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[Analysis] Small fixes

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Janis Hutz
2026-01-31 13:25:57 +01:00
parent 091b1738a2
commit 457597ea53
3 changed files with 4 additions and 2 deletions
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semester3/analysis-ii/cheat-sheet-jh/analysis-ii-cheat-sheet.pdf
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semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/02_differential.tex
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@@ -42,6 +42,7 @@ we can compute the partial derivative using the chain rule as follows:
$\frac{\partial g}{\partial \phi} = \frac{\partial g}{\partial x} \cdot \frac{\partial x}{\partial \phi} $\frac{\partial g}{\partial \phi} = \frac{\partial g}{\partial x} \cdot \frac{\partial x}{\partial \phi}
+ \frac{\partial g}{\partial y} \cdot \frac{\partial y}{\partial \phi} + \frac{\partial g}{\partial z} \cdot \frac{\partial z}{\partial \phi}$ + \frac{\partial g}{\partial y} \cdot \frac{\partial y}{\partial \phi} + \frac{\partial g}{\partial z} \cdot \frac{\partial z}{\partial \phi}$
where all $\frac{\partial g}{\partial x}$, etc are known from the gradient and the other elements can be computed quickly from the known equations. where all $\frac{\partial g}{\partial x}$, etc are known from the gradient and the other elements can be computed quickly from the known equations.
The chain rule for higher or lower dimensional functions is as one would expect from the above formula.
Finally, evaluate $\frac{\partial g}{\partial \phi}$ at the required points and compute the result. Finally, evaluate $\frac{\partial g}{\partial \phi}$ at the required points and compute the result.

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semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/03_higher_diff.tex
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@@ -1,11 +1,12 @@
\newsectionNoPB \newsectionNoPB
\subsection{Higher derivatives} \subsection{Higher derivatives}
\shortdef $f$ is in class $C^1$ if $f$ is differentiable and all its partial derivatives are continuous. \compactdef{Class} $f$ is in class $C^1$ if $f$ is differentiable and all its partial derivatives are continuous.
$f$ is of class $C^k$ if it is differentiable and each of its partial derivatives are in $C^{k - 1}$. $f$ is of class $C^k$ if it is differentiable and each of its partial derivatives are in $C^{k - 1}$.
If $f \in C^k(X; \R^m)$ for all $k \geq 1$, then $f \in C^\infty(X; \R^m)$\\ If $f \in C^k(X; \R^m)$ for all $k \geq 1$, then $f \in C^\infty(X; \R^m)$\\
% ──────────────────────────────────────────────────────────────────── % ────────────────────────────────────────────────────────────────────
\setLabelNumber{all}{4} \setLabelNumber{all}{4}
\compactproposition{Mixed derivatives commute} $\partial_{x, y} f = \partial_{y, x}$, as well as $\partial_{x, y, z} = \partial_{x, z, y} = \ldots$, etc (all mixed derivatives commute) \compactproposition{Mixed derivatives commute} $\partial_{x, y} f = \partial_{y, x}$, as well as $\partial_{x, y, z} = \partial_{x, z, y} = \ldots$,
etc (all mixed derivatives commute).
Since we have symmetry, we can use the notation $\partial_{x_1^{m_1}, \ldots, x_n^{m_n}} f = \frac{\partial^k}{\partial x^m} f = D^m f = \partial^m f$, Since we have symmetry, we can use the notation $\partial_{x_1^{m_1}, \ldots, x_n^{m_n}} f = \frac{\partial^k}{\partial x^m} f = D^m f = \partial^m f$,
where $m = (m_1, \ldots, m_n)$ and $m_1 + \ldots + m_n = k$. where $m = (m_1, \ldots, m_n)$ and $m_1 + \ldots + m_n = k$.
There are ${n + k - 1 \choose k}$ possible values for $m$ and e.g. $(1, 1, 2)$ corresponds to the derivative $\frac{\partial^4 f}{\partial x \partial y \partial^2 z}$\\ There are ${n + k - 1 \choose k}$ possible values for $m$ and e.g. $(1, 1, 2)$ corresponds to the derivative $\frac{\partial^4 f}{\partial x \partial y \partial^2 z}$\\