This commit is contained in:
parent
985b0a9e74
commit
13d4b597ff
BIN
build/main.pdf
BIN
build/main.pdf
Binary file not shown.
File diff suppressed because one or more lines are too long
|
@ -0,0 +1,17 @@
|
|||
from math import sqrt
|
||||
|
||||
def l1(x: list[float]) -> float:
|
||||
a=0
|
||||
for i in x:
|
||||
a+= abs(i)
|
||||
return a
|
||||
|
||||
def l2(x: list[float]) -> float:
|
||||
a=0
|
||||
for i in x:
|
||||
a+= i**2
|
||||
return sqrt(a)
|
||||
|
||||
|
||||
def li(x: list[float]) -> float:
|
||||
return max(x)
|
File diff suppressed because one or more lines are too long
|
@ -0,0 +1,58 @@
|
|||
== Exercise Sheet 2 - Introduction to Multivariable Functions
|
||||
|
||||
=== Exercise 1 @Exercise[2, 1]
|
||||
#block(
|
||||
fill: luma(230),
|
||||
inset: 8pt,
|
||||
radius: 4pt,
|
||||
[
|
||||
Consider the function
|
||||
|
||||
$
|
||||
f: RR^2 -> RR \
|
||||
(x,y) |-> x^2 - y^2
|
||||
$
|
||||
|
||||
Determine the levels sets at $c = 1$ and $c = −1$. What does it look like (sketch?)
|
||||
])
|
||||
|
||||
For $c = 1$:
|
||||
$
|
||||
f(x,y) =^! 1 \
|
||||
=> x^2 - y^2 = 1 <=> y = sqrt(x^2 - 1) \
|
||||
$
|
||||
|
||||
#figure(
|
||||
image("../media/img/ex2-1a.png", height: 20%), caption: [Zeichnung von $y = sqrt(x^2-1)$],
|
||||
)
|
||||
|
||||
For $c = -1$:
|
||||
$
|
||||
f(x,y) =^! -1 \
|
||||
=> x^2 - y^2 = -1 <=> y = sqrt(x^2 + 1) \
|
||||
$
|
||||
|
||||
#figure(
|
||||
image("../media/img/ex2-1b.png", height: 20%), caption: [Zeichnung von $y = sqrt(x^2+1)$]
|
||||
)
|
||||
|
||||
#pagebreak()
|
||||
=== Exercise 2 @Exercise[2, 2]
|
||||
#block(
|
||||
fill: luma(230),
|
||||
inset: 8pt,
|
||||
radius: 4pt,
|
||||
[
|
||||
Draw the following sets (what will be domains). Recall the lecture about curves.
|
||||
|
||||
#set enum(numbering: "(a)")
|
||||
+ $ D = {(x,y) thin|thin (x-2)^2 + (y+1)^2 <= 9} $
|
||||
+ $ D = {(x,y) thin|thin 0 <= x <= 1 and x^2 < y < sqrt(x)} $
|
||||
])
|
||||
|
||||
#set enum(numbering: "(a)")
|
||||
+ $
|
||||
(x-2)^2 + (y+1)^2 &=^! 9 \
|
||||
=> (x-2)^2 + (y+1)^2 &= x^2-4x+4+y^2+2y+1 \
|
||||
=> y =
|
||||
$
|
|
@ -0,0 +1,117 @@
|
|||
== Exercise Sheet 3 - Topology of Metric Spaces
|
||||
|
||||
=== Exercise 1 @Exercise[3, 1]
|
||||
#block(
|
||||
fill: luma(230),
|
||||
inset: 8pt,
|
||||
radius: 4pt,
|
||||
[
|
||||
Let $(X, d)$ be a metric space, and let $x_1, . . . , x_4 in X$. Show that
|
||||
|
||||
$
|
||||
|d(x_1,x_2) - d(x_2,x_3)| <= d(x_1,x_3)
|
||||
$
|
||||
])
|
||||
|
||||
$
|
||||
#text("z.z.") &|d(x_1,x_2) - d(x_2,x_3)| <= d(x_1,x_3) \
|
||||
d(x_1,x_3) <= &|d(x_1,x_2) + d(x_2,x_3)| \
|
||||
<=> 0 <= &|d(x_1,x_2) + d(x_2,x_3)| - d(x_1,x_3) \
|
||||
?????????????????&???????????????
|
||||
$
|
||||
|
||||
#pagebreak()
|
||||
=== Exercise 2 @Exercise[3, 2]
|
||||
#block(
|
||||
fill: luma(230),
|
||||
inset: 8pt,
|
||||
radius: 4pt,
|
||||
[
|
||||
We define the metric
|
||||
$
|
||||
d(x, y) = arctan(|x − y|)
|
||||
$
|
||||
|
||||
on $RR$. Show that $d$ indeed defines $a$ metric, i. e. it satisfies the axioms of a metric.
