12:20 AM
@DanielFischer I don't think I've ever seen you on chat this late before.

Darn... Chris's sis is gone. I finished copying a proof to an answer so that she could read it. Too slow.
I guess I have time to add the corollary, then :-)

12:44 AM
@robjohn saw your update to the Fourier series problem. like i said in my comment there, it's surprising to me that the 'Fourier series' approach ends up seeming the most obscure
surprising in the best way, of course

@Semiclassical It does give an interesting formula for the imaginary part of $\mathrm{Li}_3$ on the unit circle, however.

exactly
and, if i remember some of those exercises in fourier analysis, other such formulaes are generated by other functions on the unit circle
which suggests there's 'more than one miracle' as far as interesting polylogarithm identities

@Semiclassical yes. I believe that the imaginary part of an analytic function on the unit circle determines the analytic function, so I don't think there will be anything other than $\mathrm{Li}_3$ that matches that Fourier Series.
Unless that uniqueness falls apart if there are singularities inside the circle...

neat. i gather that other Fourier series -> series summationsw presumably bear upon other cases like Li_5

Well... Time to take my puppy to the park. BBL
@Semiclassical It would appear so.

12:52 AM
bye. it'd be neat to find a systematic perspective on that---maybe time for a new question...
have a good walk

1:29 AM
a sign of having been trained as a physicist: i can write answers like this while being blithe about the crimes against rigor i commit in doing so

1 hour later…
2:39 AM
Can't wait for this app to come out and eliminate the need to post cellphone pictures on Math.SE

2:52 AM
I only use my mobile phone for calling, texting, and alarming, lol.
Someone was nice enough to give me 2 votes an hour ago, lol.

3:08 AM
@robjohn: Well, the wanton downvoting is continuing. Obviously, I don't care about the rep points, but (as you well know) it would be nice if the person would express a reason ... unless it's just pure revenge. Hmm ... Do I notify mods at some point? [I'm sure @Mike will say there aren't any wontons ...]

@TedShifrin Hello!

is someone in the mood for training me?

@JorgeFernández Training? Are you a Pokémon?
I only do Fire types.

I'm fire-psychic

OK, that's cool.

3:15 AM
Hi @Pedro ... You wantonly downvoting me? :D

@TedShifrin You don't need to say "obviously I don't care", LOL.

@TedShifrin WANTON sounds chinese or something.
But no, I'm too busy getting knee deep with algebra.

@PedroTamaroff It's the transliteration of yuntun for some kind of dumpling.

Which I said I wouldn't be doing this semester...
But Pete's notes are pretty darn good.

See my comment to Mike above ...
It occurs to me that dear Rene may be off suspension.

3:18 AM
@TedShifrin But I thought he is still being suspended? Hmm.

Knee deep.isn't bad, @Pedro. Chest deep, scary.

I need to prove if $K\leq G$ and $H\leq G$ with $KH=G$ then $xKx^{-1}yKy^{-1}=G$ for all $x,y\in G$

@JorgeFernández What is stopping you?

surely one of those last two K's is supposed to be an H. and clearly wlog y=e.

Yup.

3:21 AM
@anon Yes, noted that.
HARRO BTW STRANGER.

I think Ted is ignoring me, but it is OK, lol.

hello peoples of #math

@anon: I cherished your "oh" :)

(:

Not really, @Jasper. I graded exams and went to a violin concert, so it's bedtime.

3:23 AM

@TedShifrin OK, good night. I am going to sleep in half an hour at 12 noon here, lol.

how does the wlog work though?
nvm, I see how
I had noticed that for finite groups
anyhow, how may I prove that?

I have only used wlog once or twice in my entire undergrad work.

haha
@anon is there a reason you are kinder in your other aliases, or is it just me?

I'm always unkind ...

3:27 AM
@JorgeFernández Where are you stuck? What have you tried?

@JorgeFernández I have been told the opposite as well. but I don't see how I'm currently being unkind. perhaps the adjective "clearly" was overdoing, and I am not being as helpful as I could be, but I am entertaining myself with other things and only scanning the chatroom very cursorily these days.

no, I didn't mean you are being unkind, its just it sound like anon is a sharper character, while blue i more laid back and relaxed

perhaps that's true

You should see purple! :D

@Emrakul Who summoned thee?

user61230
3:32 AM
[wave] Hi, I'm an emissary of Cthulu. Nice to meet you!

oh, I haven't been acquainted with purple

@Emrakul Did Cthulu send out the ebola?

