Mathematics - 2020-11-25 (page 2 of 3)

satan 29

18:10

Where can I find a proof of the statement:

anakhro

hey hot cats

satan 29

nullspace(A)=nullspace(Reduced Row echelon Form of A)

?

anakhro

@satan29 you just have to show that the row operations preserve the nullspace.

This is actually a great exercise.

satan 29

Oh! In that case, let me get my hands dirty

anakhro

Let x be in null(A) and show that x is in null(A') where A' is obtained from A' by a row operation.

Let me know if you need a little push or hint.

satan 29

18:21

Now that i think.....its very intuitive but what I am thinking hasn't got any rigor to it

Consider the equation AX=0 as a system of linear equations'

E1=0

E2=0

..and so on

say there is a solution set x=(x1,x2...)

now, x satisfies E_{i}=0 (for i =1,,2....) so it also obviously satisfies k*E_{i}=0

for a constant k

anakhro

I think that is enough rigour for that part.

copper.hat

If $E$ is invertible then $Ax=0$ iff $EAx=0$.

anakhro

Though of course you have done only one of the three elementary row operations. So the two others remain.

One of them is stupidly simple.

satan 29

and similarly, x satisfies E_{j}=0 so x will satisfy a*(E_{i}) + (b*E_{j})=0 for constants a and b

@anakhro i believe thats the interchanging two rows thing?

anakhro

Indeed. :)

So you showed null(A) is a subset of null(A'). What about null(A') being a subset of null(A)?

satan 29

18:29

hmm

copper.hat

Why not use the ERO approach?

satan 29

i cannot directly say that , if x satisfies aE1 + bE2=0 then it satisfies E1=0 and E2=0

Ted Shifrin

Copper is here procrastinating another day! :)

user2661923

Does anyone know the current method of (on the fly) inviting a user to a math chat? My rep is over 1000 and the user I wish to invite is at 21 rep. The old meta-math articles don't seem to work anymore.

anakhro

@satan29 a hint: A' was obtained from A by a row operation...

satan 29

18:33

it has got to be just reversing what i did earlier, no? I cant seem to express it properly though

copper.hat

@TedShifrin I am an expert at procrastination. Unfortunately I don't bill those hours :-).

Good morning!

Ted Shifrin

Good morning.

copper.hat

At least in Albany California :-)

satan 29

its 12:04 AM in India lol

copper.hat

Used to be named Ocean View but there were so many Ocean View's in California the mayor at the time renamed it Albany. I prefer Ocean View albeit Bay View would be more truthful.

Ted Shifrin

18:35

Crazy half hours.

copper.hat

Nepal is even stranger :-)

Ted Shifrin

It was Albany in the 50’s when I grew up there. Bay View, maybe.

copper.hat

Wow? It must have been a vastly different place then.

anakhro

@satan29 Well in the easiest case, A' being A except with two rows swapped: how do I get A from A'?

Ted Shifrin

Grew up in Kensington.

copper.hat

18:36

Nice. Still nice.

satan 29

@anakhro swap em again?

anakhro

Yes :)

Ted Shifrin

Yup. I go and drive by the house every time I am back visiting.

copper.hat

Berkeley, Kensington & El Cerrito are all about a block from my (the bank's) house.

anakhro

But you already showed that swapping the rows doesn't change the null space!

satan 29

18:37

yes!

copper.hat

Wave when you pass Curtis & Thousand Oaks :-). Come in for a glass, hopefully Covid will have dissipated then.

Ted Shifrin

I was in upper Kensington — top of Willamette.

satan 29

the other two operations remain though

copper.hat

I pass through there a few times a week on my bike. Used to run but need a replacement hip apparently. Now there is a topic for procrastination.

Ted Shifrin

Good thing you have such well-established expertise.

copper.hat

18:39

My house was owned by a 3rd gen. member of the Jacuzzi family (of hot tub fame).

One of them dropped by last Christmas to see the old house :-).

Ted Shifrin

So you're basically on Colusa.

