Mathematics - 2022-04-22

PurpleHaze

00:35

test

name didn't change here...

Ted Shifrin

VioletHaze?

PurpleHaze

I'm DiagramChasingBlues now

Ted Shifrin

Oy.

Dia for short.

robjohn

01:13

@PurpleHaze would you like the name here?

talking to ghosts again

@leslietownes Be vewwy, vewwy quiet...

leslie townes

02:24

sssshhhh

CroCo

03:04

good morning everyone, is it possible to write V() in a single term?

where K is constant values.

Adeek

Hey @leslietownes

Hi @copper.hat

Ted Shifrin

03:34

@CroCo is $K$ symmetric?

@robjohn Orange bubbly?

Adeek

@TedShifrin hi

Ted Shifrin

Hi

Adeek

I wanted to ask you. If we have a curve $C$ in $C_1 \times C_2 \times C_3 \times C_4$ product of smooth projective curves. If we take $K_1 = Pr_{1,2,3}(C) \times C_4$. Pick a curve $C_1$ different than $C$ in $K_1$. Now consider $Z_2 = C_1 \times Pr_{2,3,4}(C_1)$. Continue this process alternating between (1,2,3) and (2,3,4). At some point we get K_n that is of the form Curve x Curve x {points} right?

this is because of the common components in the $(2,3)$ we keep intersecting different things in those components

CroCo

03:51

@TedShifrin a symmetric and positive definite matrix.

Ted Shifrin

Positive definite is not relevant. Symmetric means you get twice one term.

Adeek

@CroCo what is V here?

CroCo

Adeek

what are you trying to do ?

CroCo

@TedShifrin you suggest this ?

Koro

03:53

"The elements in $\rm \mathbb Z[x,y]$ which are units modulo (y), for example, are the polynomials in $\rm \mathbb Z[x,y]$ with constant terms $\pm 1$ and all non constant terms divisible by $y$."

Adeek

@Koro what are you working on?

Koro

How is this possible? Isn't $5y$ a non constant term divisible by y but still 5y is not a unit in $\rm \mathbb Z[x,y]/(y)$?

robjohn

@TedShifrin some orange crush

Ted Shifrin

It’s $0$ in the quotient.

Adeek

$5y$ is zero in $\mathbb{Z}[x,y]/(y)$

Ted Shifrin

03:56

Yes, CroCo.

leslie townes

koro: it doesn't have a constant term \pm 1...

Adeek

$\mathbb{Z}[x,y]/(y) \cong \mathbb{Z}[x]$

CroCo

@TedShifrin thank you. this indeed reduces it into a single term.

@Adeek designing robot's controller.

Adeek

cool

Koro

If $R$ is an integral domain (commutative ring with unity has no zero divisors) then a non constant monic polynomial $p(x)$ is irreducible in R[x] if for some proper ideal $I$, the image of p(x) in $(R/I)[x]$ can't be factored in $(R/I)[x]$.
The above is the theorem that I am trying to understand. If we take $p(x,y)=xy+x+y+1\in Z[x,y]$ then $p(x,y)=(x+1)(y+1)$ but modulo (y) the polynomial (monic) p(x,y) becomes x+1, which is irreducible in $Z[x,y]/(y)$ so doesn't that contradict the theorem?

This is the explanation to the contradiction, which I don't understand:

6 mins ago, by Koro

"The elements in $\rm \mathbb Z[x,y]$ which are units modulo (y), for example, are the polynomials in $\rm \mathbb Z[x,y]$ with constant terms $\pm 1$ and all non constant terms divisible by $y$."

Adeek

04:00

@Koro non-zero

$(x + 1) * (y + 1) = (x + 1) * 1$ in the quotient

Ted Shifrin

Offhand I don’t believe the theorem.

Koro

@Adeek yes. Is (x+1) reducible in Z[x,y]/(y)? No.

