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12:11 AM
I think I have found an interesting formula, but I am not able to prove it
$$\sum_{x=1}^{\infty}\left( (-1)^x\sum_{n=1}^x \frac{1}{n^m} \right)=\frac{\zeta (m)}{2^m}$$
Any clues guys?
Right now all I know is that $\sum_{n=1}^x \frac{1}{n^m}=\zeta(m)$
 
@Mathphile That is indeed interesting
 
Mathematica says it doesn't converge, and I'm inclined to agree
 
@Semiclassical according to my computations it converges and gives these nice closed forms too
 
and according to mine it doesn't, so
there we are
 
12:28 AM
Maybe you could try computing $$\sum_{x=1}^{1000}\left( (-1)^x\sum_{n=1}^x \frac{1}{n^2} \right)$$
 
i mean, as $x\to\infty$ one has $\sum_{n=1}^x \frac{1}{n^m}=\zeta(m)$. So as you go farther into the series, you effectively have $\zeta(m)-\zeta(m)+\zeta(m)-\zeta(m)+\cdots$
which is not convergent, since it's just Grandi's series aside from a factor of $\zeta(m)$
 
You should get a value close to $\frac{\zeta(2)}{4}$ by doing the above computation
 
i get $$\sum_{x=1}^{1000}\left[(-1)^x \sum_{n=1}^x \frac{1}{n^2}\right]\approx 0.410734$$
but
i also get $$\sum_{x=1}^{1001}\left[(-1)^x \sum_{n=1}^x \frac{1}{n^2}\right]\approx -1.2332$$
which means it's bouncing back and forth between two values. aka, doesn't converge
 
i see what you mean
 
it may converge if you restrict to the subsequence of even upper limit on $x$, or odd upper limit
there may be an interesting identity in there, in other words. but it's a bit more delicate
 
12:33 AM
$$\lim_{s\to \infty}\sum_{x=1}^{2s}\left( (-1)^x\sum_{n=1}^x \frac{1}{n^m} \right)=\frac{\zeta (m)}{2^m}$$
 
My sister just got engaged
 
which makes me think the thing to do is group up the terms of the alternating sum as $(-a_2+a_1)+(-a_4+a_3)+\cdots$
@akiva kudos to her, then
 
They just spent the past two weeks quarantining in the same half of a house
(his house)
so they better like each other
lol
 
that's sorta like an anti-honeymoon, lol
 
12:35 AM
Anti meaning before, or opposite?
 
hmm. either works, i think
 
before is ante-
 
She took an airplane from New York to Seattle so we wanted to make sure she didn't get anything on the plane
 
both ante- and anti-honeymoon
 
or otherwise bring the virus across the country
 
12:36 AM
clearly it's just a cohoneymoon
 
@Semiclassical yup but I don't see what to do after that
 
@Mathphile if you let $H_{n,m}=\sum_{k=1}^n \frac{1}{k^m}$, then what I'm proposing is to consider $\sum_{n=1}^\infty (H_{2n,m}-H_{2n-1,m})$
 
hmm alright
 
which mathematica does seem to approve of, at least on a case-by-case basis
it looks like it's true so long as $m>1$, even for non-integer values
 
12:45 AM
@Semiclassical yes
 
oh, wait. $H_{2n,m}-H_{2n-1,m}=\sum_{k=1}^{2n}\frac{1}{n^m}-\sum_{k=1}^{2n-1}\frac{1}{n^m}=\frac{1}{(2n)^m}$
So it's just $$\sum_{n=1}^\infty \frac{1}{(2n)^m}=\frac{1}{2^m}\sum_{n=1}^\infty \frac{1}{n^m}=\zeta(m)/2^n$$
 
ah yes
 
So pretty simple at base
which i guess it kind of has to be
 
0
Q: Cauchy completion of transfinite "rationals"

user76284Let the Hessenberg power $\alpha^\beta$ be the supremum of ordinals that are order-isomorphic to some well-order on the set of functions $\beta \rightarrow \alpha$ of finite support that extends the following partial order: $$f \leq g \leftrightarrow \forall x \in \beta : f(x) \leq g(x)$$ Foll...

