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AbstractAlgebraLearner
AbstractAlgebraLearner
12:00 AM
Can you briefly explain G. Group?
I know field of fractions
I know a Grothendieck group is like $\Bbb{Z} $ from $\Bbb{N}$ or adding in inverses minimally to input to get a group, right?
And so how did you take G. Group of semiring, the monoid $\cdot$?
 
user76284
user76284
Yeah, you take pairs of elements of the original monoid.
 
AbstractAlgebraLearner
AbstractAlgebraLearner
I see, then free group on their direct product
then quotient by certain elements
sort of like tensoring
 
user76284
user76284
The first set of equations in my post.
(a, b) = (c, d) iff a + d = b + c and so forth.
where (a, b) "means" or is interpreted as a - b.
 
AbstractAlgebraLearner
AbstractAlgebraLearner
Why don't you just attach a sign flag to everything like you do in CS
data structures
$\operatorname{sgn}(x) \in \Bbb{Z}_2$
 
user76284
user76284
The Grothendieck group construction is more natural.
With the sign construction you get signed zeros.
 
AbstractAlgebraLearner
AbstractAlgebraLearner
12:05 AM
They say though in the first answer that you can't do it as you've posted
Is that correct?
in regards to the constraints you've given
 
user76284
user76284
That's something else, that's regarding extending from $\mathbb{Q}$ to $\mathbb{R}$.
 
AbstractAlgebraLearner
AbstractAlgebraLearner
Oh, Ic
 
user76284
user76284
Everything works fine up to $\mathbb{Q}_\text{Ord}$.
 
AbstractAlgebraLearner
AbstractAlgebraLearner
I see now
So $(\Bbb{N} \to \Bbb{Z} \to \Bbb{Q})[X]$ all are related and workable with your problem?
By injective inclusions?
 
user76284
user76284
I only have to deal with $\mathbb{N}$ and correspondingly $\mathbb{N}_\text{Ord}$ at the moment.
 
AbstractAlgebraLearner
AbstractAlgebraLearner
12:07 AM
Except with substript $\text{ord}$ meaning it's a similar construction to naturals to ints to rationsl
Yes, but then you don't have a group to work with
 
user76284
user76284
I'm just trying to figure out how to define the exponentiation for ordinals.
 
AbstractAlgebraLearner
AbstractAlgebraLearner
you want the group!
 
user76284
user76284
No, why?
$\mathbb{N}$ is closed under exponentiation.
 
AbstractAlgebraLearner
AbstractAlgebraLearner
So then you can do $e^{X} = $ taylor's analogous formula
 
user76284
user76284
I guess that's one way you could do it, but I don't have to.
I can keep everything in $\mathbb{N}$.
There will also be problems with that definition.
 
AbstractAlgebraLearner
AbstractAlgebraLearner
12:08 AM
Okay, so mult is defined could you restate it though?
Here for the record
 
user76284
user76284
You'd have to define something like transfinite series.
For addition and multiplication the Hessenberg sum and product suffice.
I'm trying to restate my question more precisely.
 
AbstractAlgebraLearner
AbstractAlgebraLearner
I think to define $e$ you can still use taylor's formula
It's all $+$'s!
for $e^{x}$
Now, check its properties one by one
Which ones do you have difficulty with?
Probably $e^{x + y}$
 
user76284
user76284
@AbstractAlgebraLearner There's 2 problems: First, I don't even have $\mathbb{R}_\text{Ord}$, since my attempted constructions failed (hence the answers to the question).
 
AbstractAlgebraLearner
AbstractAlgebraLearner
Yes, but you have approximations of $e^{x}$
Sums up to the $n$th term
which are rational
So what if you did another completion of $\Bbb{Q}_{\text{ord}}$ that says $e^x$ converges for any input.
That is in $\Bbb{Z}$ of course
In other words, take the field extension $\Bbb{Q}(\{e^x : x \in \Bbb{Z}\})$.
@user76284 since I've taken a look at yours, do you have time for glancing at mine ?
@user76284 it's really about discrete optimization: SGP
 
user76284
user76284
12:29 AM
@AbstractAlgebraLearner I don't know much about that question, unfortunately.
 
