[Analysis] Add more notes

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2026-01-28 17:26:59 +01:00
parent e8a3fcad09
commit 9ea7d3029e
4 changed files with 7 additions and 4 deletions

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@@ -40,5 +40,6 @@ $\nabla f(x_0) =
and the and the
\drmvspace\rmvspace \drmvspace\rmvspace
trace of the Jacobi Matrix, $\text{div}(f)(x_0) = \text{Tr}(J_f(x_0)) = \sum_{i = 1}^{n} \partial_{x_i} f_i(x_0)$ is called the \bi{divergence} of $f$ at $x_0$. trace of the Jacobi Matrix, $\text{div}(f)(x_0) = \text{Tr}(J_f(x_0)) = \sum_{i = 1}^{n} \partial_{x_i} f_i(x_0)$ is called the \bi{divergence} of $f$ at $x_0$.\\
The gradient is simply the transpose of the Jacobian and it points in the direction of the \bi{steepest ascent}.
\rmvspace \rmvspace

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@@ -1,4 +1,4 @@
\newsectionNoPB \newsection
\subsection{The differential} \subsection{The differential}
\setLabelNumber{all}{2} \setLabelNumber{all}{2}
\compactdef{Differentiable function} We have function $f: X \rightarrow \R^m$, linear map $u : \R^n \rightarrow \R^m$ and $x_0 \in X$. $f$ is differentiable at $x_0$ with differential $u$ if \compactdef{Differentiable function} We have function $f: X \rightarrow \R^m$, linear map $u : \R^n \rightarrow \R^m$ and $x_0 \in X$. $f$ is differentiable at $x_0$ with differential $u$ if
@@ -6,7 +6,6 @@ $\displaystyle \lim_{\elementstack{x \rightarrow x_0}{x \neq x_0}} \frac{f(x) -
We denote $\dx f(x_0) = u$. We denote $\dx f(x_0) = u$.
If $f$ is differentiable at every $x_0 \in X$, then $f$ is differentiable on $X$. If $f$ is differentiable at every $x_0 \in X$, then $f$ is differentiable on $X$.
\newpage
% ──────────────────────────────────────────────────────────────────── % ────────────────────────────────────────────────────────────────────
\stepLabelNumber{all} \stepLabelNumber{all}
\shortproposition \shortproposition

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@@ -16,7 +16,8 @@ To determine the kind of critical point, we need to determine if $H_f(x_0)$ is d
\dhrmvspace \dhrmvspace
\setLabelNumber{all}{6} \setLabelNumber{all}{6}
\compactdef{Non-degenerate critical point} If $\det(H_f(x_0)) \neq 0$ (if $H_f(x_0)$ is semi-definite, then $\det(H_f(x_0)) = 0$, thus degenerate)\\ \compactdef{Non-degenerate critical point} If $\det(H_f(x_0)) \neq 0$ (if $H_f(x_0)$ is semi-definite, then $\det(H_f(x_0)) = 0$, thus degenerate)
To figure out if a matrix is definite, we can compute the eigenvalues. $A$ is positive (negative) definite, if and only if all eigenvalues are greater (lower) than $0$. To figure out if a matrix is definite, we can compute the eigenvalues. $A$ is positive (negative) definite, if and only if all eigenvalues are greater (lower) than $0$.
$A$ is indefinite if and only if it has both positive and negative eigenvalues. $A$ is indefinite if and only if it has both positive and negative eigenvalues.
$A$ is positive (negative) semi-definite if and only if all eigenvalues are greater (lower) or equal to $0$. $A$ is positive (negative) semi-definite if and only if all eigenvalues are greater (lower) or equal to $0$.
@@ -48,3 +49,5 @@ For $2 \times 2$ matrices (i.e. 2D functions), we can use the following scheme (
(tr1) edge node [right] {$0$} (zero); (tr1) edge node [right] {$0$} (zero);
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
As in Analysis I, it is important to also check the boundaries for maximums and minimums.
For that, formulate formulas for the borders and check them for critical points.