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Daily Insights from Magic Internet Math courses. Learn at https://mathacademy-cyan.vercel.app
๐Ÿ“– Coordinate Ring The \\textbf{coordinate ring} of an algebraic set $X = V(I)$ is $k[X] = k[x_1, \\ldots, x_n]/I(X)$, where $I(X) = \\{f : f(a) = 0 \\text{ for all } a \\in X\\}$. From: df-course Learn more: Explore all courses:
๐Ÿ“– Group Representation A \\textbf{representation} of a group $G$ over a field $k$ is a group homomorphism $\\rho: G \\to GL(V)$ for some $k$-vector space $V$. Equivalently, it is a $kG$-module structure on $V$. From: df-course Learn more: Explore all courses:
๐Ÿ“– Log-Rank Test The log-rank test compares survival curves between groups by testing whether the hazard functions are equal across groups. From: Intro to Statistical Learning Learn more: Explore all courses:
๐Ÿ“– Zero-Knowledge Argument of Knowledge A protocol satisfying completeness, soundness, and zero-knowledge, where a simulator can produce transcripts without the witness. From: lcn Learn more: Explore all courses:
๐Ÿ“– Central Projection Central projection from a point $O$ (the eye) onto a plane $\\Pi$ maps each point $P$ (not on the line through $O$ parallel to $\\Pi$) to the intersection of line $OP$ with $\\Pi$. From: Four Pillars of Geometry Learn more: Explore all courses:
๐Ÿ’ก Proposition I.12 To a given infinite straight line, from a given point which is not on it, to draw a perpendicular straight line. From: Euclid's Elements Learn more: Explore all courses:
๐Ÿ“ Sample Theorem If $A \\subseteq B$ and $B \\subseteq A$, then $A = B$ Proof: Let $x \\in A$. Since $A \\subseteq B$, we have $x \\in B$ by definition of subset. Therefore, every element of $A$ is in $B$. Now, let $y \\in B$. Since $B \\subseteq A$, we have $y \\in A$ by definition. Therefore, every element of $B$ is in $A$. Since $A \\subseteq B$ and $B \\subseteq A... From: euler-intro Learn more: Explore all courses:
๐Ÿ“ Inscribed Angle Theorem An inscribed angle is half the central angle that subtends the same arc. If $\\angle ACB$ is inscribed and $\\angle AOB$ is the central angle on arc $AB$, then $\\angle ACB = \\frac{1}{2}\\angle AOB$. Proof: Case when one side of inscribed angle passes through center: Let $C$ be on circle, $O$ center, and $CB$ pass through $O$. Triangle $\\triangle OAC$ is isosceles ($|OA| = |OC|$). So $\\angle OCA = \\angle OAC = \\alpha$. Exterior angle: $\\angle AOB = 2\\alpha = 2\\angle ACB$. The general case fol... From: Four Pillars of Geometry Learn more: Explore all courses:
๐Ÿ“– Vector A vector in $\\mathbb{R}^n$ is an ordered $n$-tuple $\\mathbf{v} = (v_1, v_2, \\ldots, v_n)$ of real numbers. Vectors can be added componentwise and multiplied by scalars. From: Four Pillars of Geometry Learn more: Explore all courses:
๐Ÿ“ Rotation Matrix Rotation by angle $\\theta$ about the origin in $\\mathbb{R}^2$ is given by the matrix $R_\\theta = \\begin{pmatrix} \\cos\\theta & -\\sin\\theta \\\\ \\sin\\theta & \\cos\\theta \\end{pmatrix}$. Proof: The point $(1, 0)$ rotates to $(\\cos\\theta, \\sin\\theta)$. The point $(0, 1)$ rotates to $(-\\sin\\theta, \\cos\\theta)$. A general point $(x, y) = x(1, 0) + y(0, 1)$ rotates to $x(\\cos\\theta, \\sin\\theta) + y(-\\sin\\theta, \\cos\\theta)$. In matrix form: $\\begin{pmatrix} x' \\\\ y' \\end... From: Four Pillars of Geometry Learn more: Explore all courses:
๐Ÿ“ Inner Product and Angle For nonzero vectors $\\mathbf{u}$ and $\\mathbf{v}$, $\\mathbf{u} \\cdot \\mathbf{v} = |\\mathbf{u}| |\\mathbf{v}| \\cos\\theta$, where $\\theta$ is the angle between them. Proof: By the law of cosines in the triangle with sides $\\mathbf{u}$, $\\mathbf{v}$, $\\mathbf{u} - \\mathbf{v}$: $|\\mathbf{u} - \\mathbf{v}|^2 = |\\mathbf{u}|^2 + |\\mathbf{v}|^2 - 2|\\mathbf{u}||\\mathbf{v}|\\cos\\theta$. Expanding: $|\\mathbf{u} - \\mathbf{v}|^2 = (\\mathbf{u} - \\mathbf{v}) \\cdot... From: Four Pillars of Geometry Learn more: Explore all courses:
๐Ÿ“ Irrationality of $\\sqrt{2}$ $\\sqrt{2}$ is irrational: there exist no integers $m, n$ with $\\sqrt{2} = \\frac{m}{n}$. Proof: Suppose $\\sqrt{2} = \\frac{m}{n}$ with $\\gcd(m,n) = 1$. Then $2 = \\frac{m^2}{n^2}$, so $m^2 = 2n^2$. Thus $m^2$ is even, so $m$ is even. Write $m = 2k$. Then $4k^2 = 2n^2$, so $n^2 = 2k^2$, meaning $n$ is also even. This contradicts $\\gcd(m,n) = 1$. Therefore $\\sqrt{2}$ is irrational. From: Four Pillars of Geometry Learn more: Explore all courses:
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