Mathematics - 2017-08-07 (page 2 of 5)

Akiva Weinberger

03:00

All calculations done in Desmos

Hm. Do all cubics have 180 degree rotational symmetry about some point?

Huh, I guess they do

Neat

Secret

and I suspect that rotation point is at the inflection point

Akiva Weinberger

Yep, should be

Yeah

So you can specify a cubic, up to translations and scaling, by saying how "twisty" it is at the point of symmetry

On the other hand, up to translations and scaling, all parabolas are the same.

Secret

Are there symmetries unique to quartics and quintics?

Akiva Weinberger

And I guess quartics would need two parameters, lopsidedness and bumpiness-in-the-middle maybe

Hm that doesn't quite make sense

Whatever, they definitely would need two parameters

Secret

Hmm, these are nice findings. I wonder if these equivalence class of polynomials defined by symmetry are well studied in the literature?

Akiva Weinberger

03:10

I mean, horizontal translation just kills the $x^{n-1}$ term, vertical translation kills the constant term, and scaling kills the leading coefficient

Hm well

You would also want the ability to flip things across the x-axis, to completely kill the leading coefficient

Secret

It is always nice to find something that can help us to start tracing the behaviour of the roots of polynomials under an infinitesimal variation of one of its coefficients.

Akiva Weinberger

Otherwise up-facing parabolas and down-facing parabolas aren't equivalent

and you can't convert a down-right cubic to an up-right one

@Secret I vaguely remember there being an example somewhere that showed that they are very sensitive to small changes in the coefficients

Wilkinson's polynomial

In numerical analysis, Wilkinson's polynomial is a specific polynomial which was used by James H. Wilkinson in 1963 to illustrate a difficulty when finding the root of a polynomial: the location of the roots can be very sensitive to perturbations in the coefficients of the polynomial. The polynomial is w ( x ) = ∏ i = 1 20 ( x − i ) = ( x − 1...

> Speaking for myself I regard it as the most traumatic experience in my career as a numerical analyst. - Wilkinson, 1984

Secret

I suspect it might still be possible for rules that exists that can trace them, after all, any infinitesimal change in the coefficient must result in a continuous deformation of the polynomial in question, meaning that the roots have to trace out some continuous curve

Akiva Weinberger

The relevant section is "Conditioning of Wilkonsin's polynomial"

@Secret True

Secret

And for that to remain compatible with $Gal(S_5)$ being unsolvable, the worst case scenario is that the curve traced out by roots to only need to be some infinite combination of something I guess

Akiva Weinberger

03:20

So I guess you want $\{(a,r):f_a(r )=0\}$ or something

Hm

Say we have the polynomial $x^3+5x^2+ax+3$, as a randomly chosen example

with $a$ being a parameter for one of the variables

Maybe we just want to look at the graph of $x^3+5x^2+xy+3=0$

which solves as $y=-x^2-5x-\frac3x$

Secret

O, that's not even a polynomial, hmm...

Akiva Weinberger

Whoa that's cool looking

Graph those

Secret

But all continuous functions are computable since there are only countable many of them if I recall...

Akiva Weinberger

We instantly learn that it has one root for $a$ between $-8.27$ and $7$

At those points, it gains a double root

From $7.52$ up it has just one root

As $a$ approaches both plus and minus infinity, $0$ becomes closer and closer to becoming a root

assymptotically

Since our original polynomial ended in "+3", it never actually becomes a root

Which is good because I divided by $x$ when solving for the thing above

Secret

Interesting. I should try that out for a quintics when I get back to my comp

Akiva Weinberger

03:31

You know what, let's start over for a sec @Secret

So let's think about $x^3+5x^2+ax+3$

Divide by $x$ to get $x^2+5x+a+\frac3x$

Note that, as $a$ varies, the second graph just translates up and down

It's easy to see what happens to roots of graphs as they translate up and down

The important thing is that those two graphs have the exact same roots

So the roots of $x^3+5x^2+ax+3$ (which changes shape in weird ways as $a$ changes) can be studied by looking at the roots of $x^2+5x+a+\frac3x$ (which just translates up and down)

Similarly, if $a$ were the coefficient of $x^2$ instead, we'd divide by $x^2$

Secret

Looks sound (and we can always check the zero root case separately by plugging in zero to avoid accidental division by zero)

Akiva Weinberger

Yeah I guess that just corresponds to unioning in the y-axis sometimes

Faust7

@TedShifrin Hey do you have problem sets/hw set / midterms/ finals from that video lecture list would love to actually confirm i know my ass from a donkey's.