I think my problem is equivalent to showing $KgH=G$ for all $G\in G$
I can look at $KgH$ as a union of right cosets of $K$
and I just need to prove if there is a representative for all of them in $H$ then there is a representative for all of them in $gH$

@JorgeFernández that is a good observation. note that KgH=G can be obtained by writing g as an element of KH...

@anon well fuck
$KgH=K(kh)H=KH=G$?

3:43 AM
mmhmm

haha
I'm going to cry now

Crying is good.

that trick is actually pretty sweet though
well, it's rather straightforward, I don't know if I should call it a trick, but it's "pleasant"
I'd like to share a problem with you I enjoyed, let $G$ be a group of order $2^kl$ with $l$ odd with an element of order $2^kj$. Prove $G$ has a subgroup of index $2$.

3:59 AM
heh, the elts of order dividing |G|/2 form a subgroup
did you do it w/o classification of U(2^k)'s?

4:21 AM
yes
consider the action of $G$ on itself by left regular action. then the element of order $2^kj$ induces an odd permutation. Since every subgroup of $S_n$ which has an od permutation has exactly half of it's permutations odd we can consider the elements of $G$ that afford even permutations, these form a subgroup of the desired index.

2 hours later…
6:19 AM
@TedShifrin I will keep an eye on things, but I can't see anything unless it is bad enough. The community managers and developers are the only ones who can tell anything more.

@robjohn you have a new puppy now?

6:32 AM
@Nick Hey!

@Sawarnik Happy Diwali!

Tis tomorrow :D

But Happy Diwali to you too :D

@IceBoy No... the same one I've had for a year and a half

6:36 AM
@Nick Don't you celebrate this in Kerela?

@Sawarnik: We celebrate it under the name Deepavali

@Nick Same thing?

@Sawarnik avoid crackers? do they mean fireworks?

@robjohn Yeah.

@Sawarnik दीपावली का अर्थ है दीपों की पंक्ति। So, yes, it's the same thing.
@robjohn: It's more or less the 4th of July here.

6:40 AM
@Nick Isn't that Indpendence Day or so? :O

@Sawarnik: I mean in terms of fireworks, bud. And yes, it is the american Independance day. Welcome to Earf.

You should compare it with Christmas.
Holi and Diwali are the only two festivals I care about.
The rest are just good for holidays :D

@Sawarnik: Fireworks on christmas is not as heavy as the fireworks during the 4th of July.

@Nick Oh, acha, you re talking of fireworks.

@Sawarnik: India, Mexico and the US are the few countries that know their explosives :D

6:45 AM
@Nick Have you seen Legends of Awesomeness?

@Sawarnik: If I remember correctly, that's a Kung Fu Panda spin-off on nick.
Speaking of TV, that advertisement song "Rango Ka Mela Hai Roshni Ka Khela Hai" is stuck in my head.

@Nick Yes, it is. Have you seen it?
@Nick Never heard. I rarely watch TV shows on TV.

@Sawarnik No, not yet. Why do you ask?

Wha?

Sorry. my keyboard has some issues. It pastes whatever text I have on my clipboard whenever I press shift.

6:52 AM
lol
@Nick Some episodes are awesome :D

@Sawarnik great. any special shows your watching today? or are you going to be working on school assignments like me.

@Nick Yeah, I have to complete my book review and SST portfolio :(

@Sawarnik : Ooh, which book?

@Nick Not decided yet. But I ll choose some Sherlock story.

@Sawarnik: Try the one in which he dies. Not the best case but it was epic.

6:56 AM
No, I ll choose some real case :D
I have the only the second part right now so choices are limited.
But probably it would be Three Garridebs.

@Sawarnik: Mhh, your right. Also, um, because of your batman avatar, who would you think would win in a fight against each other, sherlock or batman?

@Nick We would team up together :P :P

@Sawarnik: lol, no, i mean batman vs sherlock. who would win?

physical fight?

eh, any fight.

7:01 AM
no tell me.

um, battle of words.

not sure. but probably shrlock then.

Actually, here's a rap battle:

lol
ok, i have to go .. bye :D
@Nick Do you want some farewell geometry questions?

bye :D

7:04 AM
ok bye :)

sure

yo

yo
YoYo!

7:35 AM
Yo :D
Yo

This room has seen more greetings than math.
and it has seen a lot of math...

Yes, within the last few minutes.

8:01 AM
@IceBoy: do you happen to know where i can find an english version of the integral kokeben

8:20 AM
@Nick There is no such version.
Integral Kokeboken is written by N3buchadnezzar and he wrote it in Norwegian only.