Maybe TMI for the web, but I know precisely.

satan 29

isnt that....too much info to be posting online?

copper.hat

Ironically my copper.hat name came from my siblings who used to make fun of my privacy concerns :-).

Ted Shifrin

Yes. Remove it.

copper.hat

A quick google search starting from my MSE bio will yield the same result :-).

But I will remove it if it makes folks uncomfortable.

anakhro

18:42

I AM SO UNCOMFORTABLE.

it's cold outside. :(((((

Ted Shifrin

Shaddup.

copper.hat

Its gone, sorry, didn't mean to induce discomfort.

Ted Shifrin

No big deal. There are crazies around here.

I'll drop by if I ever get to travel again.

anakhro

@copper.hat i was just joking. :(

copper.hat

:-) Free glass of wine, get your ears talked off :-)

satan 29

18:45

@anakhro halp... I am stuck :(

14 mins ago, by satan 29

i cannot directly say that , if x satisfies aE1 + bE2=0 then it satisfies E1=0 and E2=0

Ted Shifrin

Deal, copper :)

copper.hat

@satan29 do you see how the row operations can be written as premultiplication by particularly simple matrices?

anakhro

@satan29 how can you undo the row operations?

Ted Shifrin

Yes, reversibility is the whole point of row operations.

satan 29

how about this: since x satisfies aE1+bE2=0 for all (a,b), x MUST also satisfy the case where (a,b)= (1,0) and also the case where (a,b)=(0,1)

so we can recover E1=0 and E2=0

anakhro

18:49

Let's just make it kind of explicit what the row operation is. We have three elementary row operations: swapping rows, multiply rows by a non-zero scalar, and adding a multiple of one row to another.

You agree we are done the swapping rows case?

satan 29

yes

multiplying with constant is also obvious

anakhro

How is it obvious? Can you explicitly explain it?

satan 29

just..multiply the eqn with the multiplicative inverse?

anakhro

Great, and that again is a row operation, right?

Ted Shifrin

Just a word of warning for when you’re done. In your earlier argument, $a$ and $b$ are fixed, not arbitrary.

anakhro

18:51

So our previous direction holds for null(A') being a subset of null(A).

satan 29

yes.

@TedShifrin oh hmmmm indeed

anakhro

So now suppose I add k times row Ei of A to row Ej. So the jth row of A' is aEi + Ej.

Given A', can you now "undo" it to A again?

satan 29

subtract Ej , and multiplying by 1/a ?

anakhro

That would mean the jth row is Ei.

(in A, it is Ej)

satan 29

ok...

anakhro

19:04

What is the ith row of A'?

satan 29

Ej?

and then we swap?

anakhro

Okay, maybe this calls for some TeX

copper.hat

let $r_k$ be the $k$th row of $A$.

satan 29

okay

anakhro

Using copper.head's notation instead of Ei, Ej, etc.: $A = \begin{bmatrix}r_1\\\vdots\\r_n\end{bmatrix}$

copper.hat

19:09

What are the formulae for the rows of $A'$?

satan 29

@anakhro yes

anakhro

As copper.hat asks, what are the rows of A'?

satan 29

a1r1 +a2r2.... as the first row

b1r1+b2r2....2nd row

like this?

copper.hat

Sorry, what are you trying to do exactly?

satan 29

we are trying to prove the part where you add multiples of rows together, right>?

anakhro

19:16

Yes, so suppose we obtain A' from A by adding $k$ times $r_i$ to $r_j$.

What are the resulting rows of A'?

(keep in mind this is a single row operation, not many).

satan 29

r1....ri,(rj+kri)....rn

copper.hat

So, if $r_1,...,r_n$ are the rows of $A$ then the rows of $A'$ are something like $r_1,..., r_{j-1},r_j+ k r_i, r_{j+1},... , r_n$.

anakhro

Yes, exactly, so we get $A' = \begin{bmatrix}r_1\\\vdots\\r_j + kr_i\\\vdots\\r_n\end{bmatrix}$.

So now we want to go backwards from A' to A by using row operations.

