Mathematics - 2022-03-02

Koro

01:15

If K is a field then $K_2:= K\times K$ is not a field. This is because (1,0)*(0,1)=(0,0) hence (1,0) is a zero divisor.

Ted Shifrin

02:07

@Koro Shrug. The product of integral domains is never an integral domain.

Adeek

03:01

hi @TedShifrin

Koro

03:43

13 hours ago, by Koro

one question from an exam: K is a field and K[x]=R is a ring. Then find all maximal ideals of R/I where I=<(x-1)(x-2)>.

So we can create one homomorphism from K[x] to $K\times K$ as f(y)= (y(1),y(2))

By FIT, it follows that $R/I\sim K\times K$

So I is not a maximal ideal. I is not even prime ideal.

Now, how do I find maximal ideals of R/I?

@TedShifrin here’s the background of the question.

leslie townes

do you know the usual theorems about what ideals in R/I correspond to in R

Adeek

@leslietownes

leslie townes

and what ideals in K[x] look like when K is a field

Adeek

Ideals J in R containing I correspond to ideals of R/I

Koro

I know only that: R/I is field (integral domain) if I is maximal (prime, respectively) ideal.

If R is commutative ring with unity.

Adeek

03:50

let me check your Question koro

Koro

leslie: I know the theorem I stated in my last message.

Adeek

@Koro here is my approach

Koro

thanks Adeek.

leslie townes

do you know that K[x] is a PID

user10478

Can a computer compute the adjoint of a linear operator/boundary? I can find calculators/Mathematica code/etc. for the adjoint of a matrix, but not for an operator.

Koro

03:52

@leslietownes yes.

Adeek

@koro Ideals of R/I correspond to ideals J containing I. Now you want to find maximal ideals so you want to find the ideals J containing I so that R/J is a field.

when is R/J is a field is retty easy considering the fact that it is just K[x]

Koro

@Adeek sounds like second or 3rd isomorphism theorem.

I’ll try with that. Thanks. :)

Adeek

np

I think you will want to solve something like (x - c)g(x) = (x - 1)(x - 2) and find c

actually

I see the solution

though I think it is better for you think about it. My hints above should give you the solution.

let me know if you get stuck

Koro

Sure. Thanks.

user400188

I am having trouble proving a certain equivalence between two definitions of connected sets in point set topology.

I have asked about it before in this room, and got given a simpler problem with relative topologies instead, which I was able to solve.

However I still an unable to prove the equivalence between these two definitions:

1. $A\subset X$ is connected iff there do not exist open sets $U$ and $V$ in $X$ with $A\subset U\cup V$, $A\cap U\neq\emptyset$, $A\cap V\neq\emptyset$, and $A\cap U\cap V=\emptyset$.

2. $A\subset X$ is connected iff the only subsets of $A$ that are both open and closed are $A$ and the empty set.

I was able to prove that (1) implies (2), but not the other way around. I feel like I am missing a premises or something. $U$ and $V$ are taken from the topology of $X$, while the considerations in (2) will refer to sets in the relative topology of $A$.

I.e., something can be in the relative topology of $A$, but not in the topology of $X$, and vice versa. E.g., consider $X=(-2,2)$, $A=(0,1]$, and the metric topology on $X$. Then $(0.5,1]$ is in the topology of $A$, but not $X$, because $(0.5,1]=A\cap (0.5,2)$ and $(0.5,2)$ is in the topology of $X$.