8:51 AM
@Nick no, I don't, sorry.

9:02 AM
How can I integrate $1/z$ over a closed curve containing $0$ if $1/z$ is not continuous nor holomorphic at $0$? To use the Cauchy integral formula or the definition of the line integral, I at least need a function that is continuous. In this link <http://en.wikipedia.org/wiki/Methods_of_contour_integration#Example> they seem to ignore these concerns and just integrate anyway.

@TheSubstitute Do you mean $0$ is on the closed curve or is it in the closed curve? If it's the latter (what wiki does) then you should have no problem.

ah, thanks

No problem.

10:07 AM
@BalarkaSen $x_1=1$, $x_2=2$ and $2x_n=x_{n-1} + x_{n-2}$
Can you find the limit? :P

@Sawarnik Why should I believe that this is an interesting problem?

10:29 AM
Is there any possible "best formation" for battleships?
Is there any table of probability of someone hitting a spot?
gathered from multiple games?

@BalarkaSen: I don't care if you find the problem interesting. I'm not able to do it... at all.

Looks olympiad-ish to me. Not at all interested.

:D me neither... not today atleast.
@UserX: There are ways to always win the game. There has been research done based on probablility and there have even been mathematicians and programmers who have obsessed over the game so much that they even created their own Linear Theory of Battleships
@UserX: Whatever answer you're looking for, it is wise to first consult the enormous pile of research done on the subject.

10:47 AM
@Sawarnik Looking at $x_n - x_{n-1}$ could help.

... doh

oh right telescoping

I haven't telescoped anything in about a year...

@DanielFischer My topologies, but I have a query.
@DanielF Let $(X, d)$ be a metric space, $A \subset X$. Why can't we define $A$ to be convex by the property that for any two $x, y, \in A$, there is a $z$ which is $\neq x$ and $\neq y$ such that $d(x, y) = d(x, z) + d(z, y)$?

@BalarkaSen We could do that, but that would for example mean that $\mathbb{Q}$ is a convex subset of $\mathbb{R}$, which I find not-so-enticing.

11:07 AM
You are indeed right. Hrmph.
@DanielFischer But we can do this with the usual generalization more naturally with, say, $\mathcal{C}(X)$ for a metric space $X$.
In fact it is possible for any normed vector space over $\Bbb R$, I believe.

You could call $A$ convex if for all $x,y\in A$, for every $t\in [0,1]$, there is a $z_t\in A$ with $d(x,y) = d(x,z_t) + d(z_t,y)$ and $d(x,z_t) = t\cdot d(x,y)$. Maybe that would lead to something useful, no idea.

@DanielFischer Noted, let me think about it.

@BalarkaSen Vector spaces [or affine spaces] are the natural home of convexity.

OK, OK, I got lots of interesting stuff to think about. Thanks for your help, @DanielF. I am gonna leave chat and think about it for a bit.

Hi @userX

11:24 AM
@DanielFischer OK wait I didn't get that $d(x, z_t) = t \cdot d(x, y)$ condition.
Why are we using that?
Oh $d(z_t, y) = d(x, y) - d(x, z_t) = (1-t) \cdot d(x, y)$
I see, I see.

Balarka, Balarka.

hi
In \hyperref[NIST equation 8.20.2]{"http://dlmf.nist.gov/8.20"} what is meant by (p)_{k}

11:46 AM
hi @Sawarnik
Let $f: [0,2] \to \mathbb{R}$ be a twice differentiable with a continuous second derivative function such that the following hold;

$$\displaystyle \int_0^2 (f''(x))^2 \mathrm{d}x =\frac32 (f(0)-2f(1)+f(2))^2$$

and

$$f'' \left (\frac12 \right ) =f'' \left (\frac32 \right )$$

Find all possible $f$.
$f=ax+b, a,b \in \Bbb R$ works
but is it unique?

12:01 PM
@userX Does there exist a continuous function $f(x)$ that if $a\in\mathbb{R}$ then $f(x)-a$ has exactly 2 roots? And what about 3? :D

12:19 PM
@Sawarnik you can construct a polynomial that satisfies those.

12:52 PM

@DanielFischer A bit too late...

@JasperLoy Hey, that's the sort of question that generates millions of answers and thousands of views. Never to late to jump in.
Just got a third answer :D

I upvoted them all except my enemy's, LOL.
I have decided to get Gratzer's More Math into LaTeX and Voss's PSTricks for my LaTeX books. Of course, I have decided to use PSTricks to produce graphics.

