### A Pluto.jl notebook ###
# v0.20.21

using Markdown
using InteractiveUtils

# ╔═╡ 501f10c5-16fc-4250-91b7-405038b356c4
begin
	using PlutoTeachingTools
	using PlutoUI
	using CairoMakie
	ultramarine40 = RGBf(100/255.0, 143/255.0, 255/255.0)
	indigo50      = RGBf(120/255.0,  94/255.0, 240/255.0)
	magenta50     = RGBf(220/255.0,  38/255.0, 127/255.0)
	orange40      = RGBf(254/255.0,  97/255.0,   0/255.0)
	gold20        = RGBf(255/255.0, 176/255.0,   0/255.0)
	TableOfContents()
end

# ╔═╡ 644886f5-ff05-444c-9b2c-5f4f0c5ef37e
html"""
<h1 style="text-align:center">
	Höhere Mathematik II
</h1>
<div style="text-align:center">
	<p style="font-size: 30px; font-variant: small-caps; margin: 0px">
		Michael Schlottke-Lakemper, Manuel Torrilhon
	</p>
	<p style="font-size: 20px;">
		Universität Augsburg
	</p>
</div>
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	main {
		margin: 0 auto;
		max-width: 1000px;
    	padding-left: max(160px, 10%);
    	padding-right: max(160px, 10%);
	}
</style>
-->
"""

# ╔═╡ d638a6f8-1c7e-11f1-001e-59999a8dfd53
md"""
# 5 Analysis im ``\mathbb{R}^n``
"""

# ╔═╡ 37b665d6-1127-4dac-a03d-fd404a440d46
md"""
Wir betrachten nun Funktionen ``f: D \subset \mathbb{R}^n \to \mathbb{R}^m``
"""

# ╔═╡ 705c3e2b-f408-4506-b77d-e118b9c6b816
md"""
##### Beispiel:
1. ``n=1, m=3.\; f: [0,T] \to \mathbb{R}^3, t \mapsto \begin{pmatrix} x(t)\\y(t)\\z(t) \end{pmatrix}`` z.B. ein Pfad im Raum mit $t\;\widehat=$ Zeit.
2. ``n=3,m=1.\; f: [a,b]^3 \to \mathbb{R}, (x,y,z) \mapsto f(x,y,z)`` z.B. Temperaturfeld im Raum
3. ``n=2, m=3.\; f: [0,2\pi]^2 \to \mathbb{R}^3, (\varphi, \theta) \mapsto \begin{pmatrix} 2 \cos(\varphi) + \cos(\theta)\\ 2 \sin(\varphi) + \cos(\theta)\\ \sin(\theta) \end{pmatrix}`` Fläche eingebettet im Raum, z.B. Torus.
4. ``n=3, m=4``:
``\begin{align*}f:[0, \infty) \times [0, 2\pi] \times [0,\pi] &\to \mathbb{R}^3 \\ (r,\varphi, \theta) &\mapsto \begin{pmatrix} r \cos \varphi \cos \theta \\ r \sin \varphi \cos \theta \\ r \sin \theta \end{pmatrix}\end{align*}``
nichtlineare Koordinatentransformationen, z.B. Kugelkoordinaten.
"""

# ╔═╡ de759c8c-df0b-48e9-8df0-c8f7a8346092
md"""
## 5.1 Begrifflichkeiten (Topologie)
"""

# ╔═╡ a7adf8c8-0c4d-4749-a2e9-b3fb9e8dfcce
md"""
Wir betrachten ``\mathbb{R}^n`` mit einer Norm ``\| \cdot \|`` (z.B. 1-/2-/``\infty``-Norm).
1. ``U \subset \mathbb{R}^n`` heißt _Umgebung_ von ``x \in \mathbb{R}^n``, wenn irgendeine Kugel um $x$ zu $U$ gehört, d.h.
``\qquad \operatorname{B}_r(x) \coloneqq \{ y \in \mathbb{R}^n \mid \|x - y\| < r \} \subset U \quad \text{für ein} \quad r > 0``
2. ``V \subset \mathbb{R}^n`` heißt _offen_, falls zu jedem ``x \in V`` eine Kugel ``\operatorname{B}_r(x)`` in ``V`` liegt.
3. ``A \subset \mathbb{R}^n``heißt _abgeschlossen_, falls ``\mathbb{R}^n \setminus A`` offen ist.
4. ``B \subset \mathbb{R}^n`` heißt _beschränkt_, falls ``0 < b \in \mathbb{R}``existiert mit ``\|x\| \leq b < \infty`` für alle ``x \in B``.
5. Eine Folge ``\left( x^{(k)} \right)_{k \in \mathbb N} \subset \mathbb{R}^n`` heißt _konvergent_ gegen ``x \in \mathbb{R}^n``, falls
``\qquad \forall \varepsilon > 0 \quad \exists  k_{\varepsilon} \in \mathbb{N} \quad \|x^{(k)} - x \| \leq \varepsilon \quad k \geq k_{\varepsilon}``
"""

# ╔═╡ 1c809114-1cf1-4cdf-a52d-ffe10a0bdb45
md"""
##### Beispiel:
1. Ganz ``\mathbb{R}^n`` ist offen, also ``\emptyset = \{\}`` ist abgeschlossen.
2. ``(a,b), (a, \infty)`` sind offen, ``[a,b]`` und ``[a, \infty)`` abgeschlossen (!).
3. ``x^{(k)} = \begin{pmatrix} \frac{1}{k} \\ \frac{2k}{k+1} \end{pmatrix}`` konvergiert gegen ``x = \begin{pmatrix} 0\\2 \end{pmatrix}``
"""

# ╔═╡ cd5b92a2-78ec-4d62-beab-25a7ae1d1f74
md"""
##### Bemerkung:
1. `` A \subset \mathbb{R}^n`` abgeschlossen ``\Leftrightarrow`` Für jede konvergente Folge liegt der Grenzwert in ``A``. Das heißt, _Limes-Bildung führt nicht aus ``A`` hinaus_.
2. Über Konvergenz lässt sich keine Aussage machen.
"""

# ╔═╡ 2a00b81a-7344-4e0d-a55b-30c9bf7449bb
md"""
##### Definition:
Eine Menge ``K`` (``\infty``-dim. möglich) heißt _kompakt_, wenn zu jeder Folge ``\left( x^{(m)} \right)_{m \in \mathbb N} \subset K`` eine konvergente Teilfolge ``\left( x^{(m_j)} \right)_{m \in \mathbb N} `` existiert.
"""

# ╔═╡ 272024bf-d77d-4761-9a8b-39a0e418eb91
md"""
##### Bemerkung:
1. Kompakte Mengen sind handlich, da sie sich in mancher Hinsicht wie endliche Mengen verhalten.
2. Für ``\mathbb{R}`` sagt Bolzano/Weierstraß, dass abgeschlossene und beschränkte Intervalle kompakt sind.
"""

# ╔═╡ 86b143ec-13ab-42a3-a69f-84e6c2f9e2b8
md"""
##### Satz:
1. ``K`` kompakt ``\Rightarrow K`` abgeschlossen und beschränkt.
2. ``K \subset \mathbb{R}^n`` abgeschlossen und beschränkt ``\Rightarrow K`` kompakt. Das heißt, im Allgemeinen (``\infty``-dim.) ist Abgeschlossenheit/Beschränktheit notwendig aber nicht hinreichend für Kompaktheit. Für ``\mathbb{R}^n (n < \infty)`` ist es notwendig und hinreichend.
(ohne Beweis)
"""

# ╔═╡ fde92f57-7047-4201-9111-0fab764b9a14
md"""
##### Beispiel:
1. Sei ``X = \mathbb{R}^n`` Die Einheitskugel ``\overline{\operatorname{B}}_1(0) = \{ x \in X \mid \|x\| \leq 1 \}`` ist abgeschlossen und beschränkt, also kompakt.
2. Sei ``X = \operatorname{C}( [0,1] \to \mathbb{R})`` der Raum der stetigen Funktionen. Die Einheitskugel ``\overline{\operatorname{B}}_1(0) = \{ f \in X \mid \|f\|_{\infty} \leq 1 \}`` ist abgeschlossen und beschränkt.
Betrachte die Folge ``\left( f_n \right)_{n \in \mathbb N} \subset \overline{\operatorname{B}}_1(0)`` mit ``f_n(x) = \sin(n \pi x)``, offenbar ist
``\|f_n\|_{\infty} = \max_{x \in [0,1]} \lvert \sin(n \pi x) \rvert = 1``
aber ``(f_n)`` enthält keine konvergente Teilfolge (oszilliert beliebig stark) ``\Rightarrow \overline{\operatorname{B}}_1(0)`` ist nicht kompakt.
"""

# ╔═╡ 6bba0466-4441-45d0-aef1-10ce6df9904d
md"""
##### Definition:
Sei ``f: D \subset \mathbb{R}^n \to \mathbb{R}^m`` und ``x_0 \in D``. ``f`` heißt _stetig_ in ``x_0``, falls gilt:
``\forall \varepsilon > 0 \quad \exists \delta > 0: \quad \underbrace{\|x - x_0\|}_{\text{Norm in } \mathbb{R}^n} \leq \delta \Rightarrow \underbrace{\|f(x) - f(x_0)\|}_{\text{Norm in } \mathbb{R}^m} \leq \varepsilon``

``f`` heißt stetig in ``D``, falls ``f`` in allen ``x \in D`` stetig ist.
"""

# ╔═╡ a7d42bdd-2e33-4ed7-9fb2-ac1d07c01855
md"""
##### Bemerkung:
1. Wie in 1D gilt ``f`` stetig ``\Leftrightarrow \lim\limits_{k \to \infty} f\left(x^{(k)}\right) = f\left( \lim\limits_{k \to \infty} x^{(k)}\right)`` für alle Folgen ``\left( x^{(k)} \right)_{k \in \mathbb N} \subset \mathbb{R}^n``
2. Statt rechts/links-stetig, gibt es unendlich-viele Richtungen.
"""

# ╔═╡ bd837a2d-6f50-465d-b0f6-0a9ed78725f9
md"""
##### Beispiel:
``f: \mathbb{R}^2 \setminus \{ 0 \} \to \mathbb{R}, (x,y) \mapsto f(x,y) = \frac{2xy}{x^2 + y^2}``

Ist ``f`` stetig bei ``\begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} 0\\0 \end{pmatrix}``? Also stetig fortsetzbar? Wenn ja, mit welchem Wert? Test:
``\begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} \frac{1}{100} \\ \frac{1}{100}  \end{pmatrix}  \qquad f(x,y) = \frac{2 \cdot \frac{1}{100^2} }{\frac{1}{100^2} + \frac{1}{100^2} } = 1 \quad (!!)``

Betrachte die Folge ``\begin{pmatrix} x_n\\y_n \end{pmatrix} = \frac{1}{n} \begin{pmatrix} \cos \alpha \\ \sin \alpha \end{pmatrix}`` mit festem ``\alpha``. ``\lim\limits_{n \to \infty} \begin{pmatrix} x_n\\y_n \end{pmatrix} = \begin{pmatrix} 0\\0 \end{pmatrix}``
"""

# ╔═╡ f1d331ba-5670-4b13-ada2-5db174b5a974
RobustLocalResource("https://github.com/hpsc-lab/lecture-notes-math1_dev/raw/1dbdf8076e38678cba0e432bc7a80ecb1496cbbf/notebooks/assets/folge_in_R2.svg", "./assets/folge_in_R2.svg")

# ╔═╡ fba57cbf-7c9d-4fa3-b127-fb69942bf238
md"""
``\lim\limits_{n \to \infty} f(x_x, y_n) = \lim\limits_{n \to \infty} \frac{2 \cdot \frac{1}{n^2} \cos \alpha \sin \alpha}{\frac{1}{n^2} \cos^2 \alpha + \frac{1}{n^2} \sin^2 \alpha} = 2 \cos \alpha \sin \alpha = \cos(2 \alpha)``
``f(0,0)`` nimmt je nach Richtung alle Werte zwischen ``\pm 1`` an!
"""

# ╔═╡ d48d1091-de14-48f7-b2b5-00ff13904191
md"""
##### Satz:
Sei ``K`` kompakt und ``f: K \to W`` stetig. Dann ist auch ``f(K) \subset W`` kompakt.
"""

# ╔═╡ cd42e4ec-3dea-494e-a0d6-1cf1f364613d
md"""
##### Beweis:
Für ``\left( x^{(k)} \right)_{k \in \mathbb N} \subset K`` gibt es eine Teilfolge mit ``\lim\limits_{j \to \infty} x^{(m_j)} = a \in K``. Also für die Teilfolge von ``\left( f(x^{(m)} \right)_{m \in \mathbb N}``, dass ``	\lim\limits_{j \to \infty} f\left(x^{(m_j)}\right) \stackrel{\text{stetig}}= f\left( \lim\limits_{j \to \infty} x^{(m_j)} \right) = f(a)``
"""

# ╔═╡ 2c312344-413c-4329-969b-2ef4b9385b35
md"""
##### Satz:
Sei ``K`` kompakt und ``f:K \to \mathbb{R}`` stetig. Dann ist ``f(K)`` beschränkt und ``f`` nimmt in ``K`` ein Maximum und Minimum an.
"""

# ╔═╡ 72acb906-5813-4359-a5d2-7b4f4f965f04
md"""
##### Beweis:
Beschränktheit ist klar, denn ``f(K)`` ist kompakt. Also existiert ein Supremum ``s \coloneqq \sup \{ f(x) \mid x \in K \}`` und wir können eine Folge ``\left( x^{(k)} \right)_{k \in \mathbb N} \subset K`` finden, sodass ``	\lim\limits_{k \to \infty} f\left(x^{(k)}\right) = s``. Da ``K`` und ``f(K)`` abgeschlossen sind, gilt ``	\lim\limits_{k \to \infty} x^{(k)} = u \in K`` und ``s \in f(K)``.
"""

