightarrow6a=1,2b=1\Leftrightarrow a=\frac16,b=\frac12\\\underline{\text{Svar}}:\;y=\frac16x^3\frac12x^2+C_1x+C_0\end{align}$$
+		- 
+- **Examples**
+	- $$\begin{align}y^2y'=2xy^{1/2}\\\text{Lösn: För }y(x)\neq0,\\y^2y'=2xy^{1/2}\Leftrightarrow y^{3/2}y'=2x\\\text{Integrera m.a.p. }x,\\\frac25y^{2/5}=x^2+C\Leftrightarrow C=\frac25-1=-\frac35\\\text{Lösning är}\\y\begin{aligned}=\left(\frac52\left(x^2-\frac35\right)\right)^{2/5}\\=\left(\frac52x^2-\frac32\right)^{2/5}\end{aligned}, x^2\geq\frac35\\x\leq\sqrt{-\frac35}\text{ eller }x\geq\sqrt\frac35\end{align}$$
+	- $$\begin{align}e^{x^2}+y'e^{x^2}\times2xy=\left(e^{x^2}y\right)'\end{align}$$
+	- $$\begin{align}y'+y=2\\\text{Linjär, ordning 1}\\\text{Integrerande faktor}\\\int1dx=x+C\\\text{Vi väljer }IF=e^x\\\text{Multiplicera ekvationen med }IF\\e^xy'+x^x=2e^x\\\Leftrightarrow e^xy'+\left(e^x\right)'y=2e^x\\\Leftrightarrow\left(e^xy\right)'?2e^x\\\text{(Product regel)}\\\text{Integrera}\\e^xy=2\int e^xdx=2e^x+C\\\Leftrightarrow y=x^{-x},x\in\mathbb{R},X\in\mathbb{R}\text{ är konstant.}\end{align}$$
+	- $$\begin{align}xy'-y=x^2,x>0\\\underline{\text{Lösn}}:\text{ Linjär första ordning}\\xy'-y=x^2\Leftrightarrow y'-\frac1xy=x\\\int\left(-\frac1x\right)dx=-\ln x+C, x>0\\\end{align}$$
@@ -0,0 +1,39 @@
+**OBS Kontrolera, ALTID**
+- Definition
+	- **Def**: *En  funktion $F$ är en primär funktion till funktionen $f$ i ett intervall $I$ om $F'(x)=f(x)$ för varje $x\in{I}$*
+	- $\left.\begin{aligned}F'_1(x)=f(x)\\F'_2(x)=f(x)\end{aligned}\right\}\Rightarrow F_1(x)=F_2(x)+C,\;\;C\text{ är godtycklig konstant.}$
+	- *Beteckning: $\int{f(x)dx}=F(x)+C$ där $F$ är en partikulär primitiv funktion till $f$ och $C$ är en godtycklig konstant.*
+	- **Ex**: *Visa att $\ln\mid{x+\sqrt{x^2+a}}\mid$ är en primitiv funktion till $\frac1{\sqrt{x^2+a}}$* $$\begin{align}\frac{d}{dx}\left(\ln\left|x+\sqrt{x^2+a}\right|\right)\\=\frac1{x+\sqrt{x^2+a}}\left(\frac{d}{dx}\left(x+\sqrt{x^2+a}\right)\right)\text{ (kedjeregel)}\\=\frac1{x+\sqrt{x^2+a}}\left(1+\frac1{2\sqrt{x^2+a}}\frac{d}{dx}\left(x^2+a\right)\right)\text{ (kedjeregle, linjärtet)}\\=\frac1{x+\sqrt{x^2+a}}\left(1+\frac{\cancel2x}{\cancel2\sqrt{x^2+a}}\right)\\=\frac1{\cancel{x+\sqrt{x^2+a}}}\times\frac{\cancel{\sqrt{x^2+a}+x}}{\sqrt{x^2+a}}=\frac1{x^2+a}\\\ln\left|x+\sqrt{x^2+a}\right|\text{ är en primär funktion till }\frac1{\sqrt{x^2+a}}\end{align}$$
+- Standerd Primetiv
+	1. $f(x)=0\;\;\Rightarrow\;\;F(x)=C$
+	2. $f(x)=x^n\;\;\Rightarrow\;\;F(x)=\frac{x^{n+1}}{n+1}+C\;n\in\mathbb{Z}\setminus\{-1\}$
+	3. $f(x)=x^\alpha\;\;\Rightarrow\;\;F(x)=\frac{x^{\alpha+1}}{\alpha+1}+C\;\alpha\in\mathbb{R}\setminus\{-1\},\;x>0$
+	4. $f(x)=e^x\;\;\Rightarrow\;\;F(x)=e^x+C$
+	5. $f(x)=x^{-1}\;\;\Rightarrow\;\;F(x)=\ln\left|x\right|+C,\;x\neq0$
+	6. $f(x)=\sin x\;\;\Rightarrow\;\;F(x)=-\cos x+C$
+	7. $f(x)=\cos x\;\;\Rightarrow\;\;F(x)=\sin x+C$
