📖 Forward Stepwise Selection Start with null model, then repeatedly add the predictor that provides the greatest additional improvement to the fit, until all predictors are in the model. From: Intro to Statistical Learning Learn more:

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💡 Proposition 2.2.6 (Cancellation Law) Let $a$, $b$, $c$ be natural numbers such that $a + b = a + c$. Then $b = c$. From: tao-analysis-1 Learn more:

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📐 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: only-the-strong-survive Learn more:

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📖 Art as Spiritual Fuel Great art provides concrete images of abstract values that inspire and sustain us in the struggle to live by those values. From: Atlas Shrugged Learn more:

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💡 Proposition VI.9 From a given straight line to cut off a prescribed part. From: Euclid's Elements Learn more:

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📖 Spontaneous Order Social coordination emerging from individuals pursuing their own purposes within general rules, producing results that no one designed but serve human needs better than any conscious planning could achieve. From: The Road to Serfdom Learn more:

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An interactive course exploring F.A. Hayek

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📐 Rank-Nullity Theorem (Dimension Theorem) If $V$ is finite-dimensional, then $T(V)$ is also finite-dimensional, and: $\\dim N(T) + \\dim T(V) = \\dim V$. In other words, the nullity plus the rank of a linear transformation equals the dimension of its domain. Proof: Let $n = \\dim V$ and let $e_1, \\ldots, e_k$ be a basis for $N(T)$ where $k = \\dim N(T)$. By Theorem 1.7, these are part of some basis for $V$: $e_1, \\ldots, e_k, e_{k+1}, \\ldots, e_n$ where $k + r = n$. We show that $T(e_{k+1}), \\ldots, T(e_n)$ form a basis for $T(V)$, proving $\\dim T(V) =... From: calc2-course Learn more:

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📐 Kernel and Image are Subspaces For any linear transformation $T: V \\to W$, $\\ker(T)$ is a subspace of $V$ and $\\text{im}(T)$ is a subspace of $W$. Proof: **Kernel:** - $T(\\mathbf{0}) = \\mathbf{0}$, so $\\mathbf{0} \\in \\ker(T)$. - If $u, v \\in \\ker(T)$, then $T(u + v) = T(u) + T(v) = \\mathbf{0}$, so $u + v \\in \\ker(T)$. - If $u \\in \\ker(T)$ and $c \\in F$, then $T(cu) = cT(u) = c \\cdot \\mathbf{0} = \\mathbf{0}$, so $cu \\in \\ker(T)$.... From: Advanced Linear Algebra Learn more:

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📖 Ridge Regression Ridge regression minimizes $\\sum_{i=1}^{n}(y_i - \\beta_0 - \\sum_{j=1}^{p}\\beta_j x_{ij})^2 + \\lambda\\sum_{j=1}^{p}\\beta_j^2$ where $\\lambda \\geq 0$ is the tuning parameter controlling the shrinkage penalty. From: Intro to Statistical Learning Learn more:

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💡 Human Action Principle Human action is based on removing the largest source of dissatisfaction that can be removed at the lowest cost at a given moment. This explains why institutions will pressure executives to sell Bitcoin during adverse conditions. From: bfi Learn more:

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💡 Proposition IV.14 About a given equilateral and equiangular pentagon to circumscribe a circle. From: Euclid's Elements Learn more:

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📖 Equivalence Relation An \\textbf{equivalence relation} on a set $A$ is a relation $\\sim$ satisfying: (1) \\textbf{Reflexive:} $a \\sim a$ for all $a \\in A$; (2) \\textbf{Symmetric:} $a \\sim b$ implies $b \\sim a$; (3) \\textbf{Transitive:} $a \\sim b$ and $b \\sim c$ implies $a \\sim c$. From: df-course Learn more:

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📖 Definition 9.4.1 (Continuity at a Point) Let $X \\subseteq \\mathbb{R}$, $f : X \\to \\mathbb{R}$, and $x_0 \\in X$. We say $f$ is **continuous at $x_0$** iff for every $\\varepsilon > 0$ there exists $\\delta > 0$ such that $|f(x) - f(x_0)| \\leq \\varepsilon$ whenever $|x - x_0| < \\delta$ and $x \\in X$. From: tao-analysis-1 Learn more:

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📐 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: rothbard_controversies Learn more:

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📖 Time-to-Live (TTL) The duration in seconds that a payment registration remains active in Branta\ From: branta Learn more:

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💡 Proposition 4.4.4 (No Rational Square Root of 2) There does not exist a rational $x$ such that $x^2 = 2$. Proof: Suppose for contradiction that $x^2 = 2$ for some rational $x = p/q$ in lowest terms. Then $p^2 = 2q^2$, so $p^2$ is even, hence $p$ is even. Write $p = 2k$. Then $4k^2 = 2q^2$, so $q^2 = 2k^2$, meaning $q$ is also even. But this contradicts $p/q$ being in lowest terms. ∎ From: tao-analysis-1 Learn more:

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💡 Art Should Project the Ideal Great art shows what life could and should be at its best, not wallowing in depravity or meaningless chaos. From: Atlas Shrugged Learn more:

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📐 Subgroup Test $H \\leq G$ iff $H \\neq \\emptyset$ and for all $a, b \\in H$, $ab^{-1} \\in H$. From: intro-discrete Learn more:

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📐 Groups of Order 4 Up to isomorphism, there are exactly two groups of order 4: $\\mathbb{Z}_4$ (cyclic) and $V_4$ (Klein four-group). From: intro-discrete Learn more:

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📖 Group Homomorphism A function $\\phi: G \\to H$ between groups is a \\textbf{homomorphism} if $\\phi(ab) = \\phi(a)\\phi(b)$ for all $a, b \\in G$. It is an \\textbf{isomorphism} if it is also a bijection. From: df-course Learn more:

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