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3:42 AM
Aspyn wishes to invest an unknown sum in an account where interest is compounded continuously. Assuming Aspyn makes no additional deposits or withdrawals, what annual interest rate will allow her investment to double in exactly five years?
I interpret this as the ODE $P'(t) = rP(t)$ with the initial condition $P(5) = 2P(0)$, where $t$ is the number of years the sum has been in the account, $P(t)$ is account balance at $t$, and $r$ is a constant interest rate to be computed. I found the general solution $P(t) = Ce^{rt}$, then tried $2P(0) = Ce^{r5}$, leading to $C = 2e^{-5r}P(0)$ and the particular solution $P(t) = 2e^{rt - 5r}P(0)$.
I can solve this for $r$, but I'm not sure how to get rid of the $P(0)$ without reintroducing $C$. Am I doing this right?
4:34 AM
Not sure not really my area of focus economics. But I can introduce her well actually I think he is frightened of me now well and truly but she should try to get in touch with me ol mate Anthony, he had a lot of coin, seemed a bit creepy but yeah Anthony buck. He'll be ready to
. . . .
4:50 AM
@user10478 You can't get rid of P(0), no, but you don't need to: It's 5 years to double any initial investment.
So once you solve for $r$ you're done.
To put it another way: your formula works if P(0) is one dollar or one million dollars. after 5 years, it'll have doubled---from what to what hardly matters re: the problem (though it'd make a lot of practical difference if you were Aspyn, of course!)
Q: for each point on the curve, the line segment of the tangent line from tahe curve point to the $y-$ axis has length 1.

Math geekShow that the tractrix has the following property: for each point on the curve, the line segment of the tangent line from the curve point to the $y-$ axis has length 1. My attempt:- Let the given Tractrix can be parametrised by the formula, $$c:(0,\pi/2)\to \mathbb R^2,$$ $$c(t)=\sin(t) \hat{i}...

The attempt made by Math Geek is clear to me.
Is there any shortcut?
1 hour later…
6:23 AM
So, what kind of object would the set of all semigroup homomorphisms under function composition be? It's certainly not a magma, because there exist elements which cannot be combined with each other, which I believe disqualifies it, correct?
2 hours later…
8:03 AM
@Rithaniel they are the morphisms of a category
a category is basically like a partial monoid
Buongiorno @Alessandro
I've got a set-theoretic question: if we assume ZF + every essentially surjective fully faithful functor between small categories is an equivalence, can we prove AC?
Uh you need to explain me essentially surjective, fully faithful and equivalence :P
I explained all that in my blog :P
Alright, danke Mathein.
8:13 AM
essentially surjective means that every object in the target category is isomorphic to some object in the image. fully faithful means the map on Hom-sets is bijective for every pair of objects. $F$ is an equivalence if there's a functor $G$ that goes in the opposite direction such that $GF$ and $GF$ are isomorphic to the respective identity functors
I think I have a proof: if $(X_i)_{i \in I}$ is a non-empty family of non-empty sets, then consider the following categories: (1) $I$ as a discrete category (2) Consider the disjoint union $X$ of the sets $X_i$, each as a codiscrete category (i.e. there's a morphism between each pair of elements in the same set), then we have a functor $F:X_i \to I$ by sending each element $x_i \in X_i$ to $i$. This is essentially surjective and fully faithful,
having a quasi-inverse $G: I \to X$ is the same as a choice function
So essentially surjective + fully faithful => equivalence of categories for functors between small categories is equivalent to choice
@MatheinBoulomenos I'm confused, pick a surjective map $f\colon X\to Y$ and think about $X$ and $Y$ as trivial groupoids (the only morphisms are the identity morphisms), $f$ is now an (essentially) surjective fully faithul functor, right?
@AlessandroCodenotti no, that's not fully faithful if $f$ is not injective
since there are empty hom-sets which are mapped to non-empty hom-sets
Oh ok I see
It works out if both are codiscrete, right?
8:23 AM
if both are codiscrete, then yes, but a quasi-inverse won't necessarily be a one-sided inverse
if both are codiscrete, than even a constant functor is an equivalence of categories
Makes sense
@MatheinBoulomenos This seems to work
I think the same proof should also show that essentially surjective + fully faithful => equivalence of categories is equivalent to choice for classes over something like NBG set theory as a meta-theory
(not restricting ourselves to small categories)
@AlessandroCodenotti @MatheinBoulomenos are there classes that do not admit choice?