+	8. $f(x)=\sec^2x=1+\tan^2x\;\;\Rightarrow\;\;F(x)=\tan x+C$
+	9. $f(x)=a^x\;\;\Rightarrow\;\;F(x)=\frac{a^x}{\ln a}+C,\;a>0$
+- Regler
+	- *Låt $F$ vara så att $F'(x)=f(x)$*
+		- **Linjäritet**: $\int\left(\alpha f+\beta g\right)dx=\alpha\int gdx$
+		- **Sammansatt funktion**: $\int\left(f\left(g\left(x\right)\right)g'\left(x\right)\right)dx=F\left(g\left(x\right)\right)+C$ *I synnerhet*: $\int\left(f\left(ax+b\right)\right)dx=\frac1aF\left(ax+b\right)+C$
+		- **Divition**: $\int{\frac{f'(x)}{f(X)}dx}=\ln\left|f(x)\right|+C$
+		- **Partiell integration**: $$\begin{align}\int{\left(f\left(x\right)\right)dx}=\left(\int{fdx}\right)g\left(x\right)-\int\left(\int{fdx}\right)g'\left(x\right)dx\\=F\left(x\right)g\left(x\right)-\int{F\left(x\right)g'\left(x\right)dx}\end{align}$$
+
+| **Integral** | $\sqrt{ax+b}$   | $\sqrt{\frac{ax+b}{cx+d}}$   | $\sqrt{x^2+a}$     |
+| ------------ | --------------- | ---------------------------- | ------------------ |
+| **Utbyte**   | $t=\sqrt{ax+b}$ | $t=\sqrt{\frac{ax+b}{cx+d}}$ | $t=x+\sqrt{x^2+a}$ |
+- Regler Example $$\begin{align}\text{Låt }g(x)=y\Rightarrow dy=g'(x)dx\\\int f(g(x))g'(x)dx=\int f(y)dy\\F(x)+C=F(g(x))+C\end{align}$$
+	- **Ex** $$\begin{align}\int\frac1{x^{1/2}+x^{3/2}}dx=I\\\text{Låt }y=\sqrt{x}\Rightarrow dy=\frac1{2\sqrt{x}}dx\\I=\int\frac1{\sqrt{x}\left(1+x\right)}dx=\int\frac2{1+x}\times\frac{dx}{2\sqrt{x}}\\=\int\frac2{1+y^2}dy=2\int\frac1{1+y^2}dy\;\;\left(\frac{d}{dx}\left(\arctan x\right)=\frac1{1+x^2}\right)\\=2\arctan y+C\\=2\arctan\sqrt{x}+C,\text{ där }X\in\mathbb{R}\\\text{prof: }\left(2\arctan\sqrt{x}\right)'\\=\cancel2\times\frac1{1+\left(\sqrt{x}\right)^2}\times\frac1{\cancel2\sqrt{x}}\\=\frac1{x^{1/2}+x^{3/2}}\checkmark\end{align}$$$$\begin{align}\int\cos\left(2x+\pi\right)dx\\=\frac12\sin\left(2x+\pi\right)+C\end{align}$$$$\begin{align}I=\int\frac{\sin x}{\cos x}dx=-\int\frac{\left(\cos x\right)'}{\cos x}dx=-\ln\left|\cos x\right|+C\end{align}$$$$\begin{align}\left(F(x)g(x)\right)'=F'(x)g(x)+F(x)g'(x)\\=f(x)g(x)+F(x)+g'(x)\\F(x)g(x)=\int f(x)g(x)dx+F(x)g'(x)dx\end{align}$$$$\begin{align}\int\left(x^2-4x+5\right)\sin2xdx\\\stackrel{\text{PI}}{=}\left(\int\sin2xdx\right)\left(x^2-4x+5\right)-\int{\left(\int\sin2xdx\right)\left(x^2-4x+5\right)'dx}\\=-\frac12\left(\cos2x\right)\left(x^2-4x+5\right)+\frac12\int\left(\cos2x\right)\left(3x+4\right)dx\\\stackrel{\text{PI}}{=}-\frac12\left(x^2-4x+5\right)\cos2x+\frac12\left(\sin2x\right)\left(x-2\right)-\int\frac12\sin2xdx\\=-\frac12\left(x^2-4x+5\right)\cos2x+\frac12\left(x-2\right)+\frac14\cos2x+C\\=-\frac14\left[\left(2x^2-8x+10-1\right)\cos2x-2(x-2)\sin2x\right]+C\\=\frac{x-2}2\sin2x-\frac{2x^2-8x+9}4\cos2x+C\end{align}$$
+- **Ex kontrol** $$\begin{align}\int\left(\sin x^2\right)\left(2x\right)dx\\=-\cos x^2+C\\\text{prof: }\left(-\cos x^2+C\right)'\\=-\left(-\sin x^2\right)\left(x^2\right)'=\left(\sin x^2\right)\left(x^2\right)\end{align}$$