Adeek

why?

how can you factor (x + 1) in Z[x,y]/(y)?

Ted Shifrin

Where did you find this “theorem”?

CroCo

@TedShifrin I wish to reach to this level of confidence in math.

Ted Shifrin

04:05

Adeek, stop.

Adeek

?

Koro

It's proposition 12 in chapter 9 in Dummit and Foote's.

@Adeek I think that factoring is not possible. Because suppose that $\rm x+1= f(x) g(x)$ then degree f(x)=0 (wlog) and degree g(x)=1. It follows by comparing the coefficients that f(x)=1, which is a unit in Z[x,y]/(y).

Ted Shifrin

oh, your polynomial isn’t monic!

Adeek

yes

Ted Shifrin

No counterexample.

Adeek

04:09

No it is because he isn't applying theorem correct. He wishes to check if that polynomial is x + 1 in Z[x,y]/(y)

Ted Shifrin

Huh?

Adeek

p(x,y)=xy+x+y+1 in the quotient that becomes x + 1

not in the way he wrote

x + 1 can't be factored in (R/I)[x]

Ted Shifrin

Nonsense.

That’s what he said.

leslie townes

someone was asking a generalization of this exercise recently on main.

Adeek

wait the polynomial is monic

leslie townes

04:11

this is even less exciting than the third isomorphism theorem. lets go back to that.

Adeek

why isn't it monic

let me download Dummit

Koro

@TedShifrin thank you! But the explanation in the book didn't say that and provided a different explanation. Can I share the screenshot of that para?

leslie townes

adeek your polynomial is not a counterexample because it does not have the property that "all nonconstant terms [are] divisible by y."

Ted Shifrin

Leslie, I’m fed up with it all.

Adeek

@Koro can you give screen shot of the theorem?

Koro

04:12

@leslietownes: is it legal to share a para from the book?

@Adeek let me take lawyer advice :).

leslie townes

8

Q: Polynomial $p(x)$ is a unit (invertible) $\iff p_0 = p(0)$ is a unit & all other coef's are nilpotent

I am learning ring theory in the Dummit & Foote's Abstract Algebra, and I am doing all the exercises to get as much experience as possible... but some of them just get me stuck for hours! Like this one : Assume $R$ is a commutative ring with identity. Prove that $p(x)=a_n x^n + a_{n-1} x^{n-1} +...

Adeek

@leslietownes yes I read the theorem incorrectly

Ted Shifrin

The theorem is fine. The poly has leading coefficient $1+y$, which, last time I checked, isn’t $1$.

leslie townes

most recently asked here. math.stackexchange.com/questions/4432312/…

Adeek

wait isn't monic is the coefficient for the higher degree coefficient ?

i.e. the term $xy$ has coefficient 1

Koro

04:16

@Adeek I think you're right. The polynomial is monic.

Ted Shifrin

The ring $R$ is $\Bbb Z[y]$.

No, folks. Stop and think.

Adeek

ohhh the theorem applies in R[x]?

Ted Shifrin

Read the damn statement.

@Koro D/F are giving this example to illustrate collapsing of degree. Same thing happens if I reduce polynomials in $\Bbb Z[x]$ by a prime which kills the leading term.

Koro

Let me type the para here:

Ted Shifrin

This is not meant to be relevant to the theorem at hand. READ carefully.

I answered your question.

To say that reducing an integer poly mod $p$ sends only an irred to an irred, you need to know that reduction does not drop degree. This is the same issue.

Adeek

04:26

I guess $R = R[x]$ so the polynomial is in $R[y]$ this example $xy + x + y + 1 = (x + 1)y + x + 1$

so it is not monic

If we think about it as $R = R[y]$ then $R[x]$ the polynomial $xy + x + y + 1 = (y + 1)x + y + 1$ which isn't monic

Koro

@TedShifrin thinking about that

@Adeek: I'll share the para shortly so that the context is more clear.

