[Analysis] reorganize

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RobinB27
2026-01-02 13:05:52 +01:00
parent 04a923450a
commit 53447a0145
6 changed files with 141 additions and 139 deletions

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@@ -16,12 +16,16 @@
\section{Differential Equations} \section{Differential Equations}
\input{parts/02_diffeq.tex} \input{parts/02_diffeq.tex}
\newpage
\section{Solutions to Differential Equations}
\input{parts/03_diffeq_sol.tex}
\newpage \newpage
\section{Continuous functions in $\R^n$} \section{Continuous functions in $\R^n$}
\input{parts/03_cont.tex} \input{parts/04_cont.tex}
\newpage \newpage
\section{Differential Calculus in $\R^n$} \section{Differential Calculus in $\R^n$}
\input{parts/04_diff.tex} \input{parts/05_diff.tex}
\end{document} \end{document}

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@@ -83,139 +83,4 @@ $\S_b(F) := \{ f + f_0 \sep f \in \S(F),\ f_0 \text{ is a particular sol.} \}$\\
\remark \textbf{Applications of Linearity}\\ \remark \textbf{Applications of Linearity}\\
If $f_1$ solves $F$ for $b_1$, and $f_2$ for $b_2$: $f_1 + f_2$ solves $b_1 + b_2$. \\ If $f_1$ solves $F$ for $b_1$, and $f_2$ for $b_2$: $f_1 + f_2$ solves $b_1 + b_2$. \\
Follows from: $D(f_1) + D(f_2) = b_1 + b_2$. Follows from: $D(f_1) + D(f_2) = b_1 + b_2$.
\newpage
\subsection{Linear Solutions: First Order}
\subtext{ $I \subset \R, \quad a,b: I \to \R$ }
\textbf{Form:}
$$ y' + ay = b $$
\textbf{Approach:}
\begin{enumerate}
\item Hom. Solution $f_1$ for: $y' + ay = 0$\\
\subtext{Note that $\S$ has $\dim(\S) = 1$, so $f_1 \neq 0$ is a Basis for $\S$}
\item Part. Solution $f_0$ for $y' + ay = b$
\end{enumerate}
\textbf{Solutions:} $ f_0 + zf_1 \quad \text{ for } z \in \C $
\begin{subbox}{Explicit Homogeneous Solution}
\smalltext{$A(x)$ is a primitive of $a$, $f(x_0) = y_0$}
\begin{align*}
f_1(x) &= z \cdot \exp(-A(x)) \\
f_1(x) &= y_0 \cdot \exp(A(x_0) - a(x))
\end{align*}
\end{subbox}
Variation of Constants: Treating $z$ as $z(x)$ yields:
\begin{subbox}{Explicit Inhomogeneous Solution}
\smalltext{$A(x)$ is a primitive of $a$}
$$
f_0(x) = \underbrace{\left(\int b(x)\cdot\exp(A(x)) \right)}_{z(x)} \cdot \exp\left(-A(x)\right)
$$
\end{subbox}
\method \textbf{Educated Guess}\\
Usually, $y$ has a similar form to $b$:
\begin{tabular}{ll}
\hline
$b(x)$ & \text{Guess} \\
\hline
$a \cdot e^{\alpha x}$ & $b \cdot e^{\alpha x}$ \\
$a \cdot \sin(\beta x)$ & $c\sin(\beta x) + d\cos(\beta x)$\\
$b \cdot \cos(\beta x)$ & $c\sin(\beta x) + d\cos(\beta x)$\\
$ae^{\alpha x} \cdot \sin(\beta x)$ & $e^{\alpha x}\left(c\sin(\beta x) + d\cos(\beta x)\right)$\\
$be^{\alpha x} \cdot \cos(\beta x)$ & $e^{\alpha x}\left(c\sin(\beta x) + d\cos(\beta x)\right)$\\
$P_n(x) \cdot e^{\alpha x}$ & $R_n(x) \cdot e^{\alpha x}$\\
$P_n(x) \cdot e^{\alpha x}\sin(\beta x)$ & $e^{\alpha x}\left( R_n(x) \sin(\beta x) + S_n(x) \cos(\beta x) \right)$\\
$P_n(x) \cdot e^{\alpha x}\cos(\beta x)$ & $e^{\alpha x}\left( R_n(x) \sin(\beta x) + S_n(x) \cos(\beta x) \right)$\\
\hline
\end{tabular}
\remark If $\alpha, \beta$ are roots of $P(X)$ with multiplicity $j$, multiply guess with a $P_j(x)$.