Secret

03:50

desmos.com/calculator/nr3blbvmbn

Generic quintic shown above

Dave huff

04:14

Hey is there anyone here who is familiar and/or comfortable answering an algebriac number theory question?

ay

BAYMAX

"Just ask; don't ask to ask" , I hope some user familair in that topic will step by to help you with your question.

Dave huff

thanks fam. Okay. Here I go:

Let $p$ be a prime of the form $5k+1$. Let $\tau(\chi_{p})=\displaystyle\sum_{p \geq l \geq 1}\chi_{p}(l)\zeta^{l}_{p}$ be the Gauss sum of the Dirichlet character of order 5; that is $\chi^{5}_{p}$ is trivial while $\chi_{p}$ is not. How does the quantity $\tau^{5}(\chi_{p})$ factor as a product of irreducibles in $\mathbb{Z}(\zeta_{5})$?

I know it factors as $\pi^{e_{1}}_{1}\cdot\pi^{e_{2}}_{2}\cdot\pi^{e_{3}}_{3}\cdot\pi^{e_{4}}_{4}$, where the $e_{i}$'s sum to 10 and $e_{1}+e_{2}=e_{3}+e_{4}$, for $\bar{\pi}_{1}=\pi_{2}$ and $\bar{\pi}_{3}=\pi_{4}$. If i had at least two more linear equations, I could figure out each one. I also know that each $e_{i}$ is greater than or equal to 1.

I have no idea how to factor it, and I'm trying hard to figure out how to do so in general!

Ted Shifrin

04:41

@Faust: Sure. Just send me an email (my address is in my profile).

Faust7

Sankyuu

Excited to get to the stuff on manifolds... i think we only have one teacher who teaches it at my university and i refuse to take anther class with him i always do terrible on his finals.

Daminark

04:56

do you think there's a particular reasond why?

Faust7

his assinments always took me a alot longer than other profs but i always got like 95-100% on all of them if i put in the time

and his hw was worth 60% of your final grade

in both classes i took with him

i think i coulda done well on the finals if i had 7-10hrs to write them

but in only 3hrs i barely passes both times i took a class with him

sadly hes the only prof who teach's calculus on manifolds and manifold theory so i wont be taking eithier :(

Daminark

At least consider sitting in on the class and doing the psets if you're worried it'll mess up your grades too badly

The stuff is cool

Faust7

same reason i don't want to do my honours project on dif geo is he would be the professor id have to work with :(

Daminark

Well, even if you're having a hard time on his tests, if you're doing an honors project you wouldn't likely be doing any such time-constrained work, yeah?

Faust7

i def will and honestly i didnt end too bad in the last class i got an A- but that means i got between 50-56% on his final exam.

Daminark

05:04

I mean not all tests are built to be doable up to 90%

Faust7

well hes also a really hard prof well hard on me in particular

most of the student that went to his office hours he'd explain things even if they didnt know what what was going on... but he would make me work really hard to get even a vague insight into a question

Daminark

That may very well be a sign that he has faith in you

Faust7

not sure if he didnt like me or simply couldnt understand what i was aksing :P

Daminark

Was it that he was asking questions to try to get you to figure out the question on your own?

Faust7

yes and no he'd make me work very hard (usually he would help lead me in the direction of asking my question correctly ) but until my question was exactly correct definition wise he wasnt willing to help with the actual question at all yet i knew he understood what i meant and would help others from similar situations.

i know he was trying to help me but i have brain problems specifically with communication and dotting on the fact that i had used the wrong terminology or wording in asking my question was very frustrating.

