@AkivaWeinberger That reminds me of a motherboard, hmm...

I'm trying to find the intersection points between $x^s+y^s=1$ and $(1-x)^s+y^s=1$

these points would be algebraic right?

only if $s$ is a positive integer

where can i learn the theory of equations for elementary set theory

specifically, (A n X) U (B N X^c) = {} where A and B are fixed subsets of a universal set and X is a subset of a universal set restrained by the equation it appears in. find a solution

@Ultradark yeah if s is integer, you get a polynomial equation and such roots must be algebraic

not syre what happens if s is irrational though

@secret what if s is rational

It will be still algebraic, as you have taking roots and summing things finite number of times to get zero

It will probably be algebraic?

okay so any point can be written as $P(\Bbb A, \Bbb A)$

I guess I can answer my own question now

Thinking about how to handle if s is transcendental:

if s is transcendental, there exists some series (t) that converge to s. Hence by index laws we have:

$(1-x)^s=(1-x)^{\sum_{k=0}^{\infty}t_k}=\prod_{k=0}^{\infty} (1-x)^{t_k}$

So even if $1-x$ is an integer, depending on the details of $t_k$, the resulting product can limit to a transcendental

Suppose $(1-x)^s$ is algebraic, then there exists some $P$ such that $P((1-x)^s)=0$

moreover: $P((1-x)^s)=P(\prod_{k=0}^{\infty}(1-x)^{t_k})=0$. Ok that does not help

I'm interested in the intersection points in this space

and the relation between the points

Cannot really said much if s is irrational

too many parameters and not enough constraints

it also does not help that finite sums of algebraic is algebraic, so transcendence happens at the infinite limit

Somehow in going to the infinite limit, the difference between the series and any nth partial sum is bounded above by all conceivable rationals that is parametrised by n. If there is a way to resolve this, then we can pin down the occurrence of transcendence

really?

well I am not sure, but diophatine approximation is one of the commonly used techniques in transcendental number theory, while the other is to come up auxiliary functions so that the roots of them fall in convenient places

I was just wondering if there's anything interesting to look at regarding the intersection points for $s \in \Bbb Q \ge 1 $

Nothing much other than the intersections are all algebraic

i am not sure if anything more can be said as if your s < 1/5, then you cannot guaranteed to get surds as the roots will be beyond 5th degree and 5th degree polynomials don't necessary have surd solutions

surds?

any number that can be expressed in terms of +, x and taking nth roots

O wait, I mean constructible numbers, surds is a subset of that which has the form ${}^{n}\sqrt{x}$ for $x \in \Bbb{Q}$

I have a feeling there's something to be found...

Hey @Mathein!

I don't know enough math though

Can you call the collection of algebraic points a space?

That's only a set. A space need at least one relation to relate the points in the set in some structural fashion

do you think there's a way to do that with these points?

That equation reminds of some kind of generalised conics in $L^p$ space, thus presumably some kind of algebraic rotation can be defined on those points

but I don't know of any examples from my head which maps algebraics to algebraic in some natural fashion

mathien night have some ideas of common examples of such maps

and once the space is formalized appropriately, I wonder what kind of space it would be

I'm working on designing a drone system with some kids from school and one of the parameters of the challenge is that we can't use GPS. We decided that Light Detection and Ranging (LiDAR) was the best option and I came across this:

Is anyone familiar with these kinds of equations? I find this interesting, but am still trying to wrap my around it a bit.

Hm... well, I guess it's a little straightforward...

What $\aleph$/expression in $\aleph$ would be assigned to $C$ to denote that $\frac{2t}{C - t^2} = 0$ for any $t$?

Ok, we are getting somewhere

what's a logically equivalent statement to $\lnot P \lor Q$? $P \lor Q$?

P implies Q

$\neg P \vee Q$ is certainly not equivalent to $P \vee Q$

Whoops, I meant $\lnot P \rightarrow Q$

is that equivalent to $P \lor Q$?

Yep!

Ok, thanks :)

Because $\neg P \rightarrow Q \equiv \neg(\neg P) \lor Q \equiv P \lor Q$.

