Mathematics - 2023-01-28

robjohn

12:56 AM

@Koro that is a viewing app, not an editing app.

Xander Henderson

@robjohn Preview has some very limited annotation functionality.

geocalc33

hi

geocalc33

1:26 AM

h(x)=f(x)+ {g(x) if x<0, 0 if x>0}

so h(x)= f(x) for x>0

and h(x)=f(x)+g(x) for x<0

anak

@TedShifrin Sorry, what is (M,g) in your case? $M=S^n$ with its usual metric?

Cotton Headed Ninnymuggins

2:09 AM

How does one add curly braces underneath a term to write a note or something?

Novice

2:27 AM

tutorialspoint.com/tex_commands/underbrace.htm

Cotton Headed Ninnymuggins

Thank you

Ted Shifrin

@anak Yes. The warped product is just the Euclidean metric on $\Bbb R^{n+1}-\{0\}$.

anak

@TedShifrin Okay, maybe we aren't talking about the same thing. I am looking at $(M\times\mathbb R^+,r^2g + dr\otimes dr)$ vs. $(M\times\mathbb R^+,g + r^{-2}\,dr\otimes dr)$. You are saying one of these is isometric to the warped product?

Ted Shifrin

2:45 AM

The first is by definition a warped product, yes.

The second is isometric to the trivial product (change variables).

anak

2:57 AM

Which Ricci coefficient $R_{ij}$ of the warped product are you getting a $\pm 1$ in? I have been doing general computations without specifying $(M,g)$ explicitly, and it's all been cancelling very cleanly?

Ted Shifrin

I’m computing with moving frames. The sectional curvature of every $e_ie_j$ plane decreases by $1$. This is consistent with going from the unit sphere to Euclidean space.

anak

I am not familiar with moving frames, but I just did the computation in local coordinates. Is there a reason a local coordinates computation would fail besides typos/mistakes?

Like, is there a reason I can't do this in local coordinates?

Ted Shifrin

3:41 AM

No. Moving frames is perfect for this, though. The fact that my answer checks with the example and yours doesn’t is suggesting you messed up.

ペガサス Seiya

4:02 AM

The math ain't mathing in this annoying integral

Cotton Headed Ninnymuggins

4:20 AM

@ペガサスSeiya Calc 2 wasn't bussin. Taylor series = cool. Integral = annoying

ペガサス Seiya

4:58 AM

@CottonHeadedNinnymuggins Calc 2 was fine, not really challenging

Real Analysis is challenging though

Cotton Headed Ninnymuggins

It was the hardest of Calc 1,2,3 in my opinion. I had a better time in real analysis than Calc 2. One made me do repeated manipulation over and over while learning hardly anything of substance, and the other one taught me how to use raw mathematical definitions to prove things

Calc 2 is the only math class I've had that someone could say to me "when am I going to use this? How is this useful" and I'd respond with "Great question, I have no idea"

Koro

32

A: How can I prove formally that the projective plane is a Hausdorff space?

Let $q:S^n\to P^n$ be the quotient map, and let $u,v\in P^n$ with $u\ne v$; there are $x,y\in S^n$ such that $q^{-1}[\{u\}]=\{x,-x\}$ and $q^{-1}[\{v\}]=\{y,-y\}$. Let $\epsilon=\frac13\min\{\|x-y\|,\|x+y\|\}$, and set $$U=B(x,\epsilon)\cap S^n\quad\text{and}\quad V=B(y,\epsilon)\cap S^n\;,$$ w...

In this answer, shouldn't we have $p^{-1}(p(U))=U\cap (-U)$?

Ted Shifrin

5:25 AM

Why are you changing $q$ in Brian Scott’s proof to $p$?

Cap? No, Why do you say that? Those sets should be disjoint.

Koro

No particular reason about using p.

I'm trying to use this answer in the following case:

8 hours ago, by Koro

@TedShifrin I mixed up some things.

I want to show here that Y is Hausdorff.

Take any (x), (y) in Y, where (.) denotes the equivalence class containing . and $(x)\ne (y)$. That is, in particular we have $p_2(x)=(x), p_2(y)=(y)$.

