53,426 reputation
341109
bio website www-math.univ-poitiers.fr/…
location Poitiers, France
age 55
visits member for 3 years, 2 months
seen 3 hours ago

professor of mathematics at the Université de Poitiers (France)


3h
comment Is the product of two monotone sequences monotone?
Well "no" or "not always" don't require $2\times2$ cases. And in fact it is "no" or "not always" in all four cases, so distinguishing them is not terribly productive. If you are asking necessary and/or sufficient conditions for the product of monotone sequences to be monotone, then you should says so (and specify which you want). With a vague question you are likely to get not very useful answers, or none at all.
6h
comment Is the product of two monotone sequences monotone?
Your title and initial question suggest the answer should be too short for the lower character limit on this site (at least 30 characters). If your actual question is the one in the penultimate paragraph, then please make it clear that is the (only) question you are asking.
7h
comment What is a “supplementary subspace”?
I think "complementary subspace" (Eng.) = "sous-espace supplémentaire" (French), and this was badly translated. The highlighted statement would just be the definition of supplémentaire.
8h
comment Suppose $f : \mathbb{R}\to\mathbb{R}$ is continuous and not bounded above or below. Show $f(\mathbb{R})=\mathbb{R}$
@Sidrow: I don't understand your formulation. For any given $y$ there are $a$ with $f(a)>y$ (since $f$ is not bounded above by $y$) and $b$ with $f(b)<y$ (since $f$ is not bounded below by $y$). Now you can apply the IVT on the interval $[a,b]$ or $[b,a]$ (whichever has its bounds in the right order). That's all there is to it.
8h
comment Suppose $f : \mathbb{R}\to\mathbb{R}$ is continuous and not bounded above or below. Show $f(\mathbb{R})=\mathbb{R}$
@Quickbeam2k1: Your nitpicking is out of place for this perfectly clear natural language construct. If I would say "I did not break the object intentionally or accidentally", then nobody would construe that to mean "either I did not break the object intentionally, or I did not break the object accidentally" (which would be a tautology). Your interpretation of the title is doing just that. The obvious interpretation is: $f$ is neither bounded above nor bounded below".
10h
comment An easy example of a non-constructive proof without an obvious “fix”?
@StevenStadnicki: I agree. But continued fractions are just as discontinuous as decimals are, of course.
14h
awarded  Nice Answer
1d
comment Suppose $f : \mathbb{R}\to\mathbb{R}$ is continuous and not bounded above or below. Show $f(\mathbb{R})=\mathbb{R}$
Use the intermediate value theorem. You can do it.
1d
answered Is there a notation for being “a finite subset of”?
1d
comment Eigen value system? solution
@Omnomnomnom: The question states that $B$ is invertible. For me that is an indication the matrices (except $W$) are square.
1d
comment An easy example of a non-constructive proof without an obvious “fix”?
There is a fundamental difference between being constructively provable and having an efficient construction. In the purely finite domain dealt with in cryptography everything that exists does so constructively, by trivial brute force constructions. In other words this answer is not an answer to the question.
1d
comment An easy example of a non-constructive proof without an obvious “fix”?
@StevenStadnicki: While it is probably still true constructively that for any countable list of real numbers a real number not on that list can be produced, I do think Cantor's argument needs some modification constructively, because one cannot contruct decimals for arbitrary real numbers (that being a discontinuous operation). Cantor's argument does apply without much change to the more basic form: for any countable list of sets of natural numbers, a set of natural numbers not on that list can be produced.
1d
comment An easy example of a non-constructive proof without an obvious “fix”?
@Hanno: What you mention is an instance of the more general fact that discontinuous functions of the real numbers cannot even be defined constructively (the sign function, or the integral part, or in your example Jordan type of rank of that matrix being examples). But I don't think the statement of the IVT suffers from requiring anything discontinuous; it makes perfect sense, but just in not true constructively.
1d
revised Number of unit squares that meet a given diagonal line segment in more than one point
added 290 characters in body
1d
comment Number of unit squares that meet a given diagonal line segment in more than one point
The formula is correct, but I find the explaing text very confusing. Specialising to the example "The diagonal line will cross $8$ squares and $12$ squares, but for $4$ occasions will accomplish the transition through a corner." The sentence does not even really make sense (a transition is not a square) but I have a hard time seeing those $8$ and $12$ squares. And in the diagram the diagonal clearly goes through $5$ lattice points (corners), or $3$ if one does not count the end points (actual corners); the only way one can get $4$ is counting the end points half, or making an asymmetric choice
1d
comment Number of unit squares that meet a given diagonal line segment in more than one point
@RakeshBalhara: No, what I count is the number of passages into a new square. Since you already are in a square right from the start, you should add $1$, giving $l+b-1$ for the relatively prime case.
1d
answered Number of unit squares that meet a given diagonal line segment in more than one point
1d
revised Number of unit squares that meet a given diagonal line segment in more than one point
deleted 2 characters in body; edited title
2d
comment Simplifying a direct sum $\mathbf{3}\oplus\mathbf{3}\oplus\mathbf{2}$ etc
How do you define directs sums / tensor products of numbers?
2d
revised A linear operator $T: V \rightarrow V$ commuting with all linear operators is a scalar multiple of the identity.
Better title