diff --git a/.obsidian/community-plugins.json b/.obsidian/community-plugins.json index a37479b..e69de29 100644 --- a/.obsidian/community-plugins.json +++ b/.obsidian/community-plugins.json @@ -1,15 +0,0 @@ -[ - "obsidian-minimal-settings", - "edit-in-neovim", - "obsidian-git", - "better-export-pdf", - "templater-obsidian", - "highlightr-plugin", - "obsidian-file-color", - "darlal-switcher-plus", - "dataview", - "obsidian-vimrc-support", - "emoji-shortcodes", - "typst-mate", - "inline-math" -] \ No newline at end of file diff --git a/.obsidian/plugins/obsidian-file-color/data.json b/.obsidian/plugins/obsidian-file-color/data.json index aeb5ea2..587ba26 100644 --- a/.obsidian/plugins/obsidian-file-color/data.json +++ b/.obsidian/plugins/obsidian-file-color/data.json @@ -1,5 +1,5 @@ { - "cascadeColors": true, + "cascadeColors": false, "colorBackground": false, "palette": [], "fileColors": [] diff --git a/11 Homework/Mathe II/Homework 2.pdf b/11 Homework/Mathe II/Homework 2.pdf new file 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+--- + +=== (b) $f(x) = x^6 / (x^4 + 3x^2 + 2)$ + +Polynomdivision: +$ x^6 / (x^4 + 3x^2 + 2) = x^2 - 3 + (7x^2 + 6) / ((x^2 + 1)(x^2 + 2)) $ + +Partialbruchzerlegung des Restes: +$ (7x^2 + 6) / ((x^2 + 1)(x^2 + 2)) = 8 / (x^2 + 2) - 1 / (x^2 + 1) $ + +Stammfunktion: +$ F(x) = 1/3 x^3 - 3x - arctan(x) + 4 sqrt(2) arctan(x / sqrt(2)) + C $ + +#pagebreak() +== Problem 3: Eigenwerte + +=== (a) Berechnung der Eigenwerte + +#let id = "I" +#let det = "det" + +==== (i) $A_1 = mat(3, -1, 0; 3, -1, 0; 4, -1, -2)$ +Charakteristisches Polynom (Entwicklung nach der 3. Spalte): +$ p(lambda) = (-2-lambda) det mat(3-lambda, -1; 3, -1-lambda) $ +$ p(lambda) = (-2-lambda) (lambda^2 - 2lambda) = -lambda(lambda - 2)(lambda + 2) $ +*Eigenwerte:* $lambda_1 = 0, lambda_2 = 2, lambda_3 = -2$ + +==== (ii) $A_2 = mat(1, 0, -1; 1, 0, 2; 1, 0, 1)$ +Charakteristisches Polynom (Entwicklung nach der 2. Spalte): +$ p(lambda) = (-lambda) det mat(1-lambda, -1; 1, 1-lambda) $ +$ p(lambda) = -lambda (lambda^2 - 2lambda + 2) $ +Nullstellen via $p q$-Formel: $lambda = 1 plus.minus sqrt(-1)$ +*Eigenwerte:* $lambda_1 = 0, lambda_2 = 1 + i, lambda_3 = 1 - i$ + +--- + +=== (b) Beweis: Eigenwerte von $A B$ und $B A$ + +Sei $lambda != 0$ ein Eigenwert von $A B$. Dann existiert ein Eigenvektor $v != 0$, sodass: +$ (A B) v = lambda v $ + +Multiplikation von links mit $B$: +$ B(A B v) = B(lambda v) $ +$ (B A)(B v) = lambda (B v) $ + +Da $lambda != 0$ und $v != 0$, ist $B v != 0$. +Somit ist $B v$ ein Eigenvektor von $B A$ zum Eigenwert $lambda$. $square$ diff --git a/30 Library/Hausaufgabe02.pdf b/30 Library/Hausaufgabe02.pdf new file mode 100644 index 0000000..220eb76 Binary files /dev/null and b/30 Library/Hausaufgabe02.pdf differ diff --git a/Untitled.base b/Untitled.base new file mode 100644 index 0000000..c6ab609 --- /dev/null +++ b/Untitled.base @@ -0,0 +1,3 @@ +views: + - type: table + name: Table