Daminark

05:15

I mean, even if it's difficult, he asked the stuff from you because he thought you could handle it

Faust7

Don't get me wrong i understand the importance of what he was trying to teach me but when an assignment takes 30 hours the concepts of the class felt more important at the time

then reviewing the earlier concepts / def'ns so i could communicate the question i was actually stuck on

specially when there from diffrent courses id taken years ago ;p

Anyway i have spent most of my summer reviewing and mastering earlier concepts and definitions so that it wont happen to me again

Ted Shifrin

05:40

@Faust: I sent the stuff. I just read what you said to Demonark. It might be worth having a heart-to-heart conversation with this particular faculty member. We faculty aren't mind-readers and can't know what's going on with your particular issues; he might be sympathetic if you explained a bit, and particularly if you expressed (sincere) enthusiasm at learning stuff he teaches.

Faust7

Yeah was just going through it =)

Daminark

Oh hey, sorry I never noticed the last messages you sent, but yeah what Ted said is right

In any event, if he's being harsher with you than others it's not due to disliking you at all, if anything the opposite, so if you can find something that works, it may be good for you

And do heed Ted's advice

Ted Shifrin

interesting: I looked up the faculty. The guy who's listed on geometry/topology has answered a lot of Stack Exchange questions. I haven't interacted with him personally, but he seemed like a good dude from what I've seen.

Faust7

I loved his lectures and loved both classes he taught immensely was defiantly the most intrested student (always did the challenge problems for the assignments though usually the only one as they were always crazy hard especially in dif geo) its just there are financial reasons for me to maintain a higher gpa and his class's.

i just could never get the grades i would in others with 1/4 or even less work... ill defiantly sit in on any lectures he does on manifolds or topology but im a bit afraid to try n take a class with him again.

Ryan Budney

Ted Shifrin

Yup

Faust7

05:47

look at his rate my prof score lol lowest score ive ever seen O.o

Ted Shifrin

You still have nothing to lose (and a lot to gain) by talking with him honestly. Don't whine about grades, but explain what some of your learning issues are.

Usually RateMyProfessor is unreliable. Only students who absolutely LOVE or absolutely HATE will bother doing it.

Secret

[Experiment]

desmos.com/calculator/v5fhzkbswf

Ted Shifrin

Interestingly, a really hateful one disappeared from mine; the author (whom I can identify) must have removed it a few years ago.

Faust7

i wish i knew how to communicate with him on his level

guys really smart

and can draw a picture of anything which i reallly really like

Ted Shifrin

It's not your job to be on his level. But most teachers will listen to students who are forthcoming about issues, particularly if they can see the students really want to learn and try.

Daminark

05:50

As @Ted can confirm from my Ramsey theory lecture, pictures are my expertise

Ted Shifrin

rolls 10 1/2 eyes

@Faust: Of course I'm happy to try to help long distance, too.

Secret

Now to check whether it can help me find roots on quintics...

Ted Shifrin

heya @Tobias.

Tobias Kildetoft

@TedShifrin Hi

Daminark

Though when I actually gave the lecture to the class I did draw a picture at one point, so there's that

Hey @Tobias

Tobias Kildetoft

05:51

Hi @Daminark

Faust7

i honestly tried when i took my last class with him to explain that i had certain problems with things and that i was really interested in the subject cause

Ted Shifrin

You better make me some nice pictures for diff geo, Demonark.

I find it upsetting that a student would complain about "worst grades ever got" when they were B+ ... Most of my students were very proud of getting a B+ ...

Daminark

I will at least try. When I gave the lecture that was supposed to be on hyperbolic sets and $\epsilon$-orbits but ended up being a review of Riemannian manifolds, I did draw some pictures around

Faust7

it depends on the class i guess

Ted Shifrin

You can ask me for clues on pictures, Demonark.