AbstractAlgebraLearner
AbstractAlgebraLearner
It's actually easier to conceptualize than yours since we're dealing with the single op of concatenation of strings
 
 
1 hour later…
Stan Shunpike
Stan Shunpike
1:44 AM
@TedShifrin hey ted!
 
user574847
user574847
2:21 AM
Hello all. Is it easy to prove Riemann-Hurwitz theorem for unramified covers of compact connected orientable surfaces by other compact connected orientable surfaces?
 
user574847
user574847
3:06 AM
Nvm, yeah I proved what I wanted
 
Ted Shifrin
Ted Shifrin
Hi @Stan. Sorry, I was not here even though I was here.
@user574847: Unramified means just covering space. So it's just standard from topology (or the algebraic geometry proof works fine).
 
user574847
user574847
Yeah I just proved it by lifting a CW structure by lifting the characteristic maps
Is it easy to prove the excision property for the euler characteristic?
 
skullpatrol
skullpatrol
 
user574847
user574847
I.e. that $\chi(X)=\chi(C) + \chi(X\backslash C)$ for $C$ closed in $X$
 
Ted Shifrin
Ted Shifrin
That doesn't sound correct without hypotheses, @user574847.
 
user574847
user574847
3:20 AM
Say $X$ is a closed orientable $n$ manifold maybe
 
Ted Shifrin
Ted Shifrin
It was the closed subset $C$ that troubled me.
 
user574847
user574847
Oh
I was looking at this: www3.nd.edu/~lnicolae/EulerChar.pdf
I guess it matters what the dimension of $C$ is probably
 
Ted Shifrin
Ted Shifrin
No, apparently not. I haven't thought about these questions in a long time. I was thinking you would want some sort of neighborhood retract and then use Mayer-Vietoris. But I don't know offhand.
 
user574847
user574847
That sounds like a good idea
 
Stan Shunpike
Stan Shunpike
@TedShifrin Do I need to go through your multivariable materials before trying anything with manifolds?
I've started them and I'm just making sure u think that's what makes the most sense given where i'm at
the robot prof running that group said he thinks reading Lee's book is important for understanding robotics
which caught me off guard a bit, but that may me i need some preparatory math work before i can do the robotics
 
Ted Shifrin
Ted Shifrin
3:27 AM
Lee is well-liked, but he is wordy.
Yeah, my book (the stuff I mentioned to you weeks ago) is prerequisite. The derivative as linear map, inverse/implicit function theorems, and then you will need differential forms, most likely, for robotics.
Hey there @robjohn. I won't even ask about the skirmishes I walked in on the other day.
 
Stan Shunpike
Stan Shunpike
ok i'll start grinding on that then. Do you have a separate list of recommended problems for that? I'm resuming linear algebra this week too, so I will do both at once
 
skullpatrol
skullpatrol
\o @robjohn
 
Ted Shifrin
Ted Shifrin
You don't need to be an expert, but you definitely need to get used to thinking of the derivative as a matrix and working with it. So do exercises.
 
skullpatrol
skullpatrol
Sorry for my part in the skirmish @robjohn
I should know better :(
 
Stan Shunpike
Stan Shunpike
@TedShifrin ok i will get started
thanks a lot :) saves me a ton of time wasted spinning my wheels
i tried before to learn manifolds, so clearly i need to gain basic competency first
 
Ted Shifrin
Ted Shifrin
3:35 AM
I don't like dealing with abstract manifolds until you get used to them as they sit in $\Bbb R^n$. So at least get through that part in my book (last section of chapter 4 and then chapter 6 section 3).
Yeah, @skull, I don't see why you had to open your mouth (except you like doing that).
 
skullpatrol
skullpatrol
@StanShunpike The professor's videos will turn your spinning wheels into a drag racer!
 
robjohn
robjohn
@skullpatrol I should have made the room private.
 
skullpatrol
skullpatrol
yup, live and learn :-)
 
robjohn
robjohn
@TedShifrin it was carry-over from another room that I should have not left open.
 
Ted Shifrin
Ted Shifrin
I gathered as much. I've never had anything but pleasant dealings with AmWhy, but things are tense everywhere these days. :(
 
robjohn
robjohn
3:41 AM
@TedShifrin I'm glad that has been your experience.
 
skullpatrol
skullpatrol
::keeps mouth zipped::
 
robjohn
robjohn
@TedShifrin: I included that mitotic circle image from the other day in the answer that I was working on when I happened upon it.
@skullpatrol good job
 
skullpatrol
skullpatrol
:D
 
Ted Shifrin
Ted Shifrin
Very cool, @robjohn.
Very rare for skull to zip.
 
robjohn
robjohn
@TedShifrin it was simply supposed to be a circle with its center off the origin.
 
