1,403 reputation
214
bio website kmlinux.fjfi.cvut.cz/…
location Prague, Czech Republic
age 27
visits member for 1 year, 9 months
seen Jun 27 at 8:08

Math PhD student at Czech Technical University in Prague and at LIAFA in Paris.

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


Jun
13
comment Proof of irrationality of a series
+1 very nice, simple, straighforward yet not trivial argument :)
Jun
5
comment Prove $\sqrt6$ is irrational
sorry, corrected. And well, I say that I use stronger weapons, which can be used in a large framework, I know that simpler solutions exist.
Jun
3
comment Evaluate a limit (probably involving L'Hôpital rule)
you should use exp in the appropriate places I believe, now you mix e as a number and e(x) as a function.
Mar
31
comment How do I integrate $\frac{1}{x^6+1}$
@MorganWilde Because you have to, the space of polynomials modulo a quadratic polynomial is generated by $1,x$ and not only by $1$.
Mar
31
comment How do I integrate $\frac{1}{x^6+1}$
+1 certainly faster than my approach. It's been a while I knew these tricks, now I remember only the general techniques :-/
Mar
31
comment How do I integrate $\frac{1}{x^6+1}$
@MorganWilde No worries, I added a small tutorial.
Mar
31
comment which axiom(s) are behind the Pythagorean Theorem
@William if $A_1\wedge A_2\wedge\dots\wedge A_n \Leftrightarrow B$, then $B$ is equivalent to the system of axioms $A_1,\dots,A_n$, so I'm no quite sure what you speak to in the second part. And if $A\wedge B\Rightarrow T$ and $A\wedge C\Rightarrow T$ and $A\wedge B\not\Rightarrow C$ and $A\wedge C\not\Rightarrow T$, then you got two non-equivalent proofs of your theorem $T$. I would never "guess" which axioms are "better" in any other way than possibility to derive ones from the others. (But maybe it's just too late and I overlook some stupidity in my arguments.)
Mar
31
comment which axiom(s) are behind the Pythagorean Theorem
Equivalent = Relying the same set of axioms, usually. Therefore a proof that uses less axioms is "better", and a proof that uses different non-equivalent axioms is simply "different".
Mar
17
comment Why can't you pick socks using coin flips?
+1 for "countable vs. uncountable".
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 $\max(y_M, y(0), y(2\pi/3))$.
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
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 ;)