Ted Shifrin

Same proof. Disjoint sets on the sphere give disjoint sets of lines by taking the cones.

Koro

So from here, we have: $x\ne m y$ for any $m\in \mathbb R-\{0\}$.

Ted Shifrin

take representatives on the unit sphere.

Koro

I was thinking on straight lines through origin as follows:

Hence x, y are distinct in R^{n+1}-\{0\}$ so can be separated by disjoint open sets U and V (U contains x, V contains y).

I have shown that $p_2(U)\cap p_2(V)=\emptyset$ so I just want to show that $p_2(U)$ and $p_2(V)$ are open.

Ted Shifrin

5:38 AM

This won’t work without a ton of specifics. Do what I said and go to the sphere.

No reason those sets are disjoint. That begs the question.

ペガサス Seiya

Why's everyone still up

Ted Shifrin

<— asleep

Koro

I wanted to send an image from ipad to macbook using 'Airdrop'. It appears it does not work, seems to be just a namesake feature.

Ted Shifrin

it works for files

Koro

it sometimes works for images, sometimes it doesn't. I wanted to send the image aiding that $p_2(U)$ should be open so that I can discuss here writing how to prove it.

I have reliability issues with Apple.

Ted Shifrin

5:48 AM

Take my advice above. Your disjoint open sets may very well overlap when you project. I’m not going to saynit again.

Koro

Ted, I understand to some extent what you are saying. I'm a beginner in this subject, every statement comes as a theorem to me. But there are some things that I do find obvious but hard to prove them. I want to know your feedback on this image (whether it can help me prove p_2(U) open or not.)

(I emailed it to me from ipad and then downloaded it in macbook and then shared it because airdrop works so 'phenomenally'.)

robjohn

@Koro I send all sorts of documents: images, pdfs, text files, etc.

Koro

Macbook is so 'convenient'

@TedShifrin Ahh, I see the problem now.

The problem may be that the map $p_2\circ i$ may not be constant on $p_1^{-1}(\{y\})$ so universal property may not work the way I want it to.

Ted Shifrin

6:25 AM

Those open sets work fine, but you need to make them small and “round” enough to be sure. Suppose $y$ were “on the other side” far away.

Koro

I think the approach I tried does not work because of many reasons.

Using this, even $p_2(U)$ and $p_2(V)$ should not be disjoint.

I'm posting the details in a post shortly.

Ted Shifrin

Just choose $x$ and $y$ on the sphere as I said an hour ago.

The philosophy should be orthonormal bases are better than arbitrary bases.

Koro

yeah, I tried that. Then I though that $p_2(U)=U$

because of taking the representative on sphere.

But then I realized that $p_2(U)=U$ may not be true because we have to send a diameter to a point so in the image we get 'two spheres touching at one point'-same as what I got earlier.

0

Q: How to show that the space $Y$ obtained by identifying points on the same straight line through origin in $R^{n+1}-\{0\}$ is Hausdorff?

Background: $Y$ turns out to be projective space but I'm trying to prove this and have not proven it yet (i.e., $Y$ is homeomorphic to the space $X$ obtained by identifying antipodal points on the unit $n-$ sphere.) One of the steps in proving the homeomorphism is to prove that $Y$ is Hausdorff. ...

nickbros123

8:34 AM

Proof for Schwartz inequality for real numbers usually involves saying $ \sum (\lambda y_{i} -x_{i})^2≥0 $ and then arriving at a quadratic equation whose roots are supposed to be imaginary, hence quadratic formula proves the theorem. How did one come up with such an amazing method ?

Koro

@AlessandroCodenotti @Jakobian: can you please help me with showing that Y is Hausdorff here:

2 hours ago, by Koro

0

Q: How to show that the space $Y$ obtained by identifying points on the same straight line through origin in $R^{n+1}-\{0\}$ is Hausdorff?

