12:13 AM
How does one interpolate euler angles?
Does one just interpolate the components?

@Ultradark I have an entire discord full of people demanding you just interpolate the components.
any clue why they'd claim that?
@Ultradark if we ignore the wrapping values issue does that ever work?
just interpolating the components
@Ultradark do you know approximately what the formula for euler angle interpolation is? I don't believe its even linear which is the primary debate at hand.

12:31 AM
@anakhro Wouldn't that be Einstein rather than Eisenstein? I thought it was Einstein was a deist/atheist. Unless we're talking about an Eisenstein other than the mathematician.

12:52 AM
@s.harp I think I may try 1/2 even though having seen one of 3/4 on mathoverflow previously is scaring me off from these questions all together lol.

4

There is an interpretation of Lambda calculus that views a derivable statement $\Gamma \vdash x : A$ as the proposition $A$ with $x$ as "proof" of $A$. However, there is another interpretation in which $\Gamma \vdash x : A$ is interpreted as the statement $x \in A$. I have made my most importa...

@J.Doe oh I didn't even notice.
Thanks, I have no idea then.

1:16 AM
@MikeMiller the connection to symplectic and poisson geometry is interesting enough to me.

2:06 AM
@SimplyBeautifulArt I have not touched large cardinals formally yet, thus the only thing I can glean from that definition is B(a,b) is very huge

2:29 AM
> (insert sentence that is not directly related to academia), would you mind citing your sources
This is really a very polite way to say "I am not interested in your nonsense. Can you please shut up"
More generally, many usages of "would you mind" actually means "you must leave"
Pretentiousness culture is not just limited to the British
@ JRS

2:55 AM
@Secret be nice. >:(

Hello one and all! Is anyone here familiar with the Dormand-Prince numerical integration scheme? OR any adaptive RK method?

@Rumplestillskin nope

@TheGreatDuck Sugar!

3:30 AM
@Secret As it would turn out, my first line sums up a decent amount of large cardinal, namely Mahlos, while still being not too complicated, in my opinion.
Call $A$ stationary if every function $f$ mapping ordinals less than $\sup A$ to ordinals less than $\sup A$ has an ordinal $\alpha\in A$ such that $\alpha$ is closed over $f$.
We also restrict ourselves to $\sup A\notin A$, due to the nature of $\mathrm B(\alpha,\kappa)$.
Then the essential idea is, should such a stationary set exist, its supremum would function as a good ordinal for an ordinal collapsing function, since said ordinal collapsing function will always become constant at certain elements of $A$.
@Rumplestillskin you know people can't actually give you much feedback if you don't actually state what you want to talk about :P

3:59 AM
@SimplyBeautifulArt Did you not see the words Dormand-prince integration scheme?
It may be hard to figure out but that is what I wanted to talk about

@Rumplestillskin what about it do you want to talk about?

4:22 AM
@SimplyBeautifulArt I was wondering about the step size when you are working with a system of differential equations and whether or not you needed $n$ step sizes (for $n$ differential equations) an you adapt each step-size based off the error for each DE or whether you use some Euclidean norm for the overall error between the individual DE's. But, I've just cracked out Press - Numerical recipes and the answer is the latter.

M'kay then

2 hours later…
6:52 AM
hey
so i have a question
if you square both sides of an equation, is it still the same equation
for example we are studying about order and degree of a diff. equation
and the definition os degree is : Power od highest order derivative in a diff. equation without fraction or radical
and below that there is this example
imgur.com/a/O7M2wXC
here, to get rid of the radical sign we squared the equation
but will it still be the same equation? If the new equation we made by squaring original one have degree 2 then how can we say the original equatiin has degree 2?

By transforming the equations they in general describe different sets of solutions, all that stays is their conditional interdependence in truth value.
When you "square it" what you really do is to say "if that is true than its also true for the squared version" now since "squaring" is not invertible you cannot go back that step.

6 hours later…
1:22 PM
Problem: If $F \subseteq \Bbb{R}$ is a closed set, show that for every $\epsilon > 0$, there is a compact set $K \subseteq F$ such that $m(F \setminus K) < \epsilon$.
The case when $m(F) < \infty$ is easy: write $F = \bigcup_{n=1}^{\infty} (F \cap [-n,n])$, and then use continuity of the measure.
But what if $m(F) = \infty$? I can't figure this case out.
Wait...maybe it isn't true.
Take $F = [0,\infty)$. If it were true, then there would exist a compact set $K \subseteq F$ such that $m(F \setminus K) < 1$. Since compact sets are bounded, $m(K)< \infty$, so $m(F) < 1 + m(K)$, which is absurd.