Koro

04:26

The ideals which contain I in this case are: $\langle x-1\rangle, \langle x-2\rangle$. Suppose that M is an ideal that contains $\langle x-1\rangle$. Suppose that f(x) in M but not in <x-1>. f(x)=q(x)(x-1)+r(x), where degree r(x)=0 that is r(x) is a non zero constant. It follows that r(x)=c =f(x)-q(x)(x-1) is in M and since M is an ideal $1=c^{-1}c$ is in M. It follows that M=K[x]. So <x-1> is a maximal ideal. Similarly, <x-2> is a maximal ideal
Since every ideal of K[x] is principal ideal, suppose that there is another ideal N that contains I. N=<g(x)>, g(x) is a polynomial of minimum deg…

leslie townes

04:51

koro so what is the background vibe here? ideals of K[x] that contain <f> are of the form <p>, with p dividing f. maximal ideals are the ones where p is irreducible. so in one to one correspondence with the roots of f (ignoring multiplicity) if f splits in K.

now i'm gonna draw some riemann surfaces, and the cuspidal cubic, and get some grant money.

Koro

05:25

I'm yet to study the chapter on irreducibility. But I understood what you are saying . :)

love_sodam

07:04

I have an interesting geometry problem you people might be interested in.

What is the distance between the red dot and the red line?

Prithu biswas leftmse

09:28

What is pure math?

Jakobian

people say math is pure when it's not applied

'Pure mathematics explores the boundary of mathematics and pure reason. It has been described as "that part of mathematical activity that is done without explicit or immediate consideration of direct application," although what is "pure" in one era often becomes applied later.'

Prithu biswas leftmse

@Jakobian So pure mathematics is the study of math for the sake of studying math?

Jakobian

09:54

perhaps

Prithu biswas leftmse

13:59

Is it a misconception that "derivatives and integrals are opposite"?

anak

It's not so much as a misconception as it is a useful way of thinking about them whose limitations are not often explained.

e.g. it does not provide a one to one correspondence (one function could be the derivative of many functions!)

Prithu biswas leftmse

@anak Here is where I found that statement, (with timestamp).

Stokes' Theorem on Manifolds

►

Balarka Sen

Hi @anak

anak

@BalarkaSen HI HOW ARE YOU! LONG TIME NO TALK

Balarka Sen

I'm good, you?

anak

14:08

I think good so far, though it's the morning here so I still have plenty of time to have my day ruined.

Balarka Sen

Haha

anak

Been doing any cool/fun stuff lately?

Balarka Sen

Thinking a lot about sheaves

but not the AG kind

anak

The farming kind?

Balarka Sen

No :)

anak

14:12

What about them?

Balarka Sen

I'm trying to paste a pair of sheaves on a closed set and its complement, basically.

But the sheaves take values in topological spaces

Prithu biswas leftmse

In what sense the exterior derivative is the true derivative?

Does it mean the single variable derivative dy/dx is a lie?

anak

@BalarkaSen I am probably only familiar with sheaves of rings. So in this case since they take values in topological spaces, it's that the collection of sections at each open set that is a topological space, or something else?

Balarka Sen

Yep, thats right

anak

Is the topological space kind of uniform across all the open sets, or could it be wildly different?

Balarka Sen

14:18

Theres a technical problem with this definition because stalks are unnatural; direct limit of topological spaces are useless objects

@anak Think sheaf of sections of a fiber bundle

anak

Oh okay, fibre bundle picture makes a bit more sense.

So you are trying to fix the direct limit problem, or are there other issues you are resolving first?

Balarka Sen

nah, thats known

im just trying to paste sheaves lol

anak

Is this a technical term beyond my knowledge of sheaves, or do you just mean you are trying to impose a sheaf structure?

anak

14:33

@Prithubiswasleftmse Sorry for the delay. It means that you are placing too much value on non-mathematical statements made by some youtuber.

Prithu biswas leftmse

@anak Well I have no knowledge of differential geometry, point set topology, real-analysis , linear algebra.So I have no way of judging anything.

anak

@Prithubiswasleftmse one might ask what you seek to gain from this video then.

Prithu biswas leftmse

@anak I wanted to see what pure math is like.

Alessandro Codenotti

Where do sheaves of topological spaces come up? @Balarka

Balarka Sen

@AlessandroCodenotti h principle lol

@anak I have a sheaf over A and a sheaf over X \ A and I want to glue them to a sheaf on X, in summary

A closed

Alessandro Codenotti

14:43

@BalarkaSen I don't know what that means, but I probably don't want to know

anak

@Prithubiswasleftmse There are probably more effective ways of introducing yourself to pure math. What kind of math are you interested in (at a superficial level, I mean---so it's okay if you know nothing about it)?