Background: $Y$ turns out to be projective space but I'm trying to prove this and have not proven it yet (i.e., $Y$ is homeomorphic to the space $X$ obtained by identifying antipodal points on the unit $n-$ sphere.) One of the steps in proving the homeomorphism is to prove that $Y$ is Hausdorff. ...

general-topology algebraic-topology

@nickbros123 by experience and lot of practice.

ペガサス Seiya

9:23 AM

Optimization without calculus is fun

Thorgott

12:57 PM

again, you really do not need to prove this space is Hausdorff to show the two descriptions yield the same space

you already essentially wrote down maps in both directions that will turn out to be inverse homeomorphisms yesterday, so I'm not quite sure where the hurdle is

Orm

Hi. If someone has a strong grasp of calculus and/or abstract algebra and would be willing to casually discuss some 2nd-year university problems over Zoom for USD50/hour (paypal), send an email to bronzebear@posteo.net . No prep would be needed.

Jakobian

1:46 PM

@Koro this was resolved already right

ILikeMathematics

3:29 PM

Regarding epsilon-delta proofs of limits, if we conclude that $\delta = \frac{\epsilon}{5}$ for example, wouldn't $\delta = \frac{\epsilon}{10}$ be equally valid?
We throw some $x$ out, but since $|x - a|$ is going to be very small for $x \to a$, that shouldn't matter, right?

Xander Henderson

@ILikeMathematics Sure.

If some $\delta$ gets the job done, then any smaller $\delta$ will also get the job done.

ILikeMathematics

Thank you.

nickbros123

4:49 PM

@ペガサスSeiya i only know one other technique- getting sin and cos terms and doing optimisation on them

Obliv

5:04 PM

how does one choose a free variable in a system of linear equations?

ペガサス Seiya

5:33 PM

@Obliv I usually toss a coin and then decide it based off of that

ThunderGlove

What does x<<1 mean?

Semiclassical

x much much less than 1

typically that's understood as x>0 as well, so something small but positive

ThunderGlove

Define "much much less"

Semiclassical

it's an approximation. so for instance $(1+x)^p \approx 1+xp$ if $0<x\ll 1$

it's inherently a bit imprecise, b/c the two sides are only truly equal if $x=0$

ペガサス Seiya

In a binomial approximation, for example, being "much much smaller" means small enough to disregard it as essentially 0. At least that's what we did when learning about the law of gravitation

Semiclassical

5:37 PM

for precision you need statements like $\dfrac{1}{x}[(1+x)^p-1]\to p$ as $x\to 0$

ThunderGlove

How small so that I can disregard

Semiclassical

depends on how much error you want to tolerate

it's like asking how many grains of sand you need before it's a pile of sand. there's not going to be a hard-and-fast answer.

hence why it's language that you usually see more in physics/engineering contexts

ThunderGlove

Oh

Thanks

I expected some rigor but its fine. I understand

Semiclassical

in analysis you typically insist on more precise language. for instance, you might show that $(1+x)^p-(1+xp)$ is bounded in some fashion

(I don't remember what inequality would apply, so i can't make that statement more explicit)

Xander Henderson

@ThunderGlove "Rigor" is field specific.

ThunderGlove

5:44 PM

So true

Xander Henderson

$0 < x \ll 1$ is "rigorous enough" for most practical applications.

Semiclassical

typically in physics/engineering contexts you get statements which -could- be made rigorous with a little work

they're in some neighborhood of a rigorous statement :P

Xander Henderson

@Semiclassical With respect to which topology?

Semiclassical

siiiiigh

Xander Henderson

:P

Koro

5:55 PM

@Jakobian no, I am afraid.

But I have a better idea now, still I don't understand though how to write the proof. I'll explain: a) The picture titled 1) in a picture in my post has a subtle flaw- the image p_2(U) does not necessarily look like a dumbbell. Every line segment in U passing through the origin gets squeezed so we get something like a portion of an n sphere.

Obliv

@ペガサスSeiya if I have a 5x2 augmented matrix with 2 0s in the 2nd row, what would be the free variables?