@user193319 correct, but there is a (slightly) simpler counter-example

$F = \Bbb{R}$?

yeah

1 hour later…
2:38 PM
@Slereah hi can i ask you a question

2:52 PM
-I was wondering where he got this "3k" and this "2k" ... -I just want this information to continue my development.

I mysteriously lost 3 rep overnight

@funfun it seems they're using the similarity of the triangles ACD and AHB'
which implies that AB':AD::HB':CD

What is a bismuth function?

so |HB'|/|CD| = 2/3. Why they chose to write that as |HB'|=2k, |CD|=3k is unclear to me, but it's a valid choice

@Semiclassical someone commented " Can you construct a bismuth function $\Bbb{C} \to (0,1)+i(0,1)$" and I googled bismuth function but only could find stuff about bismuth crystals

3:03 PM
that sounds like nonsense to me
bismooth, maybe?

it was on one of my recent questions
but don't judge

note that that is indeed bismooth (as in, the function is smooth and it has a smooth inverse)

bi-smooth?

right
not great terminology imo

Hello @Semiclassical, I hope you are well.

3:10 PM
but the point is clear enough, I think: If you have a smooth function from the entire complex plane to a line segment in the complex plane, then that function can't have an inverse
hi @jasper

I have not seen bismooth before.

yeah, I'm not a fan of it

I can't understand your question, but $[0,1]^2$ is a compactification on the plane, just map the plane homemorphically to $(0,1)^2$

I might read it as bis-mooth.

@yuvrajsingh Odd but sure

3:11 PM
if you google the phrase "bismooth function" in quotations, you'll find exactly one occurence
..and that one occurrence is reuns' usage in the comment

@AlessandroCodenotti okay thanks I will look into that

so yeah, quite idiosyncratic
hmm. i may need to rewrite this recent question of mine to get more attention for it: math.stackexchange.com/questions/3330421/…

A compactification of a topological space $X$ just means an homeomorphism $p:X\to Y$ such that $p(X)$ is dense in $Y$ and $Y$ is compact. There is no reason why a continuous function $X\to Z$ should extend to a continuous function $Y\to Z$ in general though

yeah, I guess I don't need it to be continuous. I'm wondering what happens to the function when you compactify the space. Does it get warped?

It can be equivalently posed as: Given positive integers $p,q$, what is the smallest $n$ such that there exist orthogonal vectors $\vec{a},\vec{b}\in \mathbb{Q}^n$ of length $\sqrt{p},\sqrt{q}$ respectively?

3:17 PM
Unless $Y$ is the Stone-Cech compactification of $X$ and $Z$ is compact Hausdorff, then you can extend continuous maps, but that's a very hard space in general

With an obvious generalization being: Given a positive-definite matrix $A\in M_n(\mathbb{Z})$, how few rows can a rational matrix $B$ have such that $A=B^\top B$?
But that last one seems strongly dependent on the form of $A$ and I wouldn't expect a nice general answer
Actually, come to think of it, even existence isn't clear in my previous question
That is, it's not obvious that every PD matrix $A$ with integer entries must have a factorization as $A=B^\top B$ with $B$ having rational entries
If I restrict to diagonal $A$ then I think existence is guaranteed

@AlessandroCodenotti if you have a unit square in the complex plane, is it possible to then define an equivalence relation to make it a flat torus?

3:32 PM
Could anybody take a look at my question about finding the maximum of a sum of minima inside a Lebesgue integral? https://math.stackexchange.com/q/3330854/47771.
Thanks in advance to everyone who takes a look!

what is the difference between this and taking unit square in the real plane and making it a flat torus. Is the difference just that one has complex numbers on it and the other has real numbers or something?
^That's neat

1 hour later…
4:41 PM
Given $x \in M_n(\Bbb{C})$, define $||x|| := \sqrt{\frac{1}{n}tr(x^*x)}$; i.e., the normalized Hilbert-Schmidt norm. If $x,y$ are $n \times n$ unitaries, why is it true that $||xy-I|| \le ||x-I|| + ||y-I||$?
I need to show this in order to show that $x \mapsto \frac{1}{2} ||x-I||$ is a invariant length function on the $n \times n$ unitary group.