Ted Shifrin
Ted Shifrin
3:48 AM
Oh, but not the easy $r=2a\cos\theta$ or $r=2a\sin\theta$.
 
robjohn
robjohn
@TedShifrin everyone is capable of learning.
 
Ted Shifrin
Ted Shifrin
Sometimes I wonder in here.
 
skullpatrol
skullpatrol
I will try harder
 
Ted Shifrin
Ted Shifrin
I mean, everyone has bad days, but I shouldn't have to repeat things 3 times with no effect.
 
robjohn
robjohn
@TedShifrin those do end up being the $a=\pm1$ cases
$r=a\cos(\theta)+\sqrt{1-a^2\sin^2(\theta)}$
 
Ted Shifrin
Ted Shifrin
3:56 AM
Gotcha.
 
user574847
user574847
If $T^2$ is acted upon by $\Bbb Z/2\Bbb Z$ induced from the antipodal action on $S^1$, is the quotient $T^2/(\Bbb Z/2\Bbb Z)$ just $T^2$ again?
 
robjohn
robjohn
@TedShifrin the animation for that is rather boring; a circle going back and forth.
 
Ted Shifrin
Ted Shifrin
Yup, boring.
 
robjohn
robjohn
@user574847 would it be $\mathbb{P}\times\mathbb{T}$?
 
user574847
user574847
@robjohn As in $\Bbb P^1(\Bbb R)\times S^1$? What is $\Bbb T$?
 
robjohn
robjohn
4:05 AM
@user574847 I'm just guessing here. I was assuming $\mathbb{T}=S^1$
 
Ted Shifrin
Ted Shifrin
Yes, @user574847. Because the group action is orientation-preserving, so you have to ask what surface(s) the torus double covers.
 
user574847
user574847
Ah
 
robjohn
robjohn
@TedShifrin Hmm... I guess, now that I think of it, what I was thinking of as $\mathbb{P}$ is just $\mathbb{T}$.
 
user574847
user574847
Yeah I thought so @robjohn
 
Ted Shifrin
Ted Shifrin
I thought you meant projectivizing the torus, @robjohn.
 
robjohn
robjohn
4:08 AM
@user574847 so your first statement is indeed true.
@TedShifrin nothing so grand. It's been decades since I've thought about this stuff, so I'm a bit slow.
 
user574847
user574847
@TedShifrin Right, so it's orientation preserving, and the resulting quotient is then orientable? So then I can use something like Riemann-hurwitz, or whatever you want to call it
Well, first, is the cover branched, or unbranched?
I guess it's branched
at two points
 
Ted Shifrin
Ted Shifrin
You don't need Riemann-Hurwitz for plain old covering spaces. That's boring.
Whoa. Why do you say it's branched?
 
user574847
user574847
Well
$T^2\to T^2/C_2$ is a double cover, except at points which had orbit of size 1
Such as
$(1,1)$ for example
 
Ted Shifrin
Ted Shifrin
You're sending $(x,y)\rightsquigarrow (-x,-y)$, with $\|x\|=\|y\|=1$.
Huh?
There are no fixed points.
 
user574847
user574847
My action is by complex conjugation on each of the circles, which isn't rotation, which your action is I think?
Oh
I messed up my original statement sorry
 
Ted Shifrin
Ted Shifrin
4:14 AM
Grrr.
 
user574847
user574847
I wanted the induced action by complex conjugation sorry
Sorry for the confusion ted
 
Ted Shifrin
Ted Shifrin
OK, so, yeah, you'll have four fixed points.
 
user574847
user574847
At $(\pm 1,\pm 1)$
 
Ted Shifrin
Ted Shifrin
But Riemann-Hurwitz is not in play. This is not in the holomorphic world.
 
user574847
user574847
Is this still orientation preserving?
 
Ted Shifrin
Ted Shifrin
4:16 AM
Yes, product of orientation-reversing maps.
But the quotient is singular.
Notice that the quotient when you have the circle is now a closed interval.
This suggests that we get an actual filled-in square.
 
user574847
user574847
That seems reasonable
 
Ted Shifrin
Ted Shifrin
Weird.
I guess I've never encountered this before.
 
user574847
user574847
So is the euler characteristic of the closed square just $1$?
 