Koro

b) I want to prove that p_2(U), p_2(V) are open in Y: So I must prove that $p_2^{-1}(p_2(U))$ is open in R^{n+1}-{0}. This preimage can be visualized as a cone joining U to origin and then its extension in the opposite equivalent of quadrant in n+1-dimension with origin removed. This is open as the intuition tells. I think that it should be straight forward to show that these cones are open.

user123234

If I consider an element $x\in \Bbb{R}$ am I then correct if I say that every function $f\in C(\{x\})$ is constant?

Ted Shifrin

@Obliv That is not enough information. You need an echelon form of the matrix.

Obliv

idk how to do matrices in latex but it's like the first two entries on the second row would be 0s

everything else has a value

Ted Shifrin

6:05 PM

@user123234 Don't use $x$ for a fixed point. It is too confusing. You're talking about functions (forget continuous) on a one-point set?

You ignored what I said, Obliv.

What is the echelon form?

Obliv

with leading 1s*

basically the system is $x+3y-7/4 z - 5w = 11/2$ and $z-52w = 54$

Ted Shifrin

You type matrices by doing \begin{bmatrix} . & . \\ . & . \\ . & . \end{bmatrix}

So in what columns are the pivots? What are the pivot variables?

Obliv

$\begin{bmatrix} 1 & 3 & -7/4 & -5 & 11/2 \\ 0 & 0 & 1 & -52 & 54\end{bmatrix}$

Ted Shifrin

You need dollar signs, of course.

The % can't be there.

There you go.

Obliv

weirdly enough, it rendered for me even without the dollar signs

Ted Shifrin

6:08 PM

Yeah, me too.

So answer my pivot question.

user123234

@TedShifrin but how do they look like?

Obliv

I do not know what those are. The solution manual makes $y = s$ and $w = t$ but I'm not sure why.

user123234

I mean $C(A)$ are all continuous functions on $A$ why should I forget about this? @TedShifrin

Ted Shifrin

If your book defined free variables, it talked about pivots when you do row reduction.

Obliv

oh is it because $z$ is a leading variable..

Ted Shifrin

6:10 PM

The free variables are those that are not pivots.

@user123234 Because for one point it's irrelevant.

Obliv

pivots are the 1s in the matrix?

user123234

But the functions are still all constant?

Ted Shifrin

The leading ones, yes.

@user123234 If you have only one point in the domain, how can a function be non-constant?

user123234

there is no way.

Ted Shifrin

Right.

So continuity is irrelevant :)

Every function on a one-point set is continuous.

Obliv

6:13 PM

so does being able to express a system of linear equations with parameters mean it has infinite solutions?

Ted Shifrin

Provided it is consistent!

user123234

thanks a lot ted!

Ted Shifrin

What if I have the system $x+y=5$, $0=1$?

Sure thing, @user123234.

Obliv

so do we only parametrize equations of 3 or more variables?

Ted Shifrin

No.

Obliv

6:20 PM

in 3xn augmented matrices we can't have a parametrized form of $x_2$ though

it'd have to equal something

Ted Shifrin

No. Don't say "parametrized form of ..." Just say $x_2$ cannot be a free variable, if that's what you mean? Or do you mean $x_2$ is a pivot (leading) variable? I don't understand.

Obliv

Ah I think i get it, so for an n x m matrix if m > n-1 you can't parametrize the system I think

and it's consistent*

basically like for the example before it was a 5x2 system, but if it were 5x5 it wouldn't be able to have free vars?

or maybe that's totally wrong

Ted Shifrin

I still do not know what you're talking about when you say "parametrize the system." Every consistent system of equations has solutions and you can parametrize them by the free variables.

A $5\times 5$ can certainly have free variables.

Obliv

a free variable to me looks like a variable with no constraint to satisfy in the system. I imagined that if you had increasing equations then the system becomes more constrained

Ted Shifrin

Only if you have non-redundant equations. I can add the same equation over and over and over again and the solutions don't change.

Obliv

6:29 PM

is there a limit to how many non-redundant equations you can add to a system of n variables? Is it n?

Ted Shifrin

No.

Oh, non-redundant.