Most obvious thing seems to be to square both sides

Square both sides of the inequality?

right. might give you something simpler to work with

But those cross terms might complicate things, no?

they're definitely there, yeah
another thought is to write $x'=x-I, y'=y-I$
hrm, but then $x',y'$ aren't unitaries

5:12 PM
Algebra > analysis
:D
@Ultradark I think the answer to your question "turning a square into a torus" should be yes, since it's done in Topology that way.
It's really neat how they figured that stuff out and came up ultimately with the first homology theory
@Ultradark What can you tell us about the mapping from the torus to the eliptic curve graph? Is it a homeomorphism?
That's some interesting math you're into.

5:50 PM
@ShineOnYouCrazyDiamond hi
@ShineOnYouCrazyDiamond I have a question for you

1 hour later…
7:02 PM
So it seems, shapes do matter even for uncountable dimensions

@Secret There are no uncountable dimensions in that story. Just $3$
And of course shape matters for a question like that

ok sorry, I should be more precise, I mean, an infinite 3 dimensional space
This is counterintuitive even for infinity because one normally expect an unccountable number of 3D objects can be fit into a 3D space of uncountable size
But the moebius proof showed that you cannot even do that even if the uncountable cardinalities of the moebius strips you have are the same as the ambient space itself

What? You can't fit an uncountable number of solid balls into 3-space

Sanity check: $\Bbb Q(\zeta_p)$ has a quadratic subfield because $\Bbb Z/(2) \leq \Bbb Z/(p-1)$

The surprising thing here is that the Möbius strip is locally $2$-dimensional

7:08 PM
($p$ an odd prime)

@ÍgjøgnumMeg Well, because there is a subgroup of index $2$

yeah
lol

well er... regarding uncountable number of solid balls, I am thinking about balls of all sizes, so that the gaps will be arbitrarily filled in
The whole process forms a tree and thus uncountably many balls are involved, I think
The moebius proof they mentioned seemed to suggest that you cannot even pile up uncountably many moebius stripes in 3D space without intersection

yeah that seems like it would make sense
;p

But then I have not read the proof yet in detail (I got that link from h bar) so I may be grossly wrong

7:26 PM
If $f \in L^2$ then trivially $f^2 \in L^1$, right?

Another sanity check: the quadratic field inside $\Bbb Q(\zeta_p)$ has to be $\Bbb Q(\sqrt{\pm p})$ since if it's $\Bbb Q(\sqrt{\pm q})$ with $q \neq p$ prime then $q$ ramifies in that extension, and so it ramifies in $\Bbb Q(\zeta_p)$, which only happens for $p$. If $p\equiv 3 \bmod 4$ and the sign is + then $\operatorname{disc}(\Bbb Q(\sqrt{p})) = 4p$, which implies $2$ ramifies so this can't happen either, so the sign is $-$ for $p \equiv 3 \bmod 4$ and positive for $p \equiv 1 \bmod 4$.

@Secret The sizes of the balls don't matter. Better review your countability/uncountability. :)

@Ted that's what aaaaall the guys say

Hi @ÍgjøgnumMeg

Hey @Ted :P

7:35 PM
smacks @ÍgjøgnumMeg

rofl

The company I work for just released its interim Q2 report. So many unfamiliar words. But at least it has some large numbers which looks good :)

If you like numbers .... :D

Assuming non-negative $g_n$ satisfy $\int g_n \leq \frac 1n$ , can I conclude $g_n \to 0$ a.e.?

@TedShifrin Well, numbers that mean they will keep paying me are good numbers.

7:41 PM
OK, I'll stipulate that I believe you :)

@Emolga Oh I see that I cannot... So still trying to prove:
Assuming non-negative $g_n$ satisfy $\int g_n \to 0$ , can I conclude some $g_{n_k}→0$ a.e.?

