1,365 reputation
214
bio website kmlinux.fjfi.cvut.cz/…
location Prague, Czech Republic
age 26
visits member for 1 year, 6 months
seen yesterday

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

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


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
revised How do I integrate $\frac{1}{x^6+1}$
added 495 characters in body
Mar
31
answered How do I integrate $\frac{1}{x^6+1}$
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
28
revised If $\mathbb Z_m\times\mathbb Z_n$ is cyclic, then it's generated by $(\mathrm{gen}(F),\mathrm{gen}(G))$
corrected mn.gcd to mn/gcd
Mar
28
suggested suggested edit on If $\mathbb Z_m\times\mathbb Z_n$ is cyclic, then it's generated by $(\mathrm{gen}(F),\mathrm{gen}(G))$
Mar
17
comment Why can't you pick socks using coin flips?
+1 for "countable vs. uncountable".
Mar
2
revised How do I setup the lagrangian for this problem?
added 818 characters in body
Mar
2
revised How do I setup the lagrangian for this problem?
added 425 characters in body
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 $b\in\mathbb Z$