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| stats | profile views | 46 |
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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 |
awarded | Critic |
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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 |
awarded | Nice Answer |
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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. |
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Dec 16 |
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Can someone explain what this paragraph is saying more clearly? Suppose 2 divides $xy$, where $x$ and $y$ are integers. Then the prime factorisation of $xy$ contains a 2. Then either $x$ is divisible by 2, or $y$ is (or both). Because, if neither of them is divisible by 2, then neither of them has $2$ in their prime factorisation, so when you multiply them together, the prime factorisation still doesn't have a 2 in. This contradicts the fact that the prime factorisation of $xy$ does have a 2 in (because prime factorisation is unique). Now let $x = y = a$. |
