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12:00 AM
'cause $(x+1)^2-(x^2+1)=2x$, so we'd have $x$ in the thing, and then $(x+1)-x=1$
Hmmmm, well, integers are an integral domain, so none of the coefficients cancel.
With $\Bbb Z[x]$, we can still do $2(x+1)-2x=2$, so $2$'s in it
So we have $2$ and $2x$
I think we end up with all the polynomials where $f(1)$ is even?
Why do we have 2? I missed that.
$(x+1)^2-(x^2+1)$ gives us $2x$, and then $2(x+1)-2x$ gives us $2$
OK so
Okay, then I think we should be able to get 3, right? $(x+1)^3-(x^3+1)$
12:04 AM
That gives us $3x^2+3x$
And we could try subtracting $3(x^2+1)$ from that but that gives us $3x-3$
and we could try subtracting $3(x+1)$ from that but that gives us $-6$
This is definitely contained in the set of polynomials where $f(1)$ is even
(which is an ideal I think)
so I don't think we could ever get $3$
Well, this approach can be generalized to give us all even integers as constants, right?
Yeah but we have those anyway
'cause those are multiples of $2$
Ah, right
This ends up being $(x+1,2)$ I think
which I'm pretty sure is maximal
and yet isn't of the form $(M,x)$
Well, this is polynomials over $\mathbb{Z}$, and not power series
Power series are different in that everything with a unit as it's constant term is a unit.
12:08 AM
Oh right
Oh wow
So if your ideal of $\Bbb Z[[x]]$ contains $x+1$, it contains the world
So, I'm confident we won't find an ideal that isn't of the form $(M,x)$ in $R[[x]]$.
OK so say we have $\mathfrak M$
and it's not of the form $(M,x)$
But proving it . . . is an extra credit problem in the end so I might skip the extra credit problems on this week's homework. :P
12:10 AM
We want to show that it's not maximal, and we'll do it by finding something strictly containing it
Well hold on
(I joke)
Nothing in $\mathfrak M$ has a unit as a constant term, right?
Because of the thing you just mentioned
So let $M$ be the set of constant terms of everything in $\mathfrak M$
This has to be strictly smaller than $R$
Which it must be, it cannot contain any units.
12:11 AM
and it's also an ideal of $R$ as well
Can you prove that? That's where I get hung up.
Just because something doesn't contain any units doesn't make it an ideal.
Well, suppose $a_0$ is the constant term of $\mathfrak m_0$, and $a_1$ is the constant term of $\mathfrak m_1$, and suppose $r\in R$
We want to show that $a_0-a_1\in M$ and that $ra_0\in M$
Or, both together, that $a_0-ra_1\in M$
but clearly $a_0-a_1$ is the constant term of $\mathfrak m_0-\mathfrak m_1$, so it's in $M$ by the definition of $M$
@Rithaniel Sure
Right, so that's the constant term of $\mathfrak m_0-r\mathfrak m_1$
What exactly does it mean to say that $D_{2n}$ is the symmetry group of a regular $n$-gon? What does it mean to say that $\Bbb{Z}_n$ and $D_{2n}$ act on a regular $n$-gon? How does that work out? What's the set that these groups are acting on?
12:14 AM
so it's in $M$ by definition of $M$
Ah, very nice.
So the set of constant terms of things in $\mathfrak M$ is a proper ideal of $R$
Okay, you gave me the bit that I was missing: "Just use $a-rb$"
$(M,x)$ is a proper ideal of $R$
And $(M,x)$ contains $\mathfrak M$
because it's the set of all things with constant term in $M$
Furthermore, $(M,x)$ strictly contains $\mathfrak M$, since it doesn't equal $\mathfrak M$ (by hypothesis)
So $\mathfrak M$ isn't maximal (we've just found a proper ideal strictly bigger than it).
So we've proven: if $\mathfrak M$ isn't of the form $(M,x)$, then it's not a maximal ideal of $R[[x]]$.
This is why I seek help on these things.
12:17 AM
The key was that nothing in $\mathfrak M$ has unit constant term, so we can look at the set of constant terms without getting any units.
You supplied the key. Once I had that it all fell into place
@user193319 They're the group of isometries of an $n$-gon
An isometry is a rotation, reflection, translation, or combination of those - anything that doesn't change distances
Arguably you provided the key to "just show that it's an ideal."