Faust7

05:53

i have gotten some terrible grades hell i got a D in Calc 1

but i didnt exactly do anything in the class

Ted Shifrin

Perhaps he doesn't excel at reading where students are in terms of level. The stuff I've seen him answer on MSE is pretty sophisticated.

Well, anyhow, @Faust. I can't presume to know all the answers. I just hate to see you avoid stuff you're truly interested in for reasons that might be addressed.

Certainly plenty of students at UGA told other students to avoid me. Even grad students were told to avoid me by other grad students, because I didn't give automatic A's and I gave challenging courses.

Faust7

haha thanks i may try to go talk to him again in September

Daminark

Actually along those lines, so when giving a lecture I'd likely have many of the pictures prepared somehow, but when you need to generate a picture yourself, do you have any... visualization tips?

In particular, I short-circuit at rigid objects, esp in motion

Ted Shifrin

For some stuff I used Mathematica, Demonark. But for diff geo I did a lot of stuff freehand. On the other hand, animations are tremendous for a lot of things in diff geo.

Rigid objects in motion? Huh? What are you talking about?

Faust7

I dont mind working hard for a grade ill work as hard as i have to get an A in a class its just really disheartening to put 30hrs a week into a class i really enjoyed and end up with a grade that makes me uncomfortable about taking the next one down the line.

Anyway i best get back to work thanks again for the problems sets

Ted Shifrin

06:02

Sure, keep me posted, @Faust.

Blue Apple

hi is A[x] where x is indeterminated and A[y] where y is transdental over A the same thing

Daminark

So, in topology I'm somehow able to see what's happening every now and then, but somehow I've gotten the vibe that this sorta stretchy intuition is different from geometry

Tobias Kildetoft

@BlueApple Yes, at least when $A$ is commutative

Blue Apple

and when non comm, what makes them different

Secret

Ok, probably too computationally inefficient. You still need to know what the curves looks like for each piece in order to predict where the roots go correctly... hmm...

Ted Shifrin

06:04

Demonark: I referred to the physics of uniform circular motion more times than I can count when discussing curvature notions. You'll see why soon enough.

mathiu_lady

Can anyone please say 'Is R^n sigma-finite outer measure or not w.r.t. lebesgue measure'

Daminark

Ah, well, in physics I was mostly lost

Secret

This method, however, will overkill all cubics and guarentee a factorisation to be found graphically (with ever need to draw the cubic itself). To be illustrated later...

Tobias Kildetoft

@BlueApple Well, it becomes a bit more tricky to even define what the things mean (i.e. what it means to be trancendental).

Blue Apple

what if ring is non comm but indeterminate is comm

Ted Shifrin

06:07

You should know what the acceleration is for uniform circular motion, Demonark. If not, take 30 seconds to compute it.

This is an important intuition to have for curves and curves on surfaces.

Blue Apple

then i think it would make sense to talk of polynomial

Tobias Kildetoft

@BlueApple That is usually taken to be part of being an indeterminate

And indeed, then polynomials make sense

Ted Shifrin

BTW, Demonark. Do you know how to do this? (I missed the fact that the range of all the functions is $[0,1]$.)

Tobias Kildetoft

The issue arises with the term trancendental

Daminark

@mathiu_lady did you mean to just ask whether $\mathbb{R}^n$ is $\sigma$-finite wrt Lebesgue? Think of $k$-cells

mathiu_lady

06:08

@Daminark yes yes

Blue Apple

@TobiasKildetoft What the problem in previous def ie no polynomial with coeff in non comm ring has roots

Tobias Kildetoft

@BlueApple Now that you mention it, the definition is not itself a problem. But being trancendental does not imply that it commutes with everything in the original ring.

Blue Apple

@TobiasKildetoft so are trying to say that x and y differs by the fact that x commutes with ring but y mayn't

x inderteminate and y trans

Tobias Kildetoft

@BlueApple Yeah

Blue Apple

will think about it, thnks anyways

Tobias Kildetoft

06:13

The first really gives a ring of polynomials, but the other will have stuff like $yay$ where $a\in A$, and this need not be the same as $ay^2$.