Koro

@Jakobian: There is an answer to my post. I have commented there about my points of confusion.

Ted Shifrin

Yes. Right. You can have at most $n$ pivots if you have $n$ variables.

Obliv

so $x_1+x_2+...+x_n = b$ it's plain to see there exists infinite solutions to this no matter what $b$ is, but once you have multiple equations there must be consistency

and in order to tell whether the system has multiple solutions or whatever isn't there a shortcut

Ted Shifrin

Right. You need to get to an echelon form.

If the system of equations is homogeneous (all $0$'s on the right hand side), then there's no issue. It's always consistent.

user123234

6:38 PM

If we say that an operator $T$ is invertible then this means that there exists an operator $S$ s.t. $ST=TS=I$ but does this mean $S(T(x))=x$ and $T(S(y))=y$ and not multiplication right?

Ted Shifrin

Multiplication generally makes no sense.

You can interpret composition of linear functions on finite-dimensional vector spaces in terms of matrix multiplication, of course.

user123234

ah perfect thanks

Koro

@Jakobian I think that the claim that p(U) and p(V) are disjoint open should be true as this claim is in line with the answer on page no. 37 under problem 1 here: greggrant.org/armstrong.pdf

is posting link in comments in mse not allowed any more?

user123234

7:00 PM

@TedShifrin I was asking because of this question (math.stackexchange.com/questions/4627765/…). It somehow confuses me when I wanted to show that it is an inverse. since I don't have multiplication.

Node.JS

I don't understand this induction answer. math.stackexchange.com/a/4627748/49780 I wrote a comment but no response. Can someone please help?

Ted Shifrin

@user123234 I wrote a comment there for you.

user123234

should I write here with you?

or in the comments?

Ted Shifrin

If you think it will help other people who read your post, maybe there.

monoidaltransform

If $X\times Y$ is locally compact and locally connected metric spaces, then is it the case that for closed subset $N$ of $X\times Y$, one can write $N=(N_1\times Y) \cup (X\times N_2)$ where $N_1$ is closed in $X$?

user123234

7:06 PM

@TedShifrin okey I wrote there

Ted Shifrin

@monoidaltransform What you wrote makes no sense to me. And the answer, even if you fix it, is no, I'm pretty sure.

monoidaltransform

I fixed it.

Ted Shifrin

Think about random closed subspaces of $\Bbb R\times\Bbb R$. Does this make sense?

leslie townes

i don't normally recommend this kind of notational tomfoolery, but you might benefit from introducing a notation for the constant function $1$ on $C(\sigma(T))$ that is not "I", like $\mathbf{1}$, and writing something like $\phi - \lambda \mathbf{1}$ instead of $\phi - \lambda$.

you might also, somewhere in that proof, explicitly point out where you are using the hypothesis about the range of $\phi$.

Ted Shifrin

I applaud Leslie's diligence.

leslie townes

7:15 PM

my stopped clock is right once today. see you in 12 hours.

Koro

@Jakobian: my question has been answered now.

and yes the following works :-)

1 hour ago, by Koro

b) I want to prove that p_2(U), p_2(V) are open in Y: So I must prove that $p_2^{-1}(p_2(U))$ is open in R^{n+1}-{0}. This preimage can be visualized as a cone joining U to origin and then its extension in the opposite equivalent of quadrant in n+1-dimension with origin removed. This is open as the intuition tells. I think that it should be straight forward to show that these cones are open.

The word for this object is 'double cone'. Somehow the word didn't occur to me while writing the comment above 1 hr ago.

I watched the show called 'The last of us'. It's very nice :-).

leslie townes

the sum in the hypothesis 1. is probably also supposed to start at k = 0. without this, i don't see a reason why F(1) = I; F(1) could just be some self-adjoint projection that commutes with all of the F(f)'s.

ペガサス Seiya

7:31 PM

I was at the court today and I was given a severe sentence

I need to compute the inverse of 4x4 matrices. By hand

No idea if they're even invertible

leslie townes

that sucks, but if you have a few hours in a cell with nothing else to do, is not unreasonable.

copper.hat

what was the key determinant in your sentence?

just found out that my global entry card ($100, 3 mos. + interview) gets me nothing when i travel on my native country's airline. wish i figured that out before i wasted the time & money.