+- Rationella Funktioner $$\begin{align}f(x)=\frac{P(x)}{Q(x)}\text{ där }P,Q\text{ är polynomer}\\1.\;\text{ Om }grad(P)\geq grad(Q)\text{ polynomdivition }\\f(x)=\frac{P(x)}{Q(x)}=k(x)+\frac{R(x)}{Q(x)}\\\text{där }K,R\text{ är polynom, }grad(R)<grad(Q)\\2.\;\text{ Faktorisera }Q(x)\\Q(x)=c(x-a_1)(x-a_2)\dots(x^2+b_1x+d)\dots\\3.\;\text{ Partialbråkuppdelning}\\\text{Antag att}\\\frac{R(x)}{Q(x)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\dots\dots\\\text{Bestäm konstanten i HL genom att jämföra med VL}\\4.\;\text{ Integrera}\end{align}$$
+	- **Ex**: $$\begin{align}\int\frac{5x+4}{x^2+3x+2}dx=I\\\text{Lösm: Eftersom }grad(5x+4)<grad(x^2+3x+2)\text{ polinomdivision behövs inte}\\\text{Foktoresera nämnaren: }x^2+3x+2=(x+2)(x+1)\\\text{PBU: Antag att }\exists\text{ konstanten }A,B\in\mathbb{R}:\\\frac{5x+4}{x^2+3x+2}=\frac{A}{x+2}+\frac{B}{x+1}=\frac{A(x+1)+B(x+2)}{(x+2)(x+1)}\\\Rightarrow\;5x+4=A(x+1)+B(x+2)\\\text{Metod 1: Prova olika värde av }x\\x=-1:\;\;5\times(-1)+4=A\times{O}+B(-1+2)\Leftrightarrow B=-1\\x=-2:\;\;5\times(-2)+4=A\times{O}+B\times{O}\Leftrightarrow A=6\\\text{Metod 2: Jämför koefficenten}\\5x+4=A(x+1)+B(x+2)=(A+B)x+(A+2B)\\\text{Jämför koefficienter till }x^n\\\left.\begin{aligned}x^1\;\;:\;\;\;\;5=A+B\\x^0\;\;:\;\;4=A+2B\end{aligned}\right\}\Leftrightarrow\begin{aligned}A=6\\B=-1\end{aligned}\\\frac{5x+4}{x^2+3x+2}=\frac{6}{x+2}-\frac1{x+1}\\\int\frac{5x+4}{x^2+3x+2}dx=\int\left(\frac{6}{x+2}-\frac{1}{x+1}\right)dx\\=6\int\frac1{x+2}dx=\int\frac1{x+1}\\=6\ln\mid{x+2}\mid-\ln\mid{x+1}\mid+C\end{align}$$
+	- **Integral av $\frac{Ax+B}{x^2+ax+b}$** $$\frac{Ax+B}$$
+
+| Faktor i $Q(x)$       | PBU                                                             |
+| --------------------- | --------------------------------------------------------------- |
+| $x-a$                 | $\frac{A}{x-a}$                                                 |
+| $(x-a)^n$             | $\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\dots+\frac{A_n}{(x-a)^n}$ |
+| $(x^2+ax+b),\;a^2<4b$ | $\frac{Ax+B}{x^2+ax+b}$                                         |
+|                       |                                                                 |
@@ -0,0 +1,18 @@
+**Rita graf till** $f(x)=-\frac{\ln x}x$
+$$\begin{align}1.\;\;D_f=\left(0,\infty\right)\\2.\;\;\text{Lodrät asymptot }x=0\\\text{Vågrät asymtot }y=0\\3.\;\;\text{Stationära punkten:}\\f(x)=\frac{\ln x}x\\\text{derivera m.a.p. }x\\f'(x)=-\frac{x(\ln x)'-(\ln x)(x)'}{x^2}\\=-\frac{x\times\frac1x-(\ln x)(1)}{x^2}\\=\frac{\ln x-1}{x^2}\\\text{Stationär punkten uppfyller }f'(x)=0\\\Leftrightarrow\frac{\ln x-1}{x^2}=0\\\Leftrightarrow\ln x=1\\x=e\end{align}$$
+*Täkentabell*
+
+|           | $e$                           |
+| --------- | ----------------------------- |
+| $\ln x-1$ | $\;\;\;\;0\;\;+$              |