Ted Shifrin
Ted Shifrin
It's contractible.
 
user574847
user574847
4 vertices, 4 edges, 1 2-cell
True, so it's 1 that way too
 
Ted Shifrin
Ted Shifrin
4:20 AM
But, as I said, don't go anywhere near Riemann-Hurwitz.
It doesn't apply to non-holomorphic maps or to singular spaces.
 
user574847
user574847
I think you can prove that a version of it does
 
Ted Shifrin
Ted Shifrin
This thing fails to be a covering map along the (preimage of) the whole boundary of the square. This is far, far from a branched covering.
 
user574847
user574847
I.e. if you have a degree $n$ cover $X\to Y$ of finite CW complexes, then you can show that $\chi(X)=n\chi(Y)$
 
Ted Shifrin
Ted Shifrin
No, that's totally wrong.
That's for an actual covering space.
 
user574847
user574847
Yeah I was getting to that part
Then since you have excision for orientable $n$-manifolds by a finite set of points $S\subset M$, $\chi(M)= \chi(M\backslash S) + (-1)^n|S|$
For a simply branched cover of degree $d$, $X\to Y$, you have the actual cover $X\backslash p^{-1}(S) \to Y\backslash S$
Which gives you $\chi(X\backslash p^{-1}(S)) = d\chi(Y\backslash S)$. So if $X$ was genus $g$, and $Y$ was genus $h$, then we would have by excision $d(1-g_1)=2d(1-g_2)-|S|(d-1)
 
Ted Shifrin
Ted Shifrin
4:24 AM
The topology is not right. It's not a covering map over the boundary of the square.
 
user574847
user574847
$d(1-g)=2d(1-h)-|S|(d-1)$ (so here 2(1-1)=4(1-h)-4, so $h=1$?
 
Ted Shifrin
Ted Shifrin
What you're talking about is still topological 2-manifolds without boundary mapping to topological 2-manifolds without boundary.
 
user574847
user574847
Ah
That's true
 
Ted Shifrin
Ted Shifrin
So you should give serious thought to whether my square picture is correct. I'll leave you to that.
 
user574847
user574847
Sure thing
It does seem geometrically like the right picture
Do you get the same object by taking $T^2\to T^2/C_2\to (T^2/C_2)/C_2$ where in the first case you have $C_2$ act by complex conjugation on only the first factor of $S^1\times S^1$, and then in the second, on the equivalence classes of the second component
 
Stan Shunpike
Stan Shunpike
4:34 AM
Howdy
 
Knight
Knight
@TedShifrin I’m getting a discrepancy.
I want to prove $$a^3 -2 \equiv 0 \mod 3$$ is not possible
 
user574847
user574847
Well you're not in luck, since $2^3=8$ and $8 \equiv 2 \pmod 3$
Btw @Knight you only needed to check $a=0,1,2$ to check that statement
 
Knight
Knight
@user574847 Let’s begin like this $$a^3 \equiv 0 \\ a^3 \equiv 1 \\ a^3 \equiv 2 \\ a^3 \equiv -2$$
 
user574847
user574847
The heck is that
 
Knight
Knight
We have the last one, because $8 \equiv -2 $
 
user574847
user574847
4:41 AM
I just gave you one, yes
Also no
$8\equiv 2 \pmod 3$ not $1$
 
Knight
Knight
yes Sorry
 
user10478
user10478
Is it possible to type the division symbol that looks like $\sqrt()$ without the handle?
 
user574847
user574847
Note that $-2\equiv 1 \pmod 3$
 
Knight
Knight
@user574847 But we have $$-2 \equiv -2 $$ also?
 
user574847
user574847
Yes, typically you'll find that $x\equiv x$ w.r.t to any equivalence relation by reflexivity
 
Knight
Knight
4:44 AM
So, we if use $-2 \equiv -2$ then for $a^3 \equiv 2$ we will get $a^3 -2 \equiv 0$
 
user574847
user574847
What does it mean to "Use" that expression?
 
robjohn
robjohn
@Knight The thing we worked on yesterday involved $a^3\equiv2\pmod9$
is that what you're asking about?
 
