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| visits | member for | 2 years, 6 months |
| seen | Mar 10 at 15:31 | |
| stats | profile views | 68 |
My name is Paul Le Meur. I live in Bordeaux, France.
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May 24 |
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Convex functions @Egbert, $x^4$ is a strictly convex function with vanishing 2nd derivative at $0$. |
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May 23 |
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A cute geometry problem about angle trisectors. dtldarek, thanks. I think I saw it on AoPS actually. @Phira, I did not know I knew so much. :) (I did not know that interesting theorem, thx for the pointer.) |
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May 23 |
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A cute geometry problem about angle trisectors. I remember a similar problem but not where I saw it. Can you remember where you saw this one? |
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May 23 |
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Primitive recursive functions, Recursive functions and recursive set I should have said (explicitly) that at that noncomputable code of a function the operation yields a nonrecursive function. Well, all this is very informal, but I do think it is interesting -and better goes to the crux of the question than simply remarking that some recursive functions defined using minimization, i.e. not directly defined as primitive recursive, can actually be defined as primitive recursive. |
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May 23 |
answered | Primitive recursive functions, Recursive functions and recursive set |
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May 23 |
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Representations of Central Products What do you call a central product? Is it a central extension? That is, a group $E$ such that $C\subset E$ is central, and $E/C=G$ is your original group. |
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May 23 |
answered | Pen, pencils and paper to write math |
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May 22 |
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Variety vs. Manifold Also relevant is that there is no other term available than "variété" in French. I could come up with "multidim", "multiplat", "multidir", "multipart", "multigrandeur", "multicomposé(e)", "multicomp", or "pluri...". But the english language really has that power to accept such neologisms, to make them sound ok, that the french lacks. (Probably in part because english words are usually shorter and more flexible in pronunciation.) |
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May 22 |
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Variety vs. Manifold @QiaochuYuan, nitpicking, differentiable manifolds are called as you say (or "variétés différentiables"), "smooth manifolds" is usually rather "variétés lisses". But you made a very good point that there is a historical component. Also the term "algebraic manifold" is highly used (and useful) to refer to algebraic varieties which are also manifolds. |
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May 22 |
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Counting multiplicities and Bezout's theorem I think the first method to compute intersection multiplicities for all plane curves over the complex numbers was that of resultants. You take the $X$ resultant of 2 polynomials in $X$ and $Y$ which is a polynomial in $Y$, and I think the degree of this polynomial is your (total) intersection multiplicity. This is the appropriate (keeping multiplicities) projection of your intersection of the 2 varieties to 1 dimension, and the number of solutions (with multiplicities) in 1 dimension is easy to compute. en.wikipedia.org/wiki/Resultant . Wiki is not extremely helpful here. |
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May 22 |
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Concrete examples of 2-categories The link: qchu.wordpress.com/2012/02/06/… . The point of this construction is that it yields a purely categorical formulation of the center $Z(G)$ of $G$: the group of endomorphisms of the identity id$_G$. For a general 2-category the set of endomorphisms is only a monoid under vertical 2-composition. Horizontal composition coincides with the vertical one on $Z(G)$, see the blog post. This formalizes the Eckman-Hilton argument and is a major application of the construction. |
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May 22 |
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Concrete examples of 2-categories Qiaochu Yuan has another example of 2-category using groups, or a single group. If you take a group $G$ as object, morphisms its endomorphisms, and 2-morphisms from $f$ to $g$ conjugations by an element $x$ of $G$ which send $f$ to $g$, i.e. $xf(y)x^{-1}=f(y)$ for all $y\in G$. You may take any class of groups and morphisms between them to form a category this way. This construction comes from seeing groups as categories, homorphisms as functors, then "commuting" elements are natural transformations. This must make a group a 3-category in some sense. |
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May 22 |
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how to solve $aX+bX^2=e^{cX}$ Newton's method will converge for "most" values of $a,b,c$ but studying precisely each case could be time-consuming. You may start with $(a,b,c)=(1,1,1)$, the function is convex near its negative $0$, and use a homotopy method to reach all other values of $(a,b,c)$. Matlab or Mathematica will solve this for you efficiently too. I do not think there is an explicit solution in terms of roots exponentials and logarithms but I would be greatly interested if someone could point to possible proofs of this, perhaps using methods of differential algebra. |
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May 20 |
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Unique decomposition of a mapping by an equivalence relation Equality is one example of an equivalence relation. $g$ need not be injective. I think you may be assuming that the equivalence relation is defined by $x\sim y$ iff $f(x)=f(y)$ , but it is not, the implication only goes one way, to allow any element in an equivalence class, say $[x]$ to be taken to define $g([x])$. |
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May 20 |
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Unique decomposition of a mapping by an equivalence relation Are you kidding? What equivalence relation does zell "prescribe"? |
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May 19 |
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Polynomial vector space Actually André Nicolas's answer is perhaps the better way to proceed -I am referring to the comment I made on my answer. |
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May 19 |
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Polynomial vector space I guess I am being bad, doing your homework for you. :) Let's be philosophical and say it is an interesting question knowing the good from the bad in such a situation. But you may add the tag "homework" here and in the future -if I am not mistaken and this is actually homework, of course. |
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May 19 |
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Unique decomposition of a mapping by an equivalence relation @Knashik, since I take equality as equivalence relation the quotient is basically the same set. (Actually it is $\{\{0\},\{1\}\}$, with the standard construction of quotient as the set of equivalence classes, the partition induced by an equivalence relation.) |
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May 19 |
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Polynomial vector space I should have said "Jordan decomposition" instead of "Jordan form", or "Jordan-Chevalley decomposition". This refers to the sum decomposition. The "normal form" is actually writing the corresponding triangular matrix in a basis of generalized eigenvectors. |
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May 19 |
answered | Polynomial vector space |