 
1:18 AM
Would appreciate any help with this question: math.stackexchange.com/questions/3704735/…
 
I actually have to questions, one regarding my post and the solution I would like to verify since I'm studying from various sources due to all the happenings:
1
Q: Given the point $T$ and the line $p$, find the distance between their orthogonal projections onto the plane $\pi$.

Cheesecake Let $T'$ and $p'$ be the orthogonal projections of the point $T(-8,2,-3)$ and the line $$p\ldots\frac{x}4=\frac{y-4}3=\frac{z+1}{-2}$$ onto the plane $\pi\ldots x-y+3z+8=0$. Find the distance between the point $T'$ and the line $p'$. My attempt: In general, a vector perpendicular...

The other questions is: Which chat-rooms are the solution-verification questions most appropriate for?
I usually ask such questions so as to see how the things are going in my mind, but I wouldn't like to be intrusive.
Or sound as if I were promoting myself.
 
 
2 hours later…
3:53 AM
@Thorgott Thanks for your detailed response. When you say that representation theory is interdisciplinary, involving group theory and linear algebra, with applications in differential geometry or harmonic analysis, is it also interdisciplinary in the sense that it has applications to the physical sciences, say, a complex biophysics dynamical system? Or is the nature of the work exclusively proofs-based? Is any numerical work done in representation theory?
@Thorgott For instance, when I do a search for research groups at various universities, I find geometric representation theory and combinatorial representation theory groups, but nothing that intersects with science or engineering.
 
4:10 AM
@robjohn Sir when you have time please explain this answer of yours
I mean why those set of consecutive integers need to fill every class of residue classes?
 
4:36 AM
if they're in the same residue class mod n, they have to differ by a multiple of n
but how far apart can n consecutive integers be?
any two integers out of n consecutive integers, i mean
@Knight woops, forgot to ping you
 
@Semiclassical Yeah
I’m here
So, we have $n$ residue classes. One for each remainder and remainder varies from $0$ to $n-1$ , ha?
 
4:55 AM
right. the only subtle point is that the sequence of remainders may not start at 0
e.g., consider 17,18,19,20,21 mod 5. then the remainders go 2,3,4,0,1
but no two of them can have the same remainder. if they did, they'd have to differ by at least 5. but the biggest difference you can possibly have is 4
so each of them have to hit a different residue class. you've got as many integers as residue classes, so they all occur (pigeonhole principle) and that includes 0
this does have me wondering to what extent this generalizes to arithmetic progressions in general
for prime n, it probably generalizes quite nicely.
for composite n, by contrast, you can't generalize it so nicely. for instance, take n=6. then 11, 13, 15, 17, 19, 21 give remainders 5, 1, 3, 5, 1, 3. 0 doesn't show up in that case
 
5:15 AM
Okay got you! Thank you so much Semi
 
I think the main point was “that no two integers can have same remainder”
 
to put a finer point on it: no two integers in that sequence of consecutive integers can have the same remainder
if you just picked two integers out of thin air, they certainly could have the same remainder
this is why it's subtler for arithmetic progressions
 
5:53 AM
Semi, let’s we have two different integers with two different remainders $mod ~n$. That is $$a_1 = nq +r \\a_2 = nq’ + r’$$ Now, we know that $r , r’ \in [0,n-1]$, but if we add $a_1$ and $a_2$ $$ a_1 +a_2 = n(q+q’) + (r+r’)$$ but $r+r’$ can be greater than $n$ (there is a possibility). But it contradicts that any number divided by $n$ must leave the remainder less that $n$
What properties the sum of two integers will have in respect to $mod ~n$?
 