Adeek

I am off to sleep I will check it tomorrow

copper.hat

04:40

@Adeek Good night!

Adeek

you 2 nights :)

Ted Shifrin

Howdy, copper.

Koro

Example to the theorem: The polynomial $\displaystyle x^{2} +xy+1\in \mathbb{Z}[ x,y]$ is irreducible since the modulo $\displaystyle ( y)$, the polynomial becomes $\displaystyle x^{2} +1$ in $\displaystyle \mathbb{Z}[ x] ,$ which is irreducible and of the same degree.
In this type of arguments, it is necessary to be careful about 'collapsing'. For example: \ The polynomial $\displaystyle xy+x+y+1=( x+1)( y+1)$ modulo $\displaystyle ( y)$ is $\displaystyle x+1,$ which appears irreducible.
The reason for this is that non unit polynomials in Z[x,y] can reduce to units in the quotient. To take…

leslie townes

that reminds me of the time that some guy said something about collapsing

Koro

I still don't understand by the para above, why $5y$ is not a unit in Z[x,y] modulo (y).

5y is zero in Z[x,y]/(y) but by the para above, it should be a unit!

leslie townes

04:49

53 mins ago, by leslie townes

koro: it doesn't have a constant term \pm 1...

Koro

05:07

I misread the para. I thought polynomials with constant term $\pm 1$ and the polynomials whose non constant terms are divisible by $y$.

It's clear now. Thank you so much. 😊

copper.hat

@TedShifrin Hi Ted, hope all is good!

I am a terrible spellur myself, but bad spalling bothrs me.

Koro

05:43

is writing just 'Yes' in an answer to a post on mse acceptable?

just 'Yes' and nothing else.

Ted Shifrin

Rarely.

@copper.hat Did you see a FB comment of mine?!

copper.hat

06:51

@TedShifrin FB?

@Koro Are you referring to one of my comments? math.stackexchange.com/questions/4433363/…

The OP asked if the series was divergent or convergent without any attempt.

Koro

07:39

@copper.hat no. I meant posting 'Yes' in answer not in the comment section.

But it seems that your 'yes' was understood by the OP :).

LearningCHelpMeV2

10:27

is there a course in college that addresses the factorization of symmetric, cyclic, homogeneous polynomials in greater than 2 variables? if so, what course would this be under?

LearningCHelpMeV2

10:39

Example polynomial: $x(y^2-z^2)-y(x^2-z^2)+z(x^2-y^2)$

dsillman2000

14:16

Can anybody explain why $\Bbb R$ is not compact by definition, but the (closed) unit disc is compact by definition? When I see the construction that $\Bbb R$ is not compact because we can find an infinite open cover with no finite subcover, I am confused because one can do the same thing via the proper choice of pathological open cover for the closed unit disc. Likewise, one can find a finite open cover of $\Bbb R$ ($\Bbb R$ itself is open, so we can just use itself as a finite open cover).

If there's some rule that we can't use a space $X$ as its own finite open cover, then we could just split $X$ in half with overlap (i.e. $\Bbb R$ could use $(-\infty, 1)\cup(-1,\infty)$) and still find a finite open cover.

Based on this understanding (which I only gather from looking at the definition of compactness long enough), there are no non-compact spaces at all, so clearly I must be missing something... can anybody point out what I'm getting wrong?

Actually I think I could be wrong about being able to find a pathological open cover of the unit disc which challenges its compactness

Semiclassical

14:38

@dsillman2000 maybe this might help? math.stackexchange.com/questions/1982101/…

@LearningCHelpMeV2 commutative algebra, I think? i want to say it's something something Groebner bases but i'm not going to pretend i know how that works

dsillman2000

14:49

@Semiclassical Ok yes this does help. Thank you so much for the link

Koro

15:35

1

Q: Why is product of any coefficient of $p(x)\in \mathbb Q[x]$ with any coefficient of $q(x)\in \mathbb Q[x]$ is integer if $p(x)q(x)\in \mathbb Z[x]$?