Knight
Knight
@robjohn Yes sir. Today we got something different :-)
 
user574847
user574847
If $a^3-2\equiv 0\pmod 3$ then $a\equiv 2\pmod 3$ as I said before
 
Knight
Knight
So the thing I want to prove is wrong?
 
user574847
user574847
4:47 AM
Yes
That's why I said "You're not in luck"
 
robjohn
robjohn
@Knight what are you trying to prove?
 
user574847
user574847
:P
 
Knight
Knight
But see it this way, look only for positive congruences
@robjohn $a^3 -2 \equiv 0 \mod 3$ is not possible
 
Edward Evans
Edward Evans
@Knight why are you against just checking all the possible values of $a$ for one that works? $3$ is fairly small as natural numbers go
 
user574847
user574847
$\Bbb F_3$ has three elements, $0,1,2$. There is only $2^3=2$, otherwise $0^3=0$ and $1^3=1$
@Knight It literally is, as I've said thrice. Take $a=2$
 
Knight
Knight
4:49 AM
Okay
 
robjohn
robjohn
@Knight but it is... $a\equiv2\pmod3$
 
Knight
Knight
Sorry for the disturbance.
 
robjohn
robjohn
@Knight why did you think you needed to show that $a^3\equiv2\pmod3$ was impossible?
 
user574847
user574847
@Knight np, it's normal while learning
2
 
Knight
Knight
5:05 AM
@robjohn I wanted to show $$3a^3 +b^3=6$$ have no integral solution.
@user574847 Thanks for the encouraging words :)
So, what I did was $$ -b^3 = 3a^3 -6 $$ That means $3 | b^3$ well that implies $$ b= 3m$$
Substituting this into our equation we will get: $$ 3a^3 -6 = -27 m^3 \\ a^3 -2 = -9m^3$$
Well that means $$a^3 -2 \equiv 0 \mod 3$$
But since $a^3 -2 \equiv 0 \mod 3 $ is true, it tells us that the original equation can have a integer solutions
 
Edward Evans
Edward Evans
It only tells you that $a \equiv 2 \bmod 3$, which you can use
 
Knight
Knight
5:27 AM
Doesn’t that imply that the equation have an integral solution?
 
Edward Evans
Edward Evans
No
 
Knight
Knight
But I worked by assuming that integral solutions exists and found no contradiction
Please tell me Ed, I think I’m missing something
Something essential
 
Edward Evans
Edward Evans
What does $a \equiv 2 \bmod 3$ mean?
 
Knight
Knight
There exists an $a$ such that when $2$ is subtracted from it it becomes a multiple of 3
 
Edward Evans
Edward Evans
Okay so $3 \mid a - 2$
What does $3 \mid a - 2$ mean?
 
Knight
Knight
5:30 AM
$a-2$ is divisible by 3
 
Edward Evans
Edward Evans
Yeah sure, but more usefully it means that there's an integer $n$ such that $a - 2 = 3n$ right?
 
Knight
Knight
Yes
 
Edward Evans
Edward Evans
Okay so we've established that $a = 3n + 2$ and $b = 3m$ for some integers $n, m$
try plugging those in and see if you can find a contradiction
 
Knight
Knight
Okay!
I will get this:
$$27n^3 +6+54n^2 +36n =-9m^3$$
 
Edward Evans
Edward Evans
You're missing something in your expansion of $(3n+2)^3$
and also $(3m)^3 \neq 9m^3$
 
Knight
Knight
5:44 AM
We had $$a^3 -2 =9m^3$$
And I just plugged in $$(3n+2)^3 -2=9m^3$$
 
Edward Evans
Edward Evans
woops sorry, thought you plugged them straight into your original equation :)
 
Knight
Knight
:)
 
Edward Evans
Edward Evans
Anyway, what can you say about this new equation?
 
Knight
Knight
13 mins ago, by Knight
$$27n^3 +6+54n^2 +36n =-9m^3$$
Hmm... we can divide whole Eqaution by 3
 
Edward Evans
Edward Evans
Right, go ahead and do that
 
Knight
Knight
5:49 AM
$$9n^3 +18n^2 +12n +2 = -3m^2$$
 
Edward Evans
Edward Evans
$m^2$ should be an $m^3$, but yes. What happens if you reduce this guy modulo $3$?
 
Knight
Knight
$$9n^3 +18n^2 +12n +2 \equiv 0 \mod 3$$
 
robjohn
robjohn
@Knight same way... note that $3\mid b^3\implies3\mid b$, that is $b=3c$. The equation becomes $3a^3+27c^3=6$. That means $a^3+9c^3=2$ or $a^3\equiv2\mod9$
 
Edward Evans
Edward Evans
@Knight I think you can draw your own conclusion from there :P
 
Knight
Knight
@robjohn Sir can you please guide me with little more intermediate steps, I’m quite new to Modular Arithemtic
@EdwardEvans Thanks Edward
 
robjohn
robjohn
5:53 AM
@Knight can you see that $3a^3+b^3=6$ means that $3\mid b^3$?
 