$V$ having a choice function is equivalent to $V=\mathsf{HOD}$
8:38 AM
what does having a choice function mean and what is HOD?
Let $C$ be a class, a choice function is a class function $f\colon C\to\bigcup C$ such that $f(x)\in x$ for all $x\in C$
In the case of $V$ this is usually called a global choice function (global choice is often taken as an axiom in NBG)
surely the empty set...
HOD is the class of hereditarily ordinal definable sets, the construction is somewhat technical but it's an inner model
@LeakyNun The empty set is not an element of every set
so it's L?
No, L could be smaller than HOD
8:41 AM
$f(\varnothing) \in \varnothing$?
V=L implies V=HOD but the other direction does not hold
Well every nonempty set
Just like usual choice for families of sets
how do you define class?
Depends on what's the theory I guess. In ZFC you can only talk about definable classes
that's sad
in NBG a class is a collection of sets satisfying some axioms
but classes can't contain proper classes
8:45 AM
What the fuck are you dudes cooking here
No set theory in my sandwich how many times do I have to tell you
talking set theory, I guess
we're just studying discrete categories
Only continuous sheaves are allowed
we're studying set-valued sheaves on a one-point space
we're studying projective spaces over the field with one element
8:51 AM
I like F_1
It's K-theory is the sphere spectrum
F_1 is fun
Trivial things can be quite fun.
That is to say K-groups of F_1 are stable homotopy groups of spheres
@MatheinBoulomenos is espace etale a functor?
Ya it is
A map of sheaves gives a map of espace etales
8:55 AM
as objects in which category?
The right category of espace etales is probably a slice category of local homeomorphisms to X but I haven't thought this through
It's like covering spaces, more or less
yeah, exactly: there's an adjunction between the slice category $\mathbf{Top}/X$ and the category of presheaves on $X$ $\mathbf{PSh}(X)$, for one direction, you take a topological space $E \to X$ to the presheaf defined by taking preimages, for the other direction you take the espace etale.
This restricts to an equivalence of categories between etale spaces over $X$ (with local homeomorphisms as structure maps) and sheaves over $X$ and sheaves over $X$ and to an equivalence between covering spaces and locally constant sheaves
Shamelessly advertising my own question I just posted, if you want to deal with some point-set topology craziness
the counit of this adjunction is the sheafification functor
Right, you take the espace etale of the presheaf and then take the sheaf of sections to get sheafification
9:33 AM
Is the notion of a group contraction of any importance in mathematics or is it only used in physics?
9:57 AM
Hi all
Heya Meg
Hi @ÍgjøgnumMeg
So, a question: Given any function with a semigroup as it's domain, can you evaluate whether it's a semigroup homomorphism just by looking at the definition of the function?
Just read Euler‘s proof of p = a^2 + b^2 for p = 1 mod 4
so a morphism of sheaf of rings is different from a morphism of ringed spaces?
10:12 AM
morphisms of sheaf of rings have the same base space
that's stupid
for ringed spaces, you have a morphism on the spaces and on the sheaves, basically
@ÍgjøgnumMeg my favorite proof uses $\Bbb Z[i]$, how does Euler's proof work?
@Mathein you use the fact that 1, $2^{4n}$, ..., $(4n-1)^{4n}$ are all 1 mod p and then take the first difference of that sequence. Those are all divisible by p, and all factor as a difference of squares $(a^{2n} + b^{2n})(a^{2n} - b^{2n})$. Then by primality p has to divide one of those factors. There’s a Lemma that says that if (a,b)=1 then every factor of the sum of the squares of a and b is a sum of two squares, so all you need to do is show that p divides the first factor of
The difference of squares in at least one case and the Lemma gives you the proof
@ÍgjøgnumMeg did you solve any exercise on the ANT sheets I sent you?