Daminark

Okay so I'm getting that uniform circular motion should have central acceleration

Tobias Kildetoft

Unfortunately I don't have a very enlightening example at hand for this. But one example is to consider the ring $k[x]$ inside $k\langle x,y\rangle$ and the trancendental extension $k[x][y]$ which then in fact the same as $k\langle x,y\rangle$ rather than $k[x,y]$ as the notation otherwise suggests.

Daminark

Also whoa @Ted doing this without dominated convergence sounds tough. Maybe proving that it converges to something is the right idea?

And then using Arzela-Ascoli to find a uniformly converging subsequence might do it...

Ted Shifrin

06:40

@Daminark: But it does converge pointwise to $0$!

Daminark

I meant proving that the integral converges to something

Ted Shifrin

It sounds like we want Lusin's theorem, basically. Convergence will be uniform off a set of small measure, and on that set you know $0\le f\le 1$. But this small set may be a mess for Riemann integrals.

Well, I'm not going to worry about this one. Someone else will.

Daminark

Like, if we could show some sort of limit, and then have that by Arzela-Ascoli, a subsequence converges uniformly, that would work. Beyond that, it's tricky for me to see it working without some kind of measure theory

Ted Shifrin

Yeah, I dunno. Just thought you might like to think about it, but you have other things :P

Night! Talk to you tomorrow.

Daminark

See you!

Dattier

07:44

Hi,

Is every compact metric space isometric to a subset of a real vector space of finite dimension?

Give an elemantary justification

Salut Astyx, alors les concours ?

@Astyx

C'est pas grave tu feras 5/2 et tu auras le concours que tu souhaites

Bon courage pour l'année prochaine

.

$$\textbf{Is every compact metric space isometric to a subset of a real vector space of finite dimension?}$$

Secret

[Root tracing: Current progress]
Preliminary Let a real cubic be $ax^3+bx^2+cx^2+d$. Now consider the following:

\begin{align}
y & = a\\
y & = ax+b\\
y & = ax^2+bx+c
\end{align}

Let the nth zeros for step $i$ be $x_{ni}$. Note also that the value at the origin will remain unchanged unless $d$ is nonzero. Therefore $x_{1i}=0, i < 4$
Instructions:
1. Beginning: $x_{11}=0$

2. Translate $y=ax$ by $b$ units. Mark its intersection with the x-axis as $x_{22}$. If $b = 0$, then $x_{22}=0$, otherwise we should have the set $\{x_{12},x_{22}\} =\{0,x_{22}\}$

Kirill

If $(X,d)$ is a complete metric space, and $f$ is a contraction map and $(x_k)_{n \in \mathbb{N}}$ is a fixed point iteration sequence, then \[ d(x_k, x_{k+m}) \le L \left( \sum_{j=0}^{m-1} L^j\right) d(x_{k-1},x_k) \le \frac{L}{1-L} d(x_{k-1},x_k).\]
Though $L<1$, why $\sum_{j=0}^{m-1} L^j \le \frac{L}{1-L}$?

Daminark

08:00

Would you be summing from $j=0$ or $j=1$?

BAYMAX

What does it mean to be a subspace to be locally attractive ?

Tobias Kildetoft

@BAYMAX In what context?

Kirill

@Daminark it is $j=0$ in the lecture notes

Daminark

If you were doing it $j = 1$, the idea would be to say alright, if we extend from a finite geometric series to an infinite one, it'd converge to $\frac{L}{1-L}$, and these numbers are all positive so the infinite series is an upper bound

Araske

@Dattier should be able to embed it in infinite-dim real space, then argue by sequential compactness that you can choose a finite dimensional subspace

one or more of those claims might be outright lies, and you'd need to do work to prove both

Kirill

08:07

@Daminark oh you mean it is like $\sum_{k=0}^{\infty}x^k \le \frac{x}{1-x}$ if $x<1$, so the sum $\sum_{k=0}^{m-1}x^k$ is anyway $\le \sum_{k=0}^{\infty}x^k$?

