Yes, we are sadly not immortal. I’m doing tomorrow afternoon/evening to the memorial service of a friend who died at age 45 (due to previous brain damage he never told us about).

If $K/F$ is a field extension, $M$ is algebraic closure of $K$, can there exist two distinct normal closures $N, N'\subseteq M$ of $K/F$?

I'm just wondering if "minimal" can be replaced by "smallest"

my question is stupid

Isn’t $N=M$? Or are you doing inseparable things?

Oh, no, I was silly. I vote unique.

From definition $N = N'$ because they're generated by roots of the same polynomials

Well, you could consider different polynomials giving $K/F$, no?

I don't really understand

Where are you getting “the same polynomials”?

My definition of normal closure is the splitting field of minimal polynomials of $\alpha\in K$ over $F$

Ah.

I guess if you take the intersection of all Galois extensions of $F$ containing $K$ you get uniqueness, too.

I've only proved that $N$ is the unique minimal element in the poset of normal extensions contained in $M$ and containing $K$

which doesn't necessarily imply that $N$ is minimum in that poset

If $K\subseteq N'\subseteq M$ is a normal extension, then all roots of minimal polynomials of elements of $K$ are contained in $N'$

So $N\subseteq N'$ so that $N$ is also a minimum

Lots of fancy words, but what’s wrong with my saying to intersect all?

I just never thought about intersections of normal extensions being normal and I'm too tired to check this right now

Fine.

Ah I see it now. You can prove it using characterization of normal extension L/F as every irreducible polynomial over F with a root in L splits over L

That makes sense, thank you

I didn’t do much :)

The plural of "cannon" is "cannon".

That is all.

In the case of the judge, one is way too many!

that's a blast!

also a good 70s tv show starring william conrad as a private investigator in los angeles

I watched that, along with Rockford Files.

cannon also liked food and wine. did he cook? i don't remember, but it wouldn't surprise me.

william conrad also played nero wolfe in a short lived series, who also absolutely could have been a chef but generally left that to his private chef.

all of this is a short way of saying i'm hungry

loved the pi cannon, used to watch it with my mom

Can anyone help me understand lemma 1.1 of advancesincontinuousanddiscretemodels.springeropen.com/articles/…

You don't need to download the pdf just scroll down a little bit.

They call it Deformation lemma.

Okay ignore that link. I'll just state it directly from the book. Notation: $H$- HIlbert, $I$- nonlinear functional, $A_c=I^{-1}((-\infty,c] ),K_c=\{u\in H: I[u]=c,I'[u]=0\}$.

Theorem: Assume $I\in C^1(H,\mathbb R)$ satisfies Palais Smale condition. (also assume $I'$ is Lipschitz continuous on bounded sets) Suppose $K_c=\varnothing$. Then for each sufficiently small $\varepsilon>0$, there exists a constant $0<\delta<\varepsilon$ anda function $\eta\in C([0,1]\times H;H)$ such that

1) $\eta(0,u)=u$ for all u, (2) $\eta(1,u)=u$ for all $u\not\in I^{-1}[c-\varepsilon,c+\varepsilon]$, (3) $I[\eta(t,u)]\leq I[u]$, (4) $\eta(A_{c+\delta})\subset A_{c-\delta}$

I can't visualize this. can anyone give me some intuition.

Are there any (widely used) programs for drawing(?) figures in books or papers?

I don't know I asked correctly

yes but that's for diagrams isn't it?

Wdym by diagrams? It can draw a lot of things if you know how to

like commutative diagrams. Let me search what I can do more with it

Yeah that's not the only feature of it

oh I can do a lot of things with tikz lol

thanks

I'm a bit worried about my eye health these days. I usually read books using Ipad so prolonged exposure to light from an electronic device makes me worry about my eye health.

@Jakobian sorry for the direct ping, but we talked yesterday about the inequality $\frac{x^2}{\max\{x,4\}}<1$. Is it correct this inequality is equivalent to $x^2<x$ or $x^2<4$? Of course, I could use inspection, but I just want to know for future problems when the $\max$ is more involved.

Yeah

ok, thanks for confirming.

@TedShifrin As soon as I saw the reference to Cannon, my next thought was "Yeah, it's good, but Rockford was better." You have good taste, sir.

Come on, Columbo blows them out of the water.

@user858770 Peter Falk was great. James Garner was better. :P

For the proof refer to the popularity of the re-runs.

:-)

@user858770 Can't prove an opinion. Opinions are held without proof.