Prove that if $\rm p(x)$ and $\rm q(x)$ are polynomials with rational coefficients whose product $\rm f(x)g(x)$ has integer coefficients , then the product of any coefficient of $\rm g(x)$ with any coefficient of $\rm f(x)$ is an integer. I tried to prove this as follows: Let $\rm G(x):=p(x)q(x)$...

abstract-algebra ring-theory solution-verification

Can anyone please review my proof in the above link? Thanks.

@dsillman2000 the definition says that every open cover must have a finite subcover. Just because R can be used to cover R itself does not prove the compactness on its own.

leslie townes

^^

and, it's worth repeating, you actually can't find an open cover of the closed disc having no finite subcover

this isn't immediate from the definitions, but it's true

dsillman2000

@leslietownes understood. My main misunderstanding was the phrasing, b/c I thought that if any finite open cover of a set $K$ existed, then it is compact (which turns out to be a very weak criterion, because basically every set has a finite open cover)

leslie townes

oh, yeah.

unrelated, but it's always annoyed me that 'open' in 'open cover' modifies a different thing than 'finite' in 'finite subcover'

Koro

completely unrelated, but i have noted that whenever I use the tag 'proof verification', it automatically changes to 'solution verification' and no. of visitors on posts tagged with 'proof verification' is very less.

leslie townes

huh. is 'proof verification' intended for something like automated theorem proving?

Koro

15:52

Description of proof verification tag: "For questions concerning a specific proof or a specific solution, asking for verification, identifying errors, suggestions for improvement, etc."

robjohn

@leslietownes there are these things called "lives". Sometimes it might be good to get one ;-)

Koro

Description of solution verification tag: "For posts looking for feedback or verification of a proposed solution. This should not be the only tag for a question."

@leslietownes leslie, no as per the description.

leslie townes

rob: never

robjohn

@Koro it could be that they are trying to move "proof verification" to theorem proving.

leslie townes

you can take my not having a life from my cold dead hands

err

well you can't take it

Koro

15:57

@robjohn The change of tag from 'proof verification' to 'solution verification' is automatic. I have noted that whenever I have used proof verification tag.

robjohn

@leslietownes: does "red car" vs "fast car" also bother you?

dsillman2000

@robjohn We're all in the Math SE chatroom, none of us can pretend that we have lives

robjohn

@Koro I believe that is because one was made a synonym of the other.

dsillman2000

@robjohn ? these are both modifying the same object

robjohn

and it might be in preparation for repurposing one of them

leslie townes

15:59

rob: no, but i've also never had to teach definitions involving red or fast cars to people with one eye on their phones

robjohn

@dsillman2000 one modifies the paint, the other modifies the engine performance vs weight

dsillman2000

In "open cover," the adjective "open" is modifying the sets which compose the cover, while in "finite subcover," the adjective "finite" is modifying the size of the family of sets which compose the cover.

robjohn

@dsillman2000 really? is that the difference?

dsillman2000

@robjohn ok i learned this definition yesterday, don't come at me for misunderstanding it if i am

leslie townes

you've got it

speaking of cars, this morning i did try to explain to my daughter why we could drive through puddles but couldn't drive on the river, and that was hard even though she wasn't on a device at the same time

Ted Shifrin

16:03

But a river is just a big puddle?

dsillman2000

@leslietownes probably because you didn't prove the properties of the moduli space of bodies of water, parameterized by depth (of course)

Surely if you'd explained it with that level of rigour, your daughter would have understood

leslie townes

ted: this was the crux of the matter

Ted Shifrin

@dsillman2000 You are talking moduli spaces and you just learned the defn of compactness?

