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https://github.com/arnaucube/math.git
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add typos detection
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5
.github/workflows/typos.toml
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.github/workflows/typos.toml
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[default.extend-words]
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thm = "thm"
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ba = "ba"
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Strang = "Strang" # name
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Bootle = "Bootle" # name
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21
.github/workflows/typos.yml
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.github/workflows/typos.yml
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name: typos
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on:
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pull_request:
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branches: [ main ]
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types: [ready_for_review, opened, synchronize, reopened]
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push:
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branches:
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- main
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jobs:
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typos:
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if: github.event.pull_request.draft == false
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name: Spell Check with Typos
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runs-on: ubuntu-latest
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steps:
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- uses: actions/checkout@v4
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- name: Use typos with config file
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uses: crate-ci/typos@master
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with:
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config: .github/workflows/typos.toml
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@@ -174,7 +174,7 @@
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\begin{defn}{local ring}
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A \emph{local ring} has a unique maximal ideal.
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Notation: locall ring $A$, its maximal ideal $\mM$, residue field $K=A/\mM$:
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Notation: local ring $A$, its maximal ideal $\mM$, residue field $K=A/\mM$:
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$$A \supset \mM ~\text{or}~ (A, \mM) ~\text{or}~ (A, \mM, K)$$
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\end{defn}
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@@ -230,7 +230,7 @@ A subset $S \subset \Sigma$ is \emph{totally ordered} if for every pair $s_1,s_2
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Conversely:\\
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Suppose $x \not\in \mM$ for some maximal ideal $\mM$.
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Then $\mM$ and $x$ generte the unit ideal $(1)$, so that we have $u + xy = 1$ for some $u \in \mM$ and some $y \in A$.
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Then $\mM$ and $x$ generate the unit ideal $(1)$, so that we have $u + xy = 1$ for some $u \in \mM$ and some $y \in A$.
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Hence $1 -xy \in \mM$, and is therefore not a unit.
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\end{proof}
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@@ -632,7 +632,7 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi
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\begin{prop}{AM.2.8} \label{2.8}
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Let $x_i \forall i \in [n]$ be elements of $M$ whose images $\frac{M}{m M}$ from a basis of this vecctor space. Then the $x_i$ generate $M$.
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Let $x_i \forall i \in [n]$ be elements of $M$ whose images $\frac{M}{m M}$ from a basis of this vector space. Then the $x_i$ generate $M$.
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\end{prop}
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\begin{proof}
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Let $N$ submodule $M$, generated by the $x_i$.
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@@ -722,7 +722,7 @@ As in with rings, it is equivalent to say that
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\section{Exercises}
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For the exercises, I follow the assignements listed at \cite{mit-course}.
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For the exercises, I follow the assignments listed at \cite{mit-course}.
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The exercises that start with \textbf{R} are the ones from the book \cite{reid}, and the ones starting with \textbf{AM} are the ones from the book \cite{am}.
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@@ -277,7 +277,7 @@
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\end{thm}
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\begin{thm}{9.10}
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An irreducible polynomial $f \in K[t]$ ($K \subseteq \mathbb{C}$) is \emph{separable over} $K$ if it has simple zeros in $\mathbb{C}$, or equivelently, simple zeros in its splitting field.
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An irreducible polynomial $f \in K[t]$ ($K \subseteq \mathbb{C}$) is \emph{separable over} $K$ if it has simple zeros in $\mathbb{C}$, or equivalently, simple zeros in its splitting field.
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\end{thm}
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\begin{lemma}{9.13}
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@@ -809,7 +809,7 @@
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Let $f_i$ be the minimal polynimal of $\alpha_i$ over $K$.
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Then, $M \supseteq L$ is spliting field of $\Prod_{i=1}^r f_i$, since $M$ is normal enclosure of $L:K$.
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Then, $M \supseteq L$ is splitting field of $\Prod_{i=1}^r f_i$, since $M$ is normal enclosure of $L:K$.
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For every zero $\beta_{ij}$ of $f_i$ in $M$,\\
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$\exists$ an isomorphism $\sigma: K(\alpha_i) \longrightarrow K(\beta_{ij})$ by Corollary \ref{5.13}.
