Mathematics - 2021-08-22

shintuku

00:16

are there noteworthy measures other than the Lebesgue measure if I'm interested in classifying how close a function $f: X \to \mathbb{R}$ is to being surjective over an interval of its image?

shintuku

00:31

i especially want to investigate functions of type $f: \mathbb{Q} \to \mathbb{R}$ in this way, but the Lebesgue measure gives $0$ for these

Avra

How about?

$$k \le \lg n < k+1,$$

so

$$k = \lg n + \alpha,$$

where

$$0 \le \alpha < 1,$$

thus

$$k = \lfloor \lg n \rfloor.$$

Why $k=\log{n} +\alpha$? please?

I found the answer on wikipedia

Avra

01:30

What do you understand from "array in reverse sorted order" please?

Vladimir Reshetnikov

01:44

How many maximal dense chains are there in P(ℕ), partially ordered by ⊂? The question is inspired by this thread: twitter.com/JDHamkins/status/1425898421794312198?s=20, and the blogpost: jdh.hamkins.org/the-lattice-of-sets-of-natural-numbers-is-rich

P-addict

05:37

when does $f\in\mathbb{R}[x_1,\dots,x_n]$ have the property "$f(x)=0\Rightarrow g(x)=0$ implies $f\mid g$", where $x\in\mathbb{R}^n$? clearly $f$ cannot have a non-constant factor with no roots, for if $h\mid f$ has no roots then $f/h$ is $0$ exactly when $f$ is, so $f\mid f/h$, thus $h$ is constant. if $f$ is the product $P_1\dots P_k$ of irreducible non-constant polynomials, we cannot have $P_i=P_j$ for any $i\neq j$, else $f/P_i$ is $0$ exactly when $f$ is, not sure where to go from here

sorry, to be clear: the property is "'$f(x)=0\Rightarrow g(x)=0$' implies '$f\mid g$'"

and this is for any $x\in\mathbb{R}^n$, $x$ is not fixed

i suppose we also know the roots of $P_i$ cannot be a subset of those of $P_j$, $i\neq j$, but i'm not sure what else to say about $f$

shintuku

08:25

the author states the theorem: Let $X,Y$ be metric spaces, let $f$ be a function $f: X \to Y$, let $x_0$ be a point in $X$, and let $y_0 := f(x_0)$. Prove $f$ is continuous at $x_0$ if and only if, for any neighborhood $V$ of $y_0$, there exists a neighborhood $U$ of $x_0$ with $f(U) \subset V$.

doesn't the theorem need to add something like: there exists at least one neighborhood of $y_0$?

the first step of the proof requires that I say "consider an arbitrary neighborhood $V$ of $y_0$", but I don't know if it exists

shintuku

09:08

oh... the empty function is continuous lol

guess you don't need more constraints for this theorem then

shintuku

10:51

hm. apparently there's quite a few examples of degenerate continuity

geocalc33

hi

shintuku

hi

geocalc33

I was thinking about a somewhat redundant framework called multiplicative calculus in which calculus is re-written using multiplicative operators

a.k.a "Non-Newtonian" calculus

link.springer.com/article/10.1007/s10851-011-0275-1

It's basically single variable calculus re-written under the isomorphism $\exp: \Bbb R^2 \to \Bbb R^2_{\gt 0}$

but it could have applications per the link

AMDG

11:27

Is there a way to determine how a wave's frequency changes with respect to time as the input?

SAJW

11:37

@PM2Ring if the to be added term is not defined for some value of the iterator, you can bypass it still (I think). Take $\sum_{i=1}\frac{1}{i}$. I can rewrite it as $\sum_{i=0}\frac{1}{i+1}$ and get the same. But I'm not sure if that is always possible, just saying the term that gets added doesn't have to be the same.

PM 2Ring

12:15

@SAJW If you do the appropriate transformation to the expression that the Sigma is summing over, then sure, you can change the starting index to anything you want. You don't even need to add that constant. (For those wondering what we're talking about, it's this question: math.stackexchange.com/q/4229864 )

But if you can't transform the expression, then you can't just absorb the difference in a constant. Not only do you have to worry about infinite terms, you could also get complex ones, as in Ted's example:

18 hours ago, by Ted Shifrin

How about $\sum \frac 1{\sqrt{n^2+n-4}}$?