@BalarkaSen ah, I see!

Prithu biswas leftmse

@anak currently I am trying to learn single variable basic real analysis.

anak

From what book?

Balarka Sen

Here's an example where these kind of ideas are useful. Consider on the sphere $S^2$ the sheaf which associates to an open set $U$ the topological space $\mathrm{Imm}(U, \Bbb R^3)$ of immersions in $\Bbb R^3$. This is a sheaf, because $\mathrm{Imm}(U \cup V, \Bbb R^3) \cong \mathrm{Imm}(U, \Bbb R^3) \times_{\mathrm{Imm}(U \cap V, \Bbb R^3)} \mathrm{Imm}(V, \Bbb R^3)$ - a pair of immersions which agree on the intersection of their domains glue to an immersion, and this is the homeomorphism above

Prithu biswas leftmse

@anak Spivak.

anak

14:47

Decent enough choice.

Balarka Sen

Say you want to determine the homotopy type of $\mathrm{Imm}(S^2, \Bbb R^3)$. You would like to say that you understand the homotopy type of the sheaf locally, $\mathrm{Imm}(U, \Bbb R^3) \simeq V_{2, 3}$ where $U \cong D^2$ is a small disk chart on $S^2$, the homotopy equivalence given by $f \mapsto df_0$, $V_{2, 3}$ being the space of $2$-frames in $\Bbb R^3$.

So you can glue that up and understand the homotopy type of the space of global sections as well

This is how the proof of sphere eversion goes

anak

This sounds pretty interesting.

Does it appear like this in other places, too? Like can you look at the same sheaf of immersions?

Balarka Sen

other places as in

anak

Sorry, I mean is this sheaf of immersions a one trick pony for this particular case, or has it proven to be useful in other examples, too, not just on the sphere.

Balarka Sen

yeah homotopy type of $\mathrm{Imm}(M, N)$ is completely understood. nothing special about the sphere

nothing special about just smooth immersions either, you can try to do symplectic/contact immersions, ...

its a very general philosophy

anak

15:07

Are there nice expositions of the sphere eversion proof in existence?

Or just the pieces and you have to go hunting a bit?

Balarka Sen

Smale's original proof is basically this you just have to recast it in this language. The language of sheaves is developed in Gromov, PDR. He does it in full generality so you have to specialize

anak

Balarka blog post idea, I suppose.

Balarka Sen

Nah

anak

We need more expositors in math, I think.

geocalc33

15:30

@anak what kind of math are you into these days?

anak

i draw silly pictures and hope they have adequate mathematical content to get published

otherwise, i just draw more :o)

geocalc33

I do that too haha

Adeek

15:46

@Koro yes correct

Jakobian

@Prithubiswasleftmse it's true if you think about indefinite integrals

I need a hint about a topology exercise

I have a sigma-compact zero-dimensional space

I want to prove that something stronger than sigma-compactness holds

All my spaces are metrizable separable btw

Alessandro Codenotti

What's that something?

Jakobian

That there is a sequence of compact sets $(A_n)_n$ with diameters converging to $0$ such that $X$ is union of $A_n$

Alessandro Codenotti

16:03

By sigma compactness $X=\bigcup_n C_n$ for some compact $C_n$. Now I think by zero dimensionality each C_n can be split into finitely many clopen sets of diameter less than 1/n

Jakobian

Clopen in $C_n$, yeah. So we can assume WLOG that diameters of $C_n$ converge to $0$. I wonder if we can get disjointness from here somehow.

Alessandro Codenotti

You can cover C_n with finitely many pairwise disjoint clopen sets of diameter <1/n

Say you have such a cover $A_n$ of $C_n$ except the $A_n$'s are not pairwise disjoint. Replace $A_k$ with $A_k\setminus\bigcup_{i<k}A_i$

Jakobian

16:32

@AlessandroCodenotti thanks, this works.