Knight
Knight
Yes
 
robjohn
robjohn
How about if $3\mid b^3$ then $3\mid b$?
 
Knight
Knight
Yes
Yes and then $b=3c$ and then substitution
 
robjohn
robjohn
so $b=3c$ and then the equation is $3a^3+27c^2=6$, right?
 
Knight
Knight
Divide through out by 3
 
robjohn
robjohn
5:55 AM
Divide by $3$
$a^3+9c^3=2$ means $a^3\equiv2\mod 9$
 
Knight
Knight
aye
And isn’t that possible?
 
robjohn
robjohn
and we showed that was impossible yesterday
 
Knight
Knight
Yes
 
robjohn
robjohn
looking at the cubic residues mod $9$
 
Knight
Knight
So, the main step was changing the modulus from 3 to 9
 
robjohn
robjohn
5:57 AM
same as yesterday
 
Knight
Knight
Yes
 
Silent
Silent
My book says: $\int \frac{2x\,dx+2y\,dy}{x^2+y^2-3}=\ln(x^2+y^2-3)$. But when I go like $\int \frac{2x\,dx}{x^2+y^2-3}+\int \frac{2y\,dy}{x^2+y^2-3}$, I get $\ln(x^2+y^2-3)+\ln(x^2+y^2-3)$, because $\int \frac{2y\,dy}{x^2+y^2-3}=\ln(x^2+y^2-3)$
Where am I wrong?
 
robjohn
robjohn
Look at the paths you're following. In the integral with respect to $x$ you are assuming $y$ is a constant. In the integral with respect to $y$ you are assuming $x$ is a constant
Both give the difference of $\log\left(x^2+y^2-3\right)$ between points, but along paths that are restricted.
Adding both together, gives the difference between any two points in $\mathbb{R}^2$
This is a path integral.
 
Silent
Silent
Thank you so much, Sir!
 
robjohn
robjohn
Make sense?
 
Silent
Silent
6:10 AM
@robjohn Can you please elaborate this more?
 
robjohn
robjohn
$$\int_{(x_0,y_0)}^{(x_1,y_0)}\frac{2x\,\mathrm{d}x}{x^2+y^2-3}=\log\left(x^2+y_0^2-3\right){\large|}_{x_0}^{x_1}$$
 
Silent
Silent
Oh my! So, expanding like this: $\int \frac{2x\,dx}{x^2+y^2-3}+\int \frac{2y\,dy}{x^2+y^2-3}$ is not legal, right?
 
robjohn
robjohn
$$\int_{(x_1,y_0)}^{(x_1,y_2)}\frac{2y\,\mathrm{d}y}{x^2+y^2-3}=\log\left(x_1^2+y^2-3\right){\large|}_{y_0}^{y_1}$$
@Silent it is, but you have to realize that when you compute the integral like that, it depends on a particular path
Unless the form you're integrating is conservative or exact
(same thing, but talking to a physicist or a mathematician)
 
Silent
Silent
too much for me :)
 
robjohn
robjohn
you are learning about path integrals right now, correct?
 
Silent
Silent
6:18 AM
yes
 
robjohn
robjohn
This is just part of understanding path integrals.
 
Silent
Silent
But one thing is bothering me is:
 
robjohn
robjohn
Look at the two integrals I wrote above...
The integral in $x$ holds $y$ constant
and vice versa
 
Silent
Silent
@robjohn This is 'definite inegral', whereas my was indefinite or antiderivative!
@robjohn yes, i got that
 
Knight
Knight
@robjohn Sir I’m still astonished how just changing the modulus from 3 to 9 solved our problem.
 
robjohn
robjohn
6:21 AM
That is where the exact form comes in... $d\!\left(x^2+y^2-3\right)=2x\,\mathrm{d}x+2y\,\mathrm{d}y$
@Knight the cubic residues mod $3$ are complete, mod $9$ they are not.
 
Knight
Knight
Okay
 
robjohn
robjohn
$x^3\in\{0,1,2\}\pmod3$, but $x^3\in\{0,1,8\}\pmod9$
 
Knight
Knight
But if I were to consider 3 I would have lived all my life believing that that equation is solvable in integers
 
user574847
user574847
Hey again @TedShifrin. I wanted to show that for $M$ a closed orientable $n$-manifold, and $S\subset M$ a finite set of points, that $\chi(M)=\chi(M\backslash S) +(-1)^n|S|$, and I've already shown that for $X$ with two subspaces $A,B$ appropriate for Mayer-Vietoris, that $\chi(X)=\chi(A)+\chi(B)-\chi(A\cap B)$.
 