\subsection{Linear Solutions: Constant Coefficients}
\textbf{Form:}
$$
y^{(k)} + a_{k-1}y^{(k-1)} + \ldots + a_1y' + a_0y = b
$$
\subtext{Where $a_0, \ldots, a_{k-1} \in \C$ are constants, $b(x)$ is continuous.}
\subsubsection{Homogeneous Equations}
The idea is to find a Basis of $\S$:
\definition \textbf{Characteristic Polynomial} $P(X) = \prod_{i=1}^{k} (X-\alpha_i)$
\remark The unique roots $\alpha_1,\ldots,\alpha_l$ form a Basis:
$$
\text{span}(\S) = \{ x^je^{\alpha_i x} \sep i \leq l,\quad 0 \leq j \leq v_i \}
$$
\subtext{$v_1,\ldots,v_k$ are the Multiplicities of $\alpha_1,\ldots,\alpha_k$}
\remark If $\alpha_j = \beta + \gamma i \in \C$ is a root, $\bar{\alpha_j} = \beta - \gamma i$ is too.\\
To get a real-valued solution, apply:
$$
e^{\alpha_j x} = e^{\beta x}\left( \cos(\gamma x) + i \sin(\gamma x) \right)
$$
\begin{subbox}{Explicit Homogeneous Solution}
\smalltext{Using $\alpha_1,\ldots,\alpha_k$ from $P(X)$ s.t. $\alpha_i \neq \alpha_j$, $z_i \in \C$ arbitrary}
$$
f(x) = \prod_{i=1}^{k} z_i \cdot e^{\alpha_i x} \quad\text{with}\quad f^{(j)(x)} = \prod_{i=1}^{k} z_i \cdot \alpha_i^j e^{\alpha_i x}
$$
\smalltext{Multiple roots: same scheme, using the basis vectors of $\S$}
\end{subbox}
\subtext{Solutions exist $\forall Z = (z_1,\ldots,z_k)$ since that system's $\det(M_Z) \neq 0$.}
\newpage
\subsubsection{Inhomogeneous Equations}
\method \textbf{Undetermined Coefficients}: An educated guess.
\begin{enumerate}
\item $b(x) = cx^d \cdot e^{\alpha x} \implies f_p(x) = Q(x)e^{\alpha x}$\\
\subtext{$\deg(Q) \leq d + v_\alpha$, where $v_\alpha$ is $\alpha$'s multiplicity in $P(X)$}
\item $\begin{rcases*}
b(x) = cx^d \cdot \cos(\alpha x) \\
b(x) = cx^d \cdot \sin(\alpha x)
\end{rcases*} f_p = Q_1(x)\cos(\alpha x) + Q_2(x)\sin(\alpha x)$
\subtext{$\deg(Q_{1,2}) \leq d + v_\alpha$, where $v_\alpha$ is $\alpha$'s multiplicity in $P(X)$}
\end{enumerate}
\remark \textbf{Applying Linearity}\\
If $b(x) = \sum_{i=1}^{n} b_i(x)$, A solution for $b(x)$ is $f(x) = \sum_{i=1}^{n} f_i(x)$\\
\subtext{Sometimes called \textit{Superposition Principle} in this context}
\subsection{Other Methods}
\method \textbf{Change of Variable}\\
If $f(x)$ is replaced by $h(y) = f(g(y))$, then $h$ is a sol. too.\\
\subtext{Changes like $h(t) = f(e^t)$ may lead to useful properties.}
\begin{subbox}{Separation of Variables}
Form:
$$
y' = a(y)\cdot b(x)
$$
Solve using:
$$
\int \frac{1}{a(y)}\ \text{d}y = \int b(x) \dx + c