(blargh I'll get the tex right eventually, lol)

Ah, ok. That was my reasoning, but I second guessed how I reduced $\lnot (\lnot P)$

Or in other words: Permutation of sets produces nothing new

does anyone know any reference or literature on the basics of a theory of equations for sets

anyone

@Jasper Happy New Year!!! I wish you be OFF THE CHART in everything you do, in a positive sense, of course.

CHARTS

OK, I have to go now.

It's cool to see the first messages of this room

@Uticensis We're here now, you can come back

Equation of plane perpendicular to x = 0 and passing through (2, –1, 0) can be

y = -1 works right

so does z = 0 ?

For all $n \geq{1}, 4^{n}+5 \equiv{} 1^{n}+2 \equiv{} 1 + 2 \equiv{} 0 \text{ mod } 3$

How does one get from $4^{n}+5$ to $1^{n}+2$

well $4 \equiv 1$ and $5 \equiv 2$

ohhhhhh

Thank you so much!

@CaptainQuestion A line requires two equal signs. So a perpendicular to x=0 at a,b,c would be y=b, z=c.

My bad. It's a plane ye seek.

Hi, could anyone understand time complexity and basic algorithm help me this little problem?: cs.stackexchange.com/q/102209/97362

zenodo.org/record/2529103#.XCoP_u6nxdY

Happy New Year!

@AlexanderGruber Hi. Do you find it OK to post such a link at my user profile of the site, indicating the right spirit to approach the mathematical problems? youtube.com/watch?v=upwyWKzozII&t=5s

It could be inspiring, and it also reveals a true fact: the ground of mathematics is a very tough one. You want to be a super professional.

However, the clip I admit could be wrongly understood and eventually considered rude. That's why I asked.

Anyway, I prepare to celebrate the new year, but not for too long since I have a lot of research to do. I've always found these days perfect for research (toward the end of the year).

@user1357113 Yes, and for non-work hobby projects too.

@b_jonas True.

@user1357113 Well, I might not get well, but I will just keep trying. Happy New Year to you too. =)

I have born in leap day march 18.03.2000 who to calculate my age march 18 leap day in Gregorian clander please say the formula to calculate the age on leap day march 18 is leap day what is my real age?

Please help me what is my age

Please reply to my question

Please please please please

March 18 is leap day how to calculate my age

Please helpe

Is this homework @L.AravindKalaimani

There should be. I'm drawing a blank.

"The nth roots of a group element problem" perhaps?

I need to remember to use tags. As such: definition, group-theory, terminology are relevant to my previous comments.

Of course, if g=/=e, then h=/=e, so I don't know why I specified that h is not the identity.

Is there a good resource/reference for citation conventions in math papers? For example, if Smith wrote a textbook with the standard definition of a "frumious" widget, and I'm writing a paper on semiwidgets with an analogous definition, do I need to call it out explicitly, like "following section 4 of Smith, we say that a semiwidget is "frumious" if [almost identical definition]"?

Hi all, I have a bit of a silly question but

If I make a substitution inside an integral, can I add or subtract a constant times times i (the imaginary number)? Is this a legal operation

I bit like shifting the integration boundaries by a constant, except now by i times a constant

@1010011010 not always, it depends if the integrand has singularities. Are you familiar with complex lineintegrals?

@klirk yes

I am shifting inside a dilogarithm and inside a logarithm, and I would like to use the identity: wolframalpha.com/input/…

The integration boundaries are between 0 and 1, regardless, this seems to constrain what I call $\alpha$ in the link above

$\alpha$ is real in almost all cases that I could see, but may have an imaginary part of a multiple of the shift that I envisioned to apply the identity

Suppose we call the constant $c$, then I would like to shift by $ic$ for example. The denominator may have a $nic$ in it, with $n$ part of the real integers except zero (in which case the identity can be applied trivially)

@1010011010 in the complex numbers, instead of doing substitutions, we chose a different path in the complex plane along which we integrate. Using the cauchy integral formula/ the resiude theorem often such a substitution is valid, but in genereal it is not

Are the branch cuts for the complex polylogarithm identical to the monologarithm?