Not yet, I’ve been quite busy at work so I’ve been getting home and going straight to bed lol, I have a day off today which I’m spending in a different city reading Edwards‘ „genetic intro to ANT“
10:21 AM
ah nice, enjoy! I love ANT
@Mathein showing that the first factor is divisible by p in at least one case just comes from the fact that the nth difference in the sequence $\{ x^n\}$ is n!
maybe I’ll try a couple of exercises from those sheets today :)
@ÍgjøgnumMeg you seem excited
@Leaky anything to distract me from the monotony of my job is intensely exciting to me
Sitting outside Exeter cathedral doing some number theory is worthy of excitement lol
10:52 AM
@ÍgjøgnumMeg when exactly do you plan to move to Heidelberg?
11:08 AM
@Mathein 3rd of October I should think, my mother's birthday is on the 2nd, else I would've moved earlier lol
@Mathein in English one says "localisation at $S$", what does one say in German? Lokaliserung von R ____ S? Welche präposition benutzt man da hahaha
@ÍgjøgnumMeg Lokalisierung an S
Danke :)
11:24 AM
It drives me crazy that people say "localization at $\mathfrak{p}$" when it should be "localization away from $\mathfrak{p}$"
I mean, I get it, it's the stalk over $\mathfrak{p}$, so "at" makes sense. But geometrically the picture is you're inverting elements in $A \setminus \mathfrak{p}$
But you're "zooming in" on $\mathfrak{p}$ so that anything not in $\mathfrak{p}$ is invertible right?
Important math question
Should I spell it "disk" or "disc"
so you get a "zoomed" ring that tells you stuff about $\mathfrak{p}$, discarding all the other irrelevant junk
I am leaning to disk
Also hello @BalarkaSen
11:30 AM
@ÍgjøgnumMeg That's because the contravariance of the ring <-> affine scheme duality, though, innit. $A_\mathfrak{p}$ tells you stuff about the structure of $\text{Spec}\, A$ near $\mathfrak{p}$
Think about localizing at elements, or multiplicative subsets generated by them, for a change. $\Bbb Z$ localized at the multiplicative subset generated by $p$, which is $\Bbb Z[1/p]$
Tensoring with this guy, or extending scalars from $\Bbb Z$ to $\Bbb Z[1/p]$, kills $p$-torsions of abelian groups
That's why that should be "localization at $p$"
Tensoring with $\Bbb Z_{(p)}$ however kills everything but $p$-torsions :3
@Slereah Hi!
I guess you're right
it's taste tho right?
I say this because I have been localizing spaces at primes too much
Your terminology is definitely more relevant in algebraic geometry world
My only encounter with localisation has been in the context of unique prime ideal factorisation in Dedekind domains
I haven't done any alg. geo
Meh you have done more than me
i wanted to relearn localizations a bit few weeks ago so i sat down with atiyah macdonald chapter 2, went through a bit and got bored
so i read localization of spaces instead and absolutely loved it
11:43 AM
O shit thanks
Section 3.1 in that document has the proof of unique factorisation
which is nice
Nb there are mistakes because it was for my undergrad, but the essence is there :P
This is good I actually need it
I think I said at the end of the project that $\Bbb Z[\zeta_p]$ is the ring of integers of $\Bbb Q_p$ (which was almost definitely a typo but I'm sad that I missed it before my hand-in hahaha)
I want to read some algebra this semester; my ring theory course is going to be pretty droll I am sure
Also, the student journal I submitted it to required me to reformat so the font is kinda gross, it was originally just standard LaTeX font
@Balarka get sucked in by ANT
11:54 AM
Problem: Show that any group $G$ of order 108...By Sylow's theorem, I know that the number of sylow $3$-subgroups is either $1$ or $4$. But I don't know how to rule out the latter possibility. I could use a hint.
Show that any group $G$ of order 108 is what? Not simple?
haha whoops.
yes, not simple.
$G$ acts on the 3-Sylows by conjugation
That gives you a homomorphism $G \to S_4$. Can you use this?
Okay...so the homomorphism definitely can't be injective (I'm just throwing out ideas right now...need to think for a second).
12:09 PM
Okay, so the action is not faithful (just a rephrasing of non-injectivity). It looks like the action is transitive, so the orbit stabilizer theorem says $N_G(P) = P$ for every Sylow 2-subgroup $P$...Is this heading in the right direction, or am I getting sidetracked ?
the action is indeed transitive by the Sylow theorems
Ah, good. That's what I thought.