$$
\end{subbox}
\subtext{Usually $\int 1/a(y)\ \text{d}y$ can be solved directly for $\ln|a(y)|+c$.}
\subsection{Method Overview}
\begin{center}
\begin{tabular}{l|l}
\textbf{Method} & \textbf{Use case} \\
\hline
Variation of constants & LDE with $\ord(F)=1$ \\
Characteristic Polynomial & Hom. LDE w/ const. coeff. \\
Undetermined Coefficients & Inhom. LDE w/ const. coeff. \\
Separation of Variables & ODE s.t. $y' = a(y)\cdot b(x)$ \\
Change of Variables & e.g. $y' = f(ax + by + c)$ \\
\end{tabular}
\end{center}

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@@ -0,0 +1,133 @@
\subsection{Linear Solutions: First Order}
\textbf{Form:}$\quad y' + ay = b\quad $ \subtext{ $I \subset \R, \quad a,b: I \to \R$ }
\textbf{Approach:}
\begin{enumerate}
\item Hom. Solution $f_1$ for: $y' + ay = 0$\\
\subtext{Note that $\S$ has $\dim(\S) = 1$, so $f_1 \neq 0$ is a Basis for $\S$}
\item Part. Solution $f_0$ for $y' + ay = b$
\end{enumerate}
\textbf{Solutions:} $ f_0 + zf_1 \quad \text{ for } z \in \C $
\begin{subbox}{Explicit Homogeneous Solution}
\smalltext{$A(x)$ is a primitive of $a$, $f(x_0) = y_0$}
\begin{align*}
f_1(x) &= z \cdot \exp(-A(x)) \\
f_1(x) &= y_0 \cdot \exp(A(x_0) - a(x))
\end{align*}
\end{subbox}
\method \textbf{Variation of Constants}: Treating $z$ as $z(x)$ yields:
\begin{subbox}{Explicit Inhomogeneous Solution}
\smalltext{$A(x)$ is a primitive of $a$}
$$
f_0(x) = \underbrace{\left(\int b(x)\cdot\exp(A(x)) \right)}_{z(x)} \cdot \exp\left(-A(x)\right)
$$
\end{subbox}
\method \textbf{Educated Guess}\\
Usually, $y$ has a similar form to $b$:
\begin{tabular}{ll}
\hline
$b(x)$ & \text{Guess} \\
\hline
$a \cdot e^{\alpha x}$ & $b \cdot e^{\alpha x}$ \\
$a \cdot \sin(\beta x)$ & $c\sin(\beta x) + d\cos(\beta x)$\\
$b \cdot \cos(\beta x)$ & $c\sin(\beta x) + d\cos(\beta x)$\\
$ae^{\alpha x} \cdot \sin(\beta x)$ & $e^{\alpha x}\left(c\sin(\beta x) + d\cos(\beta x)\right)$\\
$be^{\alpha x} \cdot \cos(\beta x)$ & $e^{\alpha x}\left(c\sin(\beta x) + d\cos(\beta x)\right)$\\
$P_n(x) \cdot e^{\alpha x}$ & $R_n(x) \cdot e^{\alpha x}$\\
$P_n(x) \cdot e^{\alpha x}\sin(\beta x)$ & $e^{\alpha x}\left( R_n(x) \sin(\beta x) + S_n(x) \cos(\beta x) \right)$\\
$P_n(x) \cdot e^{\alpha x}\cos(\beta x)$ & $e^{\alpha x}\left( R_n(x) \sin(\beta x) + S_n(x) \cos(\beta x) \right)$\\
\hline
\end{tabular}
\remark If $\alpha, \beta$ are roots of $P(X)$ with multiplicity $j$, multiply guess with a $P_j(x)$.