 
2 hours later…
8:05 AM
@Knight I have added a detailed explanation to that answer via the Pigeonhole Principle
 
@robjohn Thank you
@skillpatrol Ah! After a long time
 
hi pal, how are you?
 
@robjohn Sir I’m finding these number theory questions quite ... because I have not been introduced to number theory ever in a formal way
@skillpatrol I’m fine pal. How are you?
 
still kicking, thanks
 
Kicking ? Didn’t get you
 
8:17 AM
it means still alive and well :-)
 
Hahahahaha
 
Hey! I need to ask some knowledgeable people here about shifting signals in matlab without having wrap-around
 
AskAway
 
Currently I have fr example a timevector stretching [-50e-15 seconds, 50e-15 seconds]. If i myself implement ifft(exp(1**omega*tau)*fft(signal)) to shift it (delay shift in fourier transform), let's say, 60e-15 seconds (further than the time axis) it wraps around to the other side.
Wait , i didn't make any sense there
i ahve a pulse centered around time 0, which is wht i am delaying
basically if i delay further than the time vector stretches, it wraps around, and i would like it to not to. I'd like it to just haev some noise low level thing there (basically nothign)
Or is this becasue the fft assumes that the signal is periodic relative to the time vector?
 
8:49 AM
can anyone suggest books to read up on math interpretations of finite element method?
I was trying to understand the answer but I dont understand some topics he's talking about like dimensions of range spaces
 
9:17 AM
@Joanna There are applications in both physics and chemistry, but I couldn't tell you about them. I've not heard of numerical aspects, though representation theory can be used for computations (I think that's how people work with the monster group).
 
Was Leonard Euler a number theorist or an analyst?
 
yes
 
or a graph theorist, or a geometer
 
@Semiclassical Can you tell me in simple terms what does a graph theorist do?
 
study graphs? i dunno, i'm not a graph theorist
 
9:28 AM
me neither, but I second that
 
Well, what was the major field of Leonard Euler? Was it Numebr Theory, or Analysis, or Geomtry?
 
in those days one person could do them all
(you forgot algebra)
 
thought about that, yeah
but i'm not sure what euler's major contributions to algebra were
 
According to Feynman it was his famous formula for e.
 
9:35 AM
eh. $e^{i\pi}+1=0$ is a bit overstated, i think. by itself, it's a curiousity. it being a special case of $e^{i\phi}$ is what makes it marvelous
exponentiation turning into rotation, bam
 
yup, that's pretty much what Feynman said :P
 
9:54 AM
I want the product boundaries to be on top and below of $\prod$ and it is putting them on the sides regardless of how many $ I enclose the expression with
i find this very upsetting and issue mods a cease and desist
fine well i guess just pout
 
$\displaystyle \prod_{k=1}^n k$? @adam
though $$\prod_{k=1}^n k$$
does the same thing
 
The thing I like about Euler was that he was not a prodigy
(Although he learned under the guidance of Jacob Bernoulli)
 
yeah
Euler just worked and worked and worked
 
I think none of Bernoulli ‘s were prodigy
Can you give me names of some mathematicians/physicists who were not very promising in their young days (young days mean up to college) but then they discovered something elegant
I know one example, Majorie Rice
(It’s not a food like Curry Rice)
 
10:20 AM
$\prod\limits_{k=1}^nk$
 
@Knight I have commented on your comment
 
@robjohn Yes, thank you so much sir
Your answer was elegant
I mean your answer didn’t require to any pen and paper work, one can think all of it in his mind
 
10:52 AM
@LeakyNun that's cool i guess
 
@robjohn morning sir
 
 
1 hour later…
11:58 AM
So I asked a question in but I did not get any appropriate answers. One meta post suggested to post link to the question in chat so heres the question, please look into it if you can math.stackexchange.com/questions/3704051/…
 