How do we "undo" it?

copper.hat

@satan29 your move :-)

satan 29

subtract kri from the jth row?

anakhro

19:21

Why a question mark? Does that seem like it would be a problem?

satan 29

no lol

anakhro

Is it a row operation?

satan 29

yes, adding (-k)ri

anakhro

And you double checked that $r_i$ was another row of A'?

satan 29

yes

anakhro

19:23

So everything checks out? Does the argument follow easily after that?

satan 29

yes, indeed :)

this was so simple man......I just wasnt thinking in terms of single operations

Mike Miller

@LukasHeger I just read your proof, thanks for writing that. It's nice.

copper.hat

When you have time, write down the form of the matrix corresponding to the above row operation.

satan 29

Thank you so much @anakhro and @copper.hat for being so patient.

copper.hat

The expression as the product of row operations is useful in many contexts so is worthwhile remembering.

@anakhro did the heavy lifting

satan 29

19:25

@copper.hat which expression?

copper.hat

the reduction of $A$ to reduced row echelon form can be done by premultiplying by a sequence of elementary row (operation) matrices.

anakhro

(an elementary row operation is an invertible linear map, so it also has the form of a matrix, which is not too hard to write down.)

Akiva Weinberger

19:37

Desmos has colors now

desmos.com/calculator/hktycfziuy

desmos.com/calculator/brbxnipnyx

Balarka Sen

cool

copper.hat

nice, but i can't say its immediately obvious to me how the colours are attached...

Akiva Weinberger

19:53

I haven't really looked at it but it's HSV rather than RGB

Or actually I think they're both available

user2103480

"When you procrastinate so much that you watch a video about what you are procrastinating"

me when I read expository texts about a topic that I should learn in detail or do exercises for

Thorgott

How do I get the bound $\vert\operatorname{SL}(2,\mathbb{Z}/N\mathbb{Z})\vert\le N^3$? A naive count only gives me the upper bound $2N^3-N^2$.

user2103480

Thorgott

Maybe I should get bounds on how many inverses an element can have in Z/NZ, but I'm not seeing anything good

user2103480

I can't help but read these in schwarzenegger-like german accents

Thorgott

20:04

ok that does make them funnier

user2103480

"commentate" is such a german error

copper.hat

doesn't he have an Austrian accent?

Balarka Sen

@Thorgott You can count cardinality of GL_2(F_q), SL_2(F_q) is the kernel of the determinant map so you just divide by p to get that

SL_2(Z/n) is a product of SL_2(F_q)'s right

user2103480

@copper.hat different dialects are everywhere in german-speaking regions, the (terrible) german accents in english don't differ too much though

copper.hat

i like the variety. if only i could speak my native language...

user2103480

20:07

may I introduce to our proudest export product, and prime german meme?

Oettinger Rede mit Untertiteln

►

Balarka Sen

SL_2(F_q) has cardinality 1/2(q^3 - q)

if I am not mistaken

So you get your bound

Much better than that

Thorgott

well, Z/n is a product of F_qs as Z-modules, so End(Z_n) is a product of End(F_q)'s as rings and they have the same invertible elements, but why does this respect determinants

Balarka Sen

yeah whatever screw it just say GL(Z/n) = prod GL(F_q) which has cardinality <= prod q^4

so at most n^4 and then uhh

yeah that doesnt quite work. interesting

Thorgott

GL(Z/n) having cardinality at most n^4 is obvious lol

but I mean, SL(Z/n) has index n

so you were right and that's it

Balarka Sen

So? lol

algebra brain

Also no man SL preserves product

its clear

Thorgott

20:11

why

Alessandro Codenotti

@user2103480 You Germans are so boring, this is what happens when an Italian politician embarasses himself speaking English at an international thing

Astyx

My ears hurt

user2103480

@AlessandroCodenotti omfg he's so f*cking italian

Balarka Sen

@Thorgott SL(R) -> SL(I) is surjective and has kernel SL(R/I) for nice rings like Euclidean domains or whatever by

row operations

elementary generation my dude

Thorgott

give a more abstract proof

Balarka Sen

20:14

so SL(Z/n) \cong prod SL(F_q) and you get exact cardinality using my argument

anakhro

@AlessandroCodenotti is there a strong expectation for Italian politicians to speak english?