1:07 PM
I just lost two points because "user was removed". Does this mean the accepted answer to one of my questions is now gone?

@MikeMiller He might have upvoted you once and downvoted you four times, for example.
@MikeMiller Accepted answers cannot be deleted except by a moderator, I believe.
I notice that often when my enemy X answers a question and someone else does with a higher number of upvotes, he will be downvoted. It is probably X that downvoted, lol.
In fact, I just saw one such downvote and how the rep of X dropped by 1, so this is very likely, lol.
This is probably the strategy of X to try to get the highest number of upvotes so that her answer will become accepted. Very cunning indeed.
To risk sounding exaggerating, I am sometimes traumatised by the actions of X, lol.

@MikeMiller Could have been that you once edited his/her question, got two points for it, and the question was deleted because of low score when the user was removed.

@DanielFischer That is very likely indeed. That would have happened before he got 2000.

1:26 PM
That makes a bit of sense.

Ladies and gentlemen, I am very excited as I only need 10 more points to reach 2000 and retire from this site. I will keep my account though as promised.

@JasperLoy Bye.

@ParthKohli I will still come to chat, lol. Hi. You should check if the dictionaries you checked yesterday are authoritative.
@ParthKohli Your take on practice yesterday is something I have never heard before. Chances are that the sources you checked are erroneous.
In American English, practice is noun and verb. In British English, practice is noun and practise is verb. Anyone who disagrees, please correct me citing your sources.

1:54 PM
Greetings

2:06 PM
In every area of life the better you become the more envious people surround you. This is also very true for mathematics.
My piece of advice for all these ones would be: don't spend your time with it, it's simply a waste of time, just try to become incredibly good, die trying it. Finally, if you cannot reach that point, appreciate those that can reach the highest peaks, admire them.
The last example that came to mind is Simona Halep that won Seren Williams, she was incredibly good. Some time ago there were some here (in my country) that talked against her when she failed, but I always put trust in her, she has that attitude of one that can even defeat gods (if you know what I mean).
World No1 Williams suffers her heaviest defeat since 1998

Greetings , The Artist is here.

@Chris'ssis You come across as narcissistic, and I say that entirely unrelated to any potential jealous that you will undoubtedly infer.
11

@Chris's sis well said. Haters gonna hate
2
Can someone take a look at this problem : math.stackexchange.com/questions/985882/…

2:26 PM
@Committingtoachallenge I was just saying some things that affected some of my math projects, I mean they were delayed to some extent (this is painful to me), but I didn't ask for your opinion, I don't care it.
I work hard to get everything in place such that I'm able to publish my book as soon as possible.

@TheArtist Welcome Artist.

@Sawarnik hello Bruce Wayne, where is your gf?

lol
:P :P

0

The plane $Π_1$ has equation $$r=i+2j+k+ \theta (2j-k) + \phi (3i+2j-2k)$$ Which is $$2x+3y+6z=14$$ The line $l$ has equation $$r = 3i + 8j + 2k + t(4i + 6j + 5k)$$ The point on $l$ where $t = λ$ is denoted by P. Find the set of values of λ for which the perpendicular...

Somebody help?
@Chris sis what's the book title/topic about ? I wish it's a topic I love, coz we are getting free copies :D
@Chris's sis sorry I forgot the "s" in tagging ^

I am given that $I(x)=\int_\frac{1}{x}^\sqrt{x} cos(t^2) dt$ and I need to find $\dfrac{dI}{dx}$.

I think I know how to do this, but i'm not sure what the best way to go about answering it formally is. For example, would the following be correct:

$\int_\frac{1}{x}^\sqrt{x} cos(t^2) dt = \int_1^\sqrt{x} cos(t^2) dt + \int_\frac{1}{x}^1 cos(t^2) dt$.

Consider $I_1^\tilda (x) = \int_1^x cos(t^2) dt, I_2^\tilda(x) = \int_x^1 cos(t^2) dt$.
Then $I(x) = I_1^\tilda(\sqrt{x}) + I_2^\tilda(\frac{1}{x})$.

2:40 PM
@Chris'ssis You don't care about your narcissistic qualities? I could easily generate a list of $30$ examples from the transcript with their classifications under narcissism within a week if you wish me to. I personally feel everyone should work on eliminating such qualities as they are hardly conducive to a healthy life(or a realistic depiction of reality). Anyway I shall sleep now, goodnight.
@Chris'ssis Very constructive
Anyway I shall sleep now. Fair-well

2:58 PM
i like my current rep score. 6321-> 6=3*2*1.