\subsection{Linear Solutions: Constant Coefficients}
\textbf{Form:}
$$
y^{(k)} + a_{k-1}y^{(k-1)} + \ldots + a_1y' + a_0y = b
$$
\subtext{Where $a_0, \ldots, a_{k-1} \in \C$ are constants, $b(x)$ is continuous.}
\subsubsection{Homogeneous Equations}
The idea is to find a Basis of $\S$:
\definition \textbf{Characteristic Polynomial} $P(X) = \prod_{i=1}^{k} (X-\alpha_i)$
\remark The unique roots $\alpha_1,\ldots,\alpha_l$ form a Basis:
$$
\text{span}(\S) = \{ x^je^{\alpha_i x} \sep i \leq l,\quad 0 \leq j \leq v_i \}
$$
\subtext{$v_1,\ldots,v_k$ are the Multiplicities of $\alpha_1,\ldots,\alpha_k$}
\remark If $\alpha_j = \beta + \gamma i \in \C$ is a root, $\bar{\alpha_j} = \beta - \gamma i$ is too.\\
To get a real-valued solution, apply:
$$
e^{\alpha_j x} = e^{\beta x}\left( \cos(\gamma x) + i \sin(\gamma x) \right)
$$
\begin{subbox}{Explicit Homogeneous Solution}
\smalltext{Using $\alpha_1,\ldots,\alpha_k$ from $P(X)$ s.t. $\alpha_i \neq \alpha_j$, $z_i \in \C$ arbitrary}
$$
f(x) = \prod_{i=1}^{k} z_i \cdot e^{\alpha_i x} \quad\text{with}\quad f^{(j)(x)} = \prod_{i=1}^{k} z_i \cdot \alpha_i^j e^{\alpha_i x}
$$
\smalltext{Multiple roots: same scheme, using the basis vectors of $\S$}
\end{subbox}
\subtext{Solutions exist $\forall Z = (z_1,\ldots,z_k)$ since that system's $\det(M_Z) \neq 0$.}
\newpage
\subsubsection{Inhomogeneous Equations}
\method \textbf{Undetermined Coefficients}: An educated guess.
\begin{enumerate}
\item $b(x) = cx^d \cdot e^{\alpha x} \implies f_p(x) = Q(x)e^{\alpha x}$\\
\subtext{$\deg(Q) \leq d + v_\alpha$, where $v_\alpha$ is $\alpha$'s multiplicity in $P(X)$}
\item $\begin{rcases*}
b(x) = cx^d \cdot \cos(\alpha x) \\
b(x) = cx^d \cdot \sin(\alpha x)
\end{rcases*} f_p = Q_1(x)\cos(\alpha x) + Q_2(x)\sin(\alpha x)$
\subtext{$\deg(Q_{1,2}) \leq d + v_\alpha$, where $v_\alpha$ is $\alpha$'s multiplicity in $P(X)$}
\end{enumerate}
\remark \textbf{Applying Linearity}\\
If $b(x) = \sum_{i=1}^{n} b_i(x)$, A solution for $b(x)$ is $f(x) = \sum_{i=1}^{n} f_i(x)$\\
\subtext{Sometimes called \textit{Superposition Principle} in this context}
\subsection{Other Methods}
\method \textbf{Change of Variable}\\
If $f(x)$ is replaced by $h(y) = f(g(y))$, then $h$ is a sol. too.\\
\subtext{Changes like $h(t) = f(e^t)$ may lead to useful properties.}
\begin{subbox}{Separation of Variables}
Form:
$$
y' = a(y)\cdot b(x)
$$
Solve using:
$$
\int \frac{1}{a(y)}\ \text{d}y = \int b(x) \dx + c
$$
\end{subbox}
\subtext{Usually $\int 1/a(y)\ \text{d}y$ can be solved directly for $\ln|a(y)|+c$.}
\subsection{Method Overview}
\begin{center}
\begin{tabular}{l|l}
\textbf{Method} & \textbf{Use case} \\
\hline
Variation of constants & LDE with $\ord(F)=1$ \\
Characteristic Polynomial & Hom. LDE w/ const. coeff. \\
Undetermined Coefficients & Inhom. LDE w/ const. coeff. \\
Separation of Variables & ODE s.t. $y' = a(y)\cdot b(x)$ \\
Change of Variables & e.g. $y' = f(ax + by + c)$ \\
\end{tabular}
\end{center}

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@@ -50,7 +50,7 @@ $\\
Partial derivatives don't provide a good approx. of $f$, unlike in the $1$-dimensional case. The \textit{differential} is a linear map which replicates this purpose in $\R^n$. Partial derivatives don't provide a good approx. of $f$, unlike in the $1$-dimensional case. The \textit{differential} is a linear map which replicates this purpose in $\R^n$.
} }
\begin{subbox}{Differentiability in $\R^n$} \begin{subbox}{Differentiability in $\R^n$ \& the Differential}
\smalltext{$X \subset \R^n \text{ open},\quad f: X \to \R^n,\quad u: \R^n \to \R^m \text{ linear map}$} \smalltext{$X \subset \R^n \text{ open},\quad f: X \to \R^n,\quad u: \R^n \to \R^m \text{ linear map}$}
$$ $$
df(x_0) := u df(x_0) := u