I am wondering, if a sequence of subsets with diameters converging to 0 is a topological invariant

Lukas Heger

@Jakobian I don't think so. Consider $(0,\frac{1}{n})$ inside $(0,1)$ and apply the homeomorphism $(0,1) \to (1,\infty), x \mapsto 1/x$

Jakobian

Yeah. Changing this example a little bit we can even have example with the subsets compact.

PDE methods

if a math theorem states : "for every $M$ we have $blahblahblah$ for some constant $C$", then does this $C$ the same for each $M$?

Semiclassical

16:53

trying to figure out if the following remark from a (physics) solution manual can be made rigorous

what they want to show is that $$\int_{-\epsilon}^{\epsilon} e^{-\beta|x|}\frac{d^2}{dx^2} e^{-\beta|x|}dx \to -2\beta$$ as $\epsilon\to 0^+$ (with $\beta>0$)

the argument they make is that $e^{-\beta|x}\to 1$ if $\epsilon\to 1$, and therefore the integral becomes $$\int_{-\epsilon}^{\epsilon} \frac{d^2}{dx^2}e^{-\beta|x|}\,dx =\left[\frac{d}{dx}e^{-|\beta|x}\right]_{-\epsilon}^\epsilon = -2\beta e^{-\beta \epsilon}\to -2\beta$$

Jakobian

@PDEmethods for every $M$ we have $M<C$ for some constant $C$

Semiclassical

should have been $\epsilon\to 0^+$ not $\epsilon \to 1$ in the last message

the more respectable version of this argument, i think, is via integration by parts

Lukas Heger

@Semiclassical what does $\frac{d^2}{dx^2} e^{-\beta|x|}$ mean?

Semiclassical

second derivative of $e^{-\beta |x|}$...which is a bit slippery given how $|x|$ isn't smooth at $x=0$

Lukas Heger

yeah

so you need distributions in first place if you want to make things precise

Semiclassical

17:07

if you do integration by parts, you have

$$\int_{-\epsilon}^\epsilon e^{-\beta |x|}\frac{d^2}{dx^2}e^{-\beta |x|}\,dx = \left[e^{-\beta |x|}\frac{d}{dx}e^{-\beta|x|}\right]_{-\epsilon}^\epsilon-\int_{-\epsilon}^\epsilon \left(\frac{d}{dx}e^{-\beta|x|}\right)^2\,dx$$

which i think (formally, at least) gets rid of the problem at $x=0$

alternatively, let $s(x)=2H(x)-1$ where $H(x)$ is the Heaviside step function with $s'(x)=2 H'(x)=2\delta(x)$. Then $|x|=x s(x)$, and $$\frac{d}{dx}e^{-\beta x s(x)}=-\beta(s(x)+2 x\delta(x))e^{-\beta x s(x)}=-\beta s(x)e^{-\beta x s(x)}$$

so the second derivative will be $$-\beta (2\delta(x))e^{-\beta x s(x)}-\beta s(x)\cdot (-\beta s(x)e^{-\beta x s(x)})=-2\beta\delta(x)+\beta^2 e^{-\beta x s(x)}$$

monoidaltransform

17:26

Rotations $\mathbb{S}^n\rightarrow \mathbb{S}^n$ have degree $1$?

leslie townes

17:41

monoidal: yes

monoidaltransform

suspending a rotation also gives us a rotation @leslietownes?