Shaun

in Helpful Commentary, 1 min ago, by Shaun

In quick succession, the following questions of mine were downvoted.

0

Q: Verifying my proof that $Y\subseteq B_X(n)$ implies $B_Y(m)\subseteq B_X(mn)$.

The Details: Let $G$ be a group and let $X\subseteq G$. Let $n\in\Bbb N$. Recall that for $H\le G$, $${\rm conj}_H(X)=\{ hxh^{-1}\mid x\in X, h\in H\}.$$ Define $$B_X(n)=\bigcup_{k=0}^n\underbrace{{\rm conj}_G(X)^{\pm 1}\dots{\rm conj}_G(X)^{\pm 1}}_{k\text{ times.}} $$ Note that the idea here is...

group-theory solution-verification

0

Q: Show, using a specific approach, that $\dim \Bbb P^n=\dim\Bbb A^n=n$.

This is Exercise 1.8.4(1) of Springer's, "Linear Algebraic Groups (Second Edition)". It is not a duplicate of The dimension of $\mathbb P^n$ is $n$ because I'm after a particular perspective; namely, the approach Springer takes (using transcendence degree and not Krull dimension). The Question: ...

abstract-algebra algebraic-geometry affine-geometry transcendence-theory

I would appreciate some feedback on them, please.

ペガサス Seiya

@leslietownes I'd rather take 100 hours of community service instead of that

I'll take driving a civic over that! Anything over that!

Shaun

I try hard when writing up my questions here. I would like to know what I'm doing wrong.

leslie townes

7:45 PM

shaun, downvotes can come for any reason or no reason at all. it's not necessarily a sign of doing anything 'wrong,' and not really something that one can avoid by following advice.

ペガサス Seiya

Imagine working really hard on a question that involved lots of MathJax, written over a few days, posting it, only for it to get down voted

That's hilarious

copper.hat

@Shaun downvoting without reason is one of the most aggravating aspects of the site gamification.

3

Shaun

@copper.hat I agree. I find it very frustrating.

copper.hat

the unexplaining downvoters will all eventually end up in some infinite tsa check line.

Shaun

From the help section of the site: "Voting up a question or answer signals to the rest of the community that a post is interesting, well-researched, and useful, while voting down a post signals the opposite: that the post contains wrong information, is poorly researched, or fails to communicate information." Do my questions contain wrong information? Are they poorly researched? Do they fail to communicate information? Honestly, it baffles me sometimes. What do people think downvotes mean?

Obliv

7:59 PM

When I'm solving a large system of linear equations is the goal to make the pivot variables of each row at the same time?

how does one choose which rows to modify and by what

leslie townes

obliv there can be a number of choices involved that are essentially arbitrary. there is no single right way to row reduce a matrix. but, to simplify things (and to ensure that one is not just randomly spinning in circles by applying row operations that undo the intended effects of other operations) most textbook and computer algorithms will, somewhat arbitrarily but not without reason, pick a particular series of steps to do it.

sometimes books kind of bungle this because they are hesitant to actually write out an algorithm as an algorithm; they'd rather write five or six examples which can look to the uninitiated like applying row operations more or less at random, in pursuit of two or three different and potentially conflicting goals.

ペガサス Seiya

Row echelon form

Its a pain, but it works

leslie townes

en.wikipedia.org/wiki/Gaussian_elimination is an example of this. it describes the general goal of row reduction, repeatedly emphasizes a small number of basic row operations, but never explicitly says how it is choosing the operations in its example. you have to infer that from what it's said about the goal.

mathworld.wolfram.com/GaussianElimination.html same problem. and i'm not picking on the web here, textbooks do it too.

