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bio website xavierm02.net
location France
age 21
visits member for 3 years, 5 months
seen 51 mins ago

Oct
18
comment Solving system of equation over distinct finite fields
Since $1\mid 2$, $GF(2)$ is a subfield of $GF(2^2)$ so you could just solve it in $GF(2^2)$. And then check that in the solution vector space, there is a vector where $y_1$ is in $GF(2)$.
Oct
15
comment Discrete mathematics proof that I have been stuck on
To go from $n$ to $n+1$, you add two parallel lines that only intersect one other line at a time. Imagine a place filled with those lines and a line being draw from farm away towards where it can intersect other lines. Each time it will intersect a line, it'll "finish" splitting a region in two. If you can see what I'm talking about, you should be able to figure out the formula. If not, try an example.
Oct
12
comment Show is a norm and a Banach space
@GiulyB : I edited your post. Please check that I didn't change its meaning.
Oct
12
revised Show is a norm and a Banach space
latex............
Oct
9
answered How to make Riemann rearrangement?
Oct
6
comment Ring subset which absorbs but is not an additive subgroup
In $2\Bbb Z$, $2\Bbb Z\setminus \{2\}$
Oct
6
awarded  Popular Question
Oct
5
comment A vector function with polynomial entries
mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html
Oct
5
comment Divisibility of a sum in a ring
Well obviously, being an integral domain isn't enough since it doesn't work in $\Bbb Z$. $2\mid 1+1 \not\implies 2 \mid 1 \land 2 \mid 1$
Sep
30
awarded  Explainer
Sep
26
comment relationship between discrete and continuous time inner product
When you take the limit of the finite dimensional one you get a countable dimension (e.g. polynomial) but your function space has an uncountable dimension. You got things backwards : when defining the Lebesgue integral, you can unify both cases by writing the second one as an integral.
Sep
22
comment How does this card trick work?
It's not a trick. It's just a bijection.
Sep
18
comment Condition for a function $f: \mathbb R \rightarrow \mathbb R$ being right or left-continuous at $a \in \mathbb R$.
$f_{\mid (-\infty,a]}:(-\infty,a] \to \Bbb R$ left-continuous at $a$ $\iff$ $f_{\mid (-\infty,a]}:(-\infty,a] \to \Bbb R$ continuous at $a$
Sep
18
comment Condition for a function $f: \mathbb R \rightarrow \mathbb R$ being right or left-continuous at $a \in \mathbb R$.
Hint: $f:\Bbb R \to \Bbb R$ left-continuous at $a$ $\iff$ $f_{\mid (-\infty,a]}:(-\infty,a] \to \Bbb R$ left-continuous at $a$ $\iff$ ...
Sep
7
comment Do the functions have monotone on $\mathbb{R}$ a vector space?
$e^x+e^{-x}$: wolframalpha.com/input/…
Sep
7
comment Do the functions have monotone on $\mathbb{R}$ a vector space?
No. It doesn't work for $f+g$.
Sep
5
comment How prove $g(x)$ is odd function :$g(x)=-g(-x)$
Well you can already set $y:=-x$ to get that their squares are equal. And then, I suppose the continuity hypothesis must be used to prove they have same sign.
Sep
4
comment self-adjoint operator without eigenvalues?
Could you please give $d$ explicitely? It shouldn't be possible.
Sep
2
comment Divergence of $\sum \frac{(-1)^n}{\sqrt{n}+(-1)^n}$
I'd assume it converges, group terms by pairs and try to bound the sum of each pair of terms.
Aug
31
comment The Shortest Distance Between 2 Points On The Earth
@Nick : Well. It seems obvious to me because rotations are isometries so moving an arbitrary point to a fixed one doesn't change the length of the straight line. But I wouldn't be able to justify that much further.