# ╔═╡ cf2d255e-f172-49c1-add1-d387e1940cb9
md"""
##### Satz:
Seien ``\|\cdot\| `` und ``||| \cdot |||`` zwei Normen auf ``\mathbb{R}^n``. Diese sind äquivalent. Das heißt es gibt ``c_1,c_2 > 0``, sodass

``\begin{align*} c_1 ||| x||| &\leq \|x\| \leq c_2 |||x||| \\ \text{bzw.} \quad \frac{1}{c_2} \|x\| &\leq |||x||| \leq \frac{1}{c_1} \|x\| \end{align*}``

für alle ``x \in \mathbb{R}^n`` gilt.
"""

# ╔═╡ ddfc97b2-20b7-4988-9d8e-985d62d030ea
md"""
##### Beweis:
Wähle ``||| \cdot ||| = \|\cdot\|_{\infty}``. Für ``x = \sum x_ie_i`` (``e_i`` Basisvektor) gilt

``	\|x\|_{\infty} \leq \sum_{i=1}^n \lvert x_i \rvert \cdot \|e_i\| _{\infty} \leq \max \{ \|e_i\| _{\infty} \} \sum \lvert x_i \rvert \leq \underbrace{\max \{ \|e_i\| _{\infty}  \} \cdot n}_{c_2} \cdot \|x\| _{\infty}``

Betrachte jetzt ``K \coloneqq \{ x \in \mathbb{R}^n \mid \|x\| _{\infty} =1 \}`` kompakt. ``\|\cdot\| `` ist stetig, also existiert ein Minimum ``c = \min_{x \in K} \|x\| > 0``. Es gilt ``	\frac{1}{\|x\| _{ \infty} } x \in K \Rightarrow \underbrace{\|\frac{1}{\|x\| _{\infty} } x  \|}_{= \frac{1}{\|x\| _{\infty} } \|x\| }  \geq c``.
"""

# ╔═╡ bb8711ab-16f4-4383-92c0-067b615f9c32
md"""
##### Bemerkung:
Im ``\mathbb{R}^n`` ist es also egal, welche Norm wir betrachten. Unterschiedliche Normen erzeugen "nur" unterschiedliche Faktoren.
"""

# ╔═╡ 7f7d0b6b-53cc-4f77-8d8d-2b563c31e2b4
md"""
##### Satz:
Für eine lineare Abbildung ``\varphi: \mathbb{R}^n \to \mathbb{R}^m`` gilt:
1. ``	\|\varphi(x)\| \leq c \|x\|`` für alle ``x \in \mathbb{R}^n`` mit festem ``c`` (hängt nur von der Norm ab).
2. ``\varphi`` ist stetig.
"""

# ╔═╡ 2cc39eb6-15e2-4906-b81d-e49cc102062e
md"""
##### Beweis:
1. Wähle ``\|\cdot\|_1`` und fixiere eine Basis, sodass ``y \in \varphi(x) \leftrightarrow y_i = \sum a_{ij} x_j``.

``\qquad\begin{align*} \|y\| _1 &= \sum_{i=1}^m \lvert y_i \rvert\\ &= \sum_{i=1}^m \lvert \sum_{j=1}^n a_{ij} x_j \rvert\\ &\leq \sum_{i=1}^m \sum_{j=1}^n \lvert a_{ij}  \rvert \cdot \lvert x_j \rvert\\ &\leq \underbrace{\max_{j=1, \dots, n} \sum_{i=1}^m \lvert a_{ij}  \rvert }_{c} \sum_{j=1}^n \lvert x_j \rvert \end{align*}``
2. Zu zeigen ist ``\forall \varepsilon > 0 \quad \exists \delta > 0, \|x - x_0\| \leq \delta \rightarrow \|\varphi(x) - \varphi(x_0) \| \leq \varepsilon``. Wähle ``\delta = \frac\varepsilon c``. Dann: ``\|\varphi(x) - \varphi(x_0)\| = \|\varphi(x - x_0)\| \leq x \|x - x_0\| \leq c \delta = \varepsilon``.
"""

# ╔═╡ f43a007e-d998-4533-90bc-69f561a76dca
md"""
## 5.2 Differentiation
Wie in 1D als Approximation:
"""

# ╔═╡ 70bd6e6d-edad-4fa5-bd88-758c6fd4b282
md"""
##### Definition:
Eine Funktion ``f: \mathbb{R}^n \to \mathbb{R}^m`` heißt in ``x_0 \in \mathbb{R}^m``(total) differenzierbar, falls ``\forall  \xi \in \mathbb{R}^n`` eine lineare Abbildung ``\varphi: \mathbb{R}^n \to \mathbb{R}^m`` existiert mit ``f(x_0 + \xi) = f(x_0) + \varphi(\xi) + r(\xi)``, wobei ``r: \mathbb{R}^n \to \mathbb{R}^m`` eine nichtlineare "Abweichung" ist, für die gilt: ``\lim_{\|\xi\|  \to 0} \frac{\|r(\xi)\| }{\|\xi\| } = 0``.

Sowohl ``\varphi`` als auch ``r`` hängen von ``x_0`` ab. Wir nennen ``\varphi = f'(x_0)`` oder ``\varphi = \text{D}f(x_0)`` die Ableitung von ``f`` bei ``x_0``.
"""

# ╔═╡ 32cf3235-e28c-42f9-ab5f-d442941d85f7
md"""
##### Bemerkung:
1. Bezüglich einer Basis entspricht ``\text{D}f(x_0)`` einer Matrix ``A = (a_{ij}) \in \mathbb{R}^{m \times n}`` und ``f,r`` haben Koordinaten ``f_i, r_i`` (``i=1,\dots,n``), sodass ``f_i(x_0 + \xi) = f_i(x_0) + 	\underbrace{\sum_{j=1}^n a_{ij} \xi_j}_{\text{lineare Approximation an } f_i} + r_i(\xi) \quad i = 1, \dots, m``.
2. Für ``n=2`` und ``m=1`` entspricht die lineare Approximation der Tangentialebene im Punkt ``x_0`` an den Graphen/"Teppich" von ``f(x,y)``.
3. Wie in 1D gilt ``\text{D}(\alpha f + \beta g) = \alpha \text{D}f + \beta \text{D}g`` (Ableitung ist linear).
4. Für ``f`` linear, das heißt ``f(x) = Ax`` mit ``A \in \mathbb{R}^{m \times n}`` unabhängig von ``x``, dann folgt, dass ``f`` differenzierbar und stetig ist und ``\text{D}f(x_0) = A``, das heißt ``r(\xi) \equiv 0``.
"""

# ╔═╡ b111ea53-1829-448b-8755-49237452ccb9
md"""
##### Berechnung der Matrix ``f'(x_0)``:
Betrachte ``\underline \xi = h \underline e_j`` mit ``\underline e_j`` Basisvektor ``\begin{pmatrix} 0\\\vdots\\0\\1\\0\\\vdots\\0 \end{pmatrix} \begin{matrix} \\\\\\ \leftarrow j \\\\\\ \\ \end{matrix}`` und ``h \in \mathbb{R}``.

``f_i(x_0 + h e_j) = f_i(x_0) + \underbrace{\varphi_i(he_j)}_{= h \varphi_i(e_j) = h a_{ij} } + r_i(he_j)``

``\Rightarrow \frac{f_i(x_0 + h e_j) - f_i(x_0)}{h} - a_{ij} = \underbrace{\frac{r_i(he_j)}{h}}_{\to 0}``

``\displaystyle\Rightarrow a_{ij} = \lim_{h \to 0} \frac{f_i(x_0 + he_j) - f_i(x_0)}{h}``
"""

# ╔═╡ 2b6a7418-25a2-425d-8c32-f59b4965020d
md"""
##### Definition:
Sei ``f: \mathbb{R}^n \to \mathbb{R}^m``. Der Grenzwert
``\lim\limits_{h \to 0} \frac{f_i(x_0 + he_j) - f_i(x_0)}{h} =: \frac{\partial f_i}{\partial x_j} ``
heißt _partielle Ableitung_. Falls der Grenzwert existiert, heißt ``f`` _partiell differenzierbar_.
"""

# ╔═╡ d2657166-50d1-4602-b8b3-30f58cae9310
md"""
##### Bemerkung:
1. Für zum Beispiel ``f(x_1,x_2,x_3)`` ist ``\frac{\partial f}{\partial x_2} = \lim_{h \to 0} \frac{f(x_1, x_2 + h, x_3) - f(x_1,x_2,x_3)}{h}``. Das heißt, ``x_1,x_3`` werden als konstant angesehen und dann nach ``x_2`` abgeleitet.
2. Für ``m=1, n >1`` heißt der Vektor ``\left( \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \dots, \frac{\partial f}{\partial x_n}\right) `` der _Gradient_ von ``f``, kurz ``\operatorname{grad} f`` oder ``\nabla f``. (sprich "Nabla")
3. Für ``m>1,n>1`` heißt die Matrix ``	\left(\frac{\partial f_i}{\partial x_j} \right) \in \mathbb{R}^{m \times n}``, das heißt

``\qquad\underbrace{\begin{pmatrix} \frac{\partial f_1}{\partial x_1} &\frac{\partial f_1}{\partial x_2} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \frac{\partial f_2}{\partial x_1} &\frac{\partial f_2}{\partial x_2} & \cdots & \frac{\partial f_2}{\partial x_n} \\ \vdots & & \ddots & \\ & & & \frac{\partial f_m}{\partial x_n} \end{pmatrix}}_{n \text{ Spalten}}\left.\begin{matrix}\\\\\\\\\\\end{matrix}\right\}\; m \text{ Zeilen}``

``\qquad``die _Jacobi-Matrix_ von ``f``, auch ``\text{D}f``.
4. Für ``m=n>1`` heißt ``\det(\text Df)`` die _Jacobi-Determinante_ von ``f``, auch ``\left\lvert \frac{\partial f_i}{\partial x_j}   \right\rvert``
"""

# ╔═╡ 0b2a502e-3fa5-49ae-b639-815d1dc86780
md"""
##### Beispiel:
1. ``f: \mathbb{R}^2 \to \mathbb{R}, (x,y) \mapsto x^2 + xy + y^2`` hat ``\operatorname{grad} f = \left( \frac{\partial f}{\partial x} , \frac{\partial f}{\partial y} \right) = (2x + y, -2y + x)``
2. ``f: \mathbb{R}^n \to \mathbb{R}, \underline x \mapsto \|x\| _2 = \sqrt{\sum_{j=1}^n x_j^2}`` hat ``	\frac{\partial f}{\partial x_i} = \frac{1}{2} \left( \sum_{j=1}^n x_j^2\right)^{- \frac{1}{2}} \cdot \frac{\partial}{\partial x_i} \left( \sum_{i=1}^{n}  x_j^2\right) = \frac{1}{2 \|x\| _2} \cdot 2x_i``, denn ``\frac{\partial}{\partial x_i}\left(x_1^2 + x_2^2 + \cdots\right) = 2x_i``. Also ist ``\operatorname{grad}f = \left( \frac{x_1}{\|x\| _2} , \frac{x_2}{\|x\| _2} , \dots, \frac{x_n}{\|x\| _2} \right) = \frac{\underline x}{\|x\| _2}`` und existiert nur für ``x \neq 0``. Für ``x=0`` ist es nicht differenzierbar.
3. ``	f: [0, \infty) \times [0, 2\pi] \times [0, \pi] \to \mathbb{R}^3, (e, \varphi, \theta) \mapsto \begin{pmatrix} r \cos \varphi \cos \theta \\ r \sin \varphi \cos \theta \\ r \sin \theta \end{pmatrix}`` hat ``\left( \frac{\partial f_i}{\partial x_j} \right) = \begin{pmatrix} \frac{\partial f_1}{\partial r} & \frac{\partial f_1}{\partial \varphi} & \frac{\partial f_1}{\partial \theta} \\ \frac{\partial f_2}{\partial r} & \frac{\partial f_2}{\partial \varphi} & \frac{\partial f_2}{\partial \theta} \\ \frac{\partial f_3}{\partial r} & \frac{\partial f_3}{\partial \varphi} & \frac{\partial f_3}{\partial \theta} \end{pmatrix} = \begin{pmatrix} \cos \varphi \cos \theta & - r \sin \varphi \cos \theta & - r \cos \varphi \sin \theta \\ \sin \varphi \cos \theta & r \cos \varphi \cos \theta & - r \sin \varphi \sin \theta \\ \sin \theta & 0 & r \cos \theta \end{pmatrix}``

``\qquad``Also 

``\qquad\begin{align*} \left\lvert \frac{\partial f_i}{\partial x_j}  \right\rvert &= \det( \cdots )\\ &= \sin \theta \left[r^2 \sin^2 \varphi \cos \theta \sin \theta + r^2 \cos^2 \varphi \sin \theta \cos \theta \right]\\ &+ r \cos \theta \left[ r^2 \cos^2 \varphi \cos^2 \theta + r \sin^2 \varphi \cos^2 \theta \right]\\ &= r^2 \cos \theta \sin^2 \theta + r^2 \cos \theta \cos^2 \theta\\ &= r^2 \cos \theta \end{align*}``
4. ``f: \mathbb{R}^2 \to \mathbb{R}, (x,y) \mapsto \begin{cases} \frac{2xy}{x^2 + y^2} &(x,y) \neq (0,0)\\ 0 &(x,y) = (0,0) \end{cases}``