 
1 hour later…
1:07 PM
so, this is slightly irritating. a certain exercise in the opening chapter of a book i'm using asks to show that countable additivity implies finite additivity
i came up with an answer pretty quick and thought that was that
but said answer hinges on the fact that the empty set has measure zero. that's something that the authors do ask you to show, but only as part of the -next- exercise
(it's specifically probability measure, for context)
So to answer the first exercise, I have to use a fact that you only prove in the second exercise. Not happy with that.
(And I can't see any way to bypass using that fact either.)
 
yeah, that's weird
I don't think you can bypass that
 
i don't think so either. suppose you've got a finite set. then the only pairwise disjoint sequence of subsets possible is one which contains infinitely-many copies of the empty set
 
Think of it this way. If $\emptyset$ had nonzero finite measure then the whole theory would be bunk, because countable union of $\emptyset$ is $\emptyset$.
 
right, that's the logic i had as well
 
there's a more convoluted option: decompose one set into a disjoint countable union of its subsets
 
1:14 PM
it doesn't even work with finite unions, of course, but with countable unions it's even more severe
 
but even that's only possible if one set is countable to begin with
 
$\sigma$-additivity is the statement that for a countable collection $A_1, A_2, \cdots$ of pairwise disjoint sets, $\sum_{i = 1}^\infty \mu(A_i)$ exists and is equal to $\mu(\cup_{i = 1}^\infty A_i)$.
The "exists" is important
 
well, you don't need that per se
 
sure, even if it's implicit
 
This forces $\mu(\emptyset) = 0$ or $\infty$.
 
1:15 PM
if it's equal to $\mu(\bigcup_{i=1}^{\infty}A_i)$, it's finite, hence exists (in the context of probability measures)
 
I'm talking general measures.
 
the sum $\sum_{i=1}^{\infty}\mu(A_i)$ always makes sense as an element of $[0,\infty]$
 
But first you have to state it exists before equating it to something
It's part of the axiom
 
the book is working specifically with $\mu:\mathcal{F}\to[0,1]$ where $\mathcal{F}$ the appropriate $\sigma$-field of subsets of a set $\Omega$
 
no, it exists as element of $[0,\infty]$ and you are postulating $\sum_{i=1}^{\infty}\mu(A_i)=\mu(\bigcup_{i=1}^{\infty}A_i)$ as an equality in $[0,\infty]$
if the RHS is finite, this equality takes place in $[0,\infty)$ and the series converges in the regular sense
 
1:17 PM
so in that case one does have $\mu(\cup_{i=1}^\infty A_i)\in [0,1]$, since $\cup_{i=1}^\infty A_i$ is certainly in $\mathcal{F}$
bleh
\cup, right
the real issue is ensuring that $\sum_{i=1}^\infty \mu(A_i)$ exists, though
 
@Thorgott It's more important to state it exists when you're working with probability measures.
There is no reason a-priori that sum of probability of a bunch of disjoint events converges
This is not a matter of definitions, it's a matter of making the intuition clear.
 
it's simply $\sum_{i=1}^{\infty}\mu(A_i)=\sup_{\substack{X\subset\mathbb{N}\\X\text{ finite}}}\sum_{i\in X}\mu(A_i)$
 
Yeah that helps nobody pedagogically and you shouldn't be allowed to write a probability textbook
:P
 
there are much better reasons for me to be disallowed from writing a probability textbook
especially the couple times I tried writing down completely formal proofs for combinatorial statements
which ended up making a simple statment take 1.5 pages
the thing is that none of the probability texts I've read ever explicitly axiomatized convergence of the series, so it was more confusing to me that way until I realized the equality always makes sense to postulate in $[0,\infty]$
 