|
||||
|
||||
Hint: You can and should use that
|
||||
|
||||
$ arctan(x + y) ≤ arctan(x) + arctan(y); x, y in RR $
|
||||
|
||||
])
|
||||
|
||||
#set enum(numbering: "(I)")
|
||||
+ $
|
||||
#text("z.z.") wide d(x,y) = 0 &<=> x=y \
|
||||
d(x,y) = 0 &<=> arctan(|x-y|) = 0 \
|
||||
&<=> |x-y| 0 \
|
||||
&<=> x-y = 0 \
|
||||
&<=> x = y space qed
|
||||
$
|
||||
+ $
|
||||
#text("z.z.") wide d(x,y) &= d(y,x) \
|
||||
d(x,y) &= arctan(|x-y|) \
|
||||
&= arctan(| -1 dot (x-y) |) \
|
||||
&= arctan(|y-x|) \
|
||||
&= d(y,x) space qed
|
||||
$
|
||||
+ $
|
||||
#text("z.z.") wide d(x , y ) &<= d(x , z) + d(z, y ) \ \
|
||||
#text("es gilt:") |x-y| &= |x-z+z-y| <= |x-z| + |z-y| \
|
||||
arctan &#text("ist streng monoton steigend.") \ \
|
||||
d(x,y) &= arctan(|x-y|) \
|
||||
&= arctan(|x-z+z-y|) \
|
||||
&<= arctan(|x-z|+|z-y|) \
|
||||
&= arctan(|x-z|) + arctan(|z-y|) #text("Zauber") \
|
||||
&= d(x,z) + d(z,y) space qed
|
||||
$
|
||||
|
||||
Somit gelten alle Eigenschaften einer Metrik für $d(x, y) = arctan(|x − y|)$ $qed$.
|
||||
|
||||
#pagebreak()
|
||||
=== Exercise 3 @Exercise[3, 3]
|
||||
#block(
|
||||
fill: luma(230),
|
||||
inset: 8pt,
|
||||
radius: 4pt,
|
||||
[
|
||||
Show that the set $U=[0, 1) subset RR$ is neither open nor closed.
|
||||
])
|
||||
|
||||
$U$ ist nicht offen, weil es keinen offenen Ball um $x=0$ mit radius $epsilon
|
||||
> 0$ geben kann, der vollständig in $U$ liegt.
|
||||
|
||||
Damit $U$ geschlossen ist, müsste $RR\\U$ offen sein. Man kann jedoch keinen
|
||||
offenen Ball um $x=1; x in RR\\U$ mit radius $epsilon >0$ legen, sodass dieser
|
||||
vollständig in $RR\\U$ liegen würde. $qed$
|
||||
|
||||
=== Exercise 4 @Exercise[3, 4]
|
||||
#block(
|
||||
fill: luma(230),
|
||||
inset: 8pt,
|
||||
radius: 4pt,
|
||||
[
|
||||
Show that the union
|
||||
$ み = union.big space U_i $
|
||||
of the sequence of open sets $U$ is open.
|
||||
])
|
||||
|
||||
@Vorlesung[3, Theorem 2.2] sagt das ist bereits wahr.
|
||||
|
||||
Trotzdem: Jeses $x$ ist auf jeden fall teil mindestens eines $U_i$. Dann existiert ein offener Ball
|
||||
$B_epsilon(x) subset U_i subset み$ für alle $x subset U_i subset み$. Somit ist $み$ offen. $qed$
|
||||
|
||||
=== Exercise 4 @Exercise[3, 5]
|
||||
#block(
|
||||
fill: luma(230),
|
||||
inset: 8pt,
|
||||
radius: 4pt,
|
||||
[
|
||||
Show that the intersection
|
||||
$ ま = sect.big space U_i $
|
||||
of the sequence of closed sets $U$ is closed.
|
||||
])
|
||||
|
||||
@Vorlesung[3, Theorem 2.2] sagt das ist bereits wahr.
|
||||
|
||||
Geschlossen bedeutet das Komplement ist offen. $overline(ま) = overline(sect.big space U_i) = union.big overline(U_i)$ Wie in Ex 4 gezeigt, ist $overline(ま)$ offen, da $overline(U_i)$ offen ist, da $U_i$ geschlossen ist. $qed$
|
|
@ -1,2 +1,5 @@
|
|||
|
||||
#include "1.typ"
|
||||
#pagebreak()
|
||||
#include "2.typ"
|
||||
#pagebreak()
|
||||
#include "3.typ"
|
||||
|
|
|
@ -32,8 +32,6 @@
|
|||
// headcolor
|
||||
#let headcolor = rgb("80b3ff")
|
||||
|
||||
#set heading(numbering: "1.")
|
||||
|
||||
// reset counter at each chapter
|
||||
#show heading.where(level:1): it => {
|
||||
counter(math.equation).update(0)
|
||||
|
@ -51,8 +49,9 @@
|
|||
#set footnote(numbering: "*")
|
||||
|
||||
#set math.equation(numbering: n => {
|
||||
let h1 = counter(heading).get().first()
|
||||
numbering("(1.1)", h1, n)
|
||||
let h1 = counter(heading.where(level: 1)).get().first()
|
||||
// let h2 = counter(heading.where(level: 2)).get().first()
|
||||
numbering("(1.1.1)", h1, n)
|
||||
})
|
||||
|
||||
#set figure(numbering: n => {
|
||||
|
|
Binary file not shown.
After Width: | Height: | Size: 13 KiB |
Binary file not shown.
After Width: | Height: | Size: 13 KiB |
|
@ -23,3 +23,9 @@
|
|||
url = "https://matheguru.com/integralrechnung/herleitung-der-stammfunktion-des-naturlichen-logarithmus.html",
|
||||
urldate = "2024-09-09",
|
||||
}
|
||||
|
||||
@online{trig-formeln,
|
||||
title = "Formelsammlung Trigonometrie",
|
||||
url = "https://de.wikipedia.org/wiki/Formelsammlung_Trigonometrie#Produkte_der_Winkelfunktionen",
|
||||
urldate = "2024-09-10",
|
||||
}
|
||||
|
|
|
@ -0,0 +1,3 @@
|
|||
== Basics & Outlook @Vorlesung[Foliensatz 2]
|
||||
|
||||
- Größtenteils Wiederholung von Analysis
|
|
@ -1 +1,2 @@
|
|||
#include "1.typ"
|
||||
#include "2.typ"
|
||||
|
|
Loading…
Reference in New Issue