``\qquad``Für ``(x,y) \neq (0,0)`` haben wir

``\qquad \begin{align*} \operatorname{grad}f &= \left( \frac{\partial f}{\partial x} , \frac{\partial f}{\partial y} \right) \\ &= \left( \frac{2y}{x^2 + y^2} - \frac{2xy}{(x^2 + y^2)^2} \cdot 2x, \frac{2x}{(x^2 + y^2)} - \frac{2xy}{(x^2 + y^2)^2} \cdot 2y\right) \\ &= \left( \frac{2yx^2 + 2y^2 - 4x^2y}{(x^2 + y^2)^2} , \frac{2x^3 + 2xy^2 - 4xy^2}{(x^2 + y^2)^2} \right) \end{align*}``

``\qquad`` Für ``(x,y) = (0,0)`` haben wir:

``\qquad \begin{align*} \left.\frac{\partial f}{\partial x} \right\vert_{(0,0)} = \lim_{h \to 0} \frac{f(0 + h,0) - f(0,0)}{h} = 0 \\ \left.\frac{\partial f}{\partial y} \right\vert_{(0,0)} = \lim_{h \to 0} \frac{f(0,0 + h) - f(0,0)}{h} = 0 \end{align*}``

``\qquad`` Aber ``f`` ist bei ``(0,0)`` nicht stetig, also existiert keine Tangentialebene. ``\Rightarrow`` nicht total differenzierbar. 
"""

# ╔═╡ 1e9ed9e7-b7a3-4f49-a2f4-a3427b1b2d3a
md"""
##### Satz:
Falls für ``f: \mathbb{R}^n \to \mathbb{R}^m`` in einer Umgebung ``U \subset \mathbb{R}^n`` von ``x_0 \in U`` die partiellen ABleitungen existieren und _stetig_ sind, so ist ``f`` in ``U`` total differenzierbar und ``\text{D}f`` in ``U`` stetig, insbesondere für ``x_0 \in U``.
"""

# ╔═╡ c42cfcc5-919a-4f1d-88c9-b1392865bc1d
md"""
##### Satz: Kettenregel
Sei ``f: \mathbb{R}^n \to \mathbb{R}^m`` differenzierbar in ``x_0 \in \mathbb{R}^n`` und ``g: \mathbb{R}^m \to \mathbb{R}^l`` differenzierbar in ``f(x_0) \in \mathbb{R}^m``. Dann ist ``g \circ f: \mathbb{R}^n \to \mathbb{R}^l`` differenzierbar in ``x_0`` mit ``\text{D}(g \circ f)(x_0) = \text{D}g(f(x_0)) \cdot \text{D}f(x_0)``

Das heißt, die Matrix ``\text{D}(g \circ f)(x_0) \in \mathbb{R}^{l \times n}`` ist das Produkt von ``\text{D}g(f(x_0)) \in \mathbb{R}^{l \times m}`` und ``\text{D}f(x_0) \in \mathbb{R}^{m \times n}``.
"""

# ╔═╡ a8ff4208-cb83-47a8-b05d-37600206ad32
md"""
##### Beispiel:
1. ``\underline x (t) = \begin{pmatrix} x_1(t) \\ x_2(t) \\x_3(t) \end{pmatrix}`` sei eine Trajektorie im Raum. ``\psi(x_1,x_2,x_3)`` eine Eigenschaft im Punkt ``(x_1,x_2,x_3)``. Wie ändert sich ``\psi`` entlang des Pfades ``\underline x(t)``?

   ``\widetilde \psi(t) = \psi(x_1(t), x_2(t), x_3(t))``

   ``\begin{align*} \frac{\text{d} \widetilde \psi}{\text{d} t}  &= \left.\frac{\partial \psi}{\partial x_1} \right\vert_{x(t)} \frac{\partial x_1}{\partial t} + \left.\frac{\partial \psi}{\partial x_2} \right\vert_{x(t)} \frac{\partial x_2}{\partial t} + \left.\frac{\partial \psi}{\partial x_3} \right\vert_{x(t)} \frac{\partial x_3}{\partial t} \\ &=\left. \left( \frac{\partial \psi}{\partial x_1}, \frac{\partial \psi}{\partial x_2}, \frac{\partial \psi}{\partial x_3}\right) \right\vert_{x(t)} \begin{pmatrix} \frac{\partial x_1}{\partial t} \\ \frac{\partial x_2}{\partial t} \\ \frac{\partial x_3}{\partial t} \end{pmatrix} \\ &= \operatorname{grad} \psi(x(t)) \cdot x'(t)\end{align*}``

2. ``f: \mathbb{R} \to \mathbb{R}^2, t \mapsto \begin{pmatrix} t^2\\ \sin t \end{pmatrix}``. ``\text Df = \begin{pmatrix} \frac{\partial f_1}{\partial t} \\ \frac{\partial f_2}{\partial t} \end{pmatrix} = \begin{pmatrix} 2t\\ \cos t \end{pmatrix}``. ``g: \mathbb{R}^2 \to \mathbb{R}^2, (x,y) \mapsto \begin{pmatrix} xy\\ \ln x + e^y \end{pmatrix}``. ``\text Dg = \begin{pmatrix} \frac{\partial g_1}{\partial x} & \frac{\partial g_1}{\partial y} \\ \frac{\partial g_2}{\partial x} & \frac{\partial g_2}{\partial y} \\ \end{pmatrix} = \begin{pmatrix} y & x \\ \frac{1}{x} & e^y \end{pmatrix}``. ``h = g \circ f: \mathbb{R} \to \mathbb{R}^2, t \mapsto g(f(t))``. ``\begin{align*} \text Dh(t) &= \begin{pmatrix} \frac{\partial h_1}{\partial t} \\ \frac{\partial h_2}{\partial t} \end{pmatrix} \\ &= \text Dg(f(t)) \text Df(t) \\ &= \left.\begin{pmatrix} \frac{\partial g_1}{\partial x} & \frac{\partial g_1}{\partial y} \\ \frac{\partial g_2}{\partial x} & \frac{\partial g_2}{\partial y} \end{pmatrix} \right\vert_{\begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} f_1(t)\\f_2(t) \end{pmatrix} }  \cdot  \begin{pmatrix} \frac{\partial f_1}{\partial t} \\ \frac{\partial f_2}{\partial t} \end{pmatrix} \\ &= \begin{pmatrix} \frac{\partial g_1}{\partial x} \frac{\partial f_1}{\partial t} + \frac{\partial g_1}{\partial y} \frac{\partial f_2}{\partial t} \\ \frac{\partial g_2}{\partial x} \frac{\partial f_1}{\partial t} + \frac{\partial g_2}{\partial y} \frac{\partial f_2}{\partial t} \\ \end{pmatrix} \\ &= \begin{pmatrix} y \cdot 2t + x \cos t \\ \frac{1}{x} 2t + e^y \cos t \end{pmatrix} \\ &= \begin{pmatrix} \sin t \cdot 2t + t^2 \cos t \\ \frac{1}{t^2} 2t + e^{\sin t} \cos t \end{pmatrix} \end{align*}``
3. Innere Energie ``U = U(T, S, N)``. Temperatur ``T(t)``, Entropie ``S(t)``, Teilchenzahl ``N(t)`` (abhängig von der Zeit ``t``). ``\leadsto U(t) = U(T(t), S(t), N(t))``. ``	\frac{\text d U}{\text d t} = \frac{\partial U}{\partial T} \frac{\text d T}{\text dt} + \frac{\partial U}{\partial S} \frac{\text d S}{\text dt} + \frac{\partial U}{\partial N} \frac{\text d N}{\text dt}``
"""

# ╔═╡ b6828b0d-31b2-44da-bf6d-14ccf7df8f8b
md"""
##### Definition:
Sei ``f: \mathbb{R}^n \to \mathbb{R}^m``, ``x \in \mathbb{R}^n`` ein Punkt und ``v \in \mathbb{R}^n`` mit ``\|v\| _2 = 1`` eine "Richtung" (Einheitsvektor). Wir nennen den Vektor ``\frac{\partial f}{\partial v} = \left(\frac{\partial f_1}{\partial v}, \frac{\partial f_2}{\partial v}, \dots, \frac{\partial f_n}{\partial v}\right)`` mit

``\frac{\partial f_i}{\partial v} (x_0) \coloneqq \lim_{h \to 0} \frac{f_1(x_0 + vh) - f_i(x_0)}{h} \qquad i = 1, \dots, m``

die _Richtungsableitung_ von ``f``.
"""

# ╔═╡ 9af0515f-5ee2-4857-8d54-f1c1b62cfcc2
md"""
##### Bemerkung:
Für ``v = e_j`` (Koordinatenrichtung) ist ``\frac{\partial f}{\partial v}`` der Vektor der partiellen Ableitung ``\frac{\partial f_i}{\partial x_j}`` (für festes ``j``).
"""

# ╔═╡ b77b4c16-5277-4bb6-80e8-01bb90a7d0b2
md"""
##### Beispiel:
Betrachte ``f: \mathbb{R}^n \to \mathbb{R}`` und ``x_0 \in \mathbb{R}^n``. In welcher Richtung ``v \in \mathbb{R}^n``, ``\|n\| = 1``, ändert sich ``f`` bei ``x_0`` am meisten? Das heißt, ``\frac{\partial f}{\partial v} (x_0)`` ist maximal? Sei ``\underline \xi(t) = \underline x_0 + t \underline v`` mit ``t \in \mathbb R`` ein Pfad in ``\mathbb R^n``. Es gilt

``\frac{\partial f}{\partial v} (x_0) = \frac{\text{d}}{\text{d}t}f(\xi(t))\big\vert_{t=0} = \operatorname{grad}f(\xi(t)) \cdot \xi'(t) \big\vert_{t=0} = \operatorname{grad}f(x_0) \cdot \underline v``

Die Richtungsableitung ist die Projektion von ``\operatorname{grad} f`` auf die Richtung ``\underline v``. Die Länge ist also maximal, dass ``v`` parallel zu ``\operatorname{grad} f`` ist.
"""

# ╔═╡ 1585111b-8ccf-41bc-9434-852a82a2f732
RobustLocalResource("https://github.com/hpsc-lab/lecture-notes-math1_dev/raw/1dbdf8076e38678cba0e432bc7a80ecb1496cbbf/notebooks/assets/gradient.svg", "./assets/gradient.svg")

# ╔═╡ bee1bef7-92e0-411a-9546-a764a88bc477
md"""
##### Mittelwertsatz:
Erinnere in 1D für ``f: [a,b] \to \mathbb R``.
1. ``\exists  \xi \in [a,b]: \qquad f(b) - f(a) = f'(\xi)(b - a)``
2. ``\exists  \xi \in [a,b]: \qquad \int_a^b f(x)\; \text dx = f(\xi)(b-a)``
"""

# ╔═╡ 10d946f4-d30e-4e04-87d4-69d3c170322b
md"""
##### Bemerkung:
1. (1.) folgt aus (2.), denn ``f(b) - f(a) \stackrel{(\ast)}= \int_a^b f'(x)\; \text dx \stackrel{(2.)}= f'(\xi)(b - a)``
2. ``(\ast)`` lässt sich auch schreiben mit ``a= x_0`` und ``b = x_0 + \xi``: ``f(x_0 + \xi) - f(x_0) = \int_{x_0}^{x_0 + \xi} f'(x) \;\text{d}x = \int_0^1 f'(x_0 + t\xi) \;\text{d}t \cdot \xi`` durch die Substitution ``x \leftrightarrow t``, ``x = x_0 + t\xi``.
3. Weder (1.) noch (2.) lassen sich im ``\mathbb R^n`` allgemein realisieren, aber ``(\ast)`` schon.
"""

# ╔═╡ e3e30523-854f-47ed-8aef-aa26a0829c4a
md"""
##### Satz:
Sei ``f: U \subset \mathbb{R}^n \to \mathbb{R}^m`` und ``x_0 \in U``, sowie ``\xi \in \mathbb R^n`` derart, dass ``x_0 + t \xi \in U`` für ``t \in [0,1]``. Dann gilt:
1. ``f(x_0 + \xi) - f(x_0) = \int_0^1 \text Df(x_0 + t \xi)\; \text dt \cdot \xi``
2. ``\|f(x_0 + \xi) - f(x_0) \| \leq M \|\xi\|`` mit ``0 < M \in \mathbb R`` wobei ``M`` nur von ``\text Df(x_0 + t\xi)`` abhängt.
"""

# ╔═╡ 9b4d1ccf-1153-45a1-b535-41e421d5c0f1
md"""
##### Bemerkung:
1. Das Integral einer Matrix erfolgt komponentenweise.
2. ``M`` entspricht der "Norm" von ``\text Df(x_0 + t \xi)``.
"""

# ╔═╡ 4f74fb37-de37-4b31-b3ae-cd94cdb14607
md"""
##### Beweis (nur von 1.):
Wir definieren ``g_i(t) = f_i(\underbrace{x_0 + t \xi}_{x(t)})`` mit ``i=1, \dots,m``. Es gilt ``g_i(1) - g_i(0) = \int_0^1 g_i(t) \;\text{d}t``. Nach der Kettenregel gilt ``g_i'(t) = \sum_{j=1}^n \frac{\partial f_i}{\partial x_j} (x(t)) \frac{\text d x_j}{\text d t} = \sum_{j=1}^n \frac{\partial f_i}{\partial x_j} (x(t)) \xi_j`` also ``\int_0^1 g'(t) \;\text{d}t = \sum_{j=1}^n \int_0^1 \frac{\partial f_i}{\partial x_j} (x_0 + t \xi) \;\text{d}t  \xi_j	\; \widehat =\; \int_0^1 \text Df(x_0 + t \xi) \;\text{d}t  \cdot \xi``
"""