I should've given honorable mention to McCloud and McMillan and Wife :P

@onepotatotwopotato you can purchase glasses that fix the issue to an extent

I put them on for prolonged gaming sessions

Youtube subtitles are weird, it wrote meromorphic functions as marijuana functions

@冥王Hades I've already worn glasses for a very long time. That's why I'm concerned.

I want to check continuity of $s(x)=\sum_{k=1}^\infty \frac{x^{2k}}{x^k+4^k}$ on $(0,2)$. Is it then enough to check uniform convergence of this series by considering an interval of the form $(0,c]$ where $0<c<2$. Because then $\left|\frac{x^{2k}}{x^k+4^k}\right|\leq \frac{c^{2k}}{4^k}=\left(\frac{c^2}{2^2}\right)^k$ and since $0<c<2 \iff 0<c^2<2^2$, we have a convergent geometric series that we can use in the Weierstrass M-test. Any objections?

I am a little unsure about the interval we need to check uniform convergence on, if it's $(0,c]$ where $0<c<2$ or even $[c,C]$, where $0<c<C<2$. I guess both will do the job.

@sunny Not reading in detail, but sure, you could do that. But, again, this will only tell you a set on which the series does converge. It doesn't really give you any information about when it does not converge. If the goal is to determine the set on which the series converges, you probably want to use a tool which does a better job of that.

@sunny What is the precise statement of the $M$ test you have in mind? It has been a while since I've thought about it, so I don't have it immediately to hand.

If the series looks like $\sum a_n(x)$, then the $M$ test works by finding a uniform bound for each $a_n$ on some set, right?

@XanderHenderson Right. You need to find a sequence $M_n$ (independent of $x$) that bounds the absolute value of your terms $a_n(x)$.

That is, if $|a_n(x)| < M_n$ (for some choice of $M_n$), and $\sum M_n$ converges, then the original series converges, too.

Now, in this case, I'm interested in the continuity of $s(x)$ on $(0,2)$, so I think I need to check uniform convergence in a neighborhood of each point there.

@sunny Seems like overkill.

The $M$ test has already given you uniform convergence to a continuous function on all of $(0,2)$.

@XanderHenderson are you claiming my function is uniformly convergent on $(0,2)$?

@sunny Sure.

The uniform limit of continuous functions is continuous, no?

If $f_n(x) = \sum_{j=1}^{n} a_j(x)$ and $f$ is the limiting function, then you have shown that $|f_n(x) - f(x)| \le \sum_{j=n+1}^{\infty} M_j$ for any $x \in (0,2)$, right? And this last sum goes to zero, since the sum of the $M_n$ converges.

@XanderHenderson I'm confused. I was claiming above it is only uniformly convergent on $(0,c]$ where $0<c<2$, but I may be wrong. I don't see how we get an upper bound (independent of $x$) for the absolute value of the terms on the interval $(0,2)$, which we need to have in the $M$-test. How would you bound $$\left|\frac{x^{2k}}{x^k+4^k}\right|$$ on $(0,2)$?

@sunny Like I said, I was not reading your work very carefully.

On whatever set you get the uniform bounds $(M_n)_n$, you get uniform convergence to a continuous function.

yes

@XanderHenderson huh?

@XanderHenderson the bound depends on $0 < c < 2$

@Jakobian I have said, multiple times, that I did not read the argument carefully.

Because I don't care to grade homework which I didn't assign.

The relevant quote is:

(and the comment above that)

Can anyone suggest a good and a basic book on numerical mathematical analysis?

@ThomasFinley What do you mean by "numerical mathematical analysis"?

Do you mean numerical analysis (which includes, for example, the study of numerical approximation methods)?

Or something else?

Also, at what level? (clever high schooler? undergraduate? graduate?)

@XanderHenderson Yes, I mean the study of numerical approximation methods like interpolation, forward interpolation, backward interpolation (Newton's method) at undergrad level

@ThomasFinley bookstore.ams.org/amstext-2

@XanderHenderson Thanks! I will surely take a look at it.

How to solve this ?

No idea. I have no knowledge of the context of the question, hence I have no idea what it is asking me to do.

@XanderHenderson Oh, no problem at all. Anyways, I am trying give more context.

This is the thing I am recently working on. I think it all seems like a 'Make an intution and let you be judged whether u r lucky ' sort of a problem

I think I should post it on the main site...

@XanderHenderson You can always just not check it then

@Jakobian And I didn't. I responded specifically to questions about the M test.

Why are you jumping down my throat about this?

quantamagazine.org/…

it's about circle packing something

Does anyone want to know how large my palace is?