I've never seen a number before
I'm an abstract math person
I only deal in greek and latin symbols

hmm good idea, I actually noticed I am making an increasing number of mistakes in most of my recent activities in various knowledge circles
It is not clear whether it is PhD stress, too many activities or something else, but anyway

It's a convex combination of all those

$\Delta^{\Lambda}$

7:48 PM
Values in a simplicial set mapped from (weird symbol)

I just put lambdas and deltas together for no reason

You're a graphomaniac like me, that's the reason

Thought that is not entirely true

It's a good thing, it makes note taking pleasurable

Because I just read some equations in my literature review which has $E^{\Delta\Delta}$ for the 2nd derivative of $E$

7:50 PM
I'm not liking the double symbols there, would have to revise it

good luck telling that chemist to review his paper lol

@TedShifrin any reason why they call them "Green's" functions? Not the Green part, that's straight forward, but the apostrophe ess part.

@ShineOnYouCrazyDiamond if you take a field on the x-axis and a field on the y-axis are the set of ordered pair points also a field?

@Ultradark no, not with componentwise ops, but if you define multiplication of pairs the same as in $\Bbb{C}$ then you get a structure that "is" $\Bbb{C}$ or atleast isomorphic to it as a ring isom.

noi8ce

7:53 PM
You can do componentwise +, then what would * be? You do (a,b)(c,d) = (ac - bd, ad + bc) or something like that. Then it is indeed "the field of complex numbers". With componentwise +, * it's only a ring.

whtacuhu mean component wise
like multiply the x coordinate by another x coordinate

(a,b)(c,d) = (ac, bd) is comonentwise
You can prove that it's not a field (no mult inverse in general)

i kne2that

Such as $(1,0)$ has none, right?

facts

7:54 PM
Cool! I like your interest in elliptic functions. I haven't gotten there yet
R x S is always a ring

nvm

when R, S are
$\Bbb{R}$ is a field, but more generally a ring

what's S

another ring
abstract = me

7:56 PM
So does the square in $\Bbb{C}$ turn into a torus some how?
Did you know that points on the planar elliptic curve form a group in some weird geometric way involving drawing tangents, chords between points, etc?

@ShineOnYouCrazyDiamond am i right in saying a subfield plotted against a subfield is a ring with multiplication wheere all elements of said fields are of the algebraic numbers in (0,1)?

You'll have to be more formal, because I'm not sure what you mean yet

okay

Also, I'm not good at ANT yet
when you say algebraic numbers I think of ANT

i dont want to bother the chat any more

7:59 PM
k
:D
Your questions seem very high quality on MSE. I would make a post

wait

Chat room?

yea

8:20 PM
I'm trying to understand whuber's comment on my answer here—
This argument cannot be correct, because the assumptions apply to two orthogonal circles in $E^3,$ but the conclusion does not hold. Somehow you need to introduce an assumption that is tantamount to the circles lying in a common plane. In other words, you also need Euclid's axiom that three points determine a plane. — whuber 2 hours ago
I'm giving an argument from a couple of simple axioms that two concentric circles with different radii cannot intersect.
whuber is saying they have a counterexample to my argument—
Whoops, I have to go.

8:31 PM
@ÍgjøgnumMeg that's my favourite argument for figuring out the quadratic subfield of $\Bbb Q(\zeta_p)$

Could someone confirm is my solution correct so far?
-1

Problem Let $V$ be finite-dimensional inner product space and $U$ it's subspace. Let's assume that $\vec{v} \in V$. Show that $\vec{v} \in U \iff \text{proj}_U(\vec{v}) = \vec{v}$ Attempt to show Let's assume subspace $U$ has orthogonal basis ($\vec{u}_1,\vec{u}_2,\vec{u}_3,\dots \vec{u}_n$) ...

Also, if someone knows why there is downvote please let me know

Also yo @MatheinBoulomenos long time no speak

8:52 PM
yo @Daminark

How's everything going?

help with problem )

@Mathein tis nice
@Mathein just having a look at washington's cyclotomic fields lol

anyone?