1:24 PM
for context, see p. 6,7 here: google.com/books/edition/…
 
It becomes natural when one stops thinking about probability or measure as set-theoretic junk and as "volume"
You have some junk of volume 1. If you pick disjoint subjunks and take their volumes and add them up you get some junk of volume <= 1
 
right
and the only way to do if you've got a finite set is if you eventually only add subjunks with zero volume
i.e. the empty set
 
the only sensible one for sure
 
@Thorgott i think this is actually not an outrageous approach eg if you read Stanley
he does things completely formally but its also not hard to read
one should formalize but not in a way that its impossible to understand
 
i guess that -maybe- the reason they organize the exercises like that is so that they can group all the deduced properties under their Proposition 4
but it makes the exercise order just kinda odd
 
1:30 PM
that book looks good
 
yeah, there's a reason i'm trying to read it pretty thoroughly
and the best compliment i can give it is to be thoroughly annoyed when i think it does something silly :P
 
Can someone please suggest me some resources (notes pdf, or a chapter from a particular book) where things like integer solutions of an Eqaution are discussed? I want to learn about things like “what are the integer values of x and y such that $$x^2 +5y= constant $$ “
 
just read measure theory and then pretend probability theory is the same
 
independence is just a nonsense idea in measure theory
 
1:34 PM
@Knight so stuff like Pell's equation?
 
thats where probability theory becomes more than measure theory
 
that, broadly speaking, goes under the rubric of Diophantine equations
 
I walk in and everyone is doing probability, I'll be back later
 
as in, there is no geometric meaning of "volume of (A \cap B) = product of volume of A and volume of B"
what the hell is that supposed to mean?
nonsense
 
@Yuvraj good morning. I just dropped in before taking the dogs out for a walk.
 
1:38 PM
lol
 
independence of rvs is just their joint pushforward measure being the tensor of their individual pushforward measures
 
@Semiclassical Yes. To be more specific I got a question saying “What is the smallest natural $n \in \mathbb{N_0} $ such that there are no integers $a,b \in\mathbb{Z}$ with $$3a^3 +b^3 =n$$?”
 
if you rephrase it by saying vol(A can B)/vol(B) = vol(A), you're saying A takes up the same proportion of volume in B as it itself takes in the whole probability space, that makes sense, you're already thinking of proportions aka probabilistically
 
@Knight oh ew
 
1:39 PM
@BalarkaSen The integral of the product of the indicator functions?
 
Robbie sir please suggest me something
 
@Thorgott Does measure theory have stuff like that?
 
i feel like i saw a problem like that recently, and specifically that it's a very very hard problem to solve if you allow both positive/negative values
 
stuff like diophantine equations? no
 
1:42 PM
Very hard? Really?
 
that's my recollection, yes. but it's possible i'm misremembering, or that the very hard prooblem is some slight variation on that
 
no clue about the specific one, but diophantine equations in general are very, very hard
 
yeah
linear diophantine equations have a general theory, but quadratic and up? yikes
i mean, there's a reason why Fermat's last theorem was so brutally hard to prove
 
I think probability theory proper probably starts when you consider large number laws/sums of independent rvs
at least I don't know a measure-theoretic perspective for that
 
see Birkhoff ergodic theorem
 
1:47 PM
things with "ergodic" in the name are scary
 
yeah thats where probability is hidden; ergodicity is not an inherently measure theoretic notion
a measure preserving transformation $T : X \to X$ is ergodic if for every $A \subset X$ such that $T^{-1}(A) = A$, $A$ is either null or full measure btw
$T$ "mixes" everything up so bad that there are no nontrivial proper $T$-invariant subset
 
yeah, I don't see why one would care about that measure-theoretically
 
yeah
 
also stochastic processes, I guess
 
@knight okay, found the source of what i was thinking of, and it's not your problem. the problem in their case was $x^3+y^3+z^3=n$ for $n=1$ to $100$
 
1:52 PM
yeah thats proper probability theory; i dunno anything about those unfortunately
 
most of the cases there can be resolved without much work
(in the sense of being able to find x,y,z or prove they don't exisit)
but the cases of n=33 and n=42 were apparently much harder, especially n=42
because...reasons?
 
