Mathematics - 2021-09-13

00:08

Just an update for anyone interested in commutative diagrams and diagram chasing software that works visually (the language is 85% visual, 15% textual as opposed to 100% textual)

It's written in python but of course will be optimized later in C++ the graph search and critical parts

shintuku

01:37

i have a function and i've gotten its critical points. if I restrict the domain, how come I need to check for other critical points along the bounds?

leslie townes

are you looking for extrema on an interval? if you don't include endpoints in your definition of 'critical points' (or separately check them after you check critical points), you might not find any, even when they exist

e.g. f(x) = x on [0,1] has a minimum at 0 and a maximum at 1 but f'(x) is never zero on [0,1] (or anywhere else)

shintuku

oh! of course

thank you

shintuku

03:16

is it possible to rotate a plane curve fixing an arbitrary point of rotation on a 2d plane so as to obtain an analytic formulation of the resulting graph?

i.e. such that the resulting graph is a set of elements satisfying some well defined equation

Koro

04:28

5

Q: Suppose $G$ is a group of order $pqr$ with $p<q<r$ and $q\nmid r-1$ for primes $p,q,r$. Prove that there is only one subgroup of order $q$.

Suppose $G$ is a group of order $pqr$ where $p,q,r$ are prime numbers such that $p<q<r$. If $q\nmid r-1$ prove that there is only one subgroup of order $q$. By the Sylow theorems we know that the number of Sylow $q$-subgroups of $G$ is in the form of $qk+1$ and has to be one of $\{1,p,r,pr\}$ b...

In the answer, why will Sylow q-subgroup be too big?

leslie townes

04:40

well under the given hypotheses, the normalizer in G of the Sylow q-subgroup of G apparently has order q (i.e. it is just the Sylow q-subgroup itself). in that first line of the solution that's what they say, anyway.

but later, it is deduced that the Sylow q-subgroup of G is apparently normal in a subgroup of G that has order qr. this meaning that everything in that subgroup of order qr is in the normalizer in G of the Sylow q-subgroup, too

so it has at least qr elements, when it's supposed to only have q elements

i should have given the sylow q-subgroup of G a name. call it H. then N_G(H), which has q elements, apparently also contains everything in a subgroup having qr elements.

Koro

05:05

@leslietownes why does it mean that subgroup of order qr is in the normalizer of q-SSG in G? I'm afraid, Leslie. I'm new to Sylow's theorems. Here is what I think: I'll try to break down the answer: If there are pr q-SSG, then we know that all q-SSG are conjugate and that $cl(q)=pr=\frac{pqr}{N(q-SSG)}$, whence N(q-SSG) is of order q

I'm using q-SSG as short for Sylow q-subgroup of G

leslie townes

lets name that subgroup of order qr K

it means that khk^{-1} is in H for every h in H and k in K

Koro

By Sylow's 3rd theorem, G must have a normal r-SSG. It follows that (r-SSG)(q-SSG) is a subgroup of G

@leslietownes ok

leslie townes

i'm not tracking the logic of the proof at all, i'm just pointing out how the statements in the solution, if true, do derive a contradiction on the size of N_G(H)

Koro

ok

leslie townes

there are probably other ways to shake it out of the sylow theorems than the way sketched in the solution

Koro

05:09

@leslietownes hmm, but still why does H contain K?

I understood that khk^{-1} is in H for every h and k as it's been shown that H is normal in K

@leslietownes I see the logic in the answer now, but I don't understand how K is contained in H:(

copper.hat

really bugs me when people downvote without explaining. i have have a bug and it is a missed opportunity to repair.

leslie townes

it isn't?

H has order q, K has order qr

Koro

So H $\subset$ K

But how come $K\subset H$? Isn't that how we're planning to draw a contradiction?

leslie townes

copper same. i'm not active enough to see a lot of that on my own answers but i don't like seeing an answer to a question that looks good to me and yet has a -1 in front of it and no comment. i wonder if i'm missing something

koro the contradiction in the answer is about |N_G(H)|

it has q elements via one route, and at least qr elements via another route, contradiction

i guess if K is a subgroup of N_G(H) and N_G(H) has q elements and thus is forced to be equal to H, this indirectly shows that K would (in the world where we are doing our proof by contradiction) must be a subset of H, but it's done via a calculation of orders

not directly trying to show that elements of K are elements of H

sylow theorems are sometimes used like this. you get your contradiction or the result you want by counting, not by reasoning directly with group elements

Koro

So N_G(H) has q elements (I understand this). So N_G(H) is cyclic. There is a subgroup of order $qr$ which has a normal subgroup of order $q$. Hence N(H) has to have atleast qr elements.

Why? Because if $A\le G$ then A is normal in G if and only if N(A)=G

@leslietownes I think I'm starting to understand it now.

Leslie, is my understanding correct though?

here

3 mins ago, by Koro

Why? Because if $A\le G$ then A is normal in G if and only if N(A)=G

leslie townes

05:23

yes

Koro

yay!

@leslietownes Thanks a lot :)

Semiclassical

@shintuku some plane curves, definitely. circles in particular :P

but surely the answer in general is "yes, if there's an analytic description of the original plane curve"

b/c rotation and translation are just coordinate transformations

PM 2Ring

Just don't expect that the rotated curve will necessarily be a function. ;)

Semiclassical

write out the new coordinates, substitute them into your original description, and you've got an analytic description in terms of the new coordinates

parametric functions are still functions :P

but yeah

doing something like $(f(t),g(t))$ isn't a problem

shintuku

05:38

thanks for the answers! is this just the change of basis stuff from standard linear algebra?

Semiclassical

yup

shintuku

neat, i'll dig in that direction. thanks!

Semiclassical

it's a transformation of the form $x'=ax+by+c$ and similarly for $y'$

shintuku

noted

Semiclassical

so you can write it as $X'=AX$ where $X=(x,y,1)^t$, $X'=(x',y',1)^t$ and a 3-by-3 matrix $A$

hence, linear algebra

the "translate and then rotate" bit means you can factorize $A$ in a particular way, but that's about the only interesting point

PM 2Ring

05:55

That 3×3 matrix is connected to homogeneous coordinates & projective geometry en.wikipedia.org/wiki/Homogeneous_coordinates

> Homogeneous coordinates are ubiquitous in computer graphics because they allow common vector operations such as translation, rotation, scaling and perspective projection to be represented as a matrix by which the vector is multiplied. By the chain rule, any sequence of such operations can be multiplied out into a single matrix, allowing simple and efficient processing

leslie townes

hah, kinda weird reversal there, attributing composition of linear maps to the chain rule and not vice versa

PM 2Ring

Kinda. I guess the point is that if you're not using projective geometry, scaling, translation, and rotation are separate process:, scalar multiplication, vector addition, and matrix multiplication. So it's rather neat that they can all be treated as matrix multiplication when you add an extra dimension.

shintuku

06:31

i just discovered \mathrm

no more ugly italicized latex

BannedUser

08:15

What is good example of algorithms with step function complexity?

step function runtime cost*

love_sodam