SAJW

So what are the rules for the starting index? Choose it, such that the expression is as simple as possible?

(but not simplier...)

PM 2Ring

$\sum_{k=0}^n f(k)$ is identical to $\sum_{k=m}^{n+m} f(k-m)$

Yes, you want $f$ to be simple. And it's traditional for the starting index to be 1, although 0 can be a sensible choice, too. Especially for people used to the zero-based indexing used in most modern programming languages.

Which gives me the perfect excuse to post this old favourite:

His books were kinda intimidating; rappelling down through his skylight seemed like the best option.

SAJW

XD

PM 2Ring

12:36

I wrote another arctan-based pi calculator today. It uses Python's decimal module, so it can produce lots of digits. It uses (approximations of) irrational arctans. We start with arctan(1), and progressively halve the angle by inverting $\tan(2x) = 2\tan(x)/(1 - \tan^2(x))$. Then we take the arctan of the small angle, and multiply it by the appropriate power of 2 to get pi.

SAJW

mmh, will $\sum_{a\in A}$ occure to me in programming? A is not a set of integers

PM 2Ring

Sure. That's a fairly common case, and easy to write in Python.

Here's my pi program. prec sets the number of digits. r controls how much work is done in the angle-halving phase vs in calculating the arctan. r must be >0, and it's best if it's <1, but you can try larger values if you like. We want the arctan phase to have more iterations than the angle-halving because the angle-halving has to do a square root on each iteration.

sagecell.sagemath.org/…

shintuku

14:02

why do people keep upvoting my questions

don't they understand i'm not qualified to have this many points

SAJW

you have an accepted answer that has 0, that's just rude

ah it's an answer to your own question héhe

AMDG

I've been messing with bits as square waves which is why I asked that question earlier, and it's really neat. desmos.com/calculator/znpma7h2lo

geocalc33

@shintuku do you know some set theory?

shintuku

yessir

AMDG

14:24

(contd.) It looks like division has a period that is directly proportional to the bit index and the divisor. If you can compute the function for one bit, then you have the function for all bits. The simplicity of them also makes them great candidates for implementation in HW I think, but I haven't contemplated yet how to implement them in the first place.

Also, @robjohn , I had the erroneous assumption that the modular decomposition of an integer would result in only bitshifts and adds, but I failed to notice that it's still requiring non powers of two, so it requires mul. However, it turns out that an alternative is to use the power of two decomposition of an integer directly because you can generalize 1/(a + b) to arbitrary numbers of terms with the same series.

So I'm still working on this, and then just yesterday, I discovered this stuff concerning division and bit square wave functions we'll call them... unless they already have a different formal name, but I wouldn't know where research has been published concerning bits as waves apart from signal processing and what little I've heard of things like FFT.

I'm probably just going to settle on using the Newton-Raphson that PM 2Ring gave me for now because I don't have the proper tools to analyze this to the extent I'd desire, and this is sufficient for immediate needs.

Not to mention, this is enough of a rabbit hole already.

PM 2Ring

I'm glad you found the Newton-Raphson reciprocal stuff useful. Messing around with that stuff motivated me to write code for the Remez algorithm that finds minimax polynomials, which is pretty amazing.

AMDG

14:44

Yeah, it was more useful than I realized after I went through the effort of inlining all the Newton Raphson iterations I needed to get to 1ulp.

It's all madds

So it ends up being 15c, not 72c

Based on what you said about 113 bits of precision, I can get 56.5 bits with four iterations and the ideal linear approximation guess you provided.

I just wish there was a way to get those few more to make it 64 bits, though that won't matter assuming I can just compute quotients with integers this way without worrying. It'll be part of the function and not as a reciprocal function. Besides, if it does work for all 64-bit integer pairs, then you can compute the reciprocal that way anyways.

I'm probably going to end up implementing this today or tomorrow, preferably today so I can get it over with and get on with my project. I've spent enough time on this for now.

In the mean time, I can rest easy knowing that there is in fact a better, more efficient method with the geometric series and bit waves.

Seriously, though, does anything like this stuff with waves show up in any literature anywhere? And is there a resource I can look at that mentions various manipulations and transforms of waves and their interactions?