?
 
Are Bala and Thor just joking about Probability theory’s relation to integer solutions or something is there really?
 
i don't think they were asserting such. the conversations just overlapped a bit
 
1:56 PM
Oh!
 
that said, there is a relationship to probability theory insofar as -finding- solutions diophantine equations go
namely, suppose the only way you know how to look for solutions is to test a bunch of integers
 
I too found something, if we have a linear Eqaution with coffexients as 1 then there are infinite integer solutions
 
in that case, you're having to search a very large solution space and hope you find it. probabillity theory certainly dictates how much luck you may need to find it in that case
yeah, linear diophantine equations do have established theory
 
Markov chains are nice
 
Yeah
 
1:58 PM
calculating hitting times/expectation by recursion
 
but anything beyond that? oof
 
really cool idea, tho that's about all I know
 
And how do we proof things like these “Prove that there are no integral $p$ and $q$ such that $$p^2 -56q^2 =105$$
Where can I find things like ^ ?
 
that's definitely in the spirit of pell's equation
which is a subject i've managed to avoid learning about
 
:-)
 
2:00 PM
it's in the realm of Olympiad math which i can't say i give a **** about
 
Aren’t these things covered in Higjer Algebra books like Hall and Knight
??
@Semiclassical I too don’t give a f*uk shit Olympiads and to any other comptetive exams. I just hate them
 
maybe? algebra books published in 1913 are generally not on my radar
 
okay... :(
 
"The more complicated equation $x^2-Dy^2=c$ can also be solved for certain values of $c$ and $D$, but the procedure is more complicated (Chrystal 1961)."
 
I have a digital copy of Chrystal
 
2:05 PM
neat
 
Thank you so much Sammy
 
2:45 PM
@robjohn no issue actually i posted a question three to four days ago but unfortunately no one has noticed.
or it was very bad to discuss
2 days ago, by Yuvraj
10 hours ago, by Yuvraj
hi guys morning/evening/afternoon/noon actually i was reading about the graohas of polynomial function $x^3$ and $x^2$ etc and they can easily be traceable ,but if i look a the $z^2$ and $z^3$ does they have any physical significance?I mean if i multiply z by z it onl gives me another complex number if it is not purely real or purely imaginary ,that is contradicting for me if i take $z$ as purely real then $z^24 surely have a physical meaning ,i am confuse in it!
 
3:15 PM
@Yuvraj I'm not exactly sure what you're asking. What is "graohas" and how is it "traceable"? Are you asking about plotting functions of complex numbers?
 
sorry sir for writing errors i mean graphs, yes sir I want to find the plane of the complex umbers
o i was thinking that can we make of plane like we draw the graph of real functions
@robjohn
 
you can make $3$-D plots of the real part or imaginary part or absolute value. Since we only have $3$ dimension to display in, we cannot plot a function $\mathbb{C}\mapsto\mathbb{C}$.
 
3:33 PM
How can I write an equation in colored form?
 
$\color{red}{\text{c}}\color{orange}{\text{o}}\color{yellow}{\text{l}}\color{green}{\text{o}}\color{blue}{\text{r}}$?
 
Do I need to remember the color codes?
 
sir can we plot like this like imaginary point on y axis and real on x axis
$\color{black}{\text{Z}}$
 
@Yuvraj You can plot complex points, but that is not a function.
 
i agree
but if the complex number is purely real
 
3:43 PM
So a function $\mathbb{R}\mapsto\mathbb{C}$ might be drawn as a curve, if you tick off the argument, or as a curve in $\mathbb{R}^3$.
@Yuvraj Well, that would be the same as plotting the real part
 
ah yes
$$ \displaystyle \lim_{x \to 1} \left( \dfrac{-ax + \sin(x-1) + a} { x + \sin(x-1) -1 } \right)^{\dfrac{1-x}{1-\sqrt x} } = \dfrac 1 4 $$
i nned to find largest value value for which the non-negative integer ( a )+1/4
i put the a=2 value in this
and a=0 for the second value
and answer is 0 which is the largest nonnegative value
i do not know where i am getting wrong
 
4:01 PM
Robbie sir if you don’t mind can I ask something now? (I asked for mind because you’re mentoring Yuvraj right now)
 
Good morning, @robjohn. Hello, @Knight.
 
