# plm

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bio website location age member for 2 years, 6 months seen Mar 10 at 15:31 profile views 68

My name is Paul Le Meur. I live in Bordeaux, France.

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 May24 comment Convex functions@Egbert, $x^4$ is a strictly convex function with vanishing 2nd derivative at $0$. May23 comment 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.) May23 comment 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? May23 comment Primitive recursive functions, Recursive functions and recursive setI 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. May23 answered Primitive recursive functions, Recursive functions and recursive set May23 comment Representations of Central ProductsWhat 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. May23 answered Pen, pencils and paper to write math May22 comment Variety vs. ManifoldAlso 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.) May22 comment 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. May22 comment Counting multiplicities and Bezout's theoremI 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. May22 comment Concrete examples of 2-categoriesThe 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. May22 comment Concrete examples of 2-categoriesQiaochu 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. May22 comment 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. May20 comment Unique decomposition of a mapping by an equivalence relationEquality 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 $g([x])$. May20 comment Unique decomposition of a mapping by an equivalence relationAre you kidding? What equivalence relation does zell "prescribe"? May19 comment Polynomial vector spaceActually André Nicolas's answer is perhaps the better way to proceed -I am referring to the comment I made on my answer. May19 comment Polynomial vector spaceI 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. May19 comment 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.) May19 comment Polynomial vector spaceI 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. May19 answered Polynomial vector space