Obliv

Oh okay thank you, I assumed there were algorithms which I understand why you wouldn't be taught them as an uninitiated so you get a better feel, but I just kept running into loops while doing these examples :\

ペガサス Seiya

@leslietownes I can't be the only one who feels like solving linear systems of equations can be done more simply using the "usual" approach, as opposed to matrices right?

Obliv

8:14 PM

tis the same approach, but without matrices you're carrying around unnecessary variables

leslie townes

obliv: there definitely are algorithms, and they're not hard, they just involve making arbitrary choices. roughly speaking, you can move left to right (ignoring columns of zeros where they appear), and top to bottom, clearing out nonzero entries in lower rows using a scaled version of the leading nonzero entry of the 'top' row. and you can swap rows only, if ever, when needed to get a nonzero entry in the upper left portion of the matrix you are working on.

you get into trouble if you begin swapping rows for other reasons, or doing anything other than swapping rows for that one reason, and adding multiples of your 'top' row to the bottom rows to clear nonzero entries out.

because then you can go in circles. you won't if you keep moving only in one direction.

there are some arbitrary choices here, like, if i don't have a nonzero number in the upper left portion of the part i'm working on, but i have multiple rows below that do, which one do i choose to swap the top row with? in specific contexts (numerical ones, or ones where staying in a particular ring/field is important) this might matter, but outside of those contexts, just pick one. pick the first one.

another arbitrary choice is whether you scale your 'top' row to have a 1 as its leading entry or not in the middle of the calculation, at the end of the calculation, or ever. again, there are reasons for caring about these things in some contexts, but in general, just pick one thing and do it every time.

so you have an algorithm and not a series of branching choices about what you're doing next.

Obliv

I think a common mistake I'm making is I'm not looking to see if each equation is non redundant, I keep expecting to get a nice system where I can get leading 1s all the way down

I also really don't like writing things out so doing everything in my head gets confusing :D

leslie townes

the other running annoyance with doing this stuff by hand is the large amount of opportunity for arithmetic errors to propagate through the calculation.

Obliv

oh yeah definitely.

leslie townes

@ペガサスSeiya my general feeling is that for doing stuff by hand with small systems (your example of 4x4 might be my informal threshold for what makes a 'large' system for hand calculation), what's simpler vs. complex is a matter of personal taste and not something that can be optimized across groups of people.

but for larger systems, it definitely helps not to have to write variables over and over.

ペガサス Seiya

8:38 PM

@leslietownes lets be real, once you've learned the approach, most of it are just tedious calculations that are better left for computers to do

leslie townes

i wonder if it's even worth learning the approach, frankly. given what hash so many books make of it.

ペガサス Seiya

I'm fine with Cramer's rule to be honest, works well enough

Don't need to worry about which rows to subtract and whatnot

ILikeMathematics

9:00 PM

Could someone please check if following proof is valid?

Proof of the uniqueness of limits.

We want to prove that if $\lim_\limits{x \to a} f(x) = l$ and $\lim_\limits{x \to a} = m$, then $l = m$.
By the epsilon delta definition of limits, for every $\varepsilon > 0$, there exists a $\delta_1 > 0$, so that whenever $0 < |x - a| < \delta_1$, we have $|f(x) - l| < \varepsilon$.
Similarly, for every $\varepsilon > 0$, there exists a $\delta_2 > 0$, so that whenever $0 < |x - a| < \delta_2$, we have $|f(x) - m| < \varepsilon$.

leslie townes

it's the right idea, it could be formatted more clearly. you're really fixing epsilon > 0 and "plugging" the same epsilon into both limit hypotheses to get your delta_1 and delta_2. this isn't clear from saying "for every epsilon > 0, one thing" followed by "for every epsilon > 0, another thing."

also, after producing your delta_1 and delta_2, you're implicitly using the fact that there are values of x for which both |f(x) - m| < e and |f(x) - l| < e hold, to deduce that |l - m| < 2e.

you could be clearer about what value or values of x are you using to do that.

relatedly, at the moment, delta_1 and delta_2 appear exactly once and are not used again later. why is that? do you need to use them later? if not, why mention them at all?

but this is absolutely the right idea and argument.

