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| visits | member for | 8 months |
| seen | 39 mins ago | |
| stats | profile views | 46 |
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Mar 21 |
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For a given metric space, how to show that $d(f(x),x) \ge \epsilon$? First of all, prove the hint (you'll need that X is compact - is this in the question?). Now, suppose there is not some e > 0 such that g(x) >= e for all x. Then, for all e' > 0, there is some x such that g(x) < e'. As g is continuous, it will attain both its infimum and its supremum... |
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Mar 21 |
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Is the free group on an empty set defined? I can't work out from the comments and postscripts whether there's still an issue here, but while the free product and the direct product match up in certain small cases, it's still far more honest to call it a free product, since it's a free product by definition and a direct product by accident. |
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Mar 21 |
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Linearly independent set can be completed to a basis $\{e_2\}$ is a linearly independent set, yes. Extensions to a basis, as you correctly guess, are usually very non-unique. Given a linearly independent set, look at the vector space W spanned by it; if this space is equal to V, then your linearly independent set is a basis; if it's not equal to V, then pick a vector in V but not in W. Notice that it's linearly independent to all of the rest. Throw it into your linearly independent set, and repeat. Can you prove that you'll eventually span all of V? (This argument only works for finite-dimensional vector spaces.) |
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Mar 21 |
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Elliptic curve question Given two points P and Q, how do I find their sum P+Q geometrically? Remember that "P is a point of order 2" means "P+P = 0", so if that's true, what geometric restriction does that place on P? |
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Mar 21 |
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Analytic Geometry question (high school level) The vertex of the parabola (in this case, (0, -4)) should be half way between the focus and the directrix, right? So let f = (0, -4-a) and d = (0, -4+a). For some value of a, the point f will be your focus, and the point d will be a point on your directrix. You don't yet know what a is, so pick a point on your parabola (e.g. (1, -3)) and work out its distance from f and its distance from the directrix. They should be equal, and this will hopefully give you an equation for a. |
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Jan 2 |
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Proving the group of homomorphisms is isomorphic to matrices What does it mean to say "f is given by A"? I think if you can answer this, then you can check easily using linearity (why is f+g linear?) that (f+g)(x) and (A+B)(x) are the same for every $x\in R^n$. |
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Jan 2 |
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Show f is uniformly continuous on $(a,b)$ if it is continuous and $\lim\limits_{x\to a^+}f(x)$ and $\lim\limits_{x\to b^-}f(x)$ exist I know what you're trying to do with the line $[a+\epsilon, b-\epsilon] \Leftrightarrow (a,b)$, but as Olivier said, it doesn't work. (Take $a = -\pi/2, b = \pi/2, f(x) = \tan(x)$ for a counterexample. Of course, the limits don't exist here as required.) Here's a hint: what do you know about continuous functions on closed intervals? Once you've answered that, can you find a large closed interval on which f is continuous? |
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Jan 2 |
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The order of the group Yes. (The smallest order of such a group is 2. There are many other bigger groups which satisfy this - for example, take a cyclic group of order 500 generated by $g$, and set $y = g^{250}$.) |
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Jan 2 |
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The order of the group (Notice - in case it's confusing - that I got $y = y$ by taking the relation $yx = x^4 y$ and setting $x = e$, as you worked out yourself a few minutes ago.) |
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Jan 2 |
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The order of the group Great. So you have a group with elements e (of order 1) and y (of order 2), with $y^2 = e$ and $y = y$. What is the smallest possible such group? |
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Jan 2 |
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The order of the group If x = e, plug it back into the relations you are given. Can you find out anything new about y? |
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Dec 23 |
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Given the result, follow de sequence Nice question! Do you know anything about "geometric series" (geometric sequences, geometric progressions)? If not, look them up, and see if you can work out how they relate to your question. |
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Dec 20 |
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Projection matrix equation What is $A^{-1}$? |
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Dec 20 |
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How do I implement an induction argument to show that we may assume that a partition $Q$ of $[a,b]\in{R}$ which refines $P$ has only one more point? There really is no (non-trivial) induction here. What you're trying to prove is "for all n, if Q is a refinement of P by n points, then (1) holds". Obviously (or as $\leq$ is transitive if you like) it suffices to prove this for n = 1 and then just repeatedly add points in. |
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Dec 20 |
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Matrix solving problem @FrederikBrinckJensen Given the question as stated, I see no reason for those plus signs to pop up. Perhaps you've written it wrongly. Can you go back a step or two and tell us how you got to this equation? Also, what makes you think y = 0.006 is correct? |
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Dec 20 |
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Matrix solving problem @FrederikBrinckJensen Is your right hand side the scalar 0 (in which case you have matrix = scalar, which is nonsense), or the matrix 0 (in which case every entry is equal to 0, in particular x = 0, y = 0, and 0.004 = 0(!))? |
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Dec 20 |
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Some arithmetic of fractional part of a irrational number Because $[\cdot]$ only returns integers. |
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Dec 20 |
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Some arithmetic of fractional part of a irrational number $[x\pm\delta]$ is an integer, so $n[x\pm\delta]$ is an integer, so $(n[x\pm\delta]) = 0$, so both claims seem to be false, unless I've misunderstood. |
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Dec 18 |
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Equality such that induction step is valid but basis is not? It's not clear to me exactly in what way you'd like the base case to fail... |
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Dec 17 |
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Find the number of homomorphisms Yes, that's exactly right. f(1) is all that matters, because 1 generates the whole group. |