Thorgott

@shintuku $C^{\ast}$ is just some infinite set, it has no reason to be in bijection with $\mathbb{N}$, but you also don't need choice to pick an element from the set. it's non-empty, so an element exists by definition.

PM 2Ring

@AMDG Well, the fundamental wave transformation is Fourier. There are more modern techniques, but all the literature on those assumes that you're thoroughly familiar with the Fourier stuff.

AMDG

Which I also learned is actually just a subset of Laplace transform today.

Which I know even less about :D

My main question is still how I can compute a function of the frequency of a wave over time.

Though I just realized that can probably be done somehow using Fourier, but I don't know enough about it to be able to use that effectively.

Multiplication produces these regular oscillations while division produces a constant increase in periodicity.

Sine appears to make a compressing/decompressing coil shape.

But they're simple, and they have regular patterns, so if we could determine how to reproduce these wave functions directly, we can compute the nth bit in constant time. It could be the first bit or the millionth. Since they could be computed individually, they can be computed in parallel.

Thorgott

15:05

@RuochanLiu a compact subspace is closed (because $\prod I$ is Hausdorff), so the only dense compact subspace of $\prod I$ is $\prod I$ itself

PM 2Ring

@AMDG Multiplication of large numbers (i.e., many thousands of digits) is often done using the DFT (discrete Fourier transform).

AMDG

I'm assuming the DFT takes a finite, discrete input and converts it to a wave-like function that can be easily manipulated.

Thorgott

@P-addict This argument does not make sense when $f$ doesn't have zeros. If $f$ doesn't have zeros, it vacuously has your property. If, on the other hand, $f$ has some zero $x$, then $T-x$ is a polynomial vanishing at $x$, so $f|T-x$ implies $f$ is constant or linear.

@shintuku Technically doesn't matter. The statement is "for any neighborhood $V$ of $y_0$,...". There is no claim to existence, it would simply be a vacuous statement if there was no neighborhood. However, there always exists at least one neighborhood, namely the entire space itself.

PM 2Ring

15:21

@AMDG Something like that... Let's say we're multiplying numbers A & B of the same length (we pad the smaller one with zeroes if they aren't). We break the numbers into digits (or larger blocks), A=(a0, a1, a2, ...), B=(b0, b1, b2, ...). Now we have to multiply each b by every a and add those products appropriately to get the final result. That process is a convolution.

But if we Fourier transform A & B to U=(u0, u1, u2, ...), V=(v0, v1, v2,...), where U & V are arrays of complex numbers with the same length as A & B, then we just need to do a pointwise multiplication, W=(u0×v0, u1×v1, u2×v2, ...). Then the inverse Fourier transform gives us the desired product.

AMDG

Well that's quite interesting.

I'll have to come back a bit later though to discuss this.

PM 2Ring

The standard multiplication is O(n^2). The pointwise multiplication is O(n). Of course, the Fourier transformations take time too. Traditionally, they're O(n^2), so you'd think there'd be no benefit to doing multiplication this way. However, it's possible to do the transforms with an O(n log n) algorithm.

There's still overheads, so there's no benefit unless n is large enough.

Peter John

16:38

Consider the sequence $\{a_n\}$ : $a_1 =3$ and $na_{n+1}-2na_n+\frac{n+2}{n+1}= 0$ for $n\geq 1$. What is an explicit formula for $a_n$ ?

user193319

16:55

Does anyone know of any virtually abelian groups? What are some examples?

Thorgott

17:35

virtually abelian groups are extensions of finite groups by abelian groups, so the split ones, i.e. semi-direct products of an abelian and a finite group, give you a large class of fairly explicit examples

P-addict

@Thorgott apologies, i was unclear - this must be true for all $x$. i'm looking for $f$ which satisfy "'$f(x)=0\Rightarrow g(x)=0$ for all $x\in\mathbb{R}^n$' implies '$f\mid g$'". in this case when $f$ has no roots, any constant function $c$ satisfies $c(x)=0$ when $f(x)=0$, so $f\mid c$ when $f$ has this property - thus non-constant $f$ without roots does not satisfy the property

SAJW

So, (a%b)%b (brackets just for no ambiguity) is useless? Or will that occur sometime?