So, if $R_{\theta}:\mathbb{S}^n\rightarrow \mathbb{S}^n$ is a rotaation, then $SR_{\theta}: \mathbb{S}^{n+1}\rightarrow \mathbb{S}^{n+1}$ is also a rotation?

yes, I think that's geometrically clear

leslie townes

i don't know, you will have to ask someone more topological than i am. it may depend on what your definition of 'rotation' is. i was assuming something like, orientation preserving bijection with whatever continuity/smoothness you usually assume.

monoidaltransform

Just rotation by some angle

amWhy

The Californians 'round here seem to arrive late in the game ;P Nearing 10:00 am, guys!! :P

At least an Arizonian has been active earlier (not hear; he lives in all chats, almost ;))

Koro

@Adeek :)

leslie townes

17:53

monoidal: i just meant, it's not clear to me which definition of 'suspension' would automatically give you something that can be 'rotated' in an ambient space. e.g. the usual definition i know does not give the suspension of the circle a smooth structure, or an 'obvious' transitive set of symmetries.

presumably this is some obvious thing to topology people that can be fixed, but it wasn't clear to me.

yes, the sun was a little late rising in california today.

monoidaltransform

i'm not thinking in terms of smooth structures, also by degree I mean the number $f_{*}:H_n(S^n)\rightarrow H_n(S^n)$ is determined by

where $f_{*}$ is the homomorphism induced on homology by a map $f:S^n\rightarrow S^n$

leslie townes

OK, well i was thinking in terms of differential topology. ask an algebraic topologist.

i think you still have the question of, what do you require of a 'rotation' if you want to conclude that a map is one. i don't think this is insurmountable, i just don't know how people usually do it.

monoidaltransform

Yeah, i'm trying to follow the reasoning of my lecture where he stated that suspending a relflection yields a relfection. Notice that $SS^n\approx S^{n+1}$

leslie townes

the candidate definition 'just rotation by some angle' seems to rely on more structure than you seem to be assuming, and maybe requires meshing with a specific definition of suspension. i dunno.

monoidaltransform

but he, just like hatcher, has written $SS^n=S^{n+1}$

So he has stated if $R_n$ denotes reflection in one coordinate, then $SR_n=R_{n+1}$

which I think is a geometrically motivated fact

rather than a formal one

M17

18:14

Does Fermat's factor analysis have an RSA mark?

Lukas Heger

@monoidaltransform every rotation is homotopic to the identity (just make the angle smaller and smaller)

so degree $1$ follows from homotopy-invariance of homology

M17

18:27

I watched a video of you and I have a query regarding Fermat factors

@TedShifrin

Ted Shifrin

Look carefully at the definition of suspension of a map, @monoidal. I do not believe the suspension of a rotation is a rotation.

@M17 what do I know about Fermat factors?

M17

I was looking for that topic on YouTube, I found a university professor’s explanation, and I watched the channel description, and I found that the description is written almost here, that he has taught for 35 years

Ted Shifrin

I do not follow.

M17

I was first asking about the channel, because you once mentioned that you were explaining on YouTube, and in your description that you are. I did the same time, I guessed it was you

Sorry if I made a mistake

Ted Shifrin

My name is clearly attached to my videos.

M17

18:35

I'm so sorry, I didn't mean that

I read your full description now I found below the youtube link it wasn't you I apologize

Is this room for academic persons only?

Leaky Nun

depends on ur definition of "academic person" :P

ypercubeᵀᴹ

@M17 the description in the top right

leslie townes

i don't think so. or at least, the implicit definition of 'academic person' is a broad one.

ypercubeᵀᴹ

> Mathematics
Associated with Math.SE; for both general discussion & math questions alike. ...

Ted Shifrin

What are Fermat factors, anyhow?

I hadn't read the blurb at the top of the chatroom. We're rarely rational? Who knew?

M17

18:41

I mean it's the university professor, the students majoring in the mathematics department, and the postgraduate students

leslie townes

participants in the chat come from all sorts of places. there might actually be fewer current students/profs than you might think.

we all, however, have one thing in common.

robjohn

Ack... The formatting of MathJax in post titles is badly wrapped again.

M17

@TedShifrin, if, pq=n

n=X^2 - Y^2

Ted Shifrin

@robjohn My growling is growing without bound.

Jakobian

@AlessandroCodenotti Actually, on the second thought, there is something missing in this argument

robjohn

18:44

@TedShifrin It was working for a while.