# ╔═╡ ab5a243d-1c44-4676-8fe5-20154926e194
md"""
## 5.3 Umkehrfunktion
Sei ``f: U \subset \mathbb{R}^n \to \mathbb{R}^n`` (``m = n``!). Wann lässt sich ``f`` umkehren? Das heißt, es existiert ``f^{-1}: W \subset \mathbb{R}^n \to \mathbb{R}^n`` mit ``f^{-1} \circ f = \operatorname{id}``.
1. Falls ``f^{-1}`` existiert, gilt ``\text D(\operatorname{id}) = I = \text D\left(f^{-1} \circ f\right) = \text D \left( f^{-1} \right) \cdot \text Df``. Das heißt, die Jacobi-Matrizen sind invertierbar.
2. In der Nähe von ``x_0`` gilt für ``x = x_0 + \xi``, dass ``y = f(\underbrace{x_0 + \xi}_{x}) = f(x_0) + \text Df(x_0) \cdot \underbrace{\xi}_{x - x_0}``, also ``x = x_0 + \text Df(x_0)^{-1}(y - f(x_0))``.
"""

# ╔═╡ 17e992a4-f9f7-44de-9272-071828dee798
md"""
##### Satz:
Sei `` f: U \subset \mathbb{R}^n \to \mathbb{R}^n`` stetig differenzierbar und für ``x_0 \in U`` sei ``\text Df(x_0)`` invertierbar (also ``\det \text Df(x_0) \neq 0``). Dann existiert eine Teilmenge ``V \subset U`` mit ``x_0 \in V`` und eine Umgebung ``V' \subset \mathbb R^n`` von ``b = f(x_0)``, sodass
1. ``f`` bildet ``V`` bijektiv auf ``V'`` ab
2. ``g \coloneqq f^{-1}: V' \to V`` ist stetig differenzierbar
3. ``\text Dg(b) = (\text Df(a))^{-1}``.
"""

# ╔═╡ 39f7fa27-e99e-4d36-8dd1-f8349187c507
md"""
Die Umkerhfunktion ``y = f(x) \Leftrightarrow x = f^{-1}(y)`` ist ein Spezialfall von implizit definierten Funktionen.
"""

# ╔═╡ 8862d324-44d9-4cfb-8c75-477acc41a47d
md"""
##### Definition:
Sei ``U \subset \mathbb{R}^p \times \mathbb{R}^n`` und ``F: U \to \mathbb{R}^n, (\underbrace{x}_{\in \mathbb{R}^p},\underbrace{y}_{\in \mathbb{R}^n}) \mapsto F(x,y)`` stetig differenzierbar. Die Funktion ``T: \mathbb R^n \to \mathbb R^n`` definiert durch ``x \mapsto y = T(x)`` mit ``F(x,y) = 0``, das heißt ``F(x, T(x)) = 0``, heißt _implizit definiert_.
"""

# ╔═╡ b8a523cb-1282-41e3-933c-7a55e9b58dff
md"""
##### Bemerkung:
1. Implizit definierte Funktionen müssen nicht existieren oder eindeutig sein.
2. Für ``p = n = 1`` beschreibt ``F(x,y) = 0 \in \mathbb R`` eine "Höhenlinie" in der x-y-Ebene. ``T(x)`` beschreibt diese Höhenlinie als Funktion.
"""

# ╔═╡ 5e74f85e-95af-42a1-82ff-b36f8f1728cf
md"""
##### Beispiel:
1. ``F(x,y) = x^2 + y^2 - 1``. ``(x,y) \in \mathbb R^2``. ``F: \mathbb R^2 \to \mathbb R``.

$(RobustLocalResource("https://github.com/hpsc-lab/lecture-notes-math1_dev/raw/1dbdf8076e38678cba0e432bc7a80ecb1496cbbf/notebooks/assets/implizite_funktionen.svg", "./assets/implizite_funktionen.svg"))

``\qquad F(x, T(x)) = 0``. ``T(x) =`` ? ``x^2 + T(x)^2 - 1 = 0``, also ``T(x)^2 = 1 - x^2``:

``\qquad T(x) = \pm \sqrt{1 - x^2}``. Nicht eindeutig! Fixiere ``F(x_0, y_0)`` mit ``F(x_0, y_0) = 0``.

``\qquad \begin{align*} y = T(x) \text{ bei } &(x_0, y_0) = (0,-1) ? \quad T(x) = - \sqrt{1 - x^2} \text{ eindeutig} \\ y = T(x) \text{ bei } &(x_0, y_0) = (1,0) ? \quad T(x) \text{ existiert nicht!} \\ \end{align*}``
2. Eine Umkehrfunktion lässt sich schreiben als ``F\left(f^{-1}(y), x\right) = 0`` mit ``F(x,y) = y - f(x)``, denn ``F\left(f^{-1}(y), y\right) = y - f\left(f^{-1}(y)\right) = 0 ``
"""

# ╔═╡ 538b1d6d-710d-4a1d-8f25-9d732f23a5cd
md"""
##### Bemerkung:
Wie bei der Umkehrfunktion lässt sich eine Aussage über die Ableitung von ``T(x)`` machen. Schreibe ``x = \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix}, T = \begin{pmatrix} T_1 \\ \vdots \\ T_n \end{pmatrix}`` und ``T(x) = \begin{pmatrix} y_1 \\ \vdots \\ y_n \end{pmatrix}``. ``\frac{\partial}{\partial x_j} F_i(x, T(x)) = 0`` mit ``i = 1, \dots, n``, ``j = 1, \dots, p``.

``\frac{\partial F_i}{\partial x_j} (x, T(x)) + \sum_{k=1}^n \frac{\partial F_i}{\partial y_k} (x, T(x)) \frac{\partial T_k}{\partial x_j} (x) =0``

Wir schreiben ``	\text D_xF = \left( \frac{\partial F_i}{\partial x_j} \right) \in \mathbb{R}^{n \times p}``, ``	\text D_yF = \left(\frac{\partial F_i}{\partial y_k} \right) \in \mathbb{R}^{n \times n}`` und damit ``\text D_xF(x, T(x)) + \text D_yF(x, T(x)) \text DT(x) = 0 \Leftrightarrow \text DT(x) = - \left[\text D_yF(x, T(x)\right]^{-1} \text D_xF(x, T(x))``

Ein invertierbares ``\text D_yF`` ist also notwendig für ein differenzierbares ``T(x)``.
"""

# ╔═╡ 3cb7186c-9ca2-4413-b369-214b1497fa4e
md"""
##### Beispiel:
``F(x,y) = x^2 + y^2 - 1``. ``\begin{pmatrix} x\\y \end{pmatrix} \in \mathbb R^2``. ``F: \mathbb R^2 \to \mathbb R``. ``\text D_xF = \frac{\partial F}{\partial x} = 2x, \text D_yF = \frac{\partial F}{\partial y} = 2y, F'(x) = - \frac{2x}{2y} = - \frac{x}{T(x)} ``. Für ``T(x) = \sqrt{1 - x^2}`` gilt ``T'(x) = -\frac{x}{\sqrt{1 - x^2}}``. Aber bei ``(x,y) = (1,0)`` ist ``T'(x) = - \frac xy \to \infty ``. Das geht nicht.
"""

# ╔═╡ 275c2528-42b0-4fe0-8bc2-2d3fa12a964c
md"""
Für den Beweis zur Existenz von impliziten Funktionen benötigen wir:
"""

# ╔═╡ 5a7851ba-70cc-4f0f-96fa-ac7088b787cf
md"""
### 5.3.1 Banach'scher Fixpunktsatz
"""

# ╔═╡ d1837cc0-a22c-46de-97d3-27ac30877eff
md"""
##### Satz:
Sei ``E \subset \mathbb R^n`` vollständig und für ``\Phi: E \to \mathbb R^n`` gelte
1. ``\Phi(E) \subset E`` (``\Phi`` führt nicht aus ``E`` heraus, "Selbstabbildung").
2. ``\|\Phi(x) - \Phi(y)\| \leq K \|x - y\|`` mit ``K < 1`` für alle ``x,y \in E`` ("kontraktion").
dann existiert ein eindeutiger _Fixpunkt_ ``x^\ast \in E`` mit ``x^\ast = \Phi(x^\ast)``.
"""

# ╔═╡ 14433d47-3ceb-42b7-80d8-259ed768cdd8
RobustLocalResource("https://github.com/hpsc-lab/lecture-notes-math1_dev/raw/1dbdf8076e38678cba0e432bc7a80ecb1496cbbf/notebooks/assets/banach.svg","./assets/banach.svg")

# ╔═╡ 3ccdaadf-d754-4d36-89bf-57da67cae148
md"""
##### Beweis:
Wir betrachten die Folge ``x^{(n+1)} = \Phi(x^{(n)})``, ``n=0,1,\dots`` mit Startwert ``x^{(0)} \in E`` und zeigen ``x^{(n)}`` ist Cauchy-Folge. Also existiert der Grenzwert ``x^\ast = \lim_{n \to \infty} x^{(n)}`` (Vollständigkeit). Cauchy: ``\forall \varepsilon \exists n_0 \in \mathbb{N}: \quad \|x^{(m)} - x^{(n)}\| \leq \varepsilon \quad m,n > n_0``.

``	\|x^{(n+1)} - x^{(n)}\| = \|\Phi(x^{(n)}) - \Phi(x^{(n-1})\| \leq K \|x^{(n)} - x^{(n-1)}\| \leq \cdots \leq K^n \underbrace{\|x^{(1)}- x^{(0)}\|}_{e_0}``

Dann mit ``n < m``:

``\begin{align*} \|x^{(m)} - x^{(n)}\| &= \|x^{(m)} - x^{(m - 1)} + x^{(m - 1)} - x^{(n)}\| \\ &\leq \underbrace{\|x^{(m)} - x^{(m-1)}\|}_{\leq K^{m-1} e_0} + \|x^{(m  - 1} - x^{(n)}\| \\ &\leq (K^{m-1} + K^{m-2})e_0 + \|x^{(m-2)} - x^{(n)}\|  \\ &\; \vdots \\ &\leq (K^{m-1} + K^{m-2} + \dots + K^n) e_0 \end{align*}``

Und es gilt ``K^{m-1} + K^{m-2} + \dots + K^n \leq \sum_{i=n}^{\infty}K^i = \sum_{i=0}^{\infty}K^i - \sum_{i=0}^{n-1} K^i = \frac{K^n}{1 - K}`` (geoemtrische Reihe).

Wähle ``\varepsilon, n_0`` so, dass ``	\frac{K^{n_0}}{1 - K}  \leq \varepsilon \Rightarrow \|x^{(m)} - x^{(n)}\| \leq \varepsilon \quad m > n > n_0``

Ist ``x^\ast`` ein Fixpunkt? ``	x^\ast = \lim_{n \to \infty} x^{(n + 1)} = \lim_{n \to \infty} \Phi\left(x^{(n)}\right) = \Phi(x^\ast)``, denn ``\Phi`` ist lipschitz-stetig.

Ist ``x^\ast`` eindeutig? Seien ``x^\ast, \tilde x \in E`` zwei Fixpunkte (unterschiedlich).

``0 < \|x^\ast - \tilde x\| = \|\Phi(x^\ast) - \Phi(\tilde x)\| \leq K \|x^\ast - \tilde x\| < \|x^\ast - \tilde x\|``

Das ist natürlich nicht möglich.
"""

# ╔═╡ 3c7a0c3e-9cf9-475c-b7b9-58f3e87a4565
md"""
##### Satz: Impliziter Funktionensatz
Sei ``F: U \subset \mathbb R^p \times \mathbb R^n \to \mathbb R^n`` stetig differenzierbar und für ``(x_0, y_0) \in U`` gelte ``\text D_yF(x_0,y_0) \in \mathbb R^{n \times n}`` ist invertierbar. Dann gibt es Umgebungen ``\text B_r(x_0)`` und ``\text B_\rho(y_0)`` mit Radien ``r, \rho > 0``, sodass die impliziete Funktion ``T: \text B_r(x_0) \subset \mathbb R^n \to \text B_\rho(y_0) \subset \mathbb R^n, x \mapsto y = T(x)`` mit ``F(x,y) = 0`` existiert, sowie eindeutig und stetig differenzierbar ist.
"""

# ╔═╡ e6d21947-f7ba-4619-838c-0d7bb06d0c9c
md"""
##### Beweis:
Zur Einfachheit ``(x_0, y_0) = (0,0)``. Wir kürzen ab: ``B \coloneqq \text D_yF(0,0) = \text{const.}`` und definieren für ``x \in \mathbb R^p`` fest ``\Phi_x: \mathbb R^n \to \mathbb R^n, y \mapsto y - B^{-1}\cdot F(x,y)`` (Den Parameter ``x`` lassen wir eventuell weg). Es gilt ``F(x,y) = 0 \Leftrightarrow \Phi_x(y) = y``, also ``y`` ein Fixpunkt von ``\Phi_x`` ist. Wir benutzen den Banach'schen Fixpunktsatz! Zu zeigen ist für eine Umgebung ``U`` von ``0 = y_0``:
1. ``\|\Phi_x(\hat y) - \Phi_x(y)\|  \leq K \|\hat y - y\| \quad y, \hat y \in U``
2. ``\Phi_x(U) \subset U``
Zu 1. verwenden wir den Mittelwertsatz: ``\Phi_x(\underbrace{y + \xi}_{\hat y}) - \Phi_x(y) = \int_0^1	\text Dy \Phi(y + t \xi) \;\text{d}t  \cdot \underbrace{\xi}_{= \hat y - y}``, oder komponentenweise ``\Phi_{x,i}(y + \xi) - \Phi_{x,i} (y) = \sum_{k=1}^n \int_0^1 \frac{\partial \Phi_i}{\partial y_k} (y + t \xi) \;\text{d}t (\hat y_k - y)``. Also ``	\max_{i = 1, \dots, n} \lvert \Phi_i( \hat y ) - \Phi_i(y) \rvert \leq \left(\sum_{k=1}^n \int_0^1 \left\lvert \frac{\partial \Phi_i}{\partial y_k} (y + t \xi)\right\rvert \;\text{d}t \right) \max_{j = 1, \dots, n} \lvert \hat y_i - y_i \rvert``.