(a mod b) mod b

Thorgott

17:55

@P-addict ah, ok. so your argument shows that $f$ is a product of linear factors, say $f=\prod_{i=1}^n(T-x_i)^{k_i}$ with $x_1,\dots,x_n$ pairwise distinct and $k_1,\dots,k_n\ge1$. Then $g=\prod_{i=1}^n(T-x_i)$ satisfies the condition, so $f\mid g$, which forces $k_1=\dots=k_n=1$, i.e. $f$ has to be product of distinct linear factors. This condition is sufficient too.

ah, I forget a possible multiplicative constant, but that's it

P-addict

18:10

@Thorgott i'm ashamed to admit my knowledge of multivariable polynomials is quite weak... i don't think i see how to show $f$ splits into linear factors. the issue is i think i'm missing a statement of the form "if $f(x)=0$ then $P\mid f$" for some polynomial $P$ depending on $x$ like we have in the single variable case. e.g. $x_1-x_2$ has root $(1,1)$, while $x_1-1$ also has root $(1,1)$, but neither divides the other

(ofc what i just gave isn't a counterexample to the claim $f$ splits into linear factors)

Derivative

18:30

Why is $\mathbb{R}^2$ with usual addition and scalar multiplication $\lambda(x,y)=(\lambda x, y)$ not a vector space?

I thought I verified all the axioms but I didn't find out why it isn't

Ted Shifrin

@Derivative: Simpler. Is $\Bbb R$ with $\lambda y = y$ for all $\lambda$ a vector space?

There's one of the axioms that pretty obviously fails. Which one?

Thorgott

sorry, I had overlooked that you wanted multivariate polynomials. what I said above applies only to univariate ones

Derivative

@TedShifrin number 6 $(\lambda+\mu)v=\lambda v+\mu v$

Thorgott

the question in the multivariate setting is probably very hard

Derivative

I see it now

Ted Shifrin

18:35

@P-addict Does it help that multivariable polynomials still form a unique factorization domain?

Right @Derivative

Thorgott

you have to analyze the extent to which the Nullstellensatz fails over $\mathbb{R}$

Derivative

When am I supposed to get used to all these definitions?

Xander Henderson

@Derivative Eventually.

You gain familiarity through practice.

Personally, I tend not to feel like I really grok a topic until I have to use the results of that topic in something which is important to me.

P-addict

@TedShifrin yeah, i have been trying to use this, and i can write $f$ as the product of irreducibles up to scalar multiplication which i think is the same as what i did before after analyzing how roots of $f$ must behave but i'm not really sure where to continue from what i had last

Xander Henderson

And it wasn't until a year after I finished my masters thesis that I really felt like I understood the Assouad dimension (the primary topic of my thesis).

P-addict

18:40

@Thorgott can you elaborate on this a bit? i'm pretty unfamiliar with the Nullstellensatz

Derivative

@XanderHenderson tbh I'm looking forward to this linear algebra class so I can understand the thing Axler does to minimize $\int_{-\pi}^\pi |p(x)-\sin(x)|dx$ over polynomials $p$ with bounded degree. So hopefully that will help eventually right

P-addict

perhaps i should give my original problem - it's possible i have overlooked something and there is actually a simpler solution which doesn't require solving the problem i stated. for an $\mathbb{R}$-algebra $\mathcal{F}$ we let $|\mathcal{F}|$ be the set of surjective morphisms $\mathcal{F}\rightarrow\mathbb{R}$, and we say $\mathcal{F}$ is geometric when $\bigcap_{p\in|\mathcal{F}|}\ker p=\{0\}$. i want to determine when $\mathbb{R}[x_1,\dots,x_n]/f^k\mathbb{R}[x_1,\dots,x_n]$ is geometric

i figure nilpotent elements necessarily map to $0$ and thus are always in $\ker p$, so $0$ must be the only nilpotent element. hence i believe we need $k=1$ - if $k>1$ then $[f]\in\mathbb{R}[x_1,\dots,x_n]/f^k\mathbb{R}[x_1,\dots,x_n]$ is nonzero and nilpotent. furthermore i think as in the one variable case we can argue that morphisms on $\mathbb{R}[x_1,\dots,x_n]/f^k\mathbb{R}[x_1,\dots,x_n]$ are evaluations of $\mathbb{R}[x_1,\dots,x_n]$ at roots of $f^k$...?