Daminark

As it stands that's false though

Either you have to start $k=1$, or you'll get $\frac{1}{1-x}$ and not $\frac{x}{1-x}$

Turns out there's a more recent proof which is slick

Araske

what exactly is being asked

BAYMAX

I

Read this in an article on synchrony

Kirill

@Daminark $\frac{1}{1-x}-1 = \frac{1-1+x}{1-x}=\frac{x}{1-x}$. Got it.

Daminark

So $d(x,y) \le d(x,f(x)) + Ld(x,y) + d(y,f(y))$

(By triangle inequality and then use the contraction)

BAYMAX

08:14

Synchronization (identical)

Daminark

So this gives $d(x,y) \le \frac{d(x,f(x)) + d(y,f(y))}{1-L}$

BAYMAX

@

Daminark

Now, let $x = T^n(z)$ and $y = T^m(z)$

Tobias Kildetoft

@BAYMAX I don't know what synchronization is about. What sort of object is this that we are taking a subspace of?

Araske

is this not just "geometric series does the right thing on arbitrary unital algebras?"

am I misreading this?

does this not just follow from differentiation?

Tobias Kildetoft

08:17

@Araske How do geometric series make sense on arbitrary unital algebras? How does differentiation?

BAYMAX

en.m.wikipedia.org/wiki/Synchronization_of_chaos

Daminark

Then $d(T^n(z),T^m(z)) \le \frac{d(T^{n+1}(z),T^n(z)) + d(T^{m+1}(z),T^m(z))}{1-L} \le \frac{L^n + L^m}{1-L}d(T(z),z)$

BAYMAX

Under identical synchronization x(t)=y(t)

Daminark

But that makes it Cauchy. Boom

Kirill

what are $T, x,y,z$?, @Daminark

Araske

08:19

@Tobias the geometric series holds on unital (Banach) algebras; we're asking about the derivative of a contraction mapping

Daminark

$T$ was meant to be $f$

Tobias Kildetoft

@Araske There is a long way from "arbitrary" to "Banach"

Daminark

$x$ and $y$ are arbitrary, and this is showing that $\{T^n(z)\}$ is Cauchy

@Araske well, we're not trying to differentiate anything right now, we're just proving the fixed point

We're doing Lipschitz continuous functions from a complete metric space to itself

Araske

@daminark oh wait derp I see

Daminark

So the derivative doesn't even make sense unless you first embed into a normed space

And then it's hazy af since that embedding is not canonical at all

Lolol

Araske

08:21

see above: very confused as to what the question being asked was

Daminark

Ah

Araske

I thought L was a matrix

Daminark

Oh kek

Kirill

@Araske then $L<1$ would have a mysterious interpretation :)

Daminark

@Kirill "Fix $\epsilon \in \mathbb{F}^{n\times n}$ and let $\delta < \epsilon$"

Araske

08:24

...I need to stop staring at Banach algebras

Daminark

Prob

Turns out there are parts of math which aren't harmonic/functional analysis, you should check them out at some point after the REU :P

Or backtracking a few days

That works too

Tobias Kildetoft

@Daminark You mean that there are parts of math that are those things?

Araske

yeah actually

Daminark

Nah, Araske has been doing working on the stuff for the REU and the ontology of the universe seems like it's basically morphing into unital Banach algebras

Araske

@TobiasKildetoft the reports of analysis' death were greatly exaggerated

Kirill

08:28

@Daminark I do not even understand how $\epsilon$ and $\delta$ should look like here :) $\epsilon$ being a matrix with the entries in $\mathbb{F}$? What is $\delta<\epsilon$ then? Referring determinants, entries, a matrix norm, or how? How you compare the matrices?

Araske

matrix norm probs

Daminark

@Kirill I was memeing with that one, in response to what you said

Araske

maybe a C* norm-completion of $L^1$

who knows really

Tobias Kildetoft

@Daminark I think that was too complicated to be a meme

(also, when did meme turn into a verb?)