Astyx

Some guy made a video of the french equivalent youtu.be/AxkBqW4dK-E

Alessandro Codenotti

@Thorgott Follows easily by cocompleting the appropriate category

Thorgott

thanks, that's much clearer

Balarka Sen

Also SL is q^3 - q, PSL is SL/2, of course

So N^3 prod_{q|N} (1 - 1/q) is the cardinality of SL_2(Z/N)

Number theorists come at me

Alessandro Codenotti

20:15

@Astyx lol

Some guy made a techno remix of the rector of my uni awarding degrees

Astyx

You can be do what we want to do

Balarka Sen

@BalarkaSen I meant Euclidean rings not domains

I is a proper ideal

@BalarkaSen *q^2

user486313

Hi @MatsGranvik

Mats Granvik

Hi

Mike Miller

20:35

@BalarkaSen This is totally wrong man

PSL is of course SL/(1+(-1)^{q+1})

Balarka Sen

Lol

I lol'd irl

Balarka Sen

20:48

Silent Hill 1-4 soundtracks are fantastic

brilliant games, wish i had the time and patience to set up and play SH2 on PC

Alessandro Codenotti

Never been a fan of the games

The Kojima/Del Toro one looked super promising though

Balarka Sen

yeah

thatd be too high end for me to play anyway lol

i wasted my childhood man, should have played way more games

Alessandro Codenotti

I waste my childhood, should have read more Lurie and less novels

Balarka Sen

lol

Alessandro Codenotti

@BalarkaSen do you know Return of the Obra Dinn? A fellow PhD student suggested it to me and it looks great from the little I've seen (but I'm trying not to see much because I prefer playing games blind)

copper.hat

20:53

once controllers had more than 3 buttons games became too complex for me

Balarka Sen

@AlessandroCodenotti nope

Semiclassical

tfw you're grading HW and someone did the right problems in the wrong chapter

so close and yet so far

Balarka Sen

ohnooo

Semiclassical

21:10

@balarka i'd pity them more but

so many of these students just end up copying from the solutions manual

geocalc33

can someone help me understand this? $\frac{dxdy}{xy(x-1)(y-1)}$

Semiclassical

that's...not a question

geocalc33

I mean this, $ds^2=\frac{dxdy}{xy(x-1)(y-1)}$

AttractorNotStrangeAtAll

Hello everyone. I would like to know what you guys think of "Do every proof yourself first".

Any of you have studied as this during undergrad or grad school?

Semiclassical

@geocalc33 once again, that's not a question.

geocalc33

21:15

I mean this metric $ds^2=\frac{dxdy}{xy(x-1)(y-1)}.$ I'm trying to understand it better

when x=1 and y=1 then it's undefined

and when x=0 and y=0 it's undefined

Semiclassical

that seems pretty dubious as a metric

pretty sure a valid first fundamental form has $ds^2=E dx^2+2F\,dxdy+G dy^2$ with $EG>F^2$. (maybe $\geq$, but not $F^2>EG$)

geocalc33

sorry I mean for a semi-riemannian space

Semiclassical

ah. dunno in that case

copper.hat

@AttractorNotStrangeAtAll Different folks have different approaches. My preference is to 'understand' the result first and often the proofs will follow from this 'understanding'. However you can spend many unproductive hours trying to reproduce results that took other years to develop without having a road map at least.

AttractorNotStrangeAtAll

21:33

Hmmm, so maybe some drawing, intuitive reasoning, then the proof will probably easily follow? Unless it is dependent on some crazy trick obsly

Semiclassical

well

if it took years for someone to come up with the argument, it may take years to come up with it again

AttractorNotStrangeAtAll

But the difference is that you know it's true, right? But I get your point

Semiclassical

the point in my book is that you never try to learn anything "from scratch". you always make use of the scaffolding that's been left behind

obviously, the more you can reconstruct without relying on said scaffolding, the better

but there's always a limit to that.