Knight
Knight
So, can you give me some Guru’s advice for future ?
 
robjohn
robjohn
6:25 AM
@Knight you should have tried to solve it and seen the difficulty
 
user574847
user574847
But I seem to get a different result. I use the manifold structure to pull back balls around each point in $S$, and take the union of all these 'balls in $M$' and call it $A$. I then take smaller closed balls in side each of these balls, containing the points of $S$, and take them back into $M$, take their complement, and call it $B$.

So in short, I have $M$ is the union 1) of an open subspace consisting of a bunch of 'balls' around the points of $S$, and 2) the complement of the union of a bunch of closed balls, contained in the open balls of (1).
 
Knight
Knight
@robjohn Actaully, I followed the same method which was used in proving this $$The~equation~ 15x^2 -7y^2 =9 ~have~no~solution~in~integers$$
 
user574847
user574847
$A\cap B$ should be homeomorphic to the disjoint union of $|S|$ copies of $S^{n-1}\times \Bbb R$, and $A$ is homotopy equivalent to the disjoint union of $|S|$-points
So I should get $\chi(M)=|S|+\chi(M\backslash S)-(-1)^{n-1}|S|= |S|+\chi(M\backslash S)+(-1)^n|S|$
 
Ted Shifrin
Ted Shifrin
That sounds right.
 
user574847
user574847
My problem is that what I get out must be wrong, since when $n$ is odd, it means that I can puncture my thing without changing the euler class
 
robjohn
robjohn
6:33 AM
@Knight There, you show that $x=3u$ and $y=3v$ and then your equation devolves to $15u^2-7v^2=1$
Look at that mod $3$
$2v^2\equiv1\pmod3$ has no solution
 
Knight
Knight
Yes and I tried the same reasoning in our cubic equation.
 
Ted Shifrin
Ted Shifrin
Oh. Duh. Your intersection term is wrong.
 
user574847
user574847
Is it?
oh
Ohhh
 
robjohn
robjohn
@Knight the quadratic residues are not complete mod $3$
 
Ted Shifrin
Ted Shifrin
Yup.
 
user574847
user574847
6:35 AM
Thanks :D
 
Ted Shifrin
Ted Shifrin
You're welcome.
 
Knight
Knight
@robjohn What are quadratic residues? Can you please please write that sentence in simple words (Sorry for being too demanding)
 
robjohn
robjohn
quadratic residues are the set of residue classes (remainders) gotten out of $x^2$ mod $3$
$0^2\equiv0$, $1^2\equiv1$, and $2^2\equiv1$ mod $3$
So the quadratic residues mod $3$ are $0$ and $1$
 
Knight
Knight
Okay, but how is it significant
 
robjohn
robjohn
$2$ is not a quadratic residue mod $3$ so $2v^2\equiv1\pmod3$ implies $v^2\equiv2\pmod3$
which is impossible
 
Knight
Knight
6:41 AM
Yes.
Sir, I want to know why $3$ Didn’t work in our cubic Eqaution proof? Why do you had to pick 9?
 
robjohn
robjohn
@Knight Because the cubic residues mod $3$ are $\{0,1,2\}$, that is, ALL the residue classes
 
Knight
Knight
@robjohn Oh Okay! Means working with 3 couldn’t give us anything ha?
 
robjohn
robjohn
you won't get a contradiction comparing cubic residues mod $3$
 
Knight
Knight
We should pick something whose cubic residues isn’t a complete residue, am I right?
 
robjohn
robjohn
yes. We were naturally lead to $9$ anyway
 
Knight
Knight
6:45 AM
Thank you so much sir.
 
SK19
SK19
7:15 AM
Hi, does anyone know if aimspress.com/journal/Math is a legit journal? I found nothing to the contrary so far, besides maybe the board members coming from all over the world.
Ah, wait, nevermind "AIMS Mathematics will charge a 800 USD Article Processing Charge (APC) for accepted papers after peer review."
 
robjohn
robjohn
7:37 AM
@SK19 I guess it can be open access if the writers are paying $800 to publish their own article.
 