Eyy @ÍgjøgnumMeg

9:06 PM
Hey @ÍgjøgnumMeg
@Daminark pretty well, thanks. Working on my undergrad thesis (I want to finish my undergrad by next spring)

@MatheinBoulomenos hi

hi @LeakyNun

I've been procrastinating from maths
wie sagt man "procrastinate" auf deutsch?

prokrastinieren

cool
and which proposition do you use

9:10 PM
Right, so, I wrote an argument, from a couple of simple axioms, that two concentric circles with different radii cannot intersect at a point. @whuber said that "This argument cannot be correct, because the assumptions apply to two orthogonal circles in $E^3,$ but the conclusion does not hold."
Anyone know what they might mean by that?
What's $E^3$? Is it $\mathbb{R}^3$?

yeah $E^3$ is just $\Bbb R^3$
$E$ stands for Euclidean

All right, I just posted a comment asking for clarification.

@LeakyNun you just use a direct object, no preposition

@MatheinBoulomenos konnten sie mir einen beweis geben?

@LeakyNun does the first example sentence here count? de.wiktionary.org/wiki/prokrastinieren

9:17 PM
@TannerSwett it's not really clear what 'concentric' is supposed to mean for circles in E^3

@MatheinBoulomenos oops ich habe beispiel gemeint

Hey there, @Leaky

hi @Ted

hi @Mathein

9:18 PM
I don't even know how to use the word in English

can't you use "procrastinate" transitively in English as well?

@TedShifrin how do you say "I am supposed to do this assignment but I am procrastinating" concisely?

I'm not doing my work.

in German do you just say "ich prokrastiniere diese Zuordnung"
@TedShifrin I mean, using the word "procrastinate"

9:21 PM
I suppose you can procrastinate on something ...

@Mathein nice, good luck! And hey Ted!
And Leaky and Semi
I just finished my topology qual

Hey there, Demonark.
Any interesting questions?

Hey @TedShifrin how does the orthonormal basis for all V help in this case?

One which my argument was needlessly complicated on but by the looks of it (thankfully) correct: If $X$ and $Y$ are simply connected CW complexes such that $X\times \mathbb{RP}^m$ is homotopy equivalent to $Y\times \mathbb{RP}^n$, then $m=n$

@Mathein @Leaky dictionary.com says procrastinate can be transitive, but I don't think I've ever heard/said it.

9:23 PM
Turns out the correct way to do it is using the cohomology ring but uh, this week in things I don't know :(

@LeakyNun if you want to use an object, you can use the synonymous "aufschieben", I think "prokrastinieren" is mostly used intransitively in German as well

separable?

@Tuki: Because then you write $v\in V$ as a linear combination of all the basis vectors. The projection maps to $\sum (v\cdot e_i)e_i$, where the sum is over the basis vectors for $U$ only. But then how can the projection equal the original vector?

auf is always separable right
man I don't remember these stuffs anymore

yes, aufschieben is separable

9:25 PM
@Daminark I was about to say what if $X = \Bbb{RP}^n$ and $Y = \Bbb{RP}^m$

Demonark: You need the Künneth theorem for homology or for cohomology. I'm not sure you need the ring structure on the cohomology, but maybe I'm being dumb.

What I did was say that if $f:X\times \mathbb{RP}^m \to Y\times \mathbb{RP}^n$ is our homotopy equivalence, it lifts to a map between the universal covers $\tilde{f}:X\times S^m \to Y\times S^n$.
$\tilde{f}$ should also be a homotopy equivalence by Whitehead's theorem

I never thought about if auf is always separable

Oh, I missed the different $X$ and $Y$. So, yeah, I guess cohomology will be easier because then we know $H^1(X) = H^1(Y)=0$.
Hmmm, why does the h.e. lift?

> Prefixes that are always inseparable are: be-, emp-, ent-, er-, ge-, miss-, ver-, & zer-. The most common of these is ver-.

> These [separable prefixes] are most frequently prepositions (e.g, ab, an, auf, aus, bei, mit, nach, statt, vor, zu) or adverbs (e.g., fort, los, nieder, vorbei, weg, zurück, zusammen). However, nouns and adjectives can also serve as separable prefixes (e.g., teilnehmen, festhalten).

9:27 PM
You project $X\times S^m \to X\times \mathbb{RP}^m$, then push across $f$
That's a map from a simply connected space to $Y\times \mathbb{RP}^n$, so that lifts to the universal cover $Y\times S^n$

I remember uber being both separable and inseparable
ubersetzen was the example for both

Oh, right, the composite is trivial on $\pi_1$ even though the original map wouldn't be.

so universal cover is a functor!

