gnometorule
• Member for 10 years, 1 month
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• Santa Monica, CA

## 130 Answers

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26 votes

You're self-motivated; the kids you mention as a general rule pushed by parents trying to compensate for what they feel is a lack in their lives. Some will excel; some will flame out as a lot that ...

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20 votes

Hint: note that $f \ast \alpha := \gamma$ is a path from $x$ to $y$, and calculate $\bar{\gamma}$ by its definition. Then you are given:   $\hat{\gamma} ([g])= \hat{\alpha} ([g])$,   which should ...

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17 votes

I can actually relate fairly well to this. Please don't take this the wrong way, but my main advice would be: Stop worrying! By this I don't mean that you should not be concerned about your grades ...

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Accepted answer
15 votes

\begin{align} \sqrt{n^2 + n} - n & = (\sqrt{n^2 + n} - n) \cdot \frac{\sqrt{n^2 + n} + n}{\sqrt{n^2 + n} +n} \\ & = \frac{n^2+n-n^2}{\sqrt{n^2 + n} + n} \\ & = \frac{1}{\frac{\sqrt{n^2 + ... View answer 15 votes Letf(x) = x, g(x) = -x$, both obviously invertible. Then$(f+g)(x) == 0$, which is not invertible. View answer 13 votes The classic book reference is probably Polya's "How to solve it". Maybe have a look at it. View answer Accepted answer 9 votes For a full solution, proceed like this:$n=1$: $$\sum_{i=1}^1 \frac{1}{(2i-1)(2i+1)} = \frac{1}{(2-1)(2+1)} = \frac{1}{3} = \frac{1}{2 \cdot 1 +1},$$ so it holds for$n=1$. Assume next that it ... View answer Accepted answer 9 votes Use the equivalent definition of local compactness: if$x \in U \subset C$,$U$open,$C$compact, then there must be a ball$B_{\epsilon}(x)$whose closure$\bar{B}$is contained in$U. Apply this ... View answer 8 votes This popped as a newsletter question, and it seems appropriate to share personal experience. It is true that most significant math research seems to be done before 30. I remember Hirzebruch from my ... View answer Accepted answer 8 votes It could be a determinant, but more likely it is a matrix norm (in particular engineering rings a bell as it could relate to some numerical analysis; but I don't know of course for sure). View answer Accepted answer 7 votes There are no such connections. View answer Accepted answer 7 votes While Munkres is arguably self-contained, there are further topics (on top of the above mentioned) you'll soon run into: (1) Free Groups (abelian and not). (2) Commutator subgroups, and (from ... View answer 6 votes Yes, the 1 cannot vanish. In your case, using the distributive law, \begin{align} \frac{9x + 8x^2 +1}{x} & = \frac{9x}{x} + \frac{8x^2}{x} + \frac{1}{x} \\ & = 9 + 8x + \frac{1}{x} \\ & \... View answer 6 votes Your question got me reading (googling). This paper: Leibniz's Infinitesimals: Their fictionality, their modern implementations, and their foes from Berkeley to Russel and beyond (Katz & Sherry, ... View answer 5 votes I can add anecdotal evidence for my former program, which wasn't math proper but attracted a fair number of math undergrads, if not people with Ph.D.s in physics prior to even entering this particular ... View answer Accepted answer 5 votes I think your central question is how to write z asz = \sum_{j=1}^D (u_j^t z) \, u_j = \sum_{j=1}^D \langle u_j, z \rangle u_j \quad (1)$$This, you can see as follows. As you have a complete, ... View answer Accepted answer 5 votes Hint: Assume without loss of generality x < y. You need to find one interval I_1 = [a, b) in which x lies, and one I_2 = [c, d) in which y lies, such that a < b \leq c < d. These ... View answer 5 votes Elementary way: For g simply take the identity e. To find another, assume that each element h has an inverse h^{-1} that is not h (h \neq h^{-1}). Summing the elements \{h, h^{-1} \} ... View answer 5 votes In Leibnitz' case (who co-invented) by (quite wrongly) assuming that the Greek atomos (indivisible) really was that: a smallest indivisible part he called "Monaden", and which also explains that his ... View answer 5 votes It's the binomial coefficient, and you should google combination and permutation as a starter. View answer 4 votes Apply your condition f(y) < y to y:=f^{-1}(x), and you immediately get$$x = f(f^{-1}(x))<f^{-1}(x)$$View answer 4 votes Let$$f(x) := \begin{cases} 1, & x\in \mathbb{Q} \\ 0, & x\in \mathbb{R} - \mathbb{Q}, \\ \end{cases}$$a very discontinuous function. Then \epsilon := 2 will do, for any x, t, \delta >0... View answer 4 votes If you use f^{-1}(U \cup V) = f^{-1}(U) \cup f^{-1}(V), which is always true, your idea will show a contradiction in the next step. Do you see it? View answer 3 votes To reconcile this with the general case, note that each block matrix addition adds two matrices of dimension \frac{n}{2} \times \frac{n}{2}, so of \frac{n^2}{4} elements, which is the complexity ... View answer Accepted answer 3 votes No, you're entirely right, and you probably just copied it down incorrectly. Here's the one point you're currently missing:$$<x, x> = ||x||^2,$$so if it weren't the square as you say it ... View answer Accepted answer 3 votes If$$1 = u(n/d) + v(s/d) := ua + vb,$$and for some integer c you have c|a and c|b, then obviously$$c|ua+vb = 1.$$Edit (in more detail per a former comment): If c|a, then a = cx (not c ... View answer Accepted answer 3 votes It should be$$ e^{-2t}(-\sin(4t) \cdot 4)+ \cos(4t)(e^{-2t}(-2)).$$Your logic is right, but$$\frac{d}{dt}(\cos(4t)) = -4\sin(4t),$$not -4\sin(4t)\cos(4t). View answer 3 votes If x, y \in G, then xyx^{-1}y^{-1} \in C. Hence,$$Cxyx^{-1}y^{-1} = C,$$or$$Cxy = Cyx,$which means that$G/C$is abelian. Edit: corrected a typo pointed out by Loki Clock. View answer 3 votes The key advantage of Ito integrals is that they are martingales. For this, a.s. convergence would not be enough in general. You don't quite need$\mathcal{L^2}\$ convergence, but something almost as ...

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3 votes

If you just want to implement Kalman Filtering, and by this mean 'coding it up' and already have the coding skills, nothing really other than a good reference book. Personally, I learned Kalman ...

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