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This is related to something which has always annoyed me: some authors define, say, the group algebra $k[G]$ of a group $G$ as the set of functions $G \to k$ with finite support, equipped with convolu …

A thing that consistently surprises me is that many "natural" sequences $f(n)$, even apparently very complicated ones, have growth rates which can be described by elementary functions $g(n)$ (say, to …

Not a complete answer. First, here is an alternate derivation of the result in the finite-dimensional case which might be more enlightening. If $A$ is positive self-adjoint, we can write $A = \exp(L)$ …

One way is to use a combinatorial definition of the derivative. Let $A(z) = \sum a_n z^n$ be a power series. In combinatorics, where $A$ is likely to be an ordinary generating function, $a_n$ is likel …

The two-point discrete space already doesn't have a (universal) connectification, in the sense that two points don't have a coproduct in the category of connected spaces. If $X$ were such a coproduct, …

The first two examples can be described more or less uniformly. Associated to a field $F$ is the category $C_F$ of algebraic field extensions of $F$ (whose objects are morphisms $F \to E$ and whose m …

Paul Siegel mentioned that to teach undergrads the Lebesgue integral you have to spend half the semester on measure theory. This is actually not true. There is a self-contained definition of the Leb …

I haven't really thought this through, but how does one actually compute integrals? For the Riemann integral one can either prove the fundamental theorem and have a large table of derivatives handy, …

No. You can take $f, g$ to be sums of smooth bump functions with disjoint support. As long as $f, g$ each have bumps of the same size and shape you can put them in arbitrary locations (and orientati …

Just a quick observation. It is not hard to see that this is impossible if there exists some $i$ such that for all $j \ge i$, the coefficients of $p(x) r(x)^j$ are all positive after the first positi …

Yes. Suppose otherwise and let $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ be two linearly independent points on the hyperplane which do not intersect the quintic. Then some po …

The standard way of doing problems like these is to look at the coefficients of the Chebyshev polynomials. The polynomial $T_n$ of degree $n$ such that $T_n(2 \cos \theta) = 2 \cos n \theta$ has lead …

A very powerful way to estimate the growth of a recurrence is to look at the analytic properties of the generating function that it implies. In this case we should take the exponential generating fun …

One way to define the algebra of differential forms $\Omega(M)$ on a smooth manifold $M$ (as explained by John Baez's week287) is as the exterior algebra of the dual of the module of derivations on th …

I can satisfy conditions 1, 3 and almost satisfy condition 2. Letting $f_n(t) = {t+n-1 \choose n}$ we have the well-known generating function $\displaystyle \sum_{n \ge 0} f_n(t) x^n = \frac{1}{(1 - …