Andrew D. Hwang
Reputation
83/100 score
 12m answered Interchange of limits of definite integral 26m revised Found a New Golden Ratio Construction with Equilateral Triangle, Square, and Circle. Geometric/Trigonmetric proof? Fixed droll cut-and-paste error, added diagram. 10h answered Found a New Golden Ratio Construction with Equilateral Triangle, Square, and Circle. Geometric/Trigonmetric proof? 10h comment Find all $g(x)$ such that $f(g(x))=g(f(x))$ 10h comment Is $\operatorname{Mat}(2,\mathbb{R})$ isomorphic to $\operatorname{Mat}(3,\mathbb{R})$? "I cannot seem to find any bijection that will do" suggests you believe the answer is "the two are isomorphic". Does that seem reasonable intuitively...? 10h answered Is a one-dimensional vector space orthogonal? 23h answered What is the reason for why we can't have general formulas for epsilon-delta proofs 1d comment A space curve with non-vanishing curvature is planar iff its torsion is 0 One approach is to let $\nu$ denote a unit normal to the plane of $\gamma$, so that $(\gamma(t) - \gamma(t_{0})) \cdot \nu = 0$ for all $t$. Differentiating twice tells you the binormal of $\gamma$ is $\pm \nu$, hence constant. 1d answered How to properly find supremum of a function $f(x,y,z)$ on a cube $[0,1]^3$? 1d comment Did Feynman mentally compute $\sqrt[3]{1729.03}$ by linear approximation? You're asking about how a mental computation was accomplished in a scene from a fictional movie...? (Just clarifying; I'm not willing to click a youtube link to get the content of the question.) 1d comment How to properly find supremum of a function $f(x,y,z)$ on a cube $[0,1]^3$? Your first equality is false (albeit not in a way that conflicts with the stated conclusion), since (say) $f(1, 1, z) = 0$ for $0 \leq z < 1$. 1d comment A space curve with non-vanishing curvature is planar iff its torsion is 0 The opposite direction is the "easy" direction. :) At worst, run your argument backward. 1d comment Holomorphic function on a domain has a primitive In case it's helpful, I gave a geometric/analytic explanation for this particular example at Why does $z^{−1}$ not have an anti derivative?. 1d answered Holomorphic function on a domain has a primitive 1d comment What is concretely a vector field? Related: Understanding tangent vectors. (Similar questions from a variety of angles have been asked many times over the years; perusing the "related" links may be helpful.) For the record, "yes, $X(0,0,1) = (0,0,1)$ is not tangent to the unit sphere", and a vector $v$ at a point $x$ defines a directional derivative operator, $$f \mapsto \frac{d}{dt}\bigg|_{t=0} f(x + tv).$$Thus, $\partial_{x} = e_{1}$ and $\partial_{y} = e_{2}$, just as you say. (Technically, $\partial_{x}$ is the constant vector field whose value is $e_{1}$ at each point.) 1d comment Refuting the Anti-Cantor Cranks Here also you can argue contrapositively, without making the initial controversial hypothesis: Let $m$ and $n$ be arbitrary positive integers. Each of $m^{2}$ and $2n^{2}$ factors uniquely into primes. However, the exponent of $2$ in $m^{2}$ is even, while the exponent of $2$ in $2n^{2}$ is odd. Since no integer is both even and odd, the factorizations of $m^{2}$ and $2n^{2}$ are distinct, and consequently $m^{2} \neq 2n^{2}$. Needless to say, I prefer this to "descent-type" arguments. :) 2d answered Equation of a Riemann surface? 2d comment Can you use both sides of an equation to prove equality? @BLAZE: An "identity" (such as $\cos^{2} x + \sin^{2} x = 1$) is just an equation that holds for all values of one or more variables (possibly with a "small number of exceptions or restrictions"), so "yes", any proof technique that works for numerical equations also works for identities. 2d answered Unique solution for ode $y' = {\sqrt{1-y^2}}$ 2d answered Idempotent and nilpotent matrices are defined differently. Why?