There are certain rotations and reflections of a regular $n$-gon that don't change what it looks like
A rotation or reflection of a shape that doesn't change what it looks like is called a symmetry of that shape
(I'm ignoring translations because you can't translate a finite shape and have the result be identical to what you started with)
The set that they're acting on is literally the set of points in the polygon
A reflection is a map that takes in a point and returns another point, on the other side of the line of reflection
So, we view the $n$-gon as living in $\Bbb{R}^2$, and the set $D_{2n}$ acts on is the set of all ordered pairs making up the $n$-gon? So, from this perspective $D_{2n}$ is really a collection of $2 \times 2$ matrices; and when one does all the calculations, one can view it more abstractly as $D_{2n} = \langle r,s \mid ... \rangle$?
Well, technically, the $n$-gon isn't necessarily living in any mathematical space. It could be a $n$-gon on your desk in front of you.
Hmm...that's a much more natural perspective---to me at least. It just seems weird when books claim that the abstract group $D_{2n} = \langle r,s \mid r^n = s^2 = (rs)^2 = 1 \rangle$ is the symmetry group and that acts on something in $\Bbb{R}^2$. It seems more natural to start with this matrix picture of acting on points in $\Bbb{R}^2$, and then move to the abstract picture.
12:30 AM
But sure, it could be a construction of line segments in $\mathbb{R}^2$
@user193319 Yeah, you can think of $D_{2n}$ as a collection of $2\times2$ matrices
And the $n$-gon is centered at $(0,0)$?
If two groups are isomorphic then they're the same, basically, by the way
The set of those matrices is isomorphic to $\langle r,s\mid r^n=s^2=(rs)^2=1\rangle$
so they're considered the same thing
Q: Eigenvalues of product of two complex matrices

MathforjobLet $A$ be a $n\times n$ complex matrix. Assume that $A$ is self-adjoint and $B$ denote the inverse of $A+iI$, where $I$ is identity matrix of order $n\times n$. Then all eigenvalues of $(A-iI)B$ are purely imaginary real of modulus one of modulus less than one My attempt: I discarded option...

How to prove that $(A-iI_n)B$ is unitary?
When we say $D_{2n}$ we're basically talking about an isomorphism class of groups
@N.Maneesh So that's $(A-iI)(A+iI)^{-1}$?
12:39 AM
@AkivaWeinberger yes
And unitary means $UU^*=I$, right?
($U^*$ or $U^\dagger$ or whatever the symbol is)
Conjugate transpose
So multiply this by its conjugate-transpose and see if you get $I$
First we need to find out what its conjugate-transpose is
I am not getting Let me post the steps here. Please help me to complete
The conjugate-transpose of $A-iI$ is $A^*+iI=A+iI$ ('cause $A$ is self-adjoint), right?
12:42 AM
Oh Thank you since I haven't use the self-adjoint property
Remember that $(XY)^*=Y^*X^*$, as well (the transpose of a product is the product of the transposes in reverse order)
So in the end the conjugate-transpose of $(A-iI)(A+iI)^{-1}$ is $(A^*-iI)^{-1}(A^*+iI)$
Hm wait
This isn't actually working
Oh wait
@AkivaWeinberger I am also stucked here.
Do $(A-iI)$ and $(A+iI)^{-1}$ commute?
need not be
I think they do
Right, they do
In general:
If $A$ and $B$ commute, then $A$ and $B^{-1}$ commute
(proof: if $AB=BA$ then multiply on the left and right by $B^{-1}$ to get $B^{-1}A=AB^{-1}$)
Here: $A+iI$ and $A-iI$ commute
(Both products give you $A^2+I$)
12:46 AM
Okay. Thank you.
So $A+iI$ and $(A-iI)^{-1}$ commute
Now I got the answer. Really Thank you.
So we can ignore things like the transpose swapping the order
You're welcome
1:08 AM
I study all day every day.
7 hours a day, 24 days a week.
1:44 AM
the taylor series for $e^x$ converges to it "rather quickly" on the positive axis (x>0) and "rather slowly" on the negative axis (x<0)...
how do make this idea formal?
Perhaps talk about the relative sizes of the $N$ necessary before the convergence is within particular $\epsilon$?
@LeakyNun I was about to say "does it really? I'd argue the opposite, rather." Then realized I was visualizing e^-x and its series =\
was wondering this - if x is rational then can $\sum_{m} x^m$ be irrational?
@nitsua60 sad
@BAYMAX the infinite series is 1/(1-x)
1:54 AM
@BAYMAX Are the rationals closed under multiplication? How 'bout addition?
@LeakyNun do u mean $\sum \frac{1}{1-x}$?
@BAYMAX 1+x+x^2+x^3+x^4+... = 1/(1-x)
how the value is irrational then?