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@@ -1065,7 +1065,7 @@ For $n \geq 3, ~~\mathbb{D}_n \subseteq \mathbb{S}_n$ (subgroup of the Symmetry
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then, $\exists$ at least one $c$ in $(a,b)$ such that $Df(c)=0$.
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\end{thm}
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\begin{proof} (proof source: cue math website)
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Notice that when $Df(x_i)=0$ occours, is a maximum or minimum (extrema) value of $f$.
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Notice that when $Df(x_i)=0$ occurs, is a maximum or minimum (extrema) value of $f$.
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$\Longrightarrow$ if a function is continuous, it is guaranteed to have both a maximum and a minimum point in the interval.\\
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Two possibilities:
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@@ -1076,9 +1076,9 @@ For $n \geq 3, ~~\mathbb{D}_n \subseteq \mathbb{S}_n$ (subgroup of the Symmetry
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since $f$ not constant, must change directions in ordder to start and end at the same $y$-value ($f(a)=f(b)$).\\
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Thus at some point between $a$ and $b$ it will either have a minimum, maximum or both.
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\begin{enumerate}[a.]
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\item does the maximum occour at a point where $Df > 0$?\\
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\item does the maximum occur at a point where $Df > 0$?\\
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No, because if $Df > 0$, then $f$ is increasing, but it can not increase since we're at its maximum point.
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\item does the maximum occour at a point where $Df < 0$?\\
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\item does the maximum occur at a point where $Df < 0$?\\
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No, because if $Df < 0$, then $f$ is deccreasing, which means that at our left it was larger, but we're at a maximum point, so a contradiction.
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\end{enumerate}
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Same with minimus.\\
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@@ -76,7 +76,7 @@ Consider the following protocol:
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\end{enumerate}
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%/// TODO tabulate this next lines
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With high probablility, $\alpha$ will not cancel the coeffs with $deg \geq d+1$. % TODO check which is the name of this theorem or why this is true
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With high probability, $\alpha$ will not cancel the coeffs with $deg \geq d+1$. % TODO check which is the name of this theorem or why this is true
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Let $g(x)=a \cdot x^{d+1}, ~~ h(x)=b \cdot x^{d+1}$, and set $f(x) = g(x) + \alpha h(x)$.
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Imagine that P can chose $\alpha$ such that $a x^{d+1} + \alpha \cdot b x^{d+1} = 0$, then, in $f(x)$ the coefficients of degree $d+1$ would cancel.
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@@ -314,14 +314,15 @@ $$g(z) = (f(z)-f(r))\cdot (z-r)^{-1}$$
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\section{STIR (main idea)}
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\emph{Update from 2024-03-22, notes from Héctor Masip Ardevol (\href{https://hecmas.github.io/}{https://hecmas.github.io}) explanations.}
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\emph{Update from 2024-03-22, notes from Héctor Masip Ardevol\\
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(\href{https://hecmas.github.io/}{https://hecmas.github.io}) explanations.}
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\vspace{0.3cm}
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Let $p \in \mathbb{F}[x]^{<n}$.
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In FRI we decompose $p(x)$ as
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$$p(x) = p_e(x^2) + x \cdot p_o(x^2)$$
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with $p_e, p_o \in \mahtbb{F}[x]^{<n}$ containing the even and odd powers respectively.
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with $p_e, p_o \in \mathbb{F}[x]^{<n}$ containing the even and odd powers respectively.
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The next FRI polynomial is
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$$p_1(x) = p_e(x) + \alpha p_o(x)$$
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@@ -329,7 +330,7 @@ for $\alpha \in^R \mathbb{F}$.
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In STIR, this would be $q(x)=x^2$,
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$$Q(z,y) = p_e(y) + z \cdot p_o(y)$$
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and then, $p(x) = Q(x, q(x))$. And $Q$ fullfills the degree from Fact 4.6 from the STIR paper.
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and then, $p(x) = Q(x, q(x))$. And $Q$ fulfills the degree from Fact 4.6 from the STIR paper.
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We can generalize to a $q$ with bigger degree, or with another shape, and adapting $Q$ on the choice of $q$.
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