"umfahren" can mean "drive over" or "drive around" depending whether it's separable or not

which one?

9:30 PM
separable one means drive over

> Some prefixes can be used as separable prefixes or inseparable prefixes. The most common of these are durch-, über-, um-, unter-, and wider-.

So then $\tilde{f}$ should be a weak homotopy equivalence. Obviously induces an iso on $\pi_1 = 0$, and then $f$ and the covering projections induce isos on higher homotopy groups. But since $X$ and $Y$ are CW complexes, Whitehead's theorem kicks in, $\tilde{f}$ is a homotopy equivalence

so "ich fahre dich um" is bad while "ich umfahre dich" is good?

@LeakyNun exactly

Now we compute homology twice
(Everything in mod 2 coefficients)

9:31 PM
Yeah, I think this argument is unnecessarily convoluted.

@MatheinBoulomenos is wieder separable or inseparable?

Assume $m<n$, then $H_m(X\times S^m) = H_0(X) \oplus H_m(X)$ by Kunneth, also $H_m(Y\times S^n) = H_m(Y)$. So this implies that the rank of $H_m(Y)$ is 1 more than that of $H_m(X)$

@Daminark when did we assume finiteness?

Once we developed moral compasses

what

9:33 PM
@LeakyNun it's separable in all examples I can think of

@MatheinBoulomenos what are the examples you thought of xd

(I'll admit this doesn't work in the infinite-dim case, that's where the cup product is what you want, but tbh finite CW complexes are what really matter)

wiedergeben, wiederfinden, wiedergewinnen

But yeah so, for $k\le m$, you have $H_k(X\times \mathbb{RP}^m) = \bigoplus_{i=0}^k H_i(X)$, and similarly $H_k(Y\times \mathbb{RP}^n) = \bigoplus_{i=0}^k H_i(Y)$. Inductively that should imply that the ranks agree up to $m$
So yeah we're screwed

That's some fancy homology :)

9:35 PM
It's a mess but it's what I thought of and I think/hope this checks out

@Daminark what are you studying?

But yeah we'll see if I pass this qual, I think algebra went well so there's a chance I'm just done already

It's a good experience to take the quals, regardless.

Uh, I'm just starting in grad school, probably gonna end up doing something under the algebra/NT umbrella but I also like topology, some parts of analysis, and combinatorics
And etc

You want to study etc?

9:38 PM
They're all related, they use combinatorics in homology-cohomology to reason about stuf

@MatheinBoulomenos I think wiederentdecken is a counterexample
and wiederholen

But my current top 3 adviser choices work in arithmetic geo, topology of singularities, and geometric representation theory, in that order
Yeah my undergrad etc class had the best psets

@LeakyNun wiederholen yeah, but wiederentdecken not

To have adviser choices before even starting is sort of unusual.
Ordinarily, students get to know faculty in classes/seminars and figure out whom they'd be comfortable working with. It's not just field.

@MatheinBoulomenos cool

9:40 PM
actually wiederholen can be both separable and inseparable, depending on the meaning

I'm weeder holin my bowl right now
:D

It's tentative, more based on impressions from talking to some students, also former profs of mine

I'm just sayin'

As far as field in general goes, there are at least 8 people who seem fun

Just try not to make them all hate you :D

9:42 PM
That... will be tough
Fingers crossed :P

I know :)

If one is a punster as well...

Oh oh
We faculty don't like to be outdone by uppity students.

:0

@MatheinBoulomenos interesting

9:46 PM
Heya @Daminark
long time no see

I'm worried that I have to prove my English profiency for the masters program I want to join

Seriously? Why be worried? Your English is superb.

lo lwht

thanks

Do you want a letter from me?

9:48 PM
thanks, but I have to take IELTS

I, too, would offer my support, but I have no credentials to draw on

Of all things to be worried about, I would say this one is nonsensical.

@Mathein which program was it you were joining again?

I want to do ALGANT, probably

I did something sooooo stoooooopid earlier today.

9:49 PM
Ohhh I know ALGANT
nice :)

I combined my gmail mail account with the other gmail accounts on my Mac, and now MacMail is downloading 20,000+ emails.

loool ouch

oh wow

<---- for someone who claims Mac competence, so stoooooooopid.

lol

9:51 PM
So far I'm back to August 2018 and there's a long way to go.