The lack of parallelism in the adjectives comes from the prefix sub.

dsillman2000

@TedShifrin I thought you would be able to tell from the way I invoked "moduli space" that I know very little about them apart from the very basics

Ted Shifrin

Nah, I’m used to pomposity in the room.

dsillman2000

16:10

Ah, yes. Using a term I don't understand fully as a humor device. How pompous of me!

Ted Shifrin

I didn’t necessarily attribute it to you — but it exists.

Daniel Adams

these two theorems are both of Arzela-Ascoli type. paraphrasing; The first says that pointwise precompactness and equicontinuity implies precompactness :

The second seems to prove precompactness under strong-equicontinuity :

leslie townes

in my experience 9 out of 10 uses of 'moduli space' are pomposity or attempted pomposity, and your use was a rare exception

Ted Shifrin

Well, it’s not exactly your field of pomposity, leslie.

leslie townes

definitely not. that's how i can tell it's pomposity when it's directed my way

Ted Shifrin

16:17

I don’t even remember what the pre in precompact means.

leslie townes

i think it just means the closure is compact

Daniel Adams

I am trying to reconcile, that the second one is called a refined version of Arzela-Ascoli theorem, because (after getting the precompactness one can then apply the first theorem)

@leslietownes yes

sorry someone called me midway as I was typing my quesiton

Ted Shifrin

Well, now we have to debate the meaning of the adjective refined.

leslie townes

i've been called worse things while typing a question

Ted Shifrin

glares and pages munchkin

Koro

16:40

1 hour ago, by Koro

1

Q: Why is product of any coefficient of $p(x)\in \mathbb Q[x]$ with any coefficient of $q(x)\in \mathbb Q[x]$ is integer if $p(x)q(x)\in \mathbb Z[x]$?

Prove that if $\rm p(x)$ and $\rm q(x)$ are polynomials with rational coefficients whose product $\rm f(x)g(x)$ has integer coefficients , then the product of any coefficient of $\rm g(x)$ with any coefficient of $\rm f(x)$ is an integer. I tried to prove this as follows: Let $\rm G(x):=p(x)q(x)$...

abstract-algebra ring-theory solution-verification

Novice

17:04

I'm watching a video series on differential forms by mathematician Michael Penn, and he describes a one-form as a linear map $\omega \colon \mathrm T_p\mathbb R^n \to \mathbb R$ given by $\omega(\langle dx_1, \dots, dx_n \rangle) = \sum_{i = 1}^n a_i dx_i$. He's describing this as a scalar projection, but I don't fully get what he's saying at this point.

I'm trying to connect this to Professor Shifrin's videos, where he introduces the exterior algebra of exterior forms, and then introduces the algebra of differential forms as being like the exterior algebra, but with functions as coefficie…

Ted Shifrin

17:16

The issue is only a pointwise issue, but what you say Penn has written on the LHS makes zero sense. I certainly discussed in the first lecture how $(\sum a_i dx_i)(v) = a\cdot v$, where $a$ is the obvious vector.

And I discussed $dx_i(v) = v_i$ as projection, And $k$-form as giving tbe signed volume of the appropriate projection of a parallelepiped.

Novice

I guess I need to look at your first couple videos again. I probably forget some details when I bounce back and forth between topics. Thanks.

Between the point-set topology and the multilinear algebra, there are many concepts that are only half-formed in my mind, so it takes a while.

Semiclassical

17:31

that feeling when you realize exactly how much grading you'll have on the final, and how quickly you'll need to get it done

yaaaaaaay

Ted Shifrin

@Novice I found his video. I suggest you not watch him. He is not a geometer and he is saying nonsense. He is writing $dx$ as a component of a vector, which is totally confusing because it’s in the dual space.

Operator algebras … just like leslie. As I’ve always said, there’s plenty of bad stuff on the web.

Semiclassical

i should really learn differential forms properly at some point

Ted Shifrin

Well, don’t learn from that guy.