``\begin{align*} \text D_y \Phi_x(y) &= \text{id} - B^{-1} \text D_yF(x,y) \quad \text{ ist stetig} \\ \text D_y \Phi_x(0) &= \text{id} - B^{-1}B = 0\quad (x = 0) \end{align*}``

Das heißt, es existiert ein (eventuell kleiner) Radius ``\rho`` um ``y=0`` und ``r`` um ``x= 0`` so, dass ``\left\lvert \frac{\partial \Phi_i}{\partial y_k}   \right\rvert \leq \frac{1}{2n} \quad \forall y \in \text B_\rho(0)``

``\Rightarrow \|\Phi_x(\hat y) - \Phi_i(y)\|_{\infty} \leq \frac{1}{2} \|\hat y - y\| _{ \infty} \qquad K = \frac{1}{2}``

Zu 2. ist zu zeigen ``y \in \text B_\rho(0) \Rightarrow \Phi_x(y) \in \text B_\rho(0)``.

``	\|\Phi_x(y) - \Phi_x(0) + \Phi_x(0)\| \leq \|\Phi_x(y) - \Phi_x(0)\| + \|\Phi_x(0)\| \leq \frac{1}{2} \underbrace{\|y\|}_{\leq \rho} + \|\Phi_x(0)\|``

``\begin{align*} \Phi_x(0) &= B^{-1}F(x,0) \quad \text{ist stetig in } x \\ \Phi_x(0=\big\vert_{x = 0} = B^{-1}F(0,0) = 0 \end{align*}``

Wieder gibt es einen (eventuell kleinen) Radius ``r`` um ``x=0`` so, dass ``\|\Phi_x(0)\| \leq \frac{\rho}{2} \Rightarrow \|\Phi_x(y)\| \leq \frac{\rho}{2} + \frac{\rho}{2} \leq \rho``

Also existiert für ein festes ``x`` ein eindeutiger Fixpunkt ``y^\ast = \Phi_x(y^\ast)``. Wir schreiben ``y^\ast = T(x)``.
"""

# ╔═╡ 505dcf09-20de-403e-a38a-f123a2666f56
md"""
##### Beispiel:
1. ``F(x,y) = ye^{x + y} - 1``. Ist ``F(x,y) = 0`` bei ``(x,y) = (-1,1)`` nach ``y`` auflösbar, das heißt, ``y = T(x)`` existiert? Welchen Wert hat ``T'(x)``? ``\text D_yF = \frac{\partial F}{\partial y} = e^{x + y} + y e^{x + y}``, ``\frac{\partial F}{\partial y} (-1,1) = e^0 + e^0 = 2 \neq 0``. ``\Rightarrow y = T(x)`` existiert.

``\qquad T'(x)\big\vert_{x = -1} = - (\text D_yF)^{-1}\text D_xF \big\vert_{(x,y) = (-1,1)} = - \frac{\frac{\partial F}{\partial x} }{\frac{\partial F}{\partial y} } = - \frac{y \cdot e^{x + y}}{(1 + y) \cdot e^{x + y}} = - \frac{1}{2}``
2. Betrachte ``F: \mathbb R^5 \to \mathbb R^3, u \mapsto v = F(u) = Au`` mit `À \in \mathbb R^{3 \times 5}``.

``\qquad\begin{pmatrix} v_1\\v_2\\v_3 \end{pmatrix} = \begin{pmatrix} \vdots & \vdots & \vdots & \vdots & \vdots \\ a_1 & a_2 & a_3 & a_4 & a_5 \\ \vdots & \vdots & \vdots & \vdots & \vdots \end{pmatrix} \begin{pmatrix} u_1 \\ \vdots \\ u_5 \end{pmatrix} ``

``\qquad`` Unterteile ``u = \begin{pmatrix} x_1\\x_2\\y_1\\y_2\\y_3\end{pmatrix} = \begin{pmatrix} x\\y \end{pmatrix}, x \in \mathbb R^2, y \in \mathbb R^3``. Unter welchen Bedingungen ist ``F(x,y) = A \begin{pmatrix} x\\y \end{pmatrix} = 0`` nach ``y`` auflösbar?

``F(x,y) = \underbrace{\begin{pmatrix} \vdots & \vdots \\ a_1 & a_2 \\ \vdots & \vdots \end{pmatrix}}_{A_1} \begin{pmatrix} x_1 \\x_2 \end{pmatrix} + 	\begin{pmatrix} \vdots & \vdots & \vdots \\ a_3 & a_4 & a_5 \\ \vdots & \vdots & \vdots \end{pmatrix} \begin{pmatrix} y_1 \\y_2 \\y_3 \end{pmatrix} = 0``

``\qquad \Rightarrow A_2y = - A_1x``

``\qquad \Rightarrow\;\;\;\;y = - A_2^{-1}A_1x``

``\qquad`` also ``A_2 = \text D_yF`` invertierbar.
"""

# ╔═╡ 18e157e4-13a3-4af8-9d49-90b265c18ea3
md"""
## 5.4 Höhere Ableitungen
"""

# ╔═╡ a0b01c16-5a8b-4023-a3f5-018f0f110a59
md"""
Wie in 1D werden höhere partielle Ableitungen rekursiv definiert. Für ``f: \mathbb R^n \to \mathbb R^m, x \mapsto f(x)`` heißt ``\frac{\partial}{\partial x_j} \left( \frac{\partial f_k}{\partial x_i}(x) \right) = \frac{\partial^2 f_k}{\partial x_j\partial x_i}`` die zweite partielle Ableitung der Komponente ``f_k`` bzgl ``x_i`` und ``x_j``.
"""

# ╔═╡ 4a665d38-4931-415e-b6e1-2ca130155f58
md"""
##### Beispiel:
``f: \mathbb R^2 \to \mathbb R, \begin{pmatrix} x\\y \end{pmatrix} \mapsto f(x,y) = x \cos(x + y)``.

``\begin{align*} \frac{\partial f}{\partial x} &= \cos(x + y) - x \sin(x + y) \\ \frac{\partial f}{\partial y} &= - x \sin (x + y) \\ \frac{\partial^2 f}{\partial x^2} &= - \sin(x + y) - \sin(x + y) - x \cos ( x + y) \\ \frac{\partial^2 f}{\partial y^2} &= -x \cos(x + y) \\ \frac{\partial^2 f}{\partial x \partial y} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) = - \sin(x + y) - x \cos(x + y) \\ \frac{\partial^2 f}{\partial y \partial x} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) = - \sin (x + y) - x \cos(x + y) \end{align*}``

Man bemerke, dass die zweiten partiellen Ableitungen _gleich_ sind!
"""

# ╔═╡ 1d7bb1a0-b41d-4e91-93b9-ab801cd05b6e
md"""
##### Satz von Schwarz:
Falls für ``f: \mathbb R^n \to \mathbb R^m`` die zweiten Ableitungen existieren und stetig sind, so gilt für alle Indexpaare ``i,j \in \{1, \dots, n\}``:

``\frac{\partial f_k}{\partial x_i \partial x_j} = \frac{\partial f_k}{\partial x_j \partial x_i}  \qquad k = 1, \dots, m``

Das heißt, die Reihenfolge der Ableitungen spielt keine Rolle.
"""

# ╔═╡ a5817e06-c8d1-4765-8ffd-3ad7b1f31f05
md"""
##### Beispiel:
1. Für ein Vektorfeld ``a: \mathbb R^n \to \mathbb R^n`` (typischerweise ``n=3``) ist die _Divergenz_ definiert durch ``\operatorname{div} a = \sum_{i=1}^n \frac{\partial a_i}{\partial x_i}`` und der _Laplace-Operator_ eines Skalarfeldes ``\phi: \mathbb R^n \to \mathbb R`` definiert durch ``\Delta \phi = \operatorname{div}( \operatorname{grad} \phi ) = \operatorname{div} \left( \frac{\partial \phi}{\partial x_1}, \dots, \frac{\partial \phi}{\partial x_n}  \right) = \sum_{i=1}^n \frac{\partial^2 \phi}{\partial x_i^2}``
2. Speziell für ``n=3`` ist für ``f: \mathbb R^3 \to \mathbb R^3`` die _Rotation_ gegeben durch ``\operatorname{rot} f = \begin{pmatrix} \frac{\partial f_3}{\partial y} - \frac{\partial f_2}{\partial z} \\ \frac{\partial f_1}{\partial z} - \frac{\partial f_3}{\partial x} \\ \frac{\partial f_2}{\partial x} - \frac{\partial f_1}{\partial y} \end{pmatrix}``.

``\qquad`` Es gilt ``\operatorname{div}(\operatorname{rot} f) = 	\frac{\partial }{\partial x} \left( \frac{\partial f_3}{\partial y} \right) - \frac{\partial}{\partial x} \left( \frac{\partial f_2}{\partial z} \right) + \frac{\partial}{\partial y} \left( \frac{\partial f_1}{\partial z} \right) - \frac{\partial}{\partial y} \left( \frac{\partial f_3}{\partial x} \right) + \frac{\partial}{\partial z} \left( \frac{\partial f_2}{\partial x} \right) - \frac{\partial}{\partial z} \left( \frac{\partial f_1}{\partial y} \right) = 0``

``\qquad`` Analog gilt ``\operatorname{grad}(\operatorname{rot} f) = 0``.
"""

# ╔═╡ 51493505-28f5-467f-afa6-813366db705a
md"""
##### Bemerkung:
Im ``\mathbb R^3`` wird oft ``\nabla`` "Nabla" als Vektor geschrieben ``\nabla = \begin{pmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{pmatrix}`` (im Sinne eines Operators). Dann gilt

``\begin{align*} \operatorname{div} a &= \nabla \cdot a \\ \operatorname{grad} \phi &= \nabla \phi \\ \operatorname{rot} f &= \nabla \times f \end{align*}``
"""

# ╔═╡ 223b6b78-29c8-4d49-9091-ec24a13e4b2b
md"""
### 5.4.1 Taylor-Entwicklung für ``f: \mathbb R^n \to \mathbb R``
"""

# ╔═╡ 589aa6ed-4169-4096-8b98-65e8fa794f41
md"""
Wir benötigen eine kompakte Schreibweise für z.B. ``\frac{\partial^4 f}{\partial x_1 \partial x_2 \partial x_3^2} \leadsto `` Multiindexnotation. Zunächst: ``D \coloneqq (\partial_1, \partial_2, \dots, \partial_n)``, ``\partial_i \coloneqq \partial_{x_i} \coloneqq \frac{\partial}{\partial x_i}``. Multiindex ``\alpha = (\alpha_1, \alpha_2, \dots, \alpha_n) \in \mathbb N_0^n`` mit ``\lvert \alpha \rvert \coloneqq \sum_{i=1}^n \alpha_i`` und ``\alpha! \coloneqq \prod_{i=1} ^n \alpha_i!``
"""

# ╔═╡ ff9d6027-6343-46f0-991d-ee694a95bd00
md"""
##### Beispiel:
``\begin{align*} n &= 2 \qquad \alpha = (1,2) \quad |\alpha| = 3 \quad \alpha! = 1! \cdot 2! = 2 \\ n &= 3 \qquad \alpha = (2,0,3) \quad |\alpha| = 5 \quad \alpha! = 2! \cdot 3! = 12 \\ \end{align*}``
"""

# ╔═╡ 76f79c6b-4eda-45ef-b69c-3df0d5a31f23
md"""
``x^\alpha \coloneqq \prod_{i=1}^n x_i^{\alpha_i}, \quad x = \begin{pmatrix}x_1\\\vdots\\ x_n \end{pmatrix} \in \mathbb R^n``

``D^\alpha \coloneqq \left( \frac{\partial}{\partial x_1}  \right)^{\alpha_1}  \left( \frac{\partial}{\partial x_2}  \right)^{\alpha_2} \cdots \left( \frac{\partial}{\partial x_n}  \right)^{\alpha_n}  = \partial_1^{\alpha_1}\partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n}``
"""

# ╔═╡ e7fceccf-c1eb-4e68-81e0-f3229abbc533
md"""
##### Satz von Taylor:
Für ``U \subset \mathbb R^n`` sei ``f: U \subset \mathbb R^n \to \mathbb R`` (``m+1``)-mal stetig partiell differenzierbar. Dann gilt für ``x_0 \in U`` die Taylor-Entwicklung

``f(x) = \underbrace{\sum_{\lvert \alpha \rvert \leq m} \frac{1}{\alpha!} D^\alpha f(x_0)(x - x_0)^\alpha}_{\text{Taylor-Polynom vom Grad $m$},\; T_m[f]} + R_m \qquad x \in U``

mit Restglied ``R_m = \sum_{\lvert \alpha \rvert = m + 1} \frac{1}{\alpha!} D^\alpha f(x_0 + \theta(x - x_0))(x - x_0)^\alpha`` mit einem ``\theta \in (0,1)``.
"""