SK19
SK19
that sum is just too damn high
if publishing an article costs more than your average hooker, someone got their priorities wrong imo
 
robjohn
robjohn
7:52 AM
@SK19 guess I'm out of touch with that scale...
 
SK19
SK19
8:03 AM
idk. I mean, if you got money to throw away that's fine. But I would rather publish in a non open-access journal and give the 800€ to charity, you know?
for me personally, 800€ is three months of food
wait, four
four to five.
 
robjohn
robjohn
8:52 AM
@SK19 That was USD, not Euros, right?
 
 
1 hour later…
AbstractAlgebraLearner
AbstractAlgebraLearner
10:14 AM
Let $1 \leq q_1 \leq q_2 \dots \leq q_{n-1}$ be a non-decreasing sequence of prime powers $q_i = p_{\pi(i)}^{\pi(k)}$ where $\pi \in S_n = \{ \text{ perms of } (0, \dots, n-1)\}$.
The smallest grammar of $s = a^{4}$ is $2^2$ itself or alternatively $g = \{A \to BB, B \to aa\}$ which has the same size.
When I say itself I means $g = \{S \to s \}$ really
I want to prove that if for $n = 1, \dots, N-1$ all of $s = 2^{n-1}$ 's smallest grammars can be achieved by taking all of $g = \{ A \to BB, B \to a^{2^{n-2}}\}$ or $h = \{ A \to B^{2^{n-2}}, B \to a^2$ or, $g = \{ A \to B^4, B \to a^{{2^{n-4}}\} $ or $\dots$
 
mick
mick
10:48 AM
1
Q: polynomial equation $ A(x+y_1)(x+y_2)...(x+y_n) + B(x+z_1)(x+z_2)...(x+z_k) = f(x) $ ??

mickConsider given integers $A,B$ such that $AB \neq 0$. Consider a given polynomial $f(x) = a_0 + a_1 x + a_2 x^2 + ... $ of degree $n > 1$ with rational coefficients $a_i$. Now I wonder about solving the diophantine equations of type : $$ A(x+y_1)(x+y_2)...(x+y_n) + B(x+z_1)(x+z_2)...(x+z_k) = f...

number-theory polynomials diophantine-equations factoring vieta-jumping
 
Knight
Knight
11:32 AM
@robjohn Sir have you ever made a cake by yourself at home?
 
SK19
SK19
12:09 PM
@robjohn true, typo. Not much of a difference, though
 
 
1 hour later…
blademan9999
blademan9999
1:26 PM
Can anyone help solve this?
3
Q: Can you define $f(x)$ such that $2^{x}<f(f(f(x)))<2^{2^x}$?

blademan9999Can you define a real-valued function $f$ using standard arthimetical operations such that $2^{x} < f(f(f(x))) < 2^{2^x}$ for sufficiently large $x\in \mathbb{R}$? I know that the rule $f(f(x))=2^{x}$ can't be established with standard arthimetical operations, but is it possible to find a functi...

functions exponentiation problem-solving
I'm being trying to find a function for a while but I can't find one
 
robjohn
robjohn
2:04 PM
@Knight I've made one with my wife, but not by myself.
 
Charlie Shuffler
Charlie Shuffler
2:25 PM
Hi everyone
I have a stylistic question
If I approximate a cosine by its taylor expansion
should I use $\approx$?
Like: cos(x) = 1 -x^2/2 + O(x^4)
or cos(x) \approx 1 -x^2/2 + O(x^4)
does the O(x^4) term provide reason not to use \approx?
 
Knight
Knight
@robjohn You and ma’am made it together, wow! What a lovely scene it would have been
 
 
2 hours later…
Tanvir Joshi
Tanvir Joshi
4:13 PM
hi everyone i had a question on how to solve this question:heres the question:For the set of whole numbers from 1 to 20 inclusive,Tammy knows that some numbers are divisible by 5 and some numbers are odd.She is going to write each number on a different ball and place the balls in a box. If one ball is randomly selected from the box,what is the probability,to the nearest tenth,that the number written on it is divisible by 5 or is an odd number?

my work:

odd numbers=3,5,7,9 P(divisible by 5 U odd number) =P(divisible by 5)+ P(odd number)-P(divisible by 5 U odd number) 3/20+5/20-7/20 =1/20
does anyone know what the correct answer is? thats my work above
 
user10478
user10478
4:38 PM
Is it fair to say that interpreting the elements of a residue system as both numbers and congruence classes is an instance of duality, or is dropping the congruence class notation really only a notational shorthand?
 
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