AttractorNotStrangeAtAll

Right! So I should concentrate my study hours doing the proposed problems/homework

Sorry if all these questions sound a little "I don't know how to study", but I'm really learning now how to learn

Semiclassical

right. you're trusting that whoever wrote the problems/HW knows the material well enough that they can judge the right path

that gets tricky when you're reading a textbook independently, of course

which you do end up having to do at some point

AttractorNotStrangeAtAll

21:40

Sometimes I understand the theorems, I 'get' the concepts and definitions, I'm ok with the examples, but when I arrive at the exercises I feel like there's something there I don't really understand what's going on. So that's why I thought about changing my study strategy

Astyx

@Semi o/

AttractorNotStrangeAtAll

Or maybe, as I'm new to math, this all is normal and it's actually good because it means I'm being challenged

Semiclassical

@astyx \o

Astyx

how are you?

Semiclassical

alright. hopped on to write up a physics stack question (my first in years) and thought i'd stop by

Astyx

21:43

What are you doing nowadays?

Thorgott

"The Nielsen–Schreier theorem in turn implies a weaker version of the axiom of choice, for finite sets."
Is Wiki trolling? Choice for finite sets is provable in ZF already or am I tripping?

Semiclassical

mostly some teaching type work. intro physics stuff

Alessandro Codenotti

@Thorgott not for all families of finite sets, not even for countable families of finite sets (that should be equivalent to what logicians call the weak Koenig's lemma)

ZF doesn't even prove choice for families of unordered pairs iirc

Thorgott

oh, it's choice for collections of finite sets

I was thinking finite collections

Semiclassical

right now i'm working on grading HW, and it's dull as anything

Thorgott

21:46

yeah ok

Alessandro Codenotti

@Thorgott Ah I see, that's provable in ZF, you're right

copper.hat

@AttractorNotStrangeAtAll I have found that having someone to discuss theorems, etc. with to be of great value.

AttractorNotStrangeAtAll

@copper.hat with everything remote is harder, and I'm doing classes as a non-degree student, so no one is really my friend

I'll try to record some videos of myself explaining some concept or whatever, because it makes me explain things out loud and clear my thought of what's happning, Idk

With a drawing tablet, of course

Semiclassical

yeah, independent learning in a math field you're not comfortable with is tough

copper.hat

@AttractorNotStrangeAtAll I understand. I have only found one person in my career (I'm 60) who shares a similar perspective & interest in mathematics and with whom I would socialise independently of mathematical motivations.

Part of doing mathematics efficiently involves a directness that makes some uncomfortable.

geocalc33

22:01

does the sum of two metrics always result in a metric?

yes

this means the sum of 4 metrics is a metric

Semiclassical

two positive-definite metrics, i'd say yes

not sure about beyond that, though. if A and -A are both permitted as metrics, then A+(-A)=0 is degenerate

which seems dubious

geocalc33

oh I overlooked that

copper.hat

How can $-A$ be a metric?

Semiclassical

semi-Riemannian context

copper.hat

Of course, thanks.

Semiclassical

22:05

calling it a metric in that case does seem weird tho

geocalc33

well at least the ds^2 I gave isn't degenerate :)

jarlemag

Hi. Does anyone know where I could find a proof/explanation of this answer about how to Calculate 3D Vector out of two angles and vector length?
https://math.stackexchange.com/questions/1385137/calculate-3d-vector-out-of-two-angles-and-vector-length

copper.hat

@jarlemag The two angles define two planes and the intersection defines a line. Then you need a length and direction (plus or minus).

Semiclassical

the description in that question is a bit strange, i think. it seems that $\alpha,\beta$ correspond to the angles of the vector upon projecting to the $XY$ and $YZ$ planes respectively

hmm, no, that doesn't work either

copper.hat

@Semiclassical Why not?

