Just because 2 things are dual, doesn't mean they're just opposites

MOM Alok’s thinking about conic sections again. The Classical Groups The classical groups (that I was thinking about) are: Real Orthogonal Group: \(O(n, \mathbb{R})\). Real-valued matrices \(A\) such that \(A^T A = I\) where \(A^T\) is the transpose of \(A\), and \(I\) is the identity matrix. Complex Orthogonal Group: \(O(n, \mathbb{C})\)....

But this is impossible by definition The title may seem like a contradiction. How can you differentiate something that’s not even continuous? The usual definition of the derivative of a function \(f\) at a point \(a\) is given by the limit: \[f'(a) := \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}\] If \(f\) is differentiable at \(a\), then it is...

Nonstandard analysis gives a nice link between continuous and bounded functions, which are small and large-scale notions.1 Let \(X\) and \(Y\) be topological spaces. Identify them with their nonstandard extensions. A function \(f: X \rightarrow Y\) is continuous if \(x \approx x' \implies f(x) \approx f(x')\). Intuitively, infinitely close points...

Let \(H > \mathbb{Nat}\) be unlimited. Then the linear map \(T(x: \mathbb{R}^*): \mathbb{R}^* := Hx\) is discontinuous. Why, its discontinuity is equivalent to it being unbounded. This holds in general, but this example is the germ of generality. See my previous post for definitions of bounded and continuous. The map \(T\) is unbounded since it...

(written long time ago, publish or languish) These are some notes I made for Davide Radaelli for the first section of Schuller’s lectures on physics. Let’s turn Boolean algebra into something we know better: arithmetic. First we’ll set False to 0 and True to 1. To handle overflow, any arithmetic is mod 2. So even numbers are \(0\) and odd numbers...