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English teacher and current math student.


Nov
23
comment Minimal polynomials and cyclic subspaces
@JeremyDaniel: I almost said no. The eigenvectors are $v_1$ and $v_2$, and we can't get from one to the other. Then I had a look at $v_1 + v_2$; its image is linearly independent. And so if by "cyclic" we mean the whole space is generated by one vector, then I think this proves it is :). The problem is, I'm not sure why I chose that linear combination besides the fact that I've seen similar choices made before. As for why one does that, no idea. In particular, the connection between what I just did and the eigenvectors eludes me. Can you recommend a source, online or otherwise?
Nov
23
comment Minimal polynomials and cyclic subspaces
@Timbuc: Thanks for the hint. Are you saying, for example, that $V$ might be written as a direct sum of, say, two cyclic subspaces, but that one of those subspaces might be contained in the other?
Nov
23
revised Minimal polynomials and cyclic subspaces
Made headline more precise
Nov
23
asked Minimal polynomials and cyclic subspaces
Oct
21
comment If $f$ is increasing, then for all $n\in\mathbb{N}$ there exists $P_n$ : $U(f,P)-L(f,P) \leq (b-a)/n$
@Did: Second response: As someone who almost went for a linguistics degree before he went for a math degree, there's something kinda annoying about that :). (See this.) Granted, I have a constant times $n$, but it's not $n$. I assume, in general, it's impossible to do with $1\cdot n$?
Oct
21
comment If $f$ is increasing, then for all $n\in\mathbb{N}$ there exists $P_n$ : $U(f,P)-L(f,P) \leq (b-a)/n$
@Did: Thought about this on the ride home from work. First of two responses: I think I see where you're heading. Let $D = \lfloor f(b) - f(a) \rfloor + 1$. Define $P_n = \{ a + i((b-a)/Dn) : 0 \leq i \leq Dn \}$. Then everything seems to work out.
Oct
21
asked If $f$ is increasing, then for all $n\in\mathbb{N}$ there exists $P_n$ : $U(f,P)-L(f,P) \leq (b-a)/n$
Oct
19
comment For groups and orders show $\forall a,b\in G: |ab|=|ba|$ and $\forall G$ with $|G| = p^n$ for $p$ prime has a subgroup of order $p$
@user149868, Apologies, apparently I got some wires crossed when I was answering the second portion. I've gone ahead and deleted that part, since it wasn't helpful. But to answer your question: If $|G|=p$, it is indeed cyclic. If $|G|=p^n$, though, it need not be cyclic; for example, $|V_4|=2^2$ and $|D_8|=2^3$, but neither is cyclic.
Oct
19
revised For groups and orders show $\forall a,b\in G: |ab|=|ba|$ and $\forall G$ with $|G| = p^n$ for $p$ prime has a subgroup of order $p$
deleted the nonsense
Oct
16
answered For groups and orders show $\forall a,b\in G: |ab|=|ba|$ and $\forall G$ with $|G| = p^n$ for $p$ prime has a subgroup of order $p$
Sep
30
awarded  Explainer
Sep
24
awarded  Autobiographer
Sep
16
comment How to find system of equations from solution space
@almagest, I'm working on a similar problem, and I have a few questions. First, why should we expect just one equation? Second, I follow up until $8w + 2x - y = 0$; where does that come from?
Sep
3
asked Prove that $p \mapsto \inf\{d(p,s) : s \in S\}$ is uniformly continuous
Aug
24
revised Positive integer solutions to $p^2 + q^2 \leq 4^k$
added 159 characters in body
Aug
24
comment Positive integer solutions to $p^2 + q^2 \leq 4^k$
@flawr: Thank you. I'll edit my post: What you write is certainly true, and it "systematic," to use my original word; what I meant, though -- at four in the morning :) -- was: Is there a closed form for the answer?
Aug
24
comment Positive integer solutions to $p^2 + q^2 \leq 4^k$
@EwanDelanoy: Thanks for your response. I know what Riemann sums are; I've never done them, like, rigorously, though. (The problem is from Chap. 1 of Pugh's Real Mathematical Analysis.) I'm not sure it'll fly with my professor if I tie up my proof by referring to some "well-known" result :). Especially since, to be frank, I'm a little skeptical: I wouldn't know how to avoid writing $\lim_{n\rightarrow\infty} S(n) = n\cdot\infty^2$. Suggestions?
Aug
24
asked Positive integer solutions to $p^2 + q^2 \leq 4^k$
Aug
24
accepted Unit disc contains finitely many dyadic squares whose total area exceeds $\pi - \epsilon$
Aug
24
comment Unit disc contains finitely many dyadic squares whose total area exceeds $\pi - \epsilon$
Thanks, Micah. I think your wording is a bit more precise :).