M17

en.m.wikipedia.org/wiki/Fermat%27s_factorization_method

Jakobian

because a clopen set of a subspace doesn't have to be clopen

leslie townes

oh, factoring by writing an odd number as a difference of squares. i'm not a specialist but i understand that this is not a very good factoring algorithm.

the wikipedia page has links to refinements.

Ted Shifrin

Factoring as a difference of squares is of course factoring as a product of a Gaussian integer and its conjugate.

M17

Does this have anything to do with parsing q and q if primaries are at n=pq Is this the best quality algorithm at that?

Jakobian

18:47

I'll have to think of something else... eh

M17

AB=2291
A=? B=?
A,B are prime numbers
---------------------------------------
A+B=E
if A>B
A - B=R

E^2 - R^2=F
F/4=2291
F=9164
E^2 - R^2=9164
(E+R) (E - R)=9164
9164/(E - R)= (E+R)

Lukas Heger

ah, fractoring integers, fun stuff

the most efficient algorithm for factoring large numbers is the number field sieve, which uses algebraic number theory

M17

Someone told me that Fermat solves the problem of factors such as I was trying it and I didn't know about Fermat factors

monoidaltransform

@TedShifrin But the suspension of a reflection is a reflection?

on the sphere

M17

9164/(E - R)= (E+R)

I want clarification on this equation

Jakobian

18:54

@LukasHeger I thought the most efficient method was the one using Fourier transform

M17

My question is, how do I solve this with algebra in specific steps?

Ted Shifrin

Yes, presumably a reflection across the suspension of the reflecting plane. Let's start with a definition of the suspension of a map $X\to X$.

@M17 Do you not know basic high school algebra?

If you have the equation $2x=7$, is not $x=7/2$?

monoidaltransform

@TedShifrin $f:X\rightarrow X$ suspends to a map $Sf([x,t])=[f(x),t]$ where $SX=\frac{X\times I}{X\times 0, X\times 1}$

M17

I know what you mean, I want to know the value

Ted Shifrin

OK, that is what I remembered. So if $x$ is fixed by $f$, $[x,t]$ is fixed by $Sf$.

monoidaltransform

18:58

Yes

Ted Shifrin

You wrote an equation and said you want clarification, @M17. We do not understand you.

So the suspension of the reflecting plane is fixed, @monoidal.

M17

Is there a video that explains this?

monoidaltransform

Sure, what goes wrong with rotation?

M17

2x=7, x=7/2

Yes

Ted Shifrin

@M17 I have no idea. You need to learn basic high school mathematics.

@monoidal We're not done yet. I suspect we need to look at the suspension of $\Bbb R^{n+1}$ with $S^n$ sitting in it to make sense of these things.

M17

19:01

(E+R) (E - R)=9164

Ted Shifrin

@M17 Go learn basic algebra.

Stop spamming us.

M17

9164/(E - R)=(E+R)

9164/(E+R)=(E - R)

Search for what on YouTube you don't know about this problem؟

?

Ted Shifrin

Linear structure makes no sense in this topological setting, @monoidal. I guess we can think of suspending a great hypersphere and getting a great hypersphere.

M17

To I*

Learn

@TedShifrin, I don't know what I did wrong with the equation

Is it possible to specify a specific topic that I learn or watch a video about it?

I mean, I, not you, searched for this problem, I just mistyped

monoidaltransform

$SS^1$ (thinking of it as living in R^3) is just a double cone which is topologically just the 2 sphere. If you have $R(x_0,x_1)=(-x_0,-x_1)$ then $SR$ at each time $t$ reflects each point on $S^1\times t \approx S^1$ to its antipode

iyop45

19:11

Hey maybe a silly qn but, can you expand out a matrix equation like (A - I) o (A - I) where o is the hadamard product and A, I are matrices?

Ted Shifrin

@monoidal But we don't want antipodes.