+| $x^2$     | $+++$                         |
+| $f'(x)$   | $-\;0\;\;+$                   |
+| $f(x)$    | $\searrow\rightarrow\nearrow$ |
+*Enlight tabellen har $f$ en lokal minimum punkt på $\left(e,-\frac1e\right)$ Punkten är också en global minimum*
+*Graf*![[TE1.png]]
+**Visa att** $x^{\frac1x}\leq e^{\frac1e}$
+*Lösning: Från ovan:*
+$$\begin{align}-\frac{\ln x}x\geq-\frac1e\\\Leftrightarrow\frac{\ln x}x\leq\frac1e\Leftrightarrow\ln x^{\frac1x}\leq\frac1e\Leftrightarrow x^{\frac1x}\leq e^{\frac1e}\\\text{(ty ln är strängt vexande)}\end{align}$$
+**Koraste avtåndet av**: $\left(0,1\right)$ till kurvan $x^2-y^2=1$
+$$\begin{align}\text{Lösn: Avståndet av }\left(0,1\right)\text{ till en punkt }\left(x,y\right)\text{ ges av}\\d)\sqrt{\left(x-0\right)^2+\left(y-1\right)^2}=\sqrt{x^2+y^2-2y+1}\\\text{Punkten }\left(x,y\right)\text{ ligger på kurvan om }x^2-y^2=1\\\text{Avståndet av }\left(0,1\right)\text{ till }\left(x,y\right)\text{ på kurvan är }\\d=\sqrt{1+y^2+y^2-2y+1}=\sqrt{2y^2-2y+2}\\\Rightarrow d^2=2y^2-2y+2\end{align}$$
+*Notera att $d$ och $d^2$ har minimum värde på samma punkt. Definiera* $$\begin{align}f(y)=d^2=2y^2-2y+2\\\text{Derivera m.a.p. }y\\f''(y)=4>0\\\text{Stationär punkt:}\\f'(x)=0\Leftrightarrow4y-2=0\Leftrightarrow y=\frac12\\f''(\frac12)=4>0\\\text{sum: }y=\frac12\text{ ger minimum värde för }f\\\text{sum: avståndet är minst då }y=\frac12\text{ Mista avståndet är}\\d_{min}=\sqrt{s\times\left(\frac12\right)^2-\cancel{2\times\frac12}+2}=\sqrt\frac32\\\text{Närmaste punkten}\\x-\left(\frac12\right)^2=1\Leftrightarrow x^2=\frac54\Leftrightarrow x=\pm\frac{\sqrt5}2\\\text{sum: }\left(-\frac{\sqrt5}2,\frac12\right)\&\left(\frac{\sqrt5}2,\frac12\right)\\\text{Kontroll: }\sqrt{\frac52+\left(\frac12-1\right)^2}=\sqrt{\frac54+\frac14}=+\sqrt\frac32\end{align}$$
@@ -0,0 +1,38 @@
+- Radian:
+	- **Def**: *It is the SI unit for measuring angles (in the plane).*
+	- **Def**: *$1$ radian is defined as the angle subtended at the center by a circular arc of length equal to the radius*
+	- **Def**: *A general angle is measured in radians as the ration of the length an associated circular arc and the corresponding radius. That is $\theta=\frac{s}{r}\text{rad}$*
+	- **Def**: *Usually "$rad$" is omitted.*
+	- Ex: $$\begin{align}180^\circ=\pi\text{ rad}\\\frac{\pi}{3}\text{ rad}=30^\circ\\\frac{\pi}{4}\text{ rad}=45^\circ\\\frac{\pi}{3}\text{ rad}=60^\circ\\\frac{\pi}{2}\text{ rad}=90^\circ\\2\pi\text{ rad}=360^\circ\end{align}$$
+- The right angled triangle
+	- **Def**: *The trigonometric functions: *$$\begin{align}\sin\theta=\frac{\text{perpendicular}}{\text{hypotenuse}}\\\cos\theta=\frac{\text{base}}{\text{hypotenuse}}\\\tan\theta=\frac{\text{perpendicular}}{\text{base}}\end{align}$$
+	- In addition to above, $\csc\theta=\frac{1}{\sin\theta},\sec\theta=\frac{1}{\cos\theta},\cot\theta=\frac{1}{\tan\theta}$