@Semiclassical $6 = 3+2+1$ too

the joy of perfect numbers
i guess i should look forward to a score of 28147421 someday :P

@Semiclassical You need to learn to type faster for that. At your current answer rate, it'll take a looooong time.

yeah...
pretty sure that's true for every user, though, given that the highest all-time rep at this point is 292k

@Semiclassical Take Jon Skeet for an example, he's not such a lazy do-nothing as we are here.

3:08 PM
indeed. though now i'm curious how long it'd take him to get to second-perfection at his present rate

@Committingtoachallenge I have the right to talk about my work in a very appreciative way. Do you know why? Because, look, while you slept like 10-12 hours per night (maybe), in the last years I slept like 3 or 4 hours and assigned my time to work, to become very good, and I got thousand of questions and solutions, maybe over 10000 questions.
Did I say "very good"? Yeah, exactly, and you know what? I have great plans, ooo yeah, I wanna become like Ramanujan one day in terms of integrals, series and limit, and this has nothing to do with the narcissism.

@DanielFischer: Looks like he averages about 330 rep per day. Even with that, it'll still take him roughly two-and-a-half years to reach 1m rep

@Semiclassical And if there weren't a rep cap, he'd already be way over 2 million.

hmm, good point

But yeah, even then the second perfect would be some way off.

3:19 PM
an interesting asymptotic question

@Committingtoachallenge You misspelled farewell.

suppose I define the nth-perfection in the way we've been using it
how fast does the size of the nth-perfection grow?

I would be really impressed if you could get an odd perfect number as your reputation

that would be a neat little feat

@Semiclassical Pretty fast, Mersenne primes aren't too common.

3:21 PM
good point
plus the factors->digits scaling isn't smal

hello

@Committingtoachallenge and I'd really like you to be right about me, and thus the world we live in to be much better than I perceive it. Hope one day you publish a paper with some work of yours (I mean your original work) and then someone suggests some of your work comes from another paper, that you took if from there. Then you'll see exactly my point.

probably the better approach is to consider the map n->decimal representation with all factors concatenated
and worry about perfect n afterwards

@DanielF Suppose there are only finitely many. Then in some sense they don't grow very fast at all.

I have a question that I feel is super trivial - if two polynomials of x are equal then they are also equal as polynomials of arbitrary powers of x right?
i.e. f(x)=g(x) => f(x^n)=g(x^n)

3:24 PM
@Max Yes.

@DanielFischer proof?

@Committingtoachallenge One of the things I said last days that seem intriguing and called as narcissistic? Well, yeah, I said I have the best proof to Au-yeung series in the world. I mean I'm very proud about it, yeah, it's an achievement I'm not willing to hide it, I worked d*amn hard for it, I suffered to get that you know?

ah i think i see the proof. typing out questions really helps :)

@Max Two polynomials are equal if and only if they are the same polynomial.

@Max for that sake subsitute $x^n$ instead of $x$ shrugs

3:27 PM
i am talking about polynomials in arbitrary rings

What Daniel said is still true.

I should really be more careful

ok but how is that a proof?
@Studentmath well there certainly are restrictions to substitution and i don't know what they are, do you?

@Max not generally, in specific cases I can guess

substitution isn't a good idea in polynomial rings, no

3:30 PM
In this case it's not a good idea indeed, from the little I know

i guess i'd just argue it as

Hi @robjohn Do you have some idea on direct method in showing the existence of minimum in Sobolev space ?

@Max Write out the polynomials as $\sum a_ix^i$ and $\sum b_i x^i$. That the polynomials are equal means the coefficients are equal, so subbing in anything for $x$ gives two still equal polynomials.

how can i enable latex in this chat room?
it shows me the dollar sign code

See the starred top comment

3:32 PM
right, thx
@MikeMiller that is the case in a ring k[x], but not in general and not in the ring i am interested in

What?

let's clarify: which ring are you studying?

So for the question I asked i'd like a general answer, but specifically I am working with the ring Z[X]/(X^r-1,p)
Since in this ring X^r = 1, you can't just compare coefficients

how would that make proving equality harder, though. if anything i'd expect easier since taking the quotient maps zero to zero, but also maps other things to zero

@Max Why not?

3:37 PM
yes i also expect that the answer is super easy, i just don't see the answer

to me the argument would just be

Also, anything to do with coding theory?

@Studentmath Point being that $X^r$ and $1$ are distinct representations of the same element.