But the same argument should work.

monoidaltransform

yeah, that's my reasoning... the same argument should work

Ted Shifrin

@iyop45 What is Hadamard product? Entry-by-entry product?

leslie townes

iyop: yes, it satisfies the distributive law that you would expect

monoidaltransform

same thing for rotation I think

Ted Shifrin

19:12

No, @monoidal. I don't see rotation at all. .

iyop45

yea element-wise product

ok cool thanks

Ted Shifrin

@iyop Then it's just the usual algebra property for each entry, right?

iyop45

oh yea im being silly

that is true it is just element wise operations

M17

@TedShifrin, Yes, I am learning algebra, but what is the name of the topic, specifying many high school mathematics? Can this equation not be solved algebraically and knowing the value or is this a very easy question so you told me that

Another clarification in that equation E and R each have only one value

user 85795

so you have a difference of squares equals 9164?

M17

19:25

Yes

monoidaltransform

I mean nudging a point in $S^1$ nudges it by the same amount in $S^1\times t$ and in turn in $S^2$, no?

M17

(79+29)^2−(79−29)^2

Semiclassical

mathematica comes up with just two (positive integer) solutions to x^2-y^2=9164

M17

=9164

Semiclassical

Easiest way to get at that, though, is to say you want (x-y)(x+y)=9164

so work out what the factorizations of 9164 are

the only real constraint is that the sum of the two factors is 2x, so the two factors must both be even or both odd

M17

19:28

...

Lukas Heger

@Jakobian no I am certain the general number field sieve is the most efficient

Semiclassical

(also x>y)

M17

Yes

AB=2291
A=? B=?
A,B are prime numbers
---------------------------------------
A+B=E
if A>B
A - B=R

E^2 - R^2=F
F/4=2291
F=9164
E^2 - R^2=9164
(E+R) (E - R)=9164
9164/(E - R)= (E+R)

Semiclassical

9164 doesn't have a huge number of divisors so it's not too hard to just search them all

M17

This equation is part of a question, I re-sent it if you didn't see it

Semiclassical

19:33

though if you're starting from AB=2291 then i have no idea why you're doing this

M17

I know the value of the unknown

Semiclassical

you don't factor an integer by writing down algebraic equations. you test factors and seeing if they work

M17

But I want to know how to get there

Semiclassical

i mean, sqrt(2291) = 47.86, so we only have to check primes up to 47

M17

@Semiclassical, This is the question, not the solution

Semiclassical

19:36

2291 = 29 * 79 is going to be annoying to find, though.

no small factors (3, 5, 7 11, etc)

but there's nothing stopping you from testing whether 2291 is divisible by 3, by 5, etc., up to 47

it's tedious but there's nothing hard about it

M17

I wanted to get it without looking at the factors, I want by equations

Semiclassical

you can't. that's not how factorization works

M17

So that's mathematically impossible, right?

Semiclassical

factorization is an algorithmic problem, not an algebraic one

i'm perhaps being a bit strong---for instance, there is Euler's factorization method which looks awfully algebraic. but that's not a general-purpose method, and it doesn't work for 2291 = 29 * 79 (because 79 = 4*19+3)

you might find Fermat's method interesting, but note that it's ultimately not about the algebra: you actually have to put in numbers and see what works

M17

Thanks

Things became clear to me

Yes, Fermat's method is based on experience with numbers

numbers test

Adeek

19:48

hi @TedShifrin

Semiclassical

Right

It is pretty effective for 2291 tho

M17

@TedShifrin, Sorry to mention you so much, you were right

Semiclassical

54^2-2291 = 625=25^2

M17

?

Semiclassical

So you don’t need to go too deep into Fermat to discover the factorization

The idea of Fermat is that you’re looking for a value of A such that A^2-N is a perfect square

robjohn

19:51

Our neighborhood is currently under an LADWP scheduled power outage. The notice we got a few days ago said 9-12. The power went out at 11. Our cable provider says we should have service back by 1. The LADWP page says 7:30 PM. Don't know what to believe.

we'll find out in 10 minutes if 12 is accurate

Semiclassical

Smallest possible A for 2291 is 48 (since 47^2<2291). But 48^2-2291 = 13 isn’t a perfect square

So then we test 49…which also fails

leslie townes

rob: around here, the most up to date 'official' time back on always seems to be several hours after it will actually come back on. i dunno why, it's useless to plan around. maybe they want to reduce call/email volume in the event they are a little late.