+	- Pythagoras' formula: $p^2+b^2=h^2$
+	  which leads to the **trigonometric identity**: $\sin^2\theta+\cos^2\theta=1$
+	  and also $\tan^2\theta+1=\sec^2\theta$
+	- Dominains and ranges:
+		- $D_{\sin}=\mathbb{R}\;\;R_{\sin}=[-1,1]$
+		- $D_{\cos}=\mathbb{R}\;\;R_{\cos}=[-1,1]$
+		- $D_{\tan}=\mathbb{R}\setminus\{n\pi+\frac{\pi}{2}:n\in\mathbb{Z}\}\;\;R_{\tan}=(-\infty,\infty)$
+- Useful relations
+	- $\sin(-\theta)=-\sin(\text{odd}),\cos(-\theta)=\cos\theta(\text{even})$
+	- Periodicity: $\sin(\theta+2n\pi)=\sin\theta,\cos(\theta+2n\pi)=\cos\theta,\tan(\theta+n\pi)=\tan\theta$
+	- Complementary angles: $\sin(\frac{\pi}{2}-\theta)=\cos\theta,\cos(\frac{\pi}{2}-\theta)=\sin\theta$
+	- Sift by $\pi$: $\sin(\theta\pm\pi)=-\sin\theta,\cos(\theta\pm\pi=-\cos\theta$
+	- Sum of angles: $\sin(\theta+\phi)=\sin\theta\times\cos\phi+\cos\theta\times\sin\phi,\cos(\theta+\phi)=\cos\theta\times\cos\phi-\sin\theta\times\sin\phi,\tan(\theta+\phi)=\frac{\tan\theta+\tan\phi}{1-\tan\theta\times\tan\phi}$
+	- Double angle: $\sin(2\theta)=2\sin\theta\cos\theta,\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta},\cos(2\theta)=\cos^2-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta$
+	- Half angle: $2\sin^2\frac{\theta}{2}=1-\cos\theta,2\cos^2\frac{\theta}{2}=1+\cos\theta$
+- Solving trigonometric equations
+	- $\sin\theta=\sin{a}\Leftrightarrow\theta=\left\{\begin{align}a+2n\pi,n\in\mathbb{Z}\\\pi-a+2n\pi,n\in\mathbb{Z}\end{align}\right.$
+	- $\cos\theta=\cos{a}\Leftrightarrow\theta=\left\{\begin{align}a+2n\pi,n\in\mathbb{Z}\\-a+2n\pi,n\in{Z}\end{align}\right.$
+	- $\tan\theta=\tan{a}\Leftrightarrow\theta=a+n\pi,n\in\mathbb{Z}$
+	- Ex: Solve $\sin(x+\frac{\pi}{6})=\frac{\sqrt{3}}{2}$ 
+- Inverse trigonometric function
+	- **Def**: *$f(x)=\sin(x),x\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$. Then $f$ is strictly increasing on $D_f$ and hence inverible. The fuction $\arcsin$ is defined as $$\arcsin(x)=f^{-1}(x)\text{ on }D_{arcsin}=R_f=[-1,1]$$*
+	- **Similarly**: *For $g(x)=\cos(x),x\in\left[0,\pi\right]$ (which is strictly decreasing) and $h(x)=\tan(x),x\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ (which is strictly increasing), the function $\arccos$ and $\arctan$ are defined as $$\begin{align}\arccos(x)=g^{-1}(x)\text{ on }D_{\arccos}=\left[-1,1\right]\\\arctan(x)=h^{-1}(x)\text{ on }D_{\arctan}=\mathbb{R}\end{align}$$*
+	- **Note**: *That the tanges $R_{\arcsin}=R_{\arctan}=\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ whereas $R_{\arccos}=\left[0,\pi\right]$*
+- Properties
+	- **Def**: $$\begin{align}\sin(\arcsin(x))=x\forall{x}\in\left[-1,1\right]\text{ | }\arcsin(sin(x))=x\text{ if }x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\\\cos(\arccos(x))=x\forall{x}\in\left[-1,1\right]\text{ | }\arccos(\cos(x))=x\text{ if }x\in\left[0,\pi\right]\\\tan(\arctan(x))=x\forall{x}\in\mathbb{R}\text{ | }\arctan(\tan(x))=x\text{ if }x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\end{align}$$