Your comment about transitivity is correct, every two Sylow p-subgroups are conjugate. But what does $G \to S_4$ being non-injective mean?
What measures non-injectivity?
but I think you're getting sidetracked
Don't think about 2-Sylows, stick to 3-Sylows
12:11 PM
The kernel?
what do you know about the kernel of a group homomorphism?
It's normal, so we have a nontrivial normal subgroup.
Why isn't it just all of $G$?
You need a proper normal subgroup
Ah, indeed...
If it were all of $G$, then every element of $G$ would fix each sylow 2-subgroup, so the action wouldn't be transitive?
3, not 2. More fundamentally, it'd mean all those four Sylow 3-subgroups are normal.
12:14 PM
we're acting on $3$-Sylows, not $2$-Sylows right now
But you can't have multiple normal Sylow p-subgroups!!
But yes, transitivity, is the key
Sorry I meant sylow 3-subgroups. Very clever solution. Thanks for the help guys!
If $G$ is a finite group it always acts transitively on it's Sylow $p$-subgroups, so you get a permutation representation $G \to S_n$ where $n$ is the number of Sylow $p$-subgroups. This is surjective by transitivity, and it's kernel is the normalizer of a Sylow $p$-subgroup. This in particular means the normalizer has index $n$.
you can generalize and slightly strengthen this approach to show that if $G$ has a subgroup of index $n$ where $G$ does not divide $\frac{n!}{2}$, then $G$ is not simple (this applies to the normalizer of a 3-Sylow in this example which is equivalent to the given solution)
You have proven that the normalizer is normal here, which is a subgroup of index 4 in a group of order 108, i.e., a subgroup of order 27.
The philosophy is if a large order group has too few Sylow p-subgroups for some prime p, then it's not going to be simple.
Or as Mathein says it's really a divisibility thing than order-comparison because if it's simple it has to embed in $S_n$ where $n$ is the number of Sylow $p$-subgroups.
But usually that's less useful
12:22 PM
So, we have shown that a group of order 108 is not simple; it could still be the case that it has four sylow 3-subgroups, right?
@MatheinBoulomenos why is $A_n$ simple for $n \ge 5$
(i.e. what's your favourite proof)
So, it is possible for a group of order 108 to have a 4 sylow 3-subgroups?
Class equation for $A_5$ then boil $A_n$ down to $A_5$ inductively
@user193319 Yes, it is possible.
a semidirect product would do it right
say $C_{27} \rtimes C_4$
12:24 PM
Take a semidirect product of .. yeah
that doesn't work
$C_3^3 \rtimes C_4$ surely will work
$X^4-1 = (X-1)(X+1)(X^2+1)$
so take $i \in GF_9$
so e.g. $\begin{pmatrix}1&0&0\\0&0&1\\0&-1&0\end{pmatrix}$
@BalarkaSen right
no that's nonsense
$C_3^3$ would be normal
Think before typing :P
yeah like you do :P
Think? Never!
12:28 PM
ok $Aut(V_4) = S_3$ so it might work
$V_4 \rtimes C_{27}$
$\langle a,b,c,g \mid a^2=b^2=c^2=abc=1, gag^{-1} = b, gbg^{-1} = c, gcg^{-1} = a, g^{27} = 1 \rangle$
You need to make a guy of order 27 normal
no you need the 27 to be not normal
What? A group of order 108 has an order 27 normal subgroup
but if you want 4 Sylow 3-subgroups then the subgroups can't be normal
Yes, but the normalizer of that Sylow 3-subgroup is normal, which is a group of index 4
12:33 PM
the normalizer is itself
which isn't normal
I'm confused. Isn't that the kernel of the homomorphism G -> S4
@LeakyNun you can also use Iwasawa's criterion to show that $A_5$ is simple, and then do induction
Iwasawa's criterion can also be used to show the simplicity of $PSL(n,k)$ unless $n=2$ and $|k| \leq 3$
@BalarkaSen I don't think $g$ is in the kernel even
Oh it's not surjective, that's not what transitivity means.