Isn't the whole underworld your palace?

found a sequence not in the OEIS

@onepotatotwopotato The original paper is arxiv.org/abs/2307.02749 .

Quite surprising.

Surprise, surprise

It isn't really my area, but the paper seems believable on a first pass. Assuming it holds up, it should get into a good journal. That undergrad is set up for a good grad school.

good for them

we need person stocks so we can invest in them

make money on bright looking undergrads

publicly trade 80% of their personship, they're entitled to 20%

free university and no more student loans, win win (for the winners)

Ew.

Gross.

I actually want to study circle packing in some (possibly?) different settings. So that article caught my (bad) eye

but the circle packing itself seems fractal

@SoumikMukherjee exactly but look at how large my residence is

Just one of the rooms in my palace holds a universe inside

In an amalgamated free product, do the groups in the amalgamation embed into the amalgamated free product?

I know they do for regular free products.

I guess it's true

i always go for free products

The amalgamation does not create new relations only in terms of one of each group. But I guess the proof should be written using universal properties

@copper.hat I would but I need to embed a group into another $G$ such that $G$ inside this group is self-normalizing but does not always have a certain "coset covering property." $G$ is always self-normalizing in $G * H$ for any $H$, but the the "coset covering property" is sometimes satisfied :( Hence, my reason for needing amalgamated free products because I need some relations (although I can't yet determine exactly what those relations need to be).

@user193319 my apologies, i meant that as a joke, not to distract from your concern

oh, okay. I understand lol :D

That palace is up for renting if anyone’s interested

@XanderHenderson I'm sorry but I'm not actually angry, if I sound so then it's most likely because of the way I talk

it's okay that you're doing all that, it was just borderline confusing

@copper.hat saves money, if nothing else.

@robjohn saving money is good :-)

indeed

It is?

Yes. If you don't need it you can help someone else.

Does each particular solution to this equation dependent on $x$ and $t$ form an infinite dim. vector space?

anyone here know functional analysis?

What’s money? All I know about is worthless @lesliecoin.

oh wait, I remember. Solutions to ODE's are functions, solutions to PDE's are manifolds. I'm curious what a solution of the form $f_1(x,t)+f_2(x,t)+\cdot\cdot\cdot$ would form. Maybe another manifold

@copper.hat or buy more donuts

or chocolate

@geocalc33 As it stands, this is nonsense.

@TedShifrin which part is nonsense?

@geocalc33 The bit between "[O]h wait" and "manifold".

power series solutions of pde's

solutions to differential equations aren't functions?

Why do PDEs magically turn into manifolds? Yes, unless you’re discussing distributional solutions, solutions are functions, whether ODE or PDE.

oh okay i see now

Do you?

If one is discussing distributional solutions that can be viewed in 2 equivalent ways: as a surface (manifold), or as a function that changes over time.

for the appropriate dim.

function that changes over time would be the distribution function changing according to the parameter.

@TedShifrin slandering the world's first true currency is a crime on the offshore nation of leslieland

and if viewing as a surface, then you plot the distribution as function the parameter and the var.

so in this picture the surface remains 'fixed' i.e. doesn't evolve according to some D.E.

@XanderHenderson I understand that my initial statement is correct yes

but I also understand why it might be criticized

it would probably be more correct to say "function" yes

for consistency

Geocalc I don’t think you even know what distributions are.

@leslietownes So pay me to cease and desist :)

I’ll drink soup with a straw

Still ailing?

@geocalc33 huh?

@TedShifrin doing better now

Most of my normal functionality is back

your what?

though my soul is still trapped, wandering the planes of existence, in search of a new person to reincarnate into

ok

reincarnate to chuck norris

been considering algebraic number theory

clearly

lom

lol

ask me for a link :)

@mick too weak for me

He isn’t worthy of my sheer elegance

Hi :) Is there a form of calculus with some sort of measure theory where one can move both towards and away from something simultaneously?

Perhaps like the direct product of two metrics or something . . .

in which you move away in one but towards in another . .

That feels like cheating though.

Consider, maybe, $y=1-x$. If you start at $(0,1)$ and move down, then you both move away from and towards the origin.

But, again, that doesn't fit with what I have in mind, unless a metric of sorts can capture it using that example.

D'you think I should ask a question about it on the main site? Or would it get shot down as nonsense?

This . . .

. . . seems like a contradiction in terms, but strange things can happen and I don't know for sure.

How about on the surface of a torus?

Simpler: the surface of a sphere.

Okay, I think I'll ask on the main site.