Semiclassical

And so forth until we hit 54, which works: 54^2-2291=625=25^2

Therefore 2291=54^2-25^2=(54+25)(54-25)=79 * 29

user4539917

Where did you first find this question? @M17

robjohn

@leslietownes yeah, I assume so. It is disheartening however.

user4539917

19:59

it feels like a JEE question

leslie townes

rob: i think people wouldn't mind these things so much if the information were accurate enough that one could rely on it in planning one's day.

during our last unscheduled power outage i had some difficulty explaining to my daughter what a power outage was. i can't blame that on the power company.

Lukas Heger

@Jakobian Fourier transform is the most efficient way for multiplication not factorization

Jakobian

Ah yes. Sorry

user4539917

it appears your power of explanation was put to the test :-)

\o @copper.hat

Lukas Heger

@leslietownes start with the Dirac equation and go from there

robjohn

20:04

12 has come and gone. Now waiting to see if our cable service is back by 1 (which would imply power is back).

leslie townes

lukas: hahaha.

we begin, as we must, with the hydrogen atom.

i hope you don't have anywhere to be, this could take us right up to snack time and nap time.

robjohn

You're assuming things are cool enough for atoms to form

M17

20:20

@user4539917, I asked this question to myself and try to answer it

But it cannot be solved algebraically

Alessandro Codenotti

20:37

@Jakobian why do you need them to be clopen in the big space? They are closed in a compact, so they are compact in the big space, you never required the sets whose diameter goes to zero to be clopen, only compact

I'm using clopenness only to argue that they can be pairwise disjoint

Jakobian

21:04

Because you're covering the compact sets with something that can be a little bigger than this compact set

so the construction breaks down

you can cover the first set with clopens, take intersection, but then what about the second one

If you remove the first compact from second compact, you'll get something not necessarily compact. And if you insist on using the clopen sets which union can be a little bigger than the first compact, you might not cover enough of the second compact this way

Alessandro Codenotti

The clopen sets are clopen in the compact

The compact subspace is zero-dimensional itself

user777

@leslietownes Thank you! I just go back and forth with the definition, I understand it now.

@TedShifrin Yes, it is.

Alessandro Codenotti

21:19

Ah ok I see what you mean now, I forgot you wanted pairwise disjoint sets whose diameter goes to zero instead of just compact sets whose diameter goes to zero

copper.hat

22:01

@user4539917 what power of explanation?

AMDG

22:56

I feel kind of silly for not noticing this sooner... but everyone is aware that modular exponentiation by squaring is $O(\log n)$, right?

That implies that you can implement division in $O(\log n)$ as well given the identity of modular reduction.

So if I revisit my paper here... the infinite sum is practically inferior to first obtaining a mixed fraction from $\frac{2^{\lceil\log_2(y)\rceil}}{y}$, and then repeatedly squaring (with one or more individual multiplies for odd exponents) until the numerator is at some desired $2^x$.

In each step, all you're doing is reducing the numerator each time with a subtraction or more which is trivial, so it would still be around $O(\log n)$ as an algorithm thereabouts. Maybe something like $O(M(n) + \log n)$.

(At least specifically for quotients involving a power of two in the numerator and a non-power of two in the denominator)

For arbitrary numerators, you would need to be able to compute roots.

Either that or I just haven't thought about it long enough. Probably the latter :)

However, if we instead consider that multiplication and division by powers of $y$ in base $y$ is a trivial shift, and that, if I understand correctly, double dabble generalized to all bases is $O(n)$, then mul and div also become $O(n)$. No need for fancy FFT :P

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