Semiclassical

22:11

@copper.hat take the (X,Y,Z) point to be (1,0,0). then the projection into the YZ plane is (0,0)

whereas the conversion given in the answer would require $\alpha=\pi/2$ and $\beta=0$. that makes sense for spherical coordinates but not for the question itself

(unless, of course, the OP's description of their angles was so unclear that it was actually correct)

copper.hat

@Semiclassical yup. need more caffeine :-)

Semiclassical

hah

@jarlemag my reading of this is that, while that answer is fine for what it sets out to do, it doesn't really match the question as stated. but the lack of detail means i'm not really sure i understand what the OP asked for in the first place

user486313

hi @copper.hat

copper.hat

Hi @user131585!

user486313

does anyone read the work of Peter Lax?

jarlemag

22:16

I got a bit confused by the question (and a similar one here: stackoverflow.com/questions/30011741/…) too, but the basic question is still "how to get (unit) vector from two angles."...

Semiclassical

Yeah, something seems odd.

jarlemag

I think this would be a statement applicable to my case: "α is an angle in the XY plane (angle from the X axis), β is an angle between the XY plane and the Z axis. Find the unit vector with direction defined by the angles.

Semiclassical

@jarlemag right. That description matches the first answer.

it doesn't match the second question you linked, though, which makes me a bit dubious of the latter

manooooh

Hello chat!

user486313

Hello

Semiclassical

22:24

yeah, the second question doesn't seem to be answered correctly

@jarlemag That said, it's not too hard to validate the first question's answer in your case

First, if you project (X,Y,Z) into the XY plane, you get the point (X,Y)

manooooh

I have a silly question: If we know that $|A|\leq|B|$ for all sets $A,B$, can we find an example of $|A|>|B|$?

user486313

I'm going all-out tonight for dinner -- steak burrito, belgian fries, and ice cream ...

user486313

then, write a research statement ...

copper.hat

@manooooh I don't understand the question. What does $|A|\leq |B|$ for all sets $A,B$ mean?

Surely this means $|A$ is a constant for the collection of sets in question?

manooooh

@copper.hat oh sorry. $|A|$ stands for cardinality of the set $A$

copper.hat

22:27

I got that, that was not what I was concerned about.

Semiclassical

@manooooh that's no better. what's your domain for A,B?

i mean, |A|>=|B| is not true for arbitrary sets A,B

so asking what the consequences of it being true is...odd

manooooh

I rephrase: Given two sets $A,B$ that satisfy $|A|\leq|B|, can we also have $|A|>|B|$?

Semiclassical

Sure. Take A to be any non-empty set and B the empty set.

n>=0 and n>0

Astyx

He's not asking for that

He wants 0>n

Semiclassical

ah

manooooh

22:30

@Semiclassical sorry, I must specify that both $A,B\neq\emptyset$

Astyx

What does < mean tho?

Semiclassical

then...no? i mean, $x\leq y\leftrightarrow x\not >y$

manooooh

@Semiclassical I see, one is the negation of the other. I guessed it was a silly question. Thanks!

Astyx

It depends how you define the relations tho

Semiclassical

this is for |A|<|B| as numbers, to be clear. it's not so clear if you mean A<B

manooooh

22:32

Yes of course, $|A|$ stands for "Amount of elements in the set $A$". And $<$ is the relation "less than"

Astyx

So is |N| < |R| ?

If you're only dealing with finite sets, then this result is just the order on natural integers

If not, that definition isn't sufficient

manooooh

Let me explain with one example

I am given a formal grammar $G=(V_n,V_t,P,S)$ where $V_n$ is the set of non-terminals (variables), $V_t$ is the set of terminals (alphabet), both are finite, $P$ are the productions or grammar rules and $S$ is the start symbol

I am asked to prove or disprove $|L(G)|=|V_t^*|$, which is obviously false

I have shown that $|L(G)|\leq|V_t^*|$, where $L(G)$ and $V_t^*$ can be finite or infinite

Is this sufficient to show that $|L(G)|\not>|V_t^*|$ hence $|L(G)|\neq|V_t^*|$?

