Now I am trying to learn prolog on the side in my free time, currently stuck with something that would be simple in any other programing language. What about you?
@Balarka Tell me if I am thinking right or not... $p:X\to Y$, A subset C of X is saturated if for some $y\in Y$, $p^{-1}(y) \cap C$ is nonempty, then C contains $p^{-1}(y)$
Hello, please what can be the condition on $f_u$ to obtain the equality between $ \int_0^1 \int_0^1 G(t,s)f_u(s,u(s)) v(s) w(t) \ ds\ dt$ and $\int_0^1 \int_0^1 G(t,s)f_u(s,u(s)) w(s) v(t) \ ds\ dt$
@TedShifrin hola ted! what does paramterizing $\mathbf{a}$, the optimum, as a function $\psi(\bar{a})$, which is from $\Bbb{R}^n-1 \mapsto \Bbb{R}^{n-1}$ accomplish?
@Hippalectryon bosse dur :), tu sais pour le théorème des fonctions implicites, il faut que la dérivée par rapport à celle que tu veux résoudre soit inversible, je vois dans des bouquins l'hypothèse $f'_y\ne 0$ seulement
@Stan: No, no, remember the maximum point is parametrized by $c$ (a scalar, the budget). If we're still talking about that problem, you want to parametrize the level surface $g(x)=c$.
I need to look at the construction of algebraic closures at some point and see if I can analogize it to get the correct notion of paths/loops in galois theory of fields.
Ah, @Gato, tu ne comprends pas ce qu'on veut dire. Quand on écrit $\partial (F_1,\dots,F_n)/\partial (x_1,\dots,x_n)$, c'est le déterminant du jacobien.
Well, @Balarka, that should keep you busy for a few years.
@TedShifrin i'm sorry but have you an idea about this: what can be the condition on $f_u(s,0)$ such that the following equality holds: $$\int_0^1 \int_0^1 G(t,s)f_u(s,0) v(s) w(t) \ ds\ dt=\int_0^1 \int_0^1 G(t,s)f_u(s,0) w(s) v(t) \ ds\ dt$$
@BalarkaSen I'm on main off and on but my inbox didn't get the chat ping right away. Just checked it and came here.
And yes there are infinite Galois extensions of finite fields - take closed subgroups of the profinite integers! (basically, pick some set Q of primes, then union over all extensions of F_l with degree a product of primes in Q). Note that there are infinite algebraic extensions of finite fields that are not Galois.
also, aren't fibers just orbits of the group (when you're talking about coverings)? so, with fields, Galois actions have orbits too. I presume this doesn't satisfy something you want from the situation though.
indeed, Galois extensions of $\Bbb F_\ell$ are in bijective correspondence with supernatural numbers, formal infinite products $\prod p^{e(p)}$ of prime powers with $e:\{{\rm primes}\}\to\{0,1,2,\cdots,\}\cup\{\infty\}$ (no restrictions on support).
@robjohn I calculated the integral for the $\frac{1}{m^4}$ case ... the resulting monster expression with harmonic numbers looks difficult to simplify to the compact form ...
there must be a neat trick .. I need to think about it ...
my keyboard h and g is not working .. copy pasting everytime is painful :P
In his biography of Einstein, Carl Seelig reports: "Einstein later laughingly recounted that his dissertation was first returned by Kleiner with the comment that it was too short. After he had added a single sentence, it was accepted without further comment."
hi guys. i have a question. hopefully some of you can take a look and provide some guidance. i've already typed it out so i'm just going to paste it now.
it's kinda trivial but i'd like to know if there's maybe a more formal approach to it. it's slightly related to base 64 but really only slightly. so when converting a byte string to a base 64 representation, every 3 bytes become 4 base 64 digits. if the number of bytes isn't divisible by 3 some padding needs to be introduced -- this way the output will always be a base 64 string whose length (number of digits) is divisible by 4
i.e. byte strings of length not divisible by 3 still give base 64 strings whose length is divisible by 4. now, when calculating the size of the base 64 representation, if we assume the byte string's length is divisible by 3, the length of the base 64 string is given by $m = 4 * (n / 3)$. but if the byte string's length isn't divisible by 3, a trick like this has to be used
$m = 4 * ((n + 2) / 3)$ to ensure that a byte string of any length would always give a base 64 string of length divisible by 4. is there a name for this trick? is there some more formal way to go about this? (note: operator / is integer division, i.e. floor("x divided by y") -- this is mainly the reason i'm asking, since the floor function is involved, is there any property of it which would make this trick "more formal")
i hope it's understandable :P
an example would be this set of byte strings $\{m, ma, man\}$ which all give a base 64 string size of 4, as expected.