# ╔═╡ e4496ba3-9405-4643-8680-78ec06dfe294
md"""
##### Beispiel:
1. Betrachte ``\varphi: \mathbb R^2 \to \mathbb R, (x,y) \mapsto \varphi(x,y)`` (also ``n=2``). Das Taylor-Polynom vom Grad ``m=3`` um ``(x_0, y_0)``

``\qquad \begin{align*} \lvert \alpha \rvert &= 0 \quad \alpha = (0,0) \quad \alpha! = 1 \\ \lvert \alpha \rvert &= 1 \quad \alpha = (1,0) \quad \alpha! = 1 \qquad \alpha = (0,1) \quad \alpha! = 1\\ \lvert \alpha \rvert &= 2 \quad \alpha = (2,0),(1,1),(0,2) \qquad \alpha! = 2,1,2 \\ \lvert \alpha \rvert &= 3 \quad \alpha = (3,0),(2,1),(1,2),(0,3) \qquad \alpha! = 6,2,2,6 \end{align*}``

``\qquad \begin{align*} T_3[\varphi](x) &= \varphi(x_0, y_0) + \underbrace{(\partial^1_x \partial^2_y \varphi)(x - x_0)^1(y - y_0)^0 + (\partial_x^0 \partial_y^1 \varphi)(x - x_0)^0 (y - y_0)^1}_{\lvert \alpha \rvert = 1} \\ &+ \underbrace{(\partial^2_x\partial^0_y \varphi)(x- x_0)^2(y - y_0)^0 + (\partial^1_x\partial^1_y \varphi)(x - x_0)^1(y - y_0)^1 + (\partial^0_x\partial^2_y \varphi)(x - x_0)^0(y - y_0)^2}_{\lvert \alpha \rvert = 2} \\ &+ \underbrace{(\partial^3_x \partial^0_y \varphi)(x - x_0)^3(y- y_0)^0 + (\partial^2_x \partial^1_y)(x - x_0)^2(y - y_0)^1 + (\partial^1_x\partial^2_y \varphi)(x - x_0)^1(y - y_0)^2 + (\partial^0_x \partial^3_y \varphi)(x - x_0)^0(y - y_0)^3}_{\lvert \alpha \rvert = 3} \\ &= \varphi + \varphi_x(x - x_0) + \varphi_y(y - y_0) + \varphi_{xx} (x - x_0)^2 + \varphi_{xy} (x - x_0)(y - y_0) + \varphi_{yy} ( y - y_0)^2 \\ &+ \varphi_{xxx} (x - x_0)^3 + \varphi_{xxy} (x - x_0)^2(y - y_0) + \varphi_{xyy} (x - x_0)(y - y_0)^2 + \varphi_{yyy} (y - y_0)^3 \end{align*}``
2. Konkret: ``\varphi(x,y,z) = \frac{e^{2x}}{y+2} + (z+1)^2``

``\qquad\begin{align*} T_2[\varphi](0,0,0) =&\; \varphi(0,0,0) + \partial_x \varphi(0,0,0)x + \partial_y \varphi(0,0,0)y + \partial_z \varphi(0)z\\ &\; + \partial_{xx}\varphi(0) \frac12 x^2 + \partial_{xy}\varphi(0)xy + \partial_{xz}\varphi(0)xz\\ &\; + \partial_{yy}\varphi(0) \frac12 y^2 + \partial_{yz}\varphi(0)yz + \partial_{zz}\varphi(0) \frac12 z^2 \end{align*}``

``\qquad \begin{align*} \partial_x \varphi &= \frac{2e^{2x}}{y+2}, &\qquad \partial_y \varphi &= -\frac{e^{2x}}{(y+2)^2}, &\qquad \partial_z \varphi &= 2(z+1), \\[6pt] \partial_{xx}\varphi &= \frac{4e^{2x}}{y+2}, &\qquad \partial_{xy}\varphi &= -\frac{2e^{2x}}{(y+2)^2}, &\qquad \partial_{xz}\varphi &= 0, \\[6pt] \partial_{yy}\varphi &= \frac{2e^{2x}}{(y+2)^3}, &\qquad \partial_{yz}\varphi &= 0, &\qquad \partial_{zz}\varphi &= 2 . \end{align*}``

``\qquad \begin{align*} T_2[\varphi] &= \frac{3}{2} + x + -\frac{1}{4}y + 2z + 2 \frac{1}{2} x^2 + -\frac{1}{2}xy + 0xz + \frac{1}{4} \frac{1}{2}y^2 + 0yz + 2 \cdot \frac12 z^2 \\ &= \frac{3}{2} + \underline b^T \underline x + \frac{1}{2} \underline x^T \underline{\underline{A}}\; \underline x\end{align*}``

``\qquad \underline x = \begin{pmatrix} x\\ y\\ z \end{pmatrix}, \qquad \underline b = \begin{pmatrix} 1\\ -\frac{1}{4}\\ 2 \end{pmatrix}, \qquad A = \begin{pmatrix} 2 & -\frac{1}{2} & 0\\ -\frac{1}{2} & \frac{1}{4} & 0\\ 0 & 0 & 2 \end{pmatrix}``
"""

# ╔═╡ 22653a5a-bfd1-4991-8b6c-d1662c638625
md"""
##### Bemerkung:
1. Statt ``T_m[f](x)=\sum_{|\alpha|\le m}\frac{1}{\alpha!}D^\alpha f(x_0)(x-x_0)^\alpha`` wird oft mit ``x = x_0 + h``, ``h \in \mathbb R^n`` geschrieben und damit ``T_m[f](h)=\sum_{|\alpha|\le m}\frac{1}{\alpha!}D^\alpha f(x_0)h^\alpha``
2. Für ``\varphi : \mathbb{R}^n \to \mathbb{R}`` heißt die Matrix ``H = (a_{ij}) \in \mathbb{R}^{n \times n}`` mit ``a_{ij} = \frac{\partial^2 \varphi}{\partial x_i \partial x_j}`` _Hesse-Matrix_. ``H`` ist symmetrisch (nach Schwarz).
"""

# ╔═╡ 556dab3a-e772-4498-b9e9-a7d7ffa26aa9
md"""
### 5.4.2 Extrema in ``\mathbb R^n``
"""

# ╔═╡ 28e9cd82-80aa-4f46-ae83-34ce65c9eaea
md"""
Für ``\varphi : U \subset \mathbb{R}^n \to \mathbb{R}`` heißt ``x_0 \in U`` _lokales_ Maximum (Minimum), falls in einer Umgebung ``B_r(x_0)`` mit Radius ``r>0`` gilt: `` \varphi(x) \le \varphi(x_0) \quad \forall x \in B_r(x_0) \qquad (\text{bzw. } \varphi(x) \ge \varphi(x_0))``

Das heißt, die Funktion ``g(t) = \varphi(x + tv)`` mit einer Richtung ``v \in \mathbb R^n`` hat bei ``t = 0`` ein Extremum. ``g'(0) = 0 \Rightarrow \langle \operatorname{grad} \varphi, v \rangle = 0``.

Bedingung für Extremum: ``\operatorname{grad} \varphi (x_0) = 0``. Das ist notwendig, aber nicht hinreichend.
"""

# ╔═╡ f8148d33-4bfd-4a9b-a0bf-4382b6ff65ef
md"""
#### Exkurs
"""

# ╔═╡ 3af4708b-42fc-4428-928e-401c4a452aa8
md"""
##### Satz und Definition:
Jede Matrix $A \in \mathbb{R}^{n \times n}$ lässt sich zerlegen in einen symmetrischen Anteil `` A_{\text{sym}}=\frac{1}{2}(A+A^T) `` und einen antisymmetrischen Anteil `` A_{\text{anti}}=\frac{1}{2}(A-A^T) `` so, dass gilt `` A=A_{\text{sym}}+A_{\text{anti}}``, sowie ``A_{\text{sym}}=A_{\text{sym}}^T`` und ``A_{\text{anti}}=-A_{\text{anti}}^T``.
"""

# ╔═╡ 4e113358-5031-4cf0-aabe-a2529b34a3ff
md"""
##### Beispiel:
``A = \begin{pmatrix} 1 & 1 & 1\\ 2 & 2 & 3\\ 1 & 1 & 3 \end{pmatrix}``

``A_{\mathrm{sym}} = \frac{1}{2} \left[\begin{pmatrix} 1 & 1 & 1\\ 2 & 2 & 3\\ 1 & 1 & 3 \end{pmatrix}+\begin{pmatrix} 1 & 2 & 1\\ 1 & 2 & 1\\ 1 & 3 & 3 \end{pmatrix}\right] = \begin{pmatrix} 1 & \frac{3}{2} & 1\\ \frac{3}{2} & 2 & 2\\ 1 & 2 & 3 \end{pmatrix}``

``A_{\mathrm{anti}} =\frac{1}{2}\left[\begin{pmatrix} 1 & 1 & 1\\ 2 & 2 & 3\\ 1 & 1 & 3 \end{pmatrix}-\begin{pmatrix} 1 & 2 & 1\\ 1 & 2 & 1\\ 1 & 3 & 3 \end{pmatrix} \right] = \begin{pmatrix} 0 & -\frac{1}{2} & 0\\ \frac{1}{2} & 0 & 1\\ 0 & -1 & 0 \end{pmatrix}``
"""

# ╔═╡ 5bb4ab2e-50c0-47f6-8598-7d7d7f170c74
md"""
##### Definition:
Eine Matrix $A \in \mathbb{R}^{n\times n}$ heißt _positiv definit_, falls gilt: ``x^T A x = \langle x, Ax \rangle > 0 \quad \forall\, x \in \mathbb{R}^n \setminus \{\underline 0\}``
"""

# ╔═╡ 8e9fe001-91d2-42f1-92c9-1f9c995a2d25
md"""
##### Bemerkung:
1. Analog _negativ definit_, falls ``x^T A x < 0``, und semi-definit falls ``\leq 0`` oder ``\geq 0`` zugelassen ist.
2. Es spielt nur der symmetrische Anteil eine Rolle, denn

``\qquad \begin{align*} x^T A x &= x^T (A_{\text{sym}} + A_{\text{anti}}) x = x^T A_{\text{sym}} x +  x^T \left(\frac{1}{2}(A - A^T) \right) x \\ &= x^T A_{\text{sym}} x + \frac{1}{2} x^T A x - \frac{1}{2} \underbrace{x^T A^T x}_{(Ax)^Tx = x^TAx} \\ &= x^T A_{\text{sym}} x\end{align*}``

``\qquad`` Es genügt also, symmetrische Matrizen zu betrachten.
"""

# ╔═╡ f1710510-16a5-4f6c-b87f-20063f0280c6
md"""
##### Satz:
``A \in \mathbb{R}^{n\times n}`` symmetrisch habe Eigenwerte ``\lambda_i \in \mathbb{R}, (i=1,\dots,n)``. Dann gilt
1. ``A`` pos./neg. definit ``\Leftrightarrow \lambda_i \gtrless 0,\quad \forall i``
2. ``A`` pos./neg. semi-definit ``\Leftrightarrow \lambda_i \gtreqless 0,\quad \forall i``
3. ``A`` indefinit ``\Leftrightarrow \lambda_i > 0,\ \lambda_j < 0`` für ein Paar ``(i,j)``
"""

# ╔═╡ 97f357b0-3658-438e-a1a8-a76f1db881ae
md"""
##### Beweis:
``A`` ist diagonalisierbar mit orthonormalen Eigenvektoren, das heißt, ``D=SAS^T`` mit ``D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)`` ist möglich. Also

``x^T \underbrace{A}_{= S^TDS} x = x^T S^T D S x = (Sx)^T D S x = y^T D y \qquad \text{mit } y=Sx . ``

(``y=0 \Leftrightarrow x=0``, denn ``S`` regulär), das heißt

`` \underbrace{x^T A x > 0}_{x\neq 0} \Leftrightarrow \underbrace{y^T D y > 0}_{y\neq 0} \qquad y^T D y=\sum_{i=1}^{n}\lambda_i \underbrace{y_i^2}_{> 0} \stackrel?>0``

``\Rightarrow \lambda_i > 0`` (``i = 1, \dots, n``)
"""

# ╔═╡ 623af91e-dd92-48c0-906c-7a7381437a93
md"""
#### Zurück zu Extrema
Betrachte die Taylor-Entwicklung von ``\varphi: \mathbb{R}^n \to \mathbb{R}`` um ``x_0`` mit ``x=x_0+h``:

`` T_2[\varphi](h) = \varphi(x_0) + \langle \underbrace{\operatorname{grad}\varphi(x_0)}_{=0}, h\rangle + \frac12 \langle h, H(x_0) h\rangle \qquad \text{mit der Hesse-Matrix $H$}``

``\phantom{T_2[\varphi](h)} = \varphi(x_0) + \frac12 \underbrace{\langle h, H(x_0)h \rangle}_{\gtrless\; 0} \stackrel?\gtrless \varphi(x_0)``
"""

# ╔═╡ f0b99f33-b310-4eeb-91e4-d0e2f1883cf7
md"""
##### Satz:
Sei ``\varphi: U \subset \mathbb{R}^n \to \mathbb{R}`` zweimal stetig differenzierbar und in ``x_0\in U`` gelte ``\operatorname{grad}\varphi(x_0)=0``. Dann hat ``\varphi`` ein Minimum in ``x_0``, falls die Hesse-Matrix `` H=\left.\left(\frac{\partial^2\varphi}{\partial x_i\partial x_j}\right)\right|_{x=x_0} `` positiv definit ist.
"""

# ╔═╡ d8789cac-e7e7-4f06-bd23-ceae2359f87b
md"""
##### Bemerkung:
1. Analog Maximum ``\Leftrightarrow`` negativ definit.
2. Bei semi-definit kann keine Aussage gemacht werden (wie in 1D)
3. indefinit bedeutet Sattelpunkt.
"""

# ╔═╡ 6a896743-df9e-46fa-89ea-5acdafee0618
md"""
##### Beispiel:
``\varphi: \mathbb R^2 \to \mathbb R`` mit ``\varphi(x,y)= 3x^2 + x^3 - 3xy^2 + 2y^2``

``\operatorname{grad}\varphi = ( \partial_x \varphi, \partial_y \varphi) = \left(6x + 3x^2 - 3y^2, -6xy + 4y\right) \stackrel!= 0``

``\begin{align*} 6x + 3x^2 - 3x^2 &= 0 \\ -6xy + 4y &= 0 \Leftrightarrow (4 - 6x)y = 0 \end{align*}``

Dazu eine Fallunterscheidung:
1. ``y = 0 \Rightarrow 6x + 3x^2 = (6 + 3x)x = 0``. Die Lösungen sind ``x=0`` und ``x=-2``.
2. ``x = \frac23 \Rightarrow 6 \cdot \frac23 + 3 \cdot \frac 49 + 3y^2 = \frac{16}3 + 3y^2 = 0``. Die Lösungen sind ``y = \frac43`` und ``y = - \frac43``.