In fact it can't be
Order issues
It's a subgroup of the normalizer, i.e., a 3-Sylow that's normal. Order 9
well, it's not a 3-Sylow if it has order 9, but yeah, the intersection of all 3-Sylows (which is always normal) will have order 9 in this case
12:41 PM
Subgroup of a 3-Sylow I meant
ah I see, I misread
Nah, I was being unclear. So it's an extension 1 -> (something of order 9) -> G -> A_4 -> 1
and the rest is group cohomology
What would taking semidirect product A_4 -> Aut(Z_9) = Z_6 using the map which kills the Klein 4-group and embeds Z_3 in Z_6 look like?
no idea
12:55 PM
No wait
This is too hard for me lmao
S saturated iff R\S is a union of prime ideals?
(S multiplicatively closed)
@ÍgjøgnumMeg yes, atiyah exercise 3.7
Cool ty :)
@Leaky @Mathein I'm interested, so is that semidirect product I suggested an example where it has four Sylow 3-subgroups?
I think so but I may be wrong
The four embeddings Z_3 -> A_4 composed with A_4 -> Aut(Z_9) = Z_6 gives the four Sylow 3-subgroups I think
They are all Z_9 semidirect Z_3, where the map Z_3 -> Aut(Z_9) = Z_6 is just the embedding
1:12 PM
if you just want an example, it's easier to take the direct product $A_4 \times C_9$, though
Hah OK yeah
no wait C_9 is normal in the semidirect product, we don't need that characteristic stuff
@MatheinBoulomenos oh right lmao
It's just bloating up the usual four 3-Sylows in A_4 intersecting trivially by a Z_9, doesn't matter how you do it
Easiest to take the trivial semidirect product.
Sounds hard to classify all such extensions, though. You'd need to calculate H^2(A_4, Z_9) and H^2(A_4, Z_3 x Z_3) for all twisted coefficients, no idea how one does that.
yeah, no idea how to do that, either
1:24 PM
They're 3-torsion I guess but that's not much
1:54 PM
Is there anything that Balarka doesn’t know 😥
Group cohomology is very scary
lol I don't know much about group cohomology (or groups, for that matter) at all
my brain just thinks of them as cohomology of BG
If $R \setminus S = \bigcup \mathfrak{p}$ for some primes $\mathfrak{p}$ of $R$ then $S = \bigcap R\setminus \mathfrak{p}$, now if $xy \in S$ then $xy \in R\setminus \mathfrak{p}$ for at least one prime, so that $xy \neq 0$ in the quotient, meaning that $x \in R\setminus \mathfrak{p} \subset S$ so $S$ is saturated
pls confirm
nah I want $xy \in R\setminus \mathfrak{p}$ for aaaall the $\mathfrak{p}$
but that doesn't change anything I just used the wrong words
I think
2:10 PM
@ÍgjøgnumMeg looks good to me
Cool :)
2:35 PM
So a primitive of $\frac{1}{x\ln{x}}$ is $\ln(\ln{x})$, but while trying to evaluate the integral within 0 and 1 of the former we get an indeterminate since the latter doesn't exist there.
I'd like to show that operations on irrational numbers produce (most often) another irrational number. The approach I have in mind is to show that rational numbers make a cut $(A,B)$ and have $max(A)$ or $min(B)$, which is untrue for irrationals. And then from the bounds of these numbers reconsturct the resulting one. Is it worthwhile?
3:05 PM
Hi all
How to expand 4 summations ??
i = 1 to N
j = i+1 to N
k = j+1 to N
l = k+1 to N
3:26 PM
Dumb question: the degree of the minimal polynomial is always less than or equal to the dimension of the matrix, right?
yes, as by Caley-Hamilton the minimal polynomial divides the characteristic polynomial
You can actually cook up a crude bound for the degree of the minimal polynomial without Cayley-Hamilton by noting that the space of linear operators is n^2 dimensional, where n is the dimension of the vector space
But that's of course much worse
3:52 PM
Problem: Determine up to similarity all matrices of finite order in $GL_2(\Bbb{Q})$...If $A \in GL_2(\Bbb{Q})$ has finite order, then $A^k = I$ for some $k \in \Bbb{N}$, so its minimal polynomial $m_A$ divides $t^k - 1$. It's either a first or second degree polynomial, but I'm not sure how to proceed.
