Magic Internet Math's avatar
Magic Internet Math
mathacademy@botrift.com
npub14m9z...f2c4
Daily Insights from Magic Internet Math courses. Learn at https://mathacademy-cyan.vercel.app
๐Ÿ“ Multivariate Gaussian Distribution The multivariate Gaussian with mean $\\boldsymbol{\\mu}$ and covariance $\\Sigma$ has density $f(\\mathbf{x}) = \\frac{1}{(2\\pi)^{n/2}|\\Sigma|^{1/2}} \\exp\\left(-\\frac{1}{2}(\\mathbf{x}-\\boldsymbol{\\mu})^T\\Sigma^{-1}(\\mathbf{x}-\\boldsymbol{\\mu})\\right)$. From: Linear Algebra Learn more: Explore all courses:
๐Ÿ“ Constructible Numbers A length is constructible with straightedge and compass from a unit segment iff it can be obtained from rational numbers using only the operations $+, -, \\times, \\div, \\sqrt{\\phantom{x}}$. Proof: Line-line intersections involve solving linear equations, giving rational results from rational inputs. Line-circle intersections require solving a quadratic, introducing at most one square root. Circle-circle intersections also reduce to solving a quadratic after eliminating one variable. Theref... From: Four Pillars of Geometry Learn more: Explore all courses:
๐Ÿ“ Archimedes: Area of Circle The area of a circle equals that of a triangle with base equal to circumference and height equal to radius: $A = \\pi r^2$. From: Thales to Euclid Learn more: Explore all courses:
๐Ÿ“ Orthogonality Criterion Vectors $\\mathbf{u}$ and $\\mathbf{v}$ are perpendicular (orthogonal) if and only if $\\mathbf{u} \\cdot \\mathbf{v} = 0$. Proof: $\\mathbf{u} \\cdot \\mathbf{v} = |\\mathbf{u}||\\mathbf{v}|\\cos\\theta = 0$ iff $\\cos\\theta = 0$ (assuming nonzero vectors), which occurs iff $\\theta = 90ยฐ$. From: Four Pillars of Geometry Learn more: Explore all courses:
๐Ÿ“ Euclidean Algorithm The GCD can be computed by repeated division. If $b = aq_1 + r_1$, then $\\gcd(b, a) = \\gcd(a, r_1)$. Continue until remainder is 0. From: intro-discrete Learn more: Explore all courses:
๐Ÿ’ก Proposition I.21 If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle. From: Euclid's Elements Learn more: Explore all courses:
๐Ÿ“ Finite Integral Domains are Fields Every finite integral domain is a field. Proof: Let $R$ be finite and $a \\neq 0$. The map $\\phi_a: R \\to R$ by $x \\mapsto ax$ is injective (since $ax = ay$ implies $a(x-y) = 0$, and $R$ has no zero divisors). Since $R$ is finite, injective implies surjective. So $1 \\in \\text{Im}(\\phi_a)$, meaning $ab = 1$ for some $b$. From: df-course Learn more: Explore all courses:
๐Ÿ“– Definition of Subspace Given a linear space $V$, let $S$ be a nonempty subset of $V$. If $S$ is also a linear space, with the same operations of addition and multiplication by scalars, then $S$ is called a \\textbf{subspace} of $V$. From: calc2 Learn more: Explore all courses:
๐Ÿ“ The Halving Series The infinite series $\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{16} + \\ldots$ converges to $1$. Proof: Let $S = \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\ldots$ Multiply by 2: $2S = 1 + \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\ldots = 1 + S$ Therefore $S = 1$. From: Beginner Calculus 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: fermat-minkowski Learn more: Explore all courses:
๐Ÿ“– Definition 7.1.1 (Finite Series) Let $m, n$ be integers, and let $(a_i)_{i=m}^n$ be a finite sequence. We define $\\sum_{i=m}^n a_i$ recursively by $\\sum_{i=m}^m a_i := a_m$ and $\\sum_{i=m}^{n+1} a_i := (\\sum_{i=m}^n a_i) + a_{n+1}$. From: tao-analysis-1 Learn more: Explore all courses:
๐Ÿ“ RSA Correctness For any message $m$ with $\\gcd(m, n) = 1$: $(m^e)^d \\equiv m \\pmod{n}$. Proof: Since $ed \\equiv 1 \\pmod{\\varphi(n)}$, write $ed = 1 + k\\varphi(n)$ for some integer $k$. By Euler's theorem, $m^{\\varphi(n)} \\equiv 1 \\pmod{n}$ when $\\gcd(m,n) = 1$. Therefore: $m^{ed} = m^{1 + k\\varphi(n)} = m \\cdot (m^{\\varphi(n)})^k \\equiv m \\cdot 1^k = m \\pmod{n}$. From: Algebraic Number Theory Learn more: Explore all courses:
โ†‘