@nitsua60 yes rationals closed under multiplication! and even closed under addition
so perhaps then u mena that $\sum x^m$ cannot be irrational
how about the other way now?
if $x$ is irrational then can $\sum x^m$ is rational
2:08 AM
Also no. We know that $\sum_m x^m=\frac{1}{1-x}$. If $x$ is irrational and $y=1-x$, then $x=1-y$. So $y$ cannot be rational unless $x$ is also rational. Now, assume $\frac{1}{y}=r$ where $r$ is rational. Then $yr=1$ which is impossible unless $y$ is rational. Thus $\frac{1}{1-x}$ is irrational.
cool yeah had done!
I was thinking that even if we sum up rationals numbers then we still get irrational number
like $\sum_{n \in \Bbb{N}} \frac{1}{n^2} = \frac{\pi^2}{6}$
yeah its not a power series
but a sum
2:35 AM
(just be careful about your convergence interval)
2:53 AM
So, if $M$ is a maximal ideal in $R$ and $n\notin M$, then why is $(M,x,nx)$ not a maximal ideal in $R[[x]]$?
$$\sum_{n=-\infty}^\infty x^n = 1 + \left( \sum_{n=1}^\infty x^n \right) + \left( \sum_{n=1}^\infty (x^{-1})^n \right) = 1 + \frac{x}{1-x} + \frac{x^{-1}}{1-x^{-1}} = 1 - \frac{x}{x-1} + \frac{1}{x-1} = \frac{x-1-x+1}{x-1} = 0$$
@Rithaniel who said it isn't?
@LeakyNun heheh
Well, I'm tasked with showing that if $\mathfrak{M}=(M,x)\in R[[x]]$ with $M$ maximal in $R$, then $\mathfrak{M}$ is maximal in $R[[x]]$. However, by my understanding, $(M,x)\subsetneq (M,x,nx)$.
@TheSimpliFire Oh :D Thank you very much.
@Rithaniel then you have a quite interesting understanding
2:58 AM
Does $(x)$ necessarily contain $nx$? Because I thought $(M,x)$ was all power series in $R[[x]]$ with coefficients in $M$.
yes, (x) necessarily contains nx
Ah, wait, does $(x)$ imply a coefficient of a unit?
that is correct
Dangit, of course it does.
Today I think Google chrome was updated. So, bookmark bar only shows in new tab, and after opening a web page, bookmark bar disappears. So, in particular, I can't click on 'Start ChatJax'. What to do?
3:01 AM
That worked. Thanks
3:17 AM
i am not very comfortable with idea of inducting over $E_1$.
I mean, we are inducting over $F$ right?
4:13 AM
Reviewing maths chat history:
When Bakarka become uncommonly on, the discussion about algebraic topology dies out, only to be replaced by even more incomprehensible barrage of field extension questions
But then, my rambles are equally incomprehensible, so that serves as a good counterweight
4:25 AM
@Silent The statement being proven by induction has a universal quantifier over fields. So, it is still just induction on $n$.
4:43 AM
It is not necessary ina parallelogram that the diagonals bisect the angles right?
4:59 AM
it is
draw a picture
1 hour later…
6:28 AM
@Silent once you're comfortable with the fact that there is an extension of $F$ in which $f(X)$ has a root it shouldn't be much more difficult
also morning
6:41 AM
Is there such a thing as n choose k to the r power or is that make believe?
well n choose k is really an integer, thus there is no problem in taking powers of it
I'm on mobile so latex is a challenge but I Don't know if I got this right because it's like an n choose k but I think it's like to the r power. I could upload the screenshot
I am thinking 3 choose k for i
And n choose k to the r power
For the second part
But it says unlimited supply so I'm also thinking 3^r choose k
For the first part and then n choose k to the r for the second part
hmm... I don't quite get 7.6, so you mean let spam, chicken and mahi be s,c,m. Then they are asking for a lunch with k plates, how many kinds of s,c,m combinations for each k (allowing repetitions of some s,c,m)?
so if it is a 5 plate lunch I have something like:
6:51 AM
I guess soooo?
But it's like ... I was thinking 3 choose k due to letting n= 3
It's the only problem I'm stuck with ;/. I think observation 13 has something to do with it
I got 7.5, 7.22, and 7.23 already because I did 7.5 earlier and 7.22 and 7.23 are the classic n stairs problem
hmm, in that case, let $r$ be some large number, then you have a total of $3^r$ plates and you want to order them in $k$ plates. Thus that will be $3^r$ choose $k$, and since it stresses it is unlimited, it means $r$ has to be taken an infinite limit somehow.