So the minimal polynomial is an atmost 2 degree polynomial which divides $t^k - 1$. What's special about these polynomials over $\Bbb Q$? What does a completely factorizable minimal polynomial say about the matrix?
That it is diagonalizable...?
Yes, and in which cases does that apply in your scenario?
Yes, because the diagonalization will have the eigenvalues on its main diagonal, so I and diag(1,-1).
$t^2 + t + 1$ is a valid quadratic factor of $t^3 - 1$, say, so no.
That fellow is irreducible over $\Bbb Q$
4:01 PM
Hmm...okay. I'm still hung up on why $m_A$ equals either $t-1$ or $t^2-1$. I mean, it's intuitively clear that if it divides $t^k-1$, then should have one of the two forms; but I don't see the proof. Is it just an annoying calculation? If so, I can figure that out later.
I just gave you an example where $m_A$ is neither $t - 1$ nor $t^2 - 1$
$t^2 + t + 1$ divides $t^3 - 1$. That could potentially be a minimal polynomial of a matrix of order 3 over $\Bbb Q$
(Can you actually find an $A \in \text{GL}_2(\Bbb Q)$ whose minimal polynomial is that?)
math.stackexchange.com/questions/1090570/…. Can anyone please answer this question? It's actually a comment on an answer.
I was recently going through this notes. However I didn't understand the proof of Proposition 1.11 (page 8). Specifically the following, "But then using a compatibility of $e_1$ and $ε_1$ we get that the composite natural transformation is $e_2$ as required." I don't understand how we can get $e_2$. Since I haven't understood this part I didn't go further. Can anyone help me here?
it's actually a comment
4:46 PM
The given answer is correct. I do not understand your comment.
More precisely, $1$ is a factor of any number (obviously), whereas $x$ is not a factor (but this also isn't implied by what the answer says). Saying $x=1$ is a factor doesn't quite make sense.
@Thorgott putting x=1 give us 0, thus x-1 must also be the factor.
Isn't this factor theorem? @Thorgott
Why would putting $x=1$ yield a zero determinant?
by mistake, I input the wrong det. Please give me a second.
$$ \begin{vmatrix}
\end{vmatrix} $$
What is happening?
I cant render this?
5:03 PM
Try fixing the "x^\\"
@user170039 Write out what the natural transformations are in the composites; use Definition 1.10(1).
And similarly the "x^&"
Here i go
putting x=1 give us 0, thus x-1 must also be the factor.
Isn't this determinant zero, regardless of the value of x, y?
5:08 PM
Which also means that (x-1) divides that determinant.
Zero polynomial is divisible by any polynomial.
I am embarrassed.
Sorry for wasting your time.
Still, this is an interesting question. (I am not really sure why we can use factorization in the post you linked.)
No problem. It's always important to play around with examples to get a feeling for how these things work.
5:11 PM
I guess the asker was also somewhat confused by that - this seems basically as a follow-up question: factor theorem for multivariables.
5:26 PM
Diving multivariable polynomials is hard. The factor theorem in the form that is used still holds though. You can consider $K[X_1,...,X_n]$ as $K[X_1,...,X_{n-1}][X]$ and apply the usual factor theorem.
The trouble with this argument is that $K[X_1,\dots,X_{n-1}]$ isn't a field.
@MartinSleziak Well, still a UFD
The factor theorem works over all commutative rings.
Here we are using that if R is an UFD, so is R[X]?
@Thorgott What factor theorem is that?
5:32 PM
If $f\in R[X],r\in R$ and $f(r)=0$, then $X-r$ divides $f$.
The accepted answer in the first one is very neat.
2 hours later…
7:44 PM
I'm trying to show that any two fields with the same order are isomorphic. If I could construct a non-zero ring homomorphism, I'd be done...But this doesn't seem possible. I could use a hint.
@user193319 You need them to be finite too of course
WHoops, forgot to mention that.
the proof essentially boils down to them both being the splitting field for a certain polynomial
Woah...didn't expect that at all.
you can apply Lagrange's theorem to the multiplicative group
7:47 PM
Lagrange's theorem about orders of subgroups dividing the order of the group?
yeah, it tells you something about the order of the elements. In this case, it will lead you to a polynomial satisfied by all elements in the field
Is main down just for me?
entire stack exchange it seems
Same here
hmm, seems to be up again
7:55 PM
I see, I guess I'll just wait then
It's still very slow for me
But I managed to submit the edit I was struggling to get through
8:21 PM
So, if $E$ and $F$ are finite fields of the same order, are $f(x) = x^n - 1_F$ and $g(x) = x^n-1_E$ considered the same polynomial, where $n = |F^\times|$?