Obliv

9:20 PM

for a homogeneous coefficient matrix like $\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \end{bmatrix}$ would it be $x_1 = 0, x_2 + x_3 = 0, x_4 = t$?

or I guess it'd be $x_3 = s$ so $x_2 = -s$

for $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ would it just be $x_1 = 0 , x_2 = 0, x_3 = 0$

or are they all free variables

actually nvm they would be all free variables

leslie townes

yes. for purposes of keeping track of the number of variables involved, it's helpful to state whether the matrix you've row reduced is an 'augmented matrix' (with the right hand sides as its last column) or not. if the 2x4 and 3x3 matrices are augmented, i agree with your description of the solution set (although there are only two free variables, x_1 and x_2, in that second example).

Obliv

oh sorry they were coefficient matrices. Also didn't realize n x m meant row by column i thought it was column by row lol

ペガサス Seiya

There's math, and then there's matrices and statistics.

Tomasz Przybyło

9:37 PM

Hi guys. I am analyzing archived data of this forum. Do you know what distincts bots here ? I found only a bot "Community" :l

leslie townes

if they're just matrices of coefficients, then the form of the solution set can depend on what the right hand side is. e.g. if you have nonzero stuff on the rhs of that 3x3 system, no variables are free and the solution set is empty. and in the first one, if there is a nonzero thing b on the right hand side of the second equation, then the formula for x_2 has to change to reflect that (although x_3 is still free)

Obliv

they were homogeneous*

leslie townes

tomasz good question, i don't know of any. i'm not sure that the terms of use allow for third party bots (it would surprise me if they did), and "community" is the only one i've seen active on the main site. i understand that it is not purely automatic, but at least sometimes is an anonymizer for moderator or voter actions.

Obliv

when is polynomial interpolation ever useful? I feel like in sciences it's always better to go with linear regression or something

amWhy

@TomaszPrzybyło Hi who? Am I an honorary "guy"? :) But rest assured I am not a Bot!

ILikeMathematics

9:50 PM

Thanks, and like this, is there anything to change?

Proof of the uniqueness of limits.

We want to prove that if $\lim_\limits{x \to a} f(x) = l$ and $\lim_\limits{x \to a} f(x) = m$, then $l = m$.
By the epsilon delta definition of limits, for every $\varepsilon_1 > 0$, there exists a $\delta_1 > 0$, so that whenever $0 < |x - a| < \delta_1$, we have $|f(x) - l| < \varepsilon_1$.
Similarly, for every $\varepsilon_2 > 0$, there exists a $\delta_2 > 0$, so that whenever $0 < |x - a| < \delta_2$, we have $|f(x) - m| < \varepsilon_2$.

amWhy

@TomaszPrzybyło Just joking. :)

leslie townes

on the third-to-last line, i would begin by fixing epsilon > 0. (taking that choice of epsilon and using it as both epsilon_1 and epsilon_2 in the stuff that comes before is what gives you delta_1 and delta_2, and hence the delta in that min( ) recipe.) it feels a little out of order to put delta and delta_1 and delta_2 ahead of the thing that is used to choose them. i like the rest of it.

ILikeMathematics

10:08 PM

Thank you.

ペガサス Seiya

10:20 PM

I have a Russian computer voice stuck in my head

@TomaszPrzybyło yep, we're all a bunch of bots here

Would you believe me if I told you that a Cessna can shoot down a raptor? @XanderHenderson

amWhy

10:50 PM

@ペガサスSeiya Speak for yourself! :D I'm not a guy and I'm not a bot. Unless some one put a chip in my Covid Vaccines! NOT... Haha.

@Kevin Parent musicfans.se., and interested in group theory. Just curious about your favorite groups in music! I'm a music fan, and greatly interested in group theory. I did not know about musicfans.se!

ペガサス Seiya

Cosmic confetti

amWhy

@Kevin not to put you on the spot. I'm just curious. No need to respond.

ペガサス Seiya

I saw someone drag race a Civic with a turbocharged V6, that thing went ballistic @TedShifrin

If you ever get bored of your car, there's your option

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