1,452 reputation
317
bio website kmlinux.fjfi.cvut.cz/…
location Prague, Czech Republic
age 27
visits member for 2 years, 1 month
seen 2 days ago

Math PhD student at Czech Technical University in Prague and at LIAFA in Paris. At the same time, I'm a typesetter (and partly the copy editor) of one scientific journal (done in LaTeX, of course).

Code licence details applicable on my posts on TeX.SX.

My favourite queries on SE's Data Explorer


Mar
2
comment How do I setup the lagrangian for this problem?
@Spacey This is analytical. You always need to treat the boundary seperately. Remember that $y(x)=2x$ has to global maximum, still it has a maximum in $[0,1]$, attained at $2$. You can actually construct Lagrangians for each of the boundary point, but the domains of these are single points, so it's a non-sense to do.
Mar
2
comment How do I setup the lagrangian for this problem?
Well, you simply search (by the derivatives method) for maxima inside $(0,2\pi/3)$, let's call it $y_M$. Then the maximum you look for is simply $\max(y_M, y(0), y(2\pi/3))$.
Mar
2
answered How do I setup the lagrangian for this problem?
Feb
26
accepted Set of all $a\in\mathbb Z$ that are coprime to $b\in\mathbb Z$
Feb
26
comment Set of all $a\in\mathbb Z$ that are coprime to $b\in\mathbb Z$
That's probably right, but $b^\perp$ doesn't so much look like a set, so I converge to $(b)^\perp$ :) Thanks anyways!
Feb
26
comment Set of all $a\in\mathbb Z$ that are coprime to $b\in\mathbb Z$
You're right that introducing a new notation will likely be necessary. I use only $\gcd(a,b)=1$ or not, so I prefer to use the symbol $\perp$. Which made me thinking: Would $(b\mathbb Z)^\perp$ do the job (of course properly defined)?
Feb
26
revised Set of all $a\in\mathbb Z$ that are coprime to $b\in\mathbb Z$
added 24 characters in body
Feb
26
asked Set of all $a\in\mathbb Z$ that are coprime to $b\in\mathbb Z$
Feb
26
comment Infinite Continued Fraction Notation
However, it is pretty confusing since $\frac{(2k+1)^2}{6}\neq\frac{(4k+2)^2}{24}$.
Feb
26
comment What is $\Bbb R^{\times}$?
However, $\mathbb{R}^*$ can be used for $\mathbb{R}\cup\{\pm\infty\}$. Some people don't like $\overline{\mathbb{R}}$ for that since for them, closure of $\mathbb{R}$ is just $\mathbb{R}$.
Feb
14
comment What's the intuition behind Pythagoras' theorem?
@Sawarnik Well, it's probably quite difficult to prove it from Euclid's axioms. However, this answer gives a nice insight into the way how PT works, which is probably more important for the OP than exact definitions and proving from axioms.
Feb
13
comment Find $f\left(A\right)$ for a polynomial function of a square matrix
@inquisitor Now it looks correct ;)
Feb
13
comment Find $f\left(A\right)$ for a polynomial function of a square matrix
@inquisitor Ok, then once you know what is determinant, you will learn what is the characteristic polynomial, and then you can understand (but probably not proof ;)) the Hamilton-Cayley Theorem. Before you have that, you can verify by hand that your matrix $A$ satisfies the relation $A^2=7A-10I$.
Feb
13
comment Find $f\left(A\right)$ for a polynomial function of a square matrix
@inquisitor Then you're wrong. Matrix multiplication is not done element-wise.
Feb
13
comment Find $f\left(A\right)$ for a polynomial function of a square matrix
Needed to add, the theorem you use is called Hamilton-Cayley Theorem, and is quite non-trivial ;)
Feb
12
comment MENSA IQ Test and rules of maths
Well, you have to remember that MENSA is just a stupid company like any other company in the world, and that IQ is defined as the ability to pass IQ tests. The correllation of IQ with intelligence is unknown.
Feb
11
comment Is $7$ the only prime followed by a cube?
@Lost1 Congratulations, you've just observed that voting system of SE is screwed globally. (Still, it seems to quite well achieve to find local maxima).
Feb
6
comment Is any norm induced by some inner product?
@TobiasKildetoft Sorry, I was a bit too short-headed.
Feb
6
answered Is any norm induced by some inner product?
Jan
15
comment How to correctly do a division using a slash?
Well, the typical convention is that $a/bc=a/(bc)$ and $a/b\cdot c=(a/b)c$. But I would rely on it.