I think I need some information on 7.4, cause I am not that familar with generating functions in combinitorics rpoblems
$3^r$ choose $k$ make sense since $3^r$ is just some very large natural number
That's what I just thought for the first part
I think the second part is either n choose k to the rth power or
n^r choose k for the second part lel
7:11 AM
I think one good way to interpret "unlimited supply" is that given k plates, you have at least k of each item thus you can request as much item you want to fill in the plates
7:23 AM
No but I'm stuck in the second part of 7.6
And that's the last one woot
That way I can create my outline and reference sheet along with handouts
Oh s--- I think I got it
On 7.5 with the donuts the unlimited vegan donuts is $(1+x+x^2+...+x^r)$
So $(1+x+x^2+...+3^r)$?
Observation 12. Suppose there is 1 object of type A, 1 object of type B, 1 object of type C, and 3 of type D. Then he number of ways to pick k objects from this set is equal to the coefficient of $x^{k}$ in the polynomial $(1+x)(1+x)(1+x)(1+x+x^{2}+x^{3})$
This has to be related with the second part of 7.6
@Secret I think I got it
Ah I finally remember, the n choose k are the binomial coefficients, and hence their generating function is $(1+x)^n$. Thus in 7.6 there are $3^r$ objects in total, thus the generating function is $(1+x)^{3^r}$

and yeah
Yayyyyyy :)
7.7 is thus when there are unlimited supply of unlimited types of food, thus taking the limit $n$ of $(1+x)^{n}$ should give you the result as required for geometric series
8:13 AM
hello all
is there a hint on solving laplace transform of $\int_{x=t}^{\infty} \frac{e^{-x}}{x} dx$
Fun puzzle: find the area of the following square (credits not mine)
@ÍgjøgnumMeg Grias-di!
@Rudi heile :)
8:29 AM
@TheSimpliFire well one could intersect those two curves with two at 1/2 inflected ones ...
or maybe rather not ...
@ÍgjøgnumMeg alles OK bei Dir?
@Rudi_Birnbaum @ÍgjøgnumMeg wie spracht man "Dirichlet" aus in Deutsch?
Ganz normal :-)
that doesn't help much lol
welche "ch"?
How can I communicate that?
wie Bach?
8:43 AM
strange, Wiki says k
there is only this ch
@Rudi_Birnbaum well there is ich
Well in case he is German than no
in case the name is french maybe
he is Johann Peter Gustav Lejeune Dirichlet
8:44 AM
Lejeune is not a German name
his grandfather was from Richelette, and then arrived to Düren
in French the ch
is the German sch
so the people called his grandfather le jeune de Richelette (the young from Richelette)
like in SH-ut up
I see
when you "verballhorn" that into German anything might result
depending on the local dialect
of French or German? :P
8:47 AM
The most logic or say contemporary one would be the (German)
so like the SHut up
but then it can be also that the germans tried to immidate the frencg
immitate the French
then it COULD be also -k-
becuase the French Ch is a bit higher in pitch and more accute
Reading the name from paper and assuming its a German name however without a doubt results in the Bach ch
I would say in the "ich" ch
well its the same for us
it's different for me...?
you notice the difference
we maybe only if its confused
so yes the ich ch
wiki cites Duden
8:52 AM
on the "k"?
I heard the name once pronounced by Tao there it sounded very wrong to me
In classes when the name was mentioned I heared it with "ch"
I love how wiki contains "Johann Peter Gustav Lejeune Dirichlet", "Peter Gustav Lejeune Dirichlet", "Gustav Lejeune Dirichlet", "Lejeune Dirichlet", and "Dirichlet"
But its possible that he emphasized that pronounciation
when he was alive
that happens for many Germans with Slavic origin names
or Austrians
And here I was hoping to see a discussion of the continuum hypothesis from a quick glance at the chat
8:54 AM
that they prefer a pronounciation that comes not self explaning from the letters
@AlessandroCodenotti sorry to disappoint :P
@Rudi man spricht ja "sprechen" anders aus als "sprache"
ch spricht man anders aus lol
hast recht
ich tät "Dirichlet" so aussprechen wie "dir-ich-let"
mit ch wie in "ich"
That phonetic description seems at least to try to resemble French, are there any French+German speakers here?
While it in my eyes could be even proper French
8:58 AM
ah jo
@Rudi_Birnbaum do I count as one?
might do a degree in linguistics if my scholarship app doesn't go through
If so you should know @LeakyNun
but like it isn't a French word either?
what do you mean?