+	- **Complementary angles**: $$\arcsin(x)+\arccos(x)=\frac{pi}{2},\;\arctan(x)+\arccos(x)=\frac{\pi}{2}$$
+	- **Negatives**: *$\arcsin$ and $\arctan$ are odd functions. $$\begin{align}\arcsin(-x)=-\arcsin(x)\\\arccos(-x)=\pi-\arccos(x)\\\arctan(-x)=-\arctan(x)\end{align}$$*
+- 
@@ -0,0 +1,13 @@
+- **DEF**
+	- *I en rätviklig rektangle stämmer $\overrightarrow{AC}=\left(\overrightarrow{AB}+\overrightarrow{AD}\right)$*
+	- $\overrightarrow{u}=\left(1,2,3\right)=\left(\begin{aligned}1\\2\\3\end{aligned}\right)=\left[\begin{aligned}1\\2\\3\end{aligned}\right]$
+	- *$\mid\mid{V}\mid\mid$ Är längden av $V$*
+- **Exemple**
+	- $$\begin{align}\text{Rektangeln }A,\;B,\;C,\;D\;\text{. Låt }E\text{ Vara punkten som delar diagonalen }\overline{AC}:\text{förhållandet }1:3\\\left(\text{dvs: }\overline{AE}:\overline{EC}=1:3\right)\\\text{Benämna }\overrightarrow{a}=\overrightarrow{AB},\overrightarrow{h}=\overrightarrow{AD}\text{ Uttryc vektor }\overrightarrow{c}=\overrightarrow{EC}\text{ i termer av }\overrightarrow{a}\text{ och }\overrightarrow{h}\\\\\text{Vart ligger punkten }E\:\text{? Hur kan vi uttrycka }\overrightarrow{c}\text{ med hjälp av }\overrightarrow{a}\text{ och }\overrightarrow{h}\:\text{?}\\\overrightarrow{c}=\overrightarrow{EC}=\frac34\overrightarrow{AC}=\frac34\left(\overrightarrow{AB}+\overrightarrow{BC}\right)=\frac34\left(\overrightarrow{a}+\overrightarrow{h}\right)=\frac34\overrightarrow{a}+\frac34\overrightarrow{h}\end{align}$$
+- **Koordinatrummet $\mathbb{R}^m$**
+	- *Mängden $\mathbb{R}^m\;\left(\text{där }m\in\mathbb{N}\right)$ består av koordinattpunkter av längden $m$ vars element är reella tal. Som skalärer tas $\mathbb{R}\text{(vanliga reela tal)}$*
+	- **Hur funkar $+$ och $\times$**
+		- $$\begin{align}\overrightarrow{u}=\left(u_1,u_2,u_3,\dots,u_m\right)\in\mathbb{R}^m\\\overrightarrow{v}=\left(v_1,v_2,\dots,y_m\right)\in\mathbb{R}^m\\\lambda\in\mathbb{R}\\\\\overrightarrow{u}+\overrightarrow{v}=\left(u_1+v_1,u_2+v_2,\dots,\lambda u_m\right)\\\text{OBS: }\left(1,2\right)+\left(3,4,5\right)\Rightarrow\text{Inte Definierat}\\\\\overrightarrow{u}\times\overrightarrow{v}=\left(u_1v_1,u_2v_2\dots,u_mv_m\right)\\\begin{aligned}\overrightarrow{u}=\left(1,2,0\right)\\\overrightarrow{v}=\left(0,0,-2\right)\end{aligned}\Rightarrow\overrightarrow{u}\times\overrightarrow{v}=\left(1\times0,2\times0,0\times\left(-2\right)\right)=\left(0,0,0\right)\\\text{Man kan i normala fall inte multiplecera vektorer!}\end{align}$$
+		- $$\overrightarrow{u},\overrightarrow{v}\in\mathbb{R}^3\Rightarrow\overrightarrow{n},\overrightarrow{n}\in\mathbb{R}^3$$