I've always skipped over the branch stuff in my CA course... guess I bit myself in the ass doing it

@1010011010 Maybe you could try to see what happens if you chose a suitable contour for integration. As you want to integrate over a bounded set, you will probably get correction summands from closing the contour. This may lead to an ugly calculation. You should ask this as a question on the main site.

Perhaps I should note that by shifting my integration coordinates by the above, I obtain a real expression inside the integral, only the variable is complex

But I'm not sure whether one can treat the variable temporarily as real

I'm aware that I'll run into some ugly expressions. I expected no less given the context of my motivation for doing the integral in the first place

I'll formulate something though, thx

$t^4+4 = (t^2+2t+2)(t^2-2t+2)$

hi chat

$t^4+4 = \prod (t \pm i \pm 1)$

hi @Astyx

Hi

numbers are annoying

Are they ?

conjecture: if $\operatorname{char}(K) \nmid n$ and $\mu_n \subseteq K$, then the factors of $t^n - a \in K[X]$ all have the same degree

$\mu_n$ ?

the $n^{\text{th}}$ roots of unity

ok

actually I have a stronger conjecture

That's the same as dealing with $t^n-1$ right ?

how so

Oh yeah

I haven't done those things in a long time

Oh it's blatantly wrong actually

Duh

conjecture: if $\operatorname{char}(K) \nmid n$ and $\mu_n \subseteq K$, then for $a \in K$ there is $ef=n$ and $b \in K$ such that the factorization of $t^n-a$ into primes is $\prod_{i=1}^f (t^e - \zeta^{ie} b)$ where $\zeta$ is a primitive $n^{\text{th}}$ root of unity

for the above case $K = \Bbb Q(i)$, $n = 4$, $a=-4$, we have $e=1$ and $f=4$ and $b=1+i$, i.e. $t^4-(-4) = (t-(1+i)) (t-i(1+i)) (t-i^2(1+i)) (t-i^3(1+i))$

When $char(K) = 0$, e=1 and f=n ?

not necessarily

the characteristic has nothing to do with the factorization, I believe

another example is $K=\Bbb Q$, $n=2$, $a=2$, in which case $e=2$ and $f=1$ and $b=2$, i.e. $t^2-2 = t^2-2$

Isn't your typical algebraic field with char 0 $\Bbb C$ ?

well in $\Bbb C$ the situation is trivial

Yes

so I'm not very interested in that case

But when char is 0, are there any other cases ?

well both cases I gave you are char 0

But $\Bbb Q$ does not contain the roots of unity

it contains the second roots of unity, i.e. 1 and -1

Oh fair enough

for $K=\Bbb Q(i)$, $n=4$, $a=9$, we should have $e=2$ and $f=2$ and $b=3$, i.e. $t^4-9 = (t^2-3)(t^2-i^23)$

Conjecture: $\Phi_\Bbb s(x) \times M_\Bbb s(x) = \Psi_\Bbb s(x)$

@Astyx do you have examples for me to compute :P

or maybe I should just prove it

Not on the top of my head

I think you can take the "smallest b" such that $\exists l|n$ $b^l = a$

If that can make sense

small as in $l$ is big

yeah sure

I don't know if that's well defined though

Then it becomes kinda trivial I think

What does char(K)|n do ?

oh well

if char(K)|n then there is no primitive n-th root of unity

Why ?

I mean, I can still come up with a conjecture, but I don't want to

oh well in that case, x^n-1 = (x^(n/p)-1)^p

ok

one would let n=p^k m with p not dividing m

and then phrase the conjecture similarly

m ?

i.e. pull out all the powers of p from n

so for example n=12 and p=2 would give you m=3

ok

Oh I misread you mb

Didn't see that first m

ok

how familiar are you with integers

A bit

I'm asking again (as I haven't found anything):

Which more often than not means not at all

so let $f(t) = t^n-a = \prod_{i=1}^f f_i(t) \in K[t]$

Sorry, nevermind . . .