Daminark

Anything is a verb if you verb it

Kirill

08:30

@Daminark I didn't get this, too...:) But, the explanation! I will go further with the fixed point stuff. Thank you!

Araske

@tobias when didn't verbing things turn into a meme? ^u^

Daminark

No problem!

Kirill

btw, the glorious interpretation would be: $A=(a) \in \mathbb{F}^{1 \times 1}, B=\begin{pmatrix}a & b \\ c &d \end{pmatrix} \in \mathbb{F}^{2 \times 2}$, so $A<B$, as $1 \times 1 < 2 \times 2$, as $1<4$. Youpie!

Daminark

Lel

Balarka Sen

08:59

Hi chat

user147690

Hey @BalarkaSen.

Daminark

Oi @Balarka

Balarka Sen

@AlexClark Long time no see

user147690

It has been awhile haha.

3

user147690

Are you at uni now @BalarkaSen?

The Raiders of Las Vegas

09:02

Alex?

user147690

Hey Skull.

The Raiders of Las Vegas

:-D

Nice to see you pal.

Balarka Sen

@AlexClark Not yet!

hopefully next year

Tobias Kildetoft

@AlexClark Hi

user147690

Ahh, so long! Any uni in particular you are looking at?

user147690

09:04

@TobiasKildetoft Hey, how have you been?

Balarka Sen

yeah one in my country in particular

Tobias Kildetoft

@AlexClark Good, you?

user147690

@TobiasKildetoft I am good. Just trying to finish up my write up for Borel-Weil atm. I was actually wondering. What language do you think of algebraic groups in? I have looked at Milne, and he essentially says its a shame that people learn algebraic groups in terms of Borel's language, and then have to relearn it in terms of modern AG. What do you think?

Tobias Kildetoft

@AlexClark I think in terms of functors

user147690

So Jantzen style?

user147690

09:08

He says I can translate easily between "schemes" and schemes, where a "scheme" is defined as in Hartshorne, and a scheme is defined as a k-functor

Daminark

Hey @Alex and @Skull

user147690

Hello @Daminark

Balarka Sen

@Daminark you have an algebraic geometry friend now

Daminark

Woohoo

user147690

Did you have a different name before @Daminark?

Tobias Kildetoft

09:09

@AlexClark Yes, Jantzen is precisely the source I use almost exclusively

The Raiders of Las Vegas

Hey @Daminark

Daminark

(For general reference I've just started reading some Atiyah-Macdonald rather casually so don't expect much out of me)

Tobias Kildetoft

@AlexClark I think he has a short explanation of how one translates between the two in his book

Daminark

(I've just started learning what a prime ideal is)

@Alex no, I've always been daminark

Just a generic hi

user147690

@TobiasKildetoft He does. I just felt a little scared. He says that the correspondence he gives will translate definitions. Such as a scheme is smooth if and only if the "scheme" |X| is. I wonder if everything else comes across nicely

user147690

09:10

@Daminark Commutative algebra? Or he has another text

Daminark

Commutative algebra

user147690

@Daminark Well you almost have the set underlying an affine scheme :P

Balarka Sen

@AlexClark Dami is one of the new guys in chat.

The Raiders of Las Vegas

Did you have any more problems with that Alec user?

user147690

@TheRaidersofLasVegas Not that I am aware of :P. I think we both got over it, and never thought about it again.

The Raiders of Las Vegas

09:13

Good good

Daminark

Sounds nifty @Alex

The Raiders of Las Vegas

He had a nifty idea for a multiple book reading tracker...

user147690

@TobiasKildetoft Not that I dislike Jantzen, but are there any other books you'd recommend? I feel like it would be good to have a book that does the general Borel/Humphreys style development of the theory, but in the language Jantzen uses.

The Raiders of Las Vegas

...I think it would have been a better sell in the physics room.