@TedShifrin Hey, Ted!
 
Heya, demonic!
 
Problem Statment: Find all positive integral solutions of $$7x + 12 y =220$$
Hello Ted
The solution given in the textbook is: *Divide the equation by $7$, that is smaller coefficient, $$ x+ y +\frac{5y}{7}= 31 + \frac{3}{7} \\ x +y + \frac{ 5y-3}{7}= 31$$
 
4:07 PM
What is $\Bbb C[x_1, \cdots, x_n]/(e_1, \cdots, e_n)$ where $e_i$ are the elementary symmetric polynomials?
 
After doing this the book writes *Since $x$ and $y$ Are to be integers we must have $$ \frac{5y-3}{7}= integers$$
I want to know why he left everything else and just equated $\frac{5y-3}{7}$ to an integer?
 
Because everything else is integers.
 
For $n = 2$ this is $\Bbb C[t]/(t^2)$. For $n = 3$ already looks to be some computation: $x = -(y + z)$ so $xy + yz + xz = yz + x(y + z) = yz - (y + z)^2 = -(y^2 + z^2 + yz)$ and $xyz = yz(y + z)$
$\Bbb C[y, z]/(y^2z + yz^2, y^2 + z^2 + yz)$ huh
 
Boo, a @Balarka
@Knight: How could it not help you?
 
$z^3$ pops out
 
4:10 PM
Okay, okay .... an Eqaution having everything as an integer would give us only integral value?
 
I don't follow.
 
why would you quotient by the symmetrics
 
Thorgott asking the real question
 
@Balarka: Presumably this is related to the coordinate ring for the symmetric product?
 
Because reasons
 
4:12 PM
Ah, you're in the middle of a question, and I missed the question. No problem. I have to leave in a few minutes anyhow.
 
We have $$x + y +\frac{5y-3}{7} = 31$$ and you said x and y are already integers so we should only be concerned about that fractional part.
 
unsure, @Ted. It doesn't seem to have a very clear geometric interpretation
 
Yes, so that gives you possible values for $y$. Then you use the equation to solve for $x$ for each of those $y$ values.
@Balarka: It sure must be related somehow to the symmetric product. Something factors through something.
 
But in the original equation $$7x + 12 y = 220$$ we can assume the same thing that both terms of x and y will be integral because integer multiplied by an integer is an integer
 
You're trying to solve it, @Knight.
 
4:14 PM
@TedShifrin Yes
 
Neat result: a space is completely regular iff it embeds into a cube $[0,1]^I$
 
So if $(5y-3)/7$ has to be an integer, what are the possible values of $y$? Everything else is gone.
 
And the proof is not hard at all either
 
@TedShifrin The algebraic symmetric product would be the GIT quotient $\Bbb C^n/S_n$, whose coordinate ring is $\Bbb C[x_1, \cdots, x_n]^{S_n} = \Bbb C[e_1, \cdots, e_n]$
 
@TedShifrin Oh! I got you
 
4:16 PM
Hmm, good point, @Balarka.
 
@Thorgott Seems like a natural ring, we should know what it looks like
Okay I mean I think it's sort of clear that $(e_1, \cdots, e_n)$ is an ideal containing $(x_1^n, \cdots, x_n^n)$.
So we're definitely looking at some quotient of $\Bbb C[x_1, \cdots, x_n]/(x_1^n, \cdots, x_n^n)$.
 