which gives the question i had

Thorgott

in the usual language of affine algebraic geometry, you're asking for polynomials $f$ such that $V(f)\subseteq V(g)$ implies $(g)\subseteq(f)$ for all $g$. The former condition is equivalent to $I(V(g))\subseteq I(V(f))$. If this were to take place over an algebraically closed field, then the Nullstellensatz would tell us that $I(V(g))=\sqrt{(g)}$ and $I(V(f))=\sqrt{(f)}$, so we get $g\in\sqrt{(g)}\subseteq\sqrt{(f)}$ (and, conversely, $g\in\sqrt{(f)}$ implies $\sqrt{(g)}\subseteq\sqrt{(f)}$).
So you'd be asking for polynomials $f$ such that $g\in\sqrt{(f)}$ implies $g\in(f)$ for all $g$. I…

P-addict

is $(f)$ the ideal of multiples of $f$?

Thorgott

over $\mathbb{R}$, this simply breaks down, because your condition only cares about the vanishing set over $\mathbb{R}$. in the univariate case, you can easily classify irreducibles based on how their vanishing set changes when extending to $\mathbb{C}$ and so a similar argument can be made to work then, but in the multivariate case, this strikes me as a complicated matter

@P-addict yes

P-addict

18:55

@Thorgott okay, that makes sense

Thorgott

19:06

@P-addict a similar argument to what I gave above still works here and shows that a necessary condition is that $f^k$ be a product of distinct irreducible factors (this also forces $k=1$). then, you can reduce the question to when $f$ itself is irreducible.

P-addict

@Thorgott yeah, and i think if $f=P_1\dots P_n$, the $P_i$ pairwise distinct irreducibles then we also know each $P_i$ has a root, and $V(P_i)\not\subseteq V(P_j)$ when $i\neq j$, else $f/P_i$ is $0$ exactly when $f$ is. i don't know if this is sufficient though (and i am wondering if there's a nicer way to characterize such $f$). but i'm not sure how to use this to reduce to the case when $f$ is irreducible.

i guess that second condition implies the first

Thorgott

the irreducible factors need not have roots

the reduction to the case $f$ irreducible follows by applying CRT and noting a product of algebras is geometric iff each factor is

P-addict

do they not? if $g\mid f$ has no roots, then i think $f/g$ is always $0$ on roots of $f$ but $[f/g]$ is not $0$ in the quotient algebra

@Thorgott ohhh, i hadn't considered this!

user10478

What does this answer (math.stackexchange.com/a/4230420/109355) mean by a solution not being "well behaved" unless $f$ is bounded? Is there something that breaks if $u$ is infinity at $r = 0$ at a mathematical level (more fundamental than the physical interpretation that temperature can't be infinite), or is Laplace's Equation just impossible to solve independent of physical interpretation?

Thorgott

oh, you mean in the geometric case

yes, having a root is a necessary condition for the quotient to be geometric

P-addict

19:31

@Thorgott i think i can see why each factor is geometric when the product is, but i'm having a harder time with the opposite direction

Thorgott

the key point is that if $A_1,\dots,A_n$ are $\mathbb{R}$-algebras, then $\mathrm{Hom}_{\mathbb{R}}(\prod_{i=1}^nA_i,\mathbb{R})\cong\coprod_{i=1}^n\mathrm{Hom}_{\mathbb{R}}(A_i,\mathbb{R})$

every algebra morphism out of a finite product factors uniquely through a projection onto one of its factors

(I should've specified finite products, I admittedly don't know what happens for infinite ones)

P-addict

yeah, i was trying to get a claim like that. i think i'm stuck on trying to construct a map $\prod_{i=1}^nA_i\rightarrow\mathbb{R}$ given maps $A_i\rightarrow\mathbb{R}$ for each $i$

Thorgott

there's no such thing

that's why this is such a curiosity, the Hom-functor takes a product to a coproduct

P-addict

oh

sorry, i'm kind of lost. does "factors uniquely through a projection" mean given $f:\prod_{i=1}^nA_i\rightarrow\mathbb{R}$ there exists a unique $f_i:A_i\rightarrow\mathbb{R}$ such that $f=f_i\circ\pi_i$, $\pi_i:\prod_{i=1}^nA_i\rightarrow A_i$ the projection?