Tobias Kildetoft

@AlexClark I am not familiar with a book that does things in terms of functors except Jantzen, but mainly because I have not needed one. They probably exist

It seems that all properties should be translatable between the two versions of "schemes"

There is a set of lecture notes by Jantzen from when he gave a course on the topic, which focus more on the case of working over a field, which avoids some of the complications that the book deals with

Secret

09:32

[Root tracing algorithm]
Ok this seemed to be not working:

$$ax^3+bx^2+cx+d = x(x(x(a)+b)+c)+d$$

The problem is that new behaviour of the roots as we move from an nth degree polynomial to an (n+1)th one is entirely controlled by the coefficients which is a constant term in each nested expression shown above.

This means, for each nested level which is a $n-k$ degree polynomial, the graph of the kth degree polynomial has to be known as the loci of the roots under vertical translation (i.e. where it will cross the x axis) is highly dependent on the shape of the kth degree polynomial

For example, suppose we are only able to draw constant functions $y=k,k\in \Bbb{R}$, then there is no hope tracing the loci of the root of $y=ax+b$ when $b$ is varied without first knowing the behaviour of $ax+b$ itself

By induction, the same argument follows up to the nth degree polynomial

Abcd

0

Q: Straight Line equation problem

Show that the equation of the line passing through $(a\cos^3\theta,a\sin^3\theta)$ and perpendicular to the limne $x\sec\theta+y\csc \theta=a$ is $x\cos\theta-y\sin\theta= a\cos2\theta$ My attempt: I converted the second line to the intercept form: $\dfrac{x}{a\cos\theta}+\dfrac...

Jasper Loy

@AlexClark Hey Alex, this is my new Math SE account, and I will try not to delete it in future. I have more things to say and I will email you.

user147690

@TobiasKildetoft Interesting. I'll see if I can find this. Thanks.

user147690

@JasperLoy Alright. I will look forward to it :).

Tobias Kildetoft

@AlexClark I doubt they are available anywhere. I can send you a copy though

user147690

09:42

@TobiasKildetoft Oh, that works too!

user147690

@TobiasKildetoft I think you have my email address perhaps?

Tobias Kildetoft

@AlexClark Not that I can see in my address book

user147690

I'll send an email to your one in a sec, so I don't get spammed from bots!

Kirill

I have to find the $\frac{\partial}{\partial x}\left( f(x,y)\right)=\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$.
1) Thinking of that as $\frac{\partial}{\partial x}\left( (x^2+y^2)^{-\frac{1}{2}}\right)$ and $\frac{\partial}{\partial a}x^a = ax^{a-1}$ I get $\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right) = -\frac{x}{\Vert (x,y) \Vert^3}$.

2) Thinking of that as $\frac{\partial}{\partial x}\left( (\sqrt{x^2+y^2})^{-1}\right)$ and $\frac{\partial}{\partial a}\sqrt{a}=\frac{1}{2\sqrt{a}}$ I get $x \cdot (x^2+y^2)$.
3) Thinking of that as $\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$ and $\frac{\partial}{\partial a}\frac{1}{a}=-\frac{1}{a^2}$ I get $-\frac{2x}{x^2+y^2}$.
Why 2) and 3) are not correct?

Secret

Hmm... $x^3+x^2+x+1 = x(x(x+1)+1)+1) = f^3(x)$ where $f = x((-)+1)$. I wonder if $f^{-1}$ exists...

$x((a_n)+1=a_{n+1}\implies x(a_n)=a_{n+1}-1 \implies a_n=\frac{a_{n+1}-1}{x},x\neq 0$

nope, that does not get anywhere except $x=\frac{\frac{\frac{-2}{x}-1}{x}-1}{x}$

Abcd

10:21

I just want to understand how can slope always be (y intercept/x intercept) irrespecitve of the inclination?

Because $tan\theta \ne tan (180-\theta)$ (wrt signs)

Secret

$-\tan \theta = \tan (180 - \theta)$, thus when the angle is reflex, the slope becomes negative

Transcript for

$Mathematics$ Mathematics