8:59 AM
the place is Richelette /riSlet/
Oh I see
yeah possibly there exists no proper pronounciation
then they verballhornen that into dirichlet
anyway is Peter pronounced as written?
sounds like PETA idk
the Verballhornung seems an integral part
@LeakyNun Depends on language level
colloqial / contemporary / salngs / dialects / age group
oder die "Hochdeutsche Bühnensprache"
Thats interestingly a normalized variant of German
(where else would you have something like this ...)
and radio/tv speakers are supposed to learn and use that
its not easy
and they usually are not able to it properly
you always here strong northern german colloquial influence
thats a result of prejudice
9:03 AM
I think the Bühnensprache is ridiculous
Since Germans have the prejudice that Hannoverians or Northerns Germans are closest to that
which is plain wrong
but however is a very strong prejudice
@LeakyNun dunno,
depends on the point of view
its however the standardized pronounciation
of course anything which is spoken written language can easily sound ridiculous
this is why Bavarians think Prussians are insane
because they speak how we only write
but usually the same bullshit
and that makes it ridiculous
how do you say "Gustav"?
is there "h" in "Johann"?
like goodstuff
without d
9:07 AM
so in the end how on earth do I say the "ch"
I would go safe and subversive "ich"
or you go smooth and adapted and use -k- or -sch
since your Asian I suppose you go for the latter ;-)
interestingly Chinese/Japanese/Korean wiki all use the "k" version in the transliteration
what did he himself say...
ja aber er ist kein Dirichlet
that guy here uses "schle" so more or less a plain french or maybe the best he can do to immitate it ...
ja, stimmt
9:19 AM
I mean I thought the name was French like a week ago
yeah :-) me until 20 min ago ...
btw his surname is Lejeune Dirichlet
yes I'm a nerd
even ..
I have no idea how names work
A linuistic characterization of what is really going on here would be quite interesting
well two words without hyphen is no normal german family name
dunno about French ...
9:23 AM
btw I don't think anyone knows where Rich(e)let(te) is...
> His paternal grandfather had come to Düren from Richelette (or more likely Richelle), a small community 5 km north east of Liège in Belgium, from which his surname "Lejeune Dirichlet" ("le jeune de Richelette", French for "the youth from Richelette") was derived.
bless icelandic naming
one person one name one word
9:27 AM
hmm it's quite close to Düren
well Liège
Who commited the crime to write "Lejeune Dirichlet" instead of "Lejeunedirichlet"
since I still don't know where Richelette is
it's also quite close to the Dutch regions of Belgium
@KarlKronenfeld, @ÍgjøgnumMeg, thank you.
9:29 AM
@LeakyNun I havent checked but its possible that these are timx places composed of two farms
@Rudi_Birnbaum I know a French surname "Lamort de Gail"
e.g. my great grandfahter was "Florian Weinmayer Holler von Obhloz"
He was a small farmer.
Holler is the name of the farmhouse
@Silent no worries
and Obholz is the "Weiler" that is composed of three farmes
and where on earth is Obholz?
9:31 AM
its a part of "Irschenberg"
not sure if it will b on the maps
then why aren't you Rudi Birnbaum Holler von Obholz?
Since he lost the farm
(and I am not the "prime" great grandson )
My grandfather had one older and many younger brothers
@LeakyNun It became a street obviously nowadays
So its very possible that this Riche* is such a tiny place not more than a conglomerate of a very few farms
9:55 AM
Dummit and Foote comments:
Since $\sqrt{-3}$ satisfies the equation $x^2+3=0$, we have $[\Bbb Q(\sqrt[3]2,\sqrt{-3}):\Bbb Q(\sqrt[3]2)]\le2$, hence degree must be $2$ since we observed above that $\Bbb Q(\sqrt[3]2)$ is not the splitting field.
I can't see why '$\Bbb Q(\sqrt[3]2)$ is not the splitting field' says that $x^2+3$ irreducible over $\Bbb Q(\sqrt[3]2)$. In fact, they showed earlier there that '$\Bbb Q(\sqrt[3]2)$ is not the splitting field for $x^3-2$'.
10:10 AM
a quadratic is reducible in your field iff it has a root
Sometimes I wish there is an easier way to determine how the roots of a n degree polynomial changes when we make an infintesimal change on one of its coefficients
Graphically, polynomials are pretty special because the elements $x^k$ for each $k$ forms a dense orthogonal basis
meaning that tweaking just one of these coefficients, and then compare that with the original, the other powers of x contributed to the graph does not change, which is why the space of polynomials behave like a vector space
but the motion of the roots are quite unpredictable with each change, at least I don't know of an analytical formula that describes them
4 hours later…
1:58 PM
@Secret it’s a certainly a perturbation theory calculation. I don’t remember the details beyond that
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