4:31 PM
$\mathbb{C}[e_1,...,e_n]$ is the natural ring here, not this
 
@Thorgott I see
 
@Thogott Why not? You look at polynomials upto symmetric polynomials. Seems fair to me!
Well, upto symmetric factors
 
@Thorgott The little bit that I'm learning about group theory seems really cool. My thesis advisor is an algebraic geometer and number theorist, so eventually I'll ask them for some advice. But I wanted to think about all this myself a bit first. Can I tell you something I notice, at least in circles around me? The ones who head down the Algebra path land up writing textbooks and are in primarily teaching careers. Is it just really hard to write papers in Algebra / pure math?
 
@Joanna Writing papers is hard in all branches of math
 
Some around me, in our dept have big-time research careers, but that's like two of them. In contrast, almost everyone in Applied Math around me is doing something research-related, with multi-year grants from the NSF, DOE, etc.
@TobiasKildetoft yeah, but is writing a paper in, say, group theory, prohibitively hard?
 
4:38 PM
Not many people write papers in say finite group theory because most of it is "done".
 
I wouldn't know anything about writing papers
 
@Joanna Not prohibitively (seeing as I have done just that)
 
@BalarkaSen Interesting, I was trying to ask that question but wasn't sure how to ...
 
@Balarka except for a comprehensive account of the things that have been done
 
@TobiasKildetoft Oo Congrats :)
 
4:40 PM
yeah they still haven't written down the proof of classification of finite simple groups have they
as in, a totally comprehensive account
 
I think there is an ongoing project for that
 
I think someone was working on it, but yeah
 
To have it all in one place written down in a somewhat readable way
 
"As of 2019, eight volumes of the second generation proof have been published (Gorenstein, Lyons & Solomon 1994, 1996, 1998, 1999, 2002, 2005, 2018a, 2018b). In 2012 Solomon estimated that the project would need another 5 volumes, but said that progress on them was slow. It is estimated that the new proof will eventually fill approximately 5,000 pages. (This length stems in part from the second generation proof being written in a more relaxed style.)"
 
@Joanna I mean to be clear group theory is a fairly vast subject. There's all kinds of other things people do; studying infinite groups (less using algebra and more using geometry, see eg. geometric group theory), representation theory (@Tobias does this I believe), ...
 
4:43 PM
@BalarkaSen Well, did :)
 
What is specifically "done" is everything that centers around finite simple groups
@TobiasKildetoft Hah. I don't value being in academia that much so I would still address you as a representation theorist though
 
Sure, I still see myself as one as well. But I don't actively do any research
 
@TobiasKildetoft Are you a professor?
@BalarkaSen How come you don't value it much?
 
I don't know if I have a coherent response. I don't like academia in general, and have trouble seeing myself as a potential academic in the future.
 
@BalarkaSen I hear ya
 
4:50 PM
can anyone recommend a good math podcast? I already listened to "A Brief History of Mathematics" by Marcus du Sautoy
 
@BalarkaSen I don't like the idea of having higher social responsibility, for instance. If I want to be provocative, and it's on social media, I'd like to do that without losing my job or being reprimanded, just like bankers, techies, artists, singers and actors can be. But, I've done some interesting research -- some call it a "great discovery", in fact, but it's early in the process in that particular question, and I actually want to move away from it, since it's gotten mired in conflicts.
So, I want to move away from Applied Math, basically -- and I'm lost and need some direction.
 
@Joanna No, I am not a professor. I used to be a postdoc, and now I am a software developer
 
@TobiasKildetoft I see :)
@TobiasKildetoft Was representation theory pretty fulfilling for you?
 
Nice
 
4:57 PM
Well yeah I mean it's kind of laughable that scientific academia is valued so much amongst the general public. Especially math, because it's -- well -- pretty useless.
It's insane that a limited number people get grants to pump out a specific number of research papers in math and others fight for that position because it's all kind of pointless anyway
 
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