Semiclassical

at present my brain has room for differentials as "things that can be integrated and which transform certain ways under change of variables" and for forms as "things which map vectors to scalars"

but i've never been able to grok how those fit together

Ted Shifrin

Map multilinearly and alternatingly … hence built precisely out of determinants, whence your transformation rule

Semiclassical

17:40

hmm

i can see that, to a point

but not enough to conclude i actually understand it

Ted Shifrin

The connection to integrals is the same jacobian determinant and the change of variables theorem.

You actually have to work stuff out in examples to understand

Semiclassical

yeah

Slereah

Hey math people

Semiclassical

i suspect i also tend to lean on the wrong examples

Slereah

Is there a way to find out all invariants of a Klein geometry/Cartan geometry?

ie distances, angles, etc etc

Novice

17:46

@TedShifrin Thanks. Guess I won't worry about those videos, at least not at this point.

Semiclassical

e.g. if someone asks me to the magnetic flux through a surface, I know how to work that example by traditional means

and thus wouldn't know where the "forms" aspect of differential forms is actually being used there

Slereah

I suspect it is possible to do so for invariance of tensors, but the "geometric" quantities, I can't seem to find anything for

Even though that's usually the starting point of every intro on Klein geometry

How would I know that a projective structure preserves colineation if I did not know it

Answer this riddle and I will grant you wishes three

Ted Shifrin

There are different notions of projective structure. You are talking about projective transformations on projective space.

Slereah

I mean more generally

Ted Shifrin

Semi … that’s just finding flux by associating a vector field to a 2-form.

Slereah

17:49

How do I know that the metric structure preserves angles, etc etc

sometimes I find little tables, like here :

But that's about it

Ted Shifrin

Structures don’t preserve anything.

Slereah

Nobody seem to ever really explain it

What the group preserve in the Klein model, if you prefer

Ted Shifrin

This table is referring to different geometric structures on $\Bbb R^n$ and not structures on general manifolds.

Semiclassical

@TedShifrin sure. my point is more that my go-to examples of "things I can calculate" probably aren't good examples

Slereah

Yes, I am aware

But still I would like to know in this case

Semiclassical

17:53

and it's hard to come up with better examples when all you have are unproductive ones :P

Ted Shifrin

If you Stay in 3D you can always do what you’re familiar with, Semi.

Semiclassical

sounds about right

i guess maybe the place it'd show us is if i wanted to have more convenient ways to, say, derive the representations of div/grad/curl/etc in spherical coordinates

Ted Shifrin

What are you calling the Klein model, @Slereah?

I showed you how to do that, Semi.

Semiclassical

i vaguely remember that, yes

something something orthonormal basis

Slereah

@TedShifrin Homogeneous space made from the quotient of a Lie group and a stabilizer

As far as I know, those objects are still invariant under gauge transformation even in a Cartan geometry?

Ted Shifrin

17:58

So totally abstract. Nothing to do with your table.

Slereah

Rotating tetrads still preserve distances and whatnot

But those lists tend to be weirdly heterogeneous

Ted Shifrin

The metric comes from an appropriate invariant metric on the Lie group. You need to do some work on the Lie algebra and the subalgebra of the stabilizer. This is totally removed from your table.

You are talking physics speak, so that removes me.

Slereah

I can get that a metric structure preserves distances since it leaves the metric tensor invariant, which is easy enough to see

But on the other hand, angles are also in there, which require dividing tensors together

How does one generate just a big list of all quantities that are invariant in those geometries

Ted Shifrin

18:23

The metric tensor gives both lengths and angles. Invariants are a difficult thing … I don’t know a crank to turn. But invariant theory (Hermann Weyl) is deep stuff.

For affine geometry, to get affine arclength requires second derivatives, not just first.

CiurkitboyN

18:45

So if something has a 7 out of 10,000 chance of happening every 2.5 seconds how often can it be expected to happen? cause my math points me to about once every 5 minutes but that sounds like too little

Sha Vuklia

Ted ;x

CiurkitboyN

is there a formula for that sort of thing?