+			- **Sats**: *Låt $\overrightarrow{m},\:\overrightarrow{n}\in\mathbb{R}^3$. Då gäller det att: $\mid\mid\overrightarrow{m}\times\overrightarrow{n}\mid\mid=\mid\mid\overrightarrow{m}\mid\mid\times\mid\mid\overrightarrow{n}\mid\mid\times\sin(\theta)$. (Där $\theta$ är vinkeln mellan $\overrightarrow{m}$ och $\overrightarrow{n}$). (Jämför: $<\overrightarrow{u},\:\overrightarrow{v}>=\mid\mid\overrightarrow{u}\mid\mid\times\mid\mid\overrightarrow{v}\mid\mid\times\cos(\theta)$)*
+			- **Prof**: *Vi börjar med: $$\begin{aligned}\mid\mid\overrightarrow{u}\times\overrightarrow{v}\mid\mid^2=<\overrightarrow{u}\times\overrightarrow{v},\:\overrightarrow{u}\times\overrightarrow{v}>\stackrel{\text{(I)}}{=}<\overrightarrow{u},\;\overrightarrow{v}\times\left(\overrightarrow{u}\times\overrightarrow{v}\right)>\stackrel{\text{(II)}}{=}<\overrightarrow{u},\;<\overrightarrow{v},\;\overrightarrow{v}>\overrightarrow{u}-<\overrightarrow{v},\;\overrightarrow{u}>\overrightarrow{v}>\\=<\overrightarrow{u},\;<\overrightarrow{v},\;\overrightarrow{v}>\overrightarrow{u}>-<\overrightarrow{u},\;<\overrightarrow{v},\;\overrightarrow{u}>\overrightarrow{v}>\\=<\overrightarrow{v},\;\overrightarrow{v}><\overrightarrow{u},\;\overrightarrow{u}>-<\overrightarrow{u},\;\overrightarrow{v}><\overrightarrow{v},\;\overrightarrow{u}>\\=\mid\mid\overrightarrow{u}\mid\mid^2\times\mid\mid\overrightarrow{v}\mid\mid^2-\left(\mid\mid\overrightarrow{u}\mid\mid\times\mid\mid\overrightarrow{v}\mid\mid\times\cos(\theta)\right)^2\\=\mid\mid\overrightarrow{u}\mid\mid^2\times\mid\mid\overrightarrow{v}\mid\mid^2\times\left(1-\cos^2(\theta)\right)\\=\mid\mid\overrightarrow{u}\mid\mid^2\times\mid\mid\overrightarrow{v}\mid\mid^2\times\sin^2(\theta)\end{aligned}$$*
Author	SHA1	Message	Date
Zacharias	6a2505c8d1	vault backup: 2026-02-26 16:22:06	2026-02-26 16:22:06 +01:00
Zacharias	847065c9f4	vault backup: 2026-02-23 16:57:23	2026-02-23 16:57:23 +01:00
Zacharias	1baf168667	vault backup: 2026-02-23 16:06:43	2026-02-23 16:06:43 +01:00
Zacharias	12fac9f8ee	vault backup: 2026-02-20 14:51:42	2026-02-20 14:51:42 +01:00
Zacharias	24190f0673	vault backup: 2026-02-20 14:05:20	2026-02-20 14:05:20 +01:00
Zacharias	e1e0eae5ad	vault backup: 2026-02-19 14:56:07	2026-02-19 14:56:07 +01:00
Zacharias	3cc168969c	vault backup: 2026-02-16 16:58:21	2026-02-16 16:58:21 +01:00
Zacharias	cb5814d1c7	vault backup: 2026-02-16 16:01:51	2026-02-16 16:01:51 +01:00
Zacharias	5ab1e30edc	vault backup: 2026-02-05 15:00:10	2026-02-05 15:00:10 +01:00
Zacharias	215b052b00	vault backup: 2026-02-05 14:02:03	2026-02-05 14:02:03 +01:00
Zacharias	f50504676f	vault backup: 2026-02-05 13:46:21	2026-02-05 13:46:21 +01:00
Zacharias	7475264267	vault backup: 2026-02-04 14:53:16	2026-02-04 14:53:16 +01:00
Zacharias	6af909a219	vault backup: 2026-02-04 14:09:52	2026-02-04 14:09:52 +01:00
Zacharias	9561d843a1	vault backup: 2026-02-04 14:09:09	2026-02-04 14:09:09 +01:00
Zacharias	9f53ff73bc	vault backup: 2026-02-04 14:06:09	2026-02-04 14:06:09 +01:00
Zacharias	d716736b43	vault backup: 2026-02-04 14:01:55	2026-02-04 14:01:55 +01:00