Definitions of formal Language and Grammar can be found at its.caltech.edu/~matilde/FormalLanguagesGNSlides.pdf

Thorgott

replace those symbols with words and then ask yourself that question again

jarlemag

@Semimclassical: I tried to solve it this way before reading the questions/answers mentioned, but it doesn't seem to give right results. If anyone can see something obviously wrong, I'd be happy to learn where I misstepped:

α is an angle in the XY plane (angle from the X axis), β is an angle between the XY plane and the Z axis. Find the corresponding unit vector v.

We know that:
x^2 + y^2 + z^2 = 1

Let f be the length of the projection of v into the XY plane. Then:

sqrt(f) = sqrt(x^2 + y^2)

Uh, I hope I remembered that trignometric identity right...

manooooh

@Thorgott yeah but some symbols need to be put there, isn't it? I mean, the generated language can be equal to the set of all words of any length (this is what $V_t^*$ means), but not always. Of course it can occur that $|L(G)|<|V_t^*|$, but my question is: How can we explain that it is impossible -no matter what grammar is- that $|L(G)|>|V_t^*|$?

So my question, again, is: Is it sufficient to prove that $|L(G)|\leq|V_t^*|$ to conclude $|L(G)|\not>|V_t^*|$?

Semiclassical

22:47

@jarlemag I agree with what you have for x,y but not for z

Thorgott

what I'm saying is that you should remind yourself of what those inequality symbols mean in plain language and then it will be very clear which parts of reasoning are sound and which aren't

jarlemag

@Semiclassical Interesting! Can you give any more clues? :-D

Thorgott

you're asking "if $L(G)$ is at most as big as $V_t^{\ast}$, can it be greater than $V_t^{\ast}$?"

Semiclassical

well, you have x^2+y^2=f^2 which is fine

but in that case, 1=x^2+y^2+z^2=f^2+z^2

so should have z=+- sqrt(1-f^2)

manooooh

@Thorgott of course not

Thorgott

22:50

indeed

@manooooh now try to see if you can figure out whether the "hence" here is reasonable or not

Semiclassical

@jarlemag so what remains is to figure out what f and z need to be to give you the desired angle beta, subject to f^2+z^2=1

jarlemag

Thanks for the help! I'll take a look.

Semiclassical

but that's not too hard either: should be be able to use trig on the right triangle formed by f,z and 1

manooooh

@Thorgott it is not reasonable. The correct should be: Since $|L(G)|\leq|V_t^*|$ but $|L(G)|\not>|V_t^*|$, we conclude $|L(G)|\neq|V_t^*|$

Now?

Shaun

Hello. I'm looking to ask a question concerning examples of finite concepts that require infinity. I'm not sure how it would be received. It might be too broad, say, or not phrased in a helpful way. Do you have any advice? What are your thoughts on some examples in the spirit of the question?

Thorgott

22:53

nope, replace inequality signs with the words "greater" and "smaller" and think

Semiclassical

that seems pretty broad yeah

Thorgott

I feel like I've seen such a question before

Semiclassical

Do you have an example in mind to start with?

Shaun

Some finitely presented groups are infinite.

Lukas Heger

@Shaun I don't know much of this, but an example that comes to mind is the methods used in the study of finite simple group of Lie type. These can be described as $G(\Bbb F_q)$ where $G$ is an algebraic group. It turns out that it's often useful to work with $G(\overline{\Bbb F_q})$ when studying those, e.g. their representations

Shaun

22:57

My motivation is to have something to point to in discussion with people skeptical of the rôle of infinity in mathematics.

Semiclassical

@Shaun i feel like the idea here would be that certain infinite structures can be described using a finite set of rules

geocalc33

f(x)=x can be described using f(x)=x but it's actually an infinite object

Semiclassical

but that is really general. for instance, a recursion relation generates an infinite sequence using a basic set of ingredients

Shaun

That's the sort of thing I have in mind, @LukasHeger; thank you.

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