I suppose $f(M)$ must be at least a submonoid of $M'$? In which case the question is whether a submonoid can have a different identity element than the monoid itself.
in the case of subgroups, at least, the answer is definitely no
(con't) I don't know how useful to learn that the function $\phi$ is separable, but it surely will help interpret the incoming future batch of data because I don't need to worry about cross terms
For all $x$, $f(x)=f(x+0)=f(x)+f(0)$. Since $f$ is additive, it follows $f$ is a homomorphism between the two monoids, thus $f(e)=e'$ where $e,e'$ are the identities of each monoid. Therefore $f(0_M)=0_{M'}$
Hi, why can we say that the determinant $\text{det}(A)$ is a polynomial in the entries of the matrix $A$? If we look at the determinant of a $3 \times 3$ matrix
There's an additional property in the case of the determinant, though, namely that each term in the expanded determinant is degree 3 (product of 3 different matrix elements).
Again, though, this shouldn't be too surprising: If you rescale a 3-by-3 matrix by a constant factor, then the determinant should rescale by that factor^3
Corollary 5: Let $\{f_n\}$ be a sequence of nonnegative integrable functions on $E$. Then
$$\lim_{n \to \infty} \int_E f_n = 0 ~~~~~~(5)$$
if and only if
$$f_n \to 0 \mbox{ in measure on } E \mbox{ and } \{f_n\} \mbox{ is uniformly integrable and tight over } E ~~~~~(6)$$
H...
Let $A$ be a ring (comm. and with $1$), and $M,N$ be $A$-modules, $I$ some submodule of $N$. Given a morphism $f:M\to N/I$, does there always exist a lift of $f$, i.e. a morphism $g:M\to N$ such that $f=i\circ g$?
For example I could set $g$ to send $m\in M$ to some element in the equivalence class of $f(m)$ in $N/I$, but I think with this arbitrary choice $g$ will not always be a morphism of $A$-modules...?
Corollary 5: Let $\{f_n\}$ be a sequence of nonnegative integrable functions on $E$. Then
$$\lim_{n \to \infty} \int_E f_n = 0 ~~~~~~(5)$$
if and only if
$$f_n \to 0 \mbox{ in measure on } E \mbox{ and } \{f_n\} \mbox{ is uniformly integrable and tight over } E ~~~~~(6)$$
H...
@cello 0celo7 didn't use any inappropriate language, you are misunderstanding. He's vocalizing that he doesn't want to help you - which is to be taken as a cue of respectful conversations to stop pinging 0celo7 on your side and drop the conversation.
Mos Eisley is currently frozen#, apparently indefinitely. I assume it has something to do with the users mentioned in @Shog9's banner, but is it possible that we be told - at least in concept - what happened in our room to make it get frozen again?
If the rest of you want to create a new room...
That's fine, we could have a discourse over that instead :) My idea is shared by many other else and lots of moderators in hbar that are actively trying to control the room (and succeeding)
If I were to suggest anyone as a more active RO it would be @Semiclassical, but of course it's a decision of himself and the other room owners here
@MikeMiller No better political position than to hand all the power to someone in your alliance and thereby gaining indirect control over the system, amiright?
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==================== Commands
//about | Let me tell you a little about myself...
//alive | Used to check if the bot is working
//appul | Apples.
//ban | Bans a user from using the bot. Only usable by hardcoded bot admins
//ban-room | Blacklists a room
//blame | No description was supplied for this command
//declare | Changes a commands status. Only commands available on the site can be edited
//doge | Such doge. Much command.
@Daminark So I need to color the integers from $1$ to $n$ such that the set $\{a,b,a+b\}$ always has more than one color, whenever $a\ne b$ and $a+b\le n$?
I want to prove that the non-zero roots of characteric polynomial of a matrix product $AB$ are the same as $BA$. I'm not given that the matrices are square, but presumably I can pad the matrices with $0s$ to get something like $[A | 0] [\frac B 0 ] \cong AB$. Can I also make a density argument by making a sequence of inv'ble matrices tending to $A$, or is that a contradiction to padding it with zeros?
Basically I'm taking $\phi$, making a linear function out of it and using it as $M$, but I'm not sure how to do that since $M$ isn't in exactly the same form as $y=mx+b$.
My social life in a nutshell, and why I am so pessmistic about socialising beyond 2020
The Predators, vicious pretentious entities, will devour me and my weirdness, turning me into a normal human being and thus eat my soul, dreams, and I will then become a lifeless cog of the society
@nitsua60 No problem, I just felt like vocalizing this because this kind of pointless flagging spree has interfered with constructive conversations in the chat (about math, say), which I participate in. Wasn't particularly trying to "lead", except perhaps ironically
Knowing the base sides and the slant heights, do we calculate the height of the pyramid by the formula $h^2=h_b^2-\left (\frac{a}{2}\right )^2$, no matter what shape of base the pyramid has, i.e. square, rectangular, right angles triangle?
Here's part of the proof of Lebesgue' theorem: "The set $E$ of points at which either $\{\psi_n(x)\}$ or $\{\phi_n(x)\}$ fail to converge to $f(x)$ has measure $0$. Let $E_0$ be the union of $E$ and the set of all the partition points in the $P_n$'s....We claim that $f$ is continuous at each point in $E-E_0$." My question: Isn't $E-E_0$ empty, since $E-E_0 = E - (E \cup \bigcup_{n=1}^\infty P_n) = \emptyset$?
@Adeek Differential forms are meant to encode area of k-planes. The derivative spits out something that encodes area of (k+1)-planes, by taking the derivative of what it does to the k-planes in the "extra direction"
I don't know that you get anything useful out of this picture tho
@BalarkaSen Yeah--it's been my experience that if the annoying behavior gets slapped immediately, rather than a flood of "outsiders" coming in, it actually dies down pretty quickly. But I know nobody comes to hang out each day thinking "hey, maybe I can call someone out correctly and get yelled at today!"
Consider the expression
$$\frac{\int_0^1 Ei(x)^5 \ln(x) dx}{\int_0^1 Ei(x)^3 \ln(x) dx} $$
A) Can we rewrite this with a single integral sign?
B) Do we have a closed form for this expression in terms of hypergeometric functions?
C) Is there a closed form without hypergeome...
Corollary 5: Let $\{f_n\}$ be a sequence of nonnegative integrable functions on $E$. Then
$$\lim_{n \to \infty} \int_E f_n = 0 ~~~~~~(5)$$
if and only if
$$f_n \to 0 \mbox{ in measure on } E \mbox{ and } \{f_n\} \mbox{ is uniformly integrable and tight over } E ~~~~~(6)$$
H...
Mos Eisley is currently frozen#, apparently indefinitely. I assume it has something to do with the users mentioned in @Shog9's banner, but is it possible that we be told - at least in concept - what happened in our room to make it get frozen again?
If the rest of you want to create a new room...
Consider the expression
$$\frac{\int_0^1 Ei(x)^5 \ln(x) dx}{\int_0^1 Ei(x)^3 \ln(x) dx} $$
A) Can we rewrite this with a single integral sign?
B) Do we have a closed form for this expression in terms of hypergeometric functions?
C) Is there a closed form without hypergeome...