Joe Shmo

does anybody know a little bit of numerical analysis? looks like the secant method is failing to converge on the right root when my function is nonnegative ($0$ in a lot of places, and then becomes positive)

suppose I'm looking for $\sup f^{-1}(0)$

$f \in C^0[0, 1]$

dsillman2000

@CiurkitboyN Yep! If an event $E$ has a probability of occurring $0 < P(E) \leq 1$ in some time unit $t$, then we can expect $E$ to occur about once every $t / P(E)$ time units.

@CiurkitboyN You should be getting about once every 59.5 minutes

CiurkitboyN

19:44

@dsillman2000 whats with all the dollar signs?

dsillman2000

@CiurkitboyN Install ChatJax

CiurkitboyN

@dsillman2000 ChatJax?

@dsillman2000 so if I had it do it like 5 times every 2.5 seconds would that mean the time would be like 11.9 minutes?

I have chatjax now

@dsillman2000 What does $P$ stand for?

robjohn

@CiurkitboyN Look at the top of the right sidebar where it says "$\LaTeX$ in chat"

CiurkitboyN

I have it now

robjohn

Good, so there are not so many dollar signs?

CiurkitboyN

19:52

yeah

robjohn

@CiurkitboyN Are you familiar with LaTeX?

There is a MathJax tutorial

CiurkitboyN

@robjohn not really

robjohn

Take a look at that page and try some of it

CiurkitboyN

Ive got to go ill be back soon

rb3652

20:38

Hi all, how do I show if 1/(2n)! is convergent or divergent?

Ohh, maybe ratio test...

leslie townes

that'll definitely do it

or randomly deduced simple inequalities. e.g. k! = 1*2*3*..*k >= 1*2*2*..*2 = 2^{k-1} and comparison with a convergent geometric series

CiurkitboyN

20:55

ehh

thanks for helping me with the probability now I just need to afk with a weight on my keyboard for awhile

Cure

22:52

Anyone knows if it is possible to reduce $1771^{151} \mod{3793}$ ?

151 is already a prime number

leslie townes

think of how frightening it would be if the answer were "no"

Cure

very

leslie townes

one way is to compute 1771 to various powers of 2 by repeated squaring and reduce the work that way. this approach does not use the fact that 151 (or the modulus 3793) is prime

one potato two potato

23:10

If I construct cantor set not by removing "middle third'' of the components but rather using strictly increasing/decreasing homeomorphism (on each ''removing'' process so they maybe different on each stage), then is it still homeomorphic to the cantor set?

Cure

@leslietownes I'm not sure if I can fully see what you mean, but from what I tried, I found that $1771 = 2^{10}+2^9+2^7+2^6+2^5+2^3+2^1+2^0$

leslie townes

oh, i meant to compute 1771^2 (by squaring), 1771^4 (by squaring again), ..., 1771^128 (by squaring 7 times), and using the fact that 151 = 128 + 16 + 4 + 2 + 1 to compute 1771^151 = 1771^128 1771^16 1771^4 1771^2 1771

same idea would work with any power of 1771; write the power in binary and use a table of precomputed squares

it's still like 10 or 11 multiplications mod 3793, but at least it isn't 151 of them

one potato two potato

23:28

Ignore my question that is not always true.

robjohn

@Cure you want $2^{10}$ to get $2^{10}$. I fixed it.

also the $1771^{151}$.

Cure

23:54

I'm getting lost among all this calculations. Does it makes sense that if I know $1771^{16}=3647$ mod 3793 then I can do $1771^{128}=(1771^16)^8 = 3647^8 = (3647^2)^4=13300609^4 = 2351^4 = (2351^2)^2=800^2 = 640000 = 2776$ using mod 3793?

@robjohn I didn't understand this. Did I type something wrong?

robjohn

@Cure do you have ChatJax installed?

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$Mathematics$ Mathematics