If not, I don't see how they could be splitting fields of a common polynomial.
@user193319 Note that the $1$ lives in their prime subfield, and those can easily be identified with each other
Hmm...All seems a little suspicious---but okay, I guess.
Speaking of suspicious, I think it's important to erase some words from the English dictionary, even if it means using additional characters, the meaning is conveyed more precisely. For example, one should no longer use the word "uniform" but rather refer to that word's meaning by using "respect costume"
the word uniform means "the same" or "unchanging", as in uniform distribution, uniformly continuous etrc
8:36 PM
the word uniform when applied to clothing then refers to the fact that the people wearing a uniform are all wearing the same thing
its not about respect
yes exactly it isn't a good word for the context that it's used for clothing
of course it is a good word
It's a uniform because a whole group is wearing it
no, you misunderstand the word thats all
you might even say
metonymy is a valid linguistic device
8:38 PM
Oh no you are right it's about pseudo respect, not actual respect, that's why costume is important, so people know it's not actual authority the individual possesses
you're only thinking about the word in one context
consider school uniforms
never have had to listen to a kid in a school uniform
Well there is a spectrum of respect our societies need to assign costumes for, that's what keeps us all in our place
gamers rise up
Can we try to keep the lunacy to a minimum (in case anyone feels like discussing something that is actually on topic)?
ah most people discuss on topic, your definition of lunacy is a little inept or naïve but ok
8:41 PM
I have a suggestion: Whats the derivative of the map $x\mapsto\exp(x) =\sum_{n=0}^\infty \frac{x^n}{n!}$ when the domain is a Banach algebra?
why don't you post it as a question then?
non-commutative of course, otherwise its just exp again
I did a couple of years ago
was reminded of it today, and since tobias suggested we talk about things that are on topic :P
that seems awkward; how do derivatives work over noncommutative spaces?
that might be too big of a hint to give but my interest is piqued
@fargle banach algebras are vector spaces, derivatives are well defined over complete vector spaces ($d_x f$ is linear part of the best affine approximation of $f$ at $x$)
oh doy
8:45 PM
:-) oh well we all have questions unanswered who knows you may get lucky and find someone cares for yours, banach algebras are just vector spaces, this is another good example of my original remark this morning actually
what might be a good example to toy with? seems like matrix spaces end up having the simple answer too
@fargle matrix spaces dont have a simple answer
at least not as far as i am aware
as for toying, the only thing i know is the form of the derivative of $x^n$
@Fargle You might be thinking of exp(tA) with respect to t for matrices
8:58 PM
yeah you are right
guys, this is a result for diagonally dominant matrices
diagonally dominant means that $\vert a_{ii}\vert^{-1}\sum_{j\neq i}\vert a_{ij}\vert<1$
and here $\rho$ is the spectral radius of the matrix (maximum of the absolute value of the eigenvalues)
does the first equality hold in general?
if so, why?
in general I mean for any matrix, not just for a diagonally dominant matrix
That equality cannot hold even for diagonally dominat matrices
hm, maybe they didn't mean the spectral radius with $\rho$
tho thus far, they always denoted spectral radius with $\rho$
I guess that's an error on my side then
definitely not, look at diagonal matrices where the expression is always zero
yea, thanks
9:35 PM
Why hello nerds
9:59 PM
Is someone of you familiar with linear (n,m)-codes?
I have a question about that
Q: Linear Codes: Generator matrix and parity-check matrix

Mary StarWe consider the following linear $(n,m)$-code $C$ : I want to determine the canonical generator matrix and the parity-check matrix. $$$$ Is the generator matrix the matrix with rows the above codes? I mean the following: \begin{equation*}G=\begin{pmatrix}0& 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 &...

2 hours later…
11:56 PM
I need a book called "A=B" someone buy it for me, im holding my breath until I get it, and if I die it will be all your fault and the story will go viral and the whole world will know what cruel monsters you are

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