Die Kandidaten sind also ``(0,0), (-2,0), \left( \frac23, \frac43 \right), \left( \frac23, - \frac43 \right)``

Die Hesse-Matrix ist ``H = \left( \frac{\partial^2 \varphi }{\partial x_i \partial x_j} \right) = \begin{pmatrix} \partial_{xx} \varphi & \partial_{xy} \varphi \\ \partial_{xy} \varphi & \partial_{yy} \varphi \end{pmatrix} = \begin{pmatrix} 6 + 6x & - 6y \\ - 6y & -6x + 4 \end{pmatrix} ``

``\begin{align*} H(0,0) &= \begin{pmatrix} 6 & 0 \\ 0 & 4 \end{pmatrix} \quad \text{EW } \lambda_1 = 6, \lambda_2 = 4 \Rightarrow \text{pos. definit: Maximum} \\ H(-2,0) &= \begin{pmatrix} -6 & 0 \\ 0 & 16 \end{pmatrix} \quad \text{EW } \lambda_1 = -6, \lambda_2 = 16 \Rightarrow \text{indefinit: Sattelpunkt} \\ H\left(\frac23, \frac43\right)  &= \begin{pmatrix} 10 & -8 \\ -8 & 0 \end{pmatrix}\end{align*}``

Das hat charakteristisches Polynom ``\det \begin{pmatrix} \lambda - 10 & 8 \\ 8 & \lambda \end{pmatrix} =(\lambda - 10)\lambda - 64 = \lambda^2 - 10 \lambda - 63``. Die Nullstellen sind ``\lambda_{1,2} = 5 \pm \sqrt{25 + 64} =  5 \pm \sqrt{89} \gtrless 0 \Rightarrow`` indefinit: Sattelpunkt.
"""

# ╔═╡ b3361b70-8d87-41b4-a0c1-672ac5395564
md"""
### 5.4.3 Extrema mit Nebenbedingung
"""

# ╔═╡ 2604872f-72c1-469c-874f-5b361c9319eb
md"""
##### Beispiel:
``\varphi:(x,y) = x^2 + y^2`` minimieren unter der Nebenbedingung ``x + 2y \stackrel!= 1`` bzw. ``g(x,y) = x + 2y \overset!= 0``
"""

# ╔═╡ 11a47cc5-a8e3-4fca-aaa3-371838fe7368
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# ╔═╡ 1df2d2f1-a270-46a6-961e-db39122807b6
md"""
Lösung: ``g(x,y) = 0 \Rightarrow y = \frac12 - \frac12 x``

``\varphi(x,y)=\varphi\left(x,\frac12-\frac12 x\right) = x^2+\left(\frac12-\frac12 x\right)^2 = x^2+\frac14 -\frac12 x + \frac14 x^2``

``\frac{\text d}{\text dx}\varphi(x,y) = 2x-\frac12 + \frac12 x = \frac52 x-\frac12 \overset!= 0 \quad\Rightarrow\quad x=\frac15 \quad\text{also}\quad y= \frac{4}{10} = \frac25``
"""

# ╔═╡ 9e76e84c-6eac-46dd-9f14-f56705d02440
md"""
##### Beispiel:
``\varphi(x,y) = 2x + y`` maximieren unter der Nebenbedingung ``x^2 + y^2 = 1`` bzw. ``g(x,y) = x^2 + y^2 - 1 \overset != 0``. ``g`` ist schwierig aufzulösen. ``y = \pm \sqrt{1 - x^2}``, wir wählen den positiven Zweig:

``\varphi(x, y(x)) = 2x + \sqrt{1 - x^2}``
"""

# ╔═╡ f4ba4d5f-aa1a-4b67-a350-60695e536a63
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# ╔═╡ f104d8a7-b1b2-488b-aac0-dc7a8ab38146
md"""
``\frac{\text d}{\text dx} \varphi(x, y(x)) = 2 - \frac{2x}{2 \sqrt{1 - x^2}} \overset != 0 \Rightarrow 2 \sqrt{1 - x^2} = x``

``4(1 - x^2) = x^2 \Leftrightarrow 4 = 5x^2 \qquad x = \pm \sqrt{\frac45}``

``y = - \sqrt{1 - x^2}`` gibt identisches Resultat. Also vier Kandidaten: ``(x,y) \in \left\{ \left( \sqrt{\frac{4}{5} } , \sqrt{\frac{1}{5} }\right), \left( - \sqrt{\frac{4}{5} } , \sqrt{\frac{1}{5} }\right), \left( \sqrt{\frac{4}{5} } , -\sqrt{\frac{1}{5} }\right), \left( -\sqrt{\frac{4}{5} } , -\sqrt{\frac{1}{5} }\right) \right\}``

Dies sind ein Minimum, ein Maximum und zwei Sattelpunkte.
"""

# ╔═╡ d9c97418-3fc5-4ac9-9550-6cf947cfa500
md"""
##### Allgemein:
``\varphi: U \subset \mathbb R^n \to \mathbb R``. Wann ist ``\varphi`` extremal unter Nebenbedingungen ``g: U \subset \mathbb R^n \to \mathbb R^m`` (``1 \leq m < n``) ``g(x) = 0`` (``m`` Gleichungen)?
"""

# ╔═╡ 6a5e2a8f-8b13-443b-b1cb-8414697eeae0
md"""
##### Bemerkung:
1. Die ``m`` Nebenbedingungen legen im Prinzip ``m`` der ``n`` Variablen in ``x \in \mathbb R^n`` fest. In den übrigen wird maximiert.
2. Bei mehreren Nebenbedingungen kann es sehr schwierig werden, nach bestimmten Variablen aufzulösen.
"""

# ╔═╡ fc7c04d4-5ac6-4ff3-8d60-3cd6e68a69ab
md"""
##### Satz:
Seien ``\varphi`` und ``g`` wie oben gegeben, ``M = \{x \in U \mid g(x) = 0\}`` die Nulsltellenmenge. Die Jacobi-Matrix ``\text Dg \in \mathbb R^{m \times n}`` habe maximalen Rang (``\operatorname{rg}(\text Dg) = m``) für alle ``x \in M``, das heißt, die Nebenbedingungen sind linear unabhängig. Die Extremalstellen von ``\varphi`` in ``M`` sind dann gegeben durch die Extremalstellen der _Lagrange-Funktion_

``\mathcal L: \mathbb R^n \times \mathbb R^m \to \mathbb R, (x, \lambda) \mapsto \mathcal L(x, \lambda) = \varphi(x) + \langle \lambda, g(x) \rangle``

wobei die _Lagrange-Multiplikatoren_ ``\lambda \in \mathbb R^m`` nur Hilfsgrößen sind.
"""

# ╔═╡ ec48d765-9866-477d-937a-b19a02ca61b9
md"""
##### Beispiel:
``\varphi(x,y)=x^2+y^2``, ``g(x,y)=x+2y-1 \overset!= 0``, ``n=2, m=1``.

``\mathcal{L}(x,y,\lambda)=x^2+y^2-\lambda(x+2y-1)`` extremal in  ``(x,y,\lambda)``

``\left.\begin{align*} \frac{\partial \mathcal{L}}{\partial x}=2x-\lambda \overset!= 0 \\ \frac{\partial \mathcal{L}}{\partial y}=2y-2\lambda\overset!= 0 \end{align*}\right\}\quad \left.\begin{matrix} \lambda &= 2x \\ \lambda &= y\end{matrix}\right \} \quad y = 2x``

``\frac{\partial \mathcal{L}}{\partial \lambda} =-(x+2y-1) \overset!= 0. \qquad (\text{NB}) ``

Es gilt dass ``\lambda=y`` und ``y=2x``. Durch Einsetzen dieses in die Nebenbedingung erhält man `` x=\frac15 \quad \text{und} \quad y=\frac25. ``

Also ist ``x + 4x = 1 \qquad x = \frac15, y = \frac25``.
"""

# ╔═╡ 87f3875b-be93-4c4c-90fb-ee633c69bdc5
md"""
##### Beweis:
Für Extrema muss gelten
`` \operatorname{grad}\mathcal{L}(x,\lambda)=0. ``

``\operatorname{grad}\mathcal{L}(x,\lambda) = (\operatorname{grad}_x \mathcal{L}, \operatorname{grad}_\lambda \mathcal{L}) = \left(\operatorname{grad}\varphi-\langle \lambda,\operatorname{grad} g\rangle,\; g(x)\right) \overset != 0``

In Komponenten:
``\frac{\partial \varphi}{\partial x_i} - \sum_{k=1}^m \lambda_k \frac{\partial g_k}{\partial x_i} =0, \qquad i=1,\dots,n \qquad`` (Bed. für ``x``)

``g_j(x)=0, \qquad j=1,\dots,m \qquad`` (NB)
"""

# ╔═╡ 8745d18c-3bb3-4ac7-8182-58dabfa868cf
md"""
##### Anschaulich ``(m=1)``:
`` \operatorname{grad}\varphi=\lambda \operatorname{grad} g, ``
das heißt, beide Gradienten zeigen in die gleiche Richtung.

Sei ``v`` die Richtung der Konturlinie ``g(x)=0`` bei ``x_0``, das heißt, ``g(x_0+t v)=0`` für ``t \in (-\varepsilon,\varepsilon). ``


Durch Ableiten folgt `` \langle \operatorname{grad} g, v\rangle =0. ``

`` \operatorname{grad} g `` steht senkrecht auf der Konturlinie und definiert damit eine "verbotene" Richtung. ``\operatorname{grad}\varphi`` muss nun nicht komplett verschwinden, sondern nur die Projektion auf ``v``, das heißt `` \langle \operatorname{grad}\varphi, v\rangle =0. ``

Änderungen von ``\varphi`` sind also nur in den "verbotenen" Richtungen
(proportional zu ``\operatorname{grad} g``) erlaubt.
"""

# ╔═╡ d696f636-1a71-4fe5-a3ff-8dd7bb93f463
md"""
##### Formal:
Wir unterteilen ``x=\begin{pmatrix}y\\ z\end{pmatrix}\in\mathbb{R}^n`` mit
``z\in\mathbb{R}^m`` und ``y\in\mathbb{R}^{n-m}`` so, dass für ``g(x)=g(y,z)`` die Jacobi-Matrix ``\text D_z g(y,z)`` in einem Punkt ``(y,z)\in M`` invertierbar ist. Mit dem Satz über implizite Funktionen existiert ein ``h:\mathbb{R}^{n-m}\to\mathbb{R}^m, y\mapsto z=h(y)``
mit ``g(y,z)=0``.

Wir maximieren ``\varphi(y,h(y))=: \widetilde{\varphi}(y)``.