Ahhh.

@Studentmath because the polynomials f(X)=X^r and g(X)=1 are equal and the coefficents of powers of X are not

3:37 PM
prove that it's true in k[x], and conclude that it must also be true in the quotient ring

@Semiclassical ah yes, that sounds reasonable

@Max Just so you know, not everyone agrees on the term "polynomial ring"... Some (me) take it to mean "ring of polynomials over another ring", some "quotient of rings in the previous sense", some require the coefficient ring to be a field, etc
Hence the confusion. Sorry!

@Max No, the polynomials aren't equal. Their residue classes in $\mathbb{Z}[X]/(X^r-1,p)$ are, that is something different.

@DanielFischer hence they are equivalent in the ring

@Max Right, but they aren't polynomials.

3:41 PM
@MikeMiller aight, i use the term because that's what we use at uni

the elements of Z[x] mod whatever are residue classes of polynomials, not polynomials

well this is just terminology bullshit, no?
i want to show that two polynomials of Z[X] are in the same residue class of Z[X]/(X^r-1,p)
and they are in the same res class iff their difference is in the res class of zero

calling distinctions in math 'terminology bullshit'? thems fighting words :)

@Max We can first factor out $p$ - that is a prime, I hope, then we have a nice polynomial ring $(\mathbb{Z}/p\mathbb{Z})[X] \cong \mathbb{Z}[X]/(p)$. Then you want to see whether $f(X) \equiv g(X) \pmod{X^r-1} \implies f(X^n) \equiv g(X^n) \pmod{X^r-1}$, where $f,g \in (\mathbb{Z}/p\mathbb{Z})[X]$.

@DanielFischer yes that's what i want to show in this particular case, but i thought it would be easy to show for any ring

3:48 PM
If in this chat robjohn hadn't been I would have probably never stayed here.

@Max Let $R$ be a commutative ring (with $1$), and $m \in R[X]$. Then you want to see whether $f(X) \equiv g(X)\pmod{(m)} \implies f(X^n) \equiv g(X^n)\pmod{(m)}$, right?

Anyway.

@Max Its simply not true for any ring. Let $A=k[X]/I$. Then your claim is true if and only if for any $f \in I$, then $f(x^n) \in I$.

@DanielFischer exactly
@MikeMiller care to elaborate?

Concretely pick $I =\langle x+1\rangle$. Then $[0]=[x+1]$ but $[0]\neq [x^2+1]$,

3:50 PM
@Max Okay, so you have $f(X) = g(X) + m(X)\cdot q(X)$, whence $f(X^n) = g(X^n) + m(X^n)\cdot q(X^n)$. For that to always hold, whatever $f,g$, you need $m(X^n) \equiv 0 \pmod{(m)}$, as Mike said.

@MikeMiller: writing $f(x)=\sum_i a_i x^i$, then $f(x^k)=\sum_i a_i x^{ik}$ would seem to be in the ring manifestly

Well, I need $k$ not of characteristic two for that example.

though that is a good counterexample

@Chris'ssis Despite what some people might say about me, I have very little confidence in my mathematical abilities and I'm always questioning myself. I wish I had just a small fraction of the confidence that you have.

@Max For $m(X) = X^r-1$, you have $m(X^n) = X^{nr} -1 \equiv 0 \pmod{(m)}$, so for that particular class of polynomials it holds.

3:52 PM
@MikeMiller ok that's neat
@DanielFischer right, let me think about that for a sec

If you don't have the $f(x^n)\in I$ condition you automatically have a counterexample like that; if you do, suppose $f \equiv g$, that is $f(x)=g(x)+i(x)$. Then $f(x^n)=g(x^n)+i(x^n)$ and by our condition $f(x^n)\equiv g(x^n)$.

hmm. that suggests a characterization problem: for what $I$ is it the case that $f(x)\in I$ iff $f(x^n)\in I$

It's true at least for the ideal under current consideration

so in this particular ring Z[x]/(x^r-1,p), x is a root of unity. is that maybe somehow the reason it works?

Not "iff", Mike.

3:56 PM
should be just $f(x)\in I\implies f(x^n)\in I$?

I just proved the biconditional. $n$ is fixed, and to clarify I meant "if and only if the thing semi classical just said"

@Semiclassical Yes, that's what we are after.

so what would be the argument that $f(x) \equiv g(x) \implies f(x^n) \equiv g(x^n)$ in my ring $Z[x]/(x^r-1,p)$?

@Max That $(X^r-1) \mid (X^{nr}-1)$.