I'm pretty sure if $b^l = c^m =a$ then there exists $d$ s.t. $d^{\lcm(l,m)}=a$, or at least I want this to be true

we observe that $f(\zeta t) = f(t)$

Yeah

i.e. $\prod_{i=1}^f f_i(t) = \prod_{i=1}^f f_i(\zeta t)$

Agreed

@Shaun So you're good ?

so $C_n$ acts on $\{1, 2, 3, \cdots, f\}$ transitively

What are those dots ?

oh no

I used $f$ twice

kill me

Didn't you call $C_n$ $\mu_n$ earlier on ?

oh well they're isomorphic... non-canonically

@LeakyNun Not before the new year.

What's $C_n$ ?

cyclic group of order $n$

Yeah right

yeah the action from $\mu_n$ is more canonical

They're the same group anyways

non-canonically

@Astyx I guess so, yeah. But if you have anything to add in the Group Theory chatroom, I'd be happy to hear it :)

@Astyx do you see how the action is transitive

What does that mean again ?

$\forall i \forall j \exists \sigma, \sigma(i)=j$

hi @AlessandroCodenotti I made a conjecture

and I'm proving it

Given a family of 2D curves, find a 3D manifold whose geodesics project to the plane curves

Like take $\sigma = j-i$

Or something

@Astyx well anyway here's the (elementary-)number-theoretic version of the conjecture: if a partition of $\{1, 2, \cdots, n\}$ into $f$ boxes is stable under $i \mapsto i+1$ (map $n$ to $1$) then the partition must be in the form $\{ \{ i + jf \mid 1 \le j \le e \} \mid 1 \le i \le f\}$

i.e. if $n=9$ and $f=3$ then it must be {{1,4,7},{2,5,8},{3,6,9}}

Do you think it's easier to prove it through number theory ?

Oh yeah it might

yes

Ok

you know, it's very intuitively obvious to me

both the conjecture and the fact that I don't want to prove it

corollary: under the above condition, $t^n-a$ is irreducible iff $t^p-a$ is irreducible for all prime factor $p$ of $n$

hey @BalarkaSen

Hey

do you think the conjecture/theorem above has a name

It's quick straightforward no ?

Seems simple

I have not the slightest clue how to prove it

the set of differences between consecutive terms in one of the sets partitionning remains the same accross the partition

Then it's not hard to prove that it's the same everywhere

Meaning all sets of the partition are of the form $\{n, n+k, n+2k, \dots\}$

hi

Hi

You're looking at partitions of Z/n which are invariant under the left regular action. Has to be cosets of a subgroup

Or something

sure

I'm too lazy to write down a proof but it's not hard

is there any literature/Information on a "theory of equations" for the algebra of sets

that you guys may know of

Anyway, I need to go

Happy new year to you guys

@Astyx joyeux nouvel an

It's a new year alright

i read "set theory and logic" by robert stoll and the author talks about such things but it's seemingly obscure and i'd like to see other literature on the topic

anyone

$$\operatorname{Hom}(\operatorname{Gal}(L/K), \mu_n) \cong (K^\times \cap L^{\times n})/K^{\times n}$$

redpilled

the left hand side clearly being finite... I do not see why the right hand side is finite

is anyone familiar with algebra of sets or is it obscure?

so the generalization is that if a partition of a group $G$ is stable under the left-regular action, then it must be a coset partition?

@BalarkaSen

Good question. You should think about it.

anyone?

$X^4-X^2+1 = (X^2+\sqrt3X+1)(X^2-\sqrt3X+1) \in \Bbb R[X]$ so $r_1 = 0$ and $r_2 = 2$

$X = \frac{\sqrt3+i}2$

$X^2 = \frac{-1+\sqrt{-3}}2$

$X^3 = i$

man this is such a s*ty basis

Happy new year everyone

Am I missing a LaTeX addon for the chat?

i think i am missing a latex add on as well

In the info panel on the side there's a bookmarklet

> Chat guidelines: tinyurl.com/hzl2955 | LATEX in chat: tinyurl.com/cfqcvpc

@Jasper I'm sure you will. Any difficulty can be surpassed, even if it's hard and often seems impossible.

@Jasper You never need confirmation you can, you have to know within yourself you can and you just do it. That's the attitude, to be extremely powerful, in all circumstances.

Always winner, always you know you'll do it no matter what.

Happy New Year, Jasper!

i wish this chat was more active

Heppy Ner Yer

sleep