Zacharias	db1643f5c9	vault backup: 2026-02-02 16:47:13	2026-02-02 16:47:13 +01:00
Zacharias	89047ac829	vault backup: 2026-02-02 16:07:18	2026-02-02 16:07:18 +01:00
Zacharias	c5fcf125a5	vault backup: 2026-01-30 19:40:30	2026-01-30 19:40:30 +01:00
Zacharias	1607e5ef22	vault backup: 2026-01-30 15:02:03	2026-01-30 15:02:03 +01:00
Zacharias	042a78753c	vault backup: 2026-01-29 15:41:23	2026-01-29 15:41:23 +01:00
Zacharias	3084db1e72	vault backup: 2026-01-26 11:53:05	2026-01-26 11:53:05 +01:00
Zacharias	f253fe796d	vault backup: 2026-01-20 09:26:43	2026-01-20 09:26:43 +01:00
Zacharias	3d1374f978	vault backup: 2025-12-10 14:50:14	2025-12-10 14:50:14 +01:00
Zacharias	507f7fb003	vault backup: 2025-12-09 14:41:50	2025-12-09 14:41:50 +01:00
Zacharias	07338a26a2	vault backup: 2025-12-08 14:59:26	2025-12-08 14:59:26 +01:00
Zacharias	ca651ab00e	vault backup: 2025-12-03 14:53:27	2025-12-03 14:53:28 +01:00
Zacharias	b4d244d347	vault backup: 2025-12-01 14:57:57	2025-12-01 14:57:57 +01:00
Zacharias	74ae29c2b4	vault backup: 2025-11-27 14:56:31	2025-11-27 14:56:32 +01:00
Zacharias	5c6c7ee11e	vault backup: 2025-11-26 16:53:02	2025-11-26 16:53:03 +01:00
Zacharias	1bd452ec8f	vault backup: 2025-11-25 16:05:22	2025-11-25 16:05:22 +01:00
Zacharias	a4931ccf37	vault backup: 2025-11-19 15:00:24	2025-11-19 15:00:24 +01:00
Zacharias	4763397f7a	vault backup: 2025-11-18 15:03:14	2025-11-18 15:03:14 +01:00
Zacharias	3d0926d746	vault backup: 2025-11-17 17:04:18	2025-11-17 17:04:18 +01:00
Zacharias	81464c72b9	vault backup: 2025-11-13 16:00:03	2025-11-13 16:00:03 +01:00
Zacharias	f718c50824	vault backup: 2025-11-13 15:59:21	2025-11-13 15:59:21 +01:00
Zacharias	8ba3e10fd0	vault backup: 2025-11-13 15:59:01	2025-11-13 15:59:01 +01:00
Zacharias	48eaaff3bb	vault backup: 2025-11-13 15:58:17	2025-11-13 15:58:17 +01:00
Zacharias	ec960e98fa	vault backup: 2025-11-13 15:01:21	2025-11-13 15:01:21 +01:00
Zacharias	d4b855f926	trying to fix merge issue	2025-11-12 17:11:34 +01:00
Zacharias	8f9dadd51c	vault backup: 2025-11-12 17:06:06	2025-11-12 17:06:06 +01:00
Zacharias	17f9f3aafa	vault backup: 2025-11-12 17:02:15	2025-11-12 17:02:15 +01:00
Zacharias	76331adc17	Merge branch 'master' of git.server.4zellen.se:Zacharias/Analys-och-Linj-r-algibra	2025-11-12 17:01:33 +01:00
Zacharias	a09b5a99c0	vault backup: 2025-11-12 17:01:15	2025-11-12 17:01:15 +01:00
Zacharias	a23a0d3bb0	vault backup: 2025-11-12 15:01:58	2025-11-12 15:01:58 +01:00
Zacharias	a3b1fbefd8	vault backup: 2025-11-11 17:01:13	2025-11-11 17:01:13 +01:00
Zacharias	4bd845de6b	vault backup: 2025-11-04 16:53:01	2025-11-04 16:53:01 +01:00
Zacharias	8e3b175d5d	vault backup: 2025-11-03 14:20:29	2025-11-03 14:20:29 +01:00
Zacharias	289f7fa403	vault backup: 2025-10-29 17:01:23	2025-10-29 17:01:23 +01:00
Zacharias	8031e75f98	vault backup: 2025-10-28 16:59:44	2025-10-28 16:59:44 +01:00