``0=\frac{\partial \widetilde{\varphi}}{\partial y_j}=\frac{\partial \varphi}{\partial y_j}+\sum_{i=1}^{m}\frac{\partial \varphi}{\partial z_i}\frac{\partial h_i}{\partial y_j},\qquad j=1,\dots,n-m``

Es gilt
``\frac{\partial h_i}{\partial y_j}=-\sum_{k=1}^{m}(\text D_z g)^{-1}_{ik}\frac{\partial g_k}{\partial y_j}.``

Also ``\text D_y h=-(D_z g)^{-1} \text D_y g.``

``\Rightarrow\frac{\partial \varphi}{\partial y_j}=\sum_{i=1}^m\frac{\partial \varphi}{\partial z_i}\sum_{k=1}^m(\text D_z g)^{-1}_{ik}\frac{\partial g_k}{\partial y_j}=\sum_{k=1}^m\underbrace{\sum_{i=1}^m\frac{\partial \varphi}{\partial z_i}(\text D_z g)^{-1}_{ik}}_{=:\lambda_k}\frac{\partial g_k}{\partial y_j} =\sum_{k=1}^m\lambda_k\frac{\partial g_k}{\partial y_j}``

Für ``\frac{\partial \varphi}{\partial z_j}`` gilt dies trivialerweise:

``
\frac{\partial \varphi}{\partial z_j}
= \sum_{k=1}^m
\lambda_k
\frac{\partial g_k}{\partial z_j}
= \sum_{i=1}^m
\sum_{k=1}^m
\frac{\partial \varphi}{\partial z_i}
\underbrace{(\text D_z g)^{-1}_{ik}
(\text D_z g)_{kj}}_{\delta_{ij}}
= \frac{\partial \varphi}{\partial z_j}.
``

Also existieren ``\lambda_k\in\mathbb{R}`` so, dass
``
\frac{\partial \varphi}{\partial x_i}
= \sum_{k=1}^m
\lambda_k
\frac{\partial g_k}{\partial x_i},
\qquad
i=1,\dots,n.
``
"""

# ╔═╡ 2cf580da-d04a-4af7-baec-36344d3954d2
md"""
##### Beispiel:
``\varphi(x,y,z)=x^2+y^2+z^2, g(x,y,z) = \begin{pmatrix} x+y+z-1\\ 2x-y-2z+1 \end{pmatrix} \overset != \begin{pmatrix}0\\0\end{pmatrix}``

``
\mathcal{L}(x,y,z,\lambda_1,\lambda_2)
=
x^2+y^2+z^2
-\lambda_1(x+y+z-1)
-\lambda_2(2x-y-2z+1).
``

``\begin{align*} \frac{\partial \mathcal{L}}{\partial x} &=2x-\lambda_1-2\lambda_2=0 \\ \frac{\partial \mathcal{L}}{\partial y} &=2y \lambda_1+\lambda_2=0 \\ \frac{\partial \mathcal{L}}{\partial z} &=2z-\lambda_1+2\lambda_2=0 \end{align*}``

Also ``2(x - y) - 3 \lambda_2 = 0`` und ``2(y - z) - \lambda_2 = 0``, es folgt ``2(x-y) - 6(y - z) = 0``

``
\frac{\partial \mathcal{L}}{\partial \lambda_1}=0,
\qquad
\frac{\partial \mathcal{L}}{\partial \lambda_2}=0
\quad\Rightarrow\quad \text{NB}
``

Zum Lösen:

``\begin{align*} x+y+z &= 1 \\ 2x-y-2z & =1 \\  2x-8y+6z &= 0 \end{align*} \quad \begin{pmatrix} x\\ y\\ z \end{pmatrix} = \begin{pmatrix} \frac{9}{13}\\ \frac{3}{13}\\ \frac{1}{13} \end{pmatrix}``
"""

# ╔═╡ 8d25c672-324c-49fd-b347-263aab0f1842
md"""
## 5.5 Mehrfach-Integrale
"""

# ╔═╡ f4b2d2dd-f527-4fc4-a942-65e7d328878f
md"""
Wir schreiben ``f:\mathbb{R}^{m + n}\to\mathbb{R}`` als `` f:\mathbb{R}^m \times \mathbb{R}^n \to \mathbb{R}, (x,y)\mapsto f(x,y)`` und suchen nach einer Formel

`` \int_{\mathbb{R}^m\times\mathbb{R}^n} f(x,y)\; \text d\lambda_{m+n} = \, ?``

``f`` sei Lebesgue-integrierbar in ``\mathbb{R}^{m+n}``
"""

# ╔═╡ c68a13ac-76c8-4787-8e49-0dd04d40ab69
md"""
##### Satz von Fubini:
Sei ``\Omega_1\subset\mathbb{R}^n``, ``\Omega_2\subset\mathbb{R}^m`` und ``\Omega=\Omega_1\times\Omega_2\subset\mathbb{R}^{n+m}`` jeweils messbar und `` f:\Omega_1\times\Omega_2\to\mathbb{R}, (x,y)\mapsto f(x,y) `` ebenfalls messbar.

Falls ``f`` Lebesgue-integrierbar in ``\Omega`` ist, dann ist die Funktion ``f(x,\cdot)`` Lebesgue-integrierbar in ``\Omega_2`` (für fast alle ``x\in\Omega_1``) und ``f(\cdot,y)`` Lebesgue-integrierbar in ``\Omega_1`` (für fast alle ``y\in\Omega_2``), und es gilt

`` \begin{align*} \int_{\Omega_1\times\Omega_2} f(x,y)\; \text d\lambda_{n+m} &= \int_{\Omega_1} \underbrace{\left( \int_{\Omega_2} f(x,y)\; \text d\lambda_m(y) \right)}_{= F(x)}\; \text d\lambda_n(x)\\ &= \int_{\Omega_2} \underbrace{\left( \int_{\Omega_1} f(x,y)\; \text d\lambda_n(x) \right)}_{=G(y)}\; \text d\lambda_m(y). \end{align*} ``
"""

# ╔═╡ 4ec262fc-00c5-470b-b7a3-f9f0e3231d4e
md"""
##### Bemerkung:
1. Im wesentlichen folgt diese Aussage aus der Eigenschaft von Quadern ``Q_1\subset\mathbb{R}^n``, ``Q_2\subset\mathbb{R}^m``: `` \lambda_{n+m}(\underbrace{Q_1\times Q_2}_{\text{Quader im } \mathbb R^{m + n}})=\lambda_n(Q_1)\lambda_m(Q_2). ``

2. Wir schreiben ``\text d\lambda_{n+m}`` auch als ``\text d(x,y) = \text dx\text dy`` sowie ``\text d \lambda_n(x)`` als ``\text dx`` und ``\text d\lambda_m(y)`` als ``\text dy``.
"""

# ╔═╡ ca2f00e8-be6f-4366-996f-3575e41820b0
md"""
##### Bemerkung:
Falls ``A\subset\mathbb{R}^{n+m}`` kein kartesisches Produkt ist, betrachten wir die sogenannten _Schnitte_

``\begin{align*} A_x &\coloneqq \{y\in\mathbb{R}^m\mid (x,y)\in A\}\subset\mathbb{R}^m \\ A_y &\coloneqq \{x\in\mathbb{R}^n\mid (x,y)\in A\}\subset\mathbb{R}^n \end{align*}``

Diese sind messbar und es gilt für ``f:A\to\mathbb{R}`` (messbar)

``\begin{align*} \int_A f(x,y)\; \text d\lambda_{n+m} &= \int_{\mathbb{R}^{n+m}} f(x,y)\,\chi_A(x,y)\; \text d\lambda_{n+m} \\ &= \int_{\mathbb{R}^n}\left(\int_{A_x} f(x,y)\; \text dy\right)\; \text dx \\ &= \int_{\mathbb{R}^m}\left(\int_{A_y} f(x,y)\; \text dx\right)\; \text dy\end{align*}``
"""

# ╔═╡ beb10bd2-c344-4401-b27d-8df5cb4560db
md"""
##### Beispiel:
`` \Omega=[0,1]\times[2,3]\subset\mathbb{R}^2, f:\mathbb{R}^2\to\mathbb{R},(x,y)\mapsto x+x^2y^2 ``

``\begin{align*} \int_{\Omega} f(x,y)\; \text d(x,y) &= \int_0^1\left(\int_2^3 (x+x^2y^2)\; \text dy\right)\; \text dx \\ \text{(Integral bzgl.} y) \quad  &=\int_0^1\left(xy+\frac13 x^2 y^3\right)\Big|_{y=2}^{y=3}\; \text dx \\ &= \int_0^1\left(3x+9x^2-2x-\frac83 x^2\right)\; \text dx\\ &=\int_0^1\left(x+\frac{19}{3}x^2\right)\; \text dx \\ &= \left(\frac12 x^2+\frac{19}{9}x^3\right)\Big|_{x=0}^{x=1} \\ &= \frac12+\frac{19}{9} = \frac{47}{18} \end{align*}``
"""

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uuid = "0c0b7dd1-d40b-584c-a123-a41640f87eec"
version = "1.0.13+0"

[[deps.Xorg_libXdmcp_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl"]
git-tree-sha1 = "52858d64353db33a56e13c341d7bf44cd0d7b309"
uuid = "a3789734-cfe1-5b06-b2d0-1dd0d9d62d05"
version = "1.1.6+0"

[[deps.Xorg_libXext_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Xorg_libX11_jll"]
git-tree-sha1 = "a4c0ee07ad36bf8bbce1c3bb52d21fb1e0b987fb"
uuid = "1082639a-0dae-5f34-9b06-72781eeb8cb3"
version = "1.3.7+0"

[[deps.Xorg_libXrender_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Xorg_libX11_jll"]
git-tree-sha1 = "7ed9347888fac59a618302ee38216dd0379c480d"
uuid = "ea2f1a96-1ddc-540d-b46f-429655e07cfa"
version = "0.9.12+0"

[[deps.Xorg_libxcb_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Xorg_libXau_jll", "Xorg_libXdmcp_jll"]
git-tree-sha1 = "bfcaf7ec088eaba362093393fe11aa141fa15422"
uuid = "c7cfdc94-dc32-55de-ac96-5a1b8d977c5b"
version = "1.17.1+0"

[[deps.Xorg_xtrans_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl"]
git-tree-sha1 = "a63799ff68005991f9d9491b6e95bd3478d783cb"
uuid = "c5fb5394-a638-5e4d-96e5-b29de1b5cf10"
version = "1.6.0+0"

[[deps.Zlib_jll]]
deps = ["Libdl"]
uuid = "83775a58-1f1d-513f-b197-d71354ab007a"
version = "1.3.1+2"

[[deps.Zstd_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl"]
git-tree-sha1 = "446b23e73536f84e8037f5dce465e92275f6a308"
uuid = "3161d3a3-bdf6-5164-811a-617609db77b4"
version = "1.5.7+1"

[[deps.isoband_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Pkg"]
git-tree-sha1 = "51b5eeb3f98367157a7a12a1fb0aa5328946c03c"
uuid = "9a68df92-36a6-505f-a73e-abb412b6bfb4"
version = "0.2.3+0"

[[deps.libaom_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl"]
git-tree-sha1 = "371cc681c00a3ccc3fbc5c0fb91f58ba9bec1ecf"
uuid = "a4ae2306-e953-59d6-aa16-d00cac43593b"
version = "3.13.1+0"

[[deps.libass_jll]]
deps = ["Artifacts", "Bzip2_jll", "FreeType2_jll", "FriBidi_jll", "HarfBuzz_jll", "JLLWrappers", "Libdl", "Zlib_jll"]
git-tree-sha1 = "125eedcb0a4a0bba65b657251ce1d27c8714e9d6"
uuid = "0ac62f75-1d6f-5e53-bd7c-93b484bb37c0"
version = "0.17.4+0"

[[deps.libblastrampoline_jll]]
deps = ["Artifacts", "Libdl"]
uuid = "8e850b90-86db-534c-a0d3-1478176c7d93"
version = "5.15.0+0"

[[deps.libfdk_aac_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl"]
git-tree-sha1 = "646634dd19587a56ee2f1199563ec056c5f228df"
uuid = "f638f0a6-7fb0-5443-88ba-1cc74229b280"
version = "2.0.4+0"

[[deps.libpng_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Zlib_jll"]
git-tree-sha1 = "07b6a107d926093898e82b3b1db657ebe33134ec"
uuid = "b53b4c65-9356-5827-b1ea-8c7a1a84506f"
version = "1.6.50+0"

[[deps.libsixel_jll]]
deps = ["Artifacts", "JLLWrappers", "JpegTurbo_jll", "Libdl", "libpng_jll"]
git-tree-sha1 = "c1733e347283df07689d71d61e14be986e49e47a"
uuid = "075b6546-f08a-558a-be8f-8157d0f608a5"
version = "1.10.5+0"

[[deps.libvorbis_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Ogg_jll"]
git-tree-sha1 = "11e1772e7f3cc987e9d3de991dd4f6b2602663a5"
uuid = "f27f6e37-5d2b-51aa-960f-b287f2bc3b7a"
version = "1.3.8+0"

[[deps.libwebp_jll]]
deps = ["Artifacts", "Giflib_jll", "JLLWrappers", "JpegTurbo_jll", "Libdl", "Libglvnd_jll", "Libtiff_jll", "libpng_jll"]
git-tree-sha1 = "4e4282c4d846e11dce56d74fa8040130b7a95cb3"
uuid = "c5f90fcd-3b7e-5836-afba-fc50a0988cb2"
version = "1.6.0+0"

[[deps.nghttp2_jll]]
deps = ["Artifacts", "Libdl"]
uuid = "8e850ede-7688-5339-a07c-302acd2aaf8d"
version = "1.64.0+1"

[[deps.oneTBB_jll]]
deps = ["Artifacts", "JLLWrappers", "LazyArtifacts", "Libdl"]
git-tree-sha1 = "d5a767a3bb77135a99e433afe0eb14cd7f6914c3"
uuid = "1317d2d5-d96f-522e-a858-c73665f53c3e"
version = "2022.0.0+0"

[[deps.p7zip_jll]]
deps = ["Artifacts", "CompilerSupportLibraries_jll", "Libdl"]
uuid = "3f19e933-33d8-53b3-aaab-bd5110c3b7a0"
version = "17.7.0+0"

[[deps.x264_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl"]
git-tree-sha1 = "14cc7083fc6dff3cc44f2bc435ee96d06ed79aa7"
uuid = "1270edf5-f2f9-52d2-97e9-ab00b5d0237a"
version = "10164.0.1+0"

[[deps.x265_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl"]
git-tree-sha1 = "e7b67590c14d487e734dcb925924c5dc43ec85f3"
uuid = "dfaa095f-4041-5dcd-9319-2fabd8486b76"
version = "4.1.0+0"
"""

# ╔═╡ Cell order:
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# ╟─8d25c672-324c-49fd-b347-263aab0f1842
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