Thorgott

19:53

yes

and the $i$ is unique too, of course

P-addict

ohhhh, i was confusing myself. so what you're saying is a morphism out of the product is secretly just determined by a morphism out of one of the algebras in the product

Thorgott

yes

Ruochan Liu

20:12

@Thorgott I am still a bit confused, how does this imply not first-countable in this situation? Thank you!

Thorgott

you said you already know $\prod I$ is not first-countable, no?

Ruochan Liu

Right

I'm trying to show there is a point with no countable local basis

That is the same definition?

Semiclassical

20:32

@TedShifrin oh, i also figured out my problem on polytopes. i was basically just misinterpreting the documentation on the program i was using: the two polytopes were affinely equivalent, but the program only had a function for testing for congruence (i.e., equivalent up to rotation and scaling). So the polytopes were affinely equivalent but not congruent.

so a very boring issue

Ted Shifrin

Ah, that makes sense.

Thorgott

yes, that is the definition

Ruochan Liu

@Thorgott Could you elaborate more on your reasoning since I still don't understand?

Semiclassical

20:48

right. in my defense i assumed that there would be a function for testing affine equivalence

and that's the only function which seemed to address that

but yeah, congruence > affine equivalence > combinatorial equivalence

Thorgott

what part needs elaboration? I'm just saying the only dense compact subspace of $\prod I$ is $\prod I$ itself

Ruochan Liu

I have a non-compact subset though

So I'm not sure if compactness is even relevant information

Thorgott

you asked whether the non-compactness condition made a difference, I was just replying to that

Semiclassical

the slightly neat part is that the affine equivalence had a simple interpretation. Suppose you have bits $x,y\in\{0,1\}$. Then the logical AND of those two bits can computed in ordinary arithmetic by doing $\frac12(x+y-(x\oplus y))$

where $x\oplus y$ is XOR, i.e., $x+y$ taken mod 2.

P-addict

21:05

@Thorgott okay, this makes sense. i suppose now if $f$ is irreducible and the quotient algebra is geometric, if $f$ has degree at least $2$, for each root $(r_1,\dots,r_n)$ of $f$ we can consider the polynomial $(x_1-r_1)\dots(x_n-r_n)$ and take the product of all such polynomials over all roots of $f$. clearly this new polynomial is $0$ exactly when $f$ is, but $f$ cannot divide it since it splits into linear factors. does this argument prove $f$ must have degree $1$?

(or $0$)

Thorgott

it doesn't

because a multivariate polynomial in general has infinitely many roots

P-addict

oooooooops

Thorgott

also, quotients like $\mathbb{R}[X_1,\dots,X_n]/(f)$ are finite-dimensional if $n=1$, but infinite-dimensional if $n>1$

these are both reasons why the multivariate case is much harder than the univariate one

P-addict

dang

Ted Shifrin

@Thorgott Indeed, algebraic varieties !

P-addict

21:10

do you happen to have any idea how to solve this? i'm stumped

Thorgott

I don't, this is way too hard for me

you could try asking on the site, maybe some algebraic geometer will know

P-addict

yeah, i'll post a question

Thorgott

let me know if you get an answer

P-addict

sure thing!

P-addict

21:43

@Thorgott hmm, why does CRT apply? we have to show if $P$ and $Q$ are distinct irreducible polynomials then $\langle P\rangle$ and $\langle Q\rangle$ are coprime, right? i'm not sure how to do this though

Thorgott

oh yeah, that's completely untrue

urgh, I'm sorry

P-addict

no worries

Thorgott

I've grown too used to my rings being 1-dimensional

robjohn

23:34

There is a two dimensional ring ;-) It is also how to square a circle

Xander Henderson

Ouch...

leslie townes

i laughed.

Xander Henderson

Though isn't that really more a group than a ring? What is the multiplication on $\mathbb{R}/\mathbb{Z}$?

robjohn

@XanderHenderson what does the ring on a finger look like?

topologically similar

Xander Henderson

@robjohn I got the joke. I just don't like it. :P

robjohn

23:36

I guess it might be considered 3 dimensional

Xander Henderson

It hurts too much.

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$Mathematics$ Mathematics