Mathematics - 2023-02-14 (page 1 of 2)

Jack Ohara

00:07

@TedShifrin Hi Ted

I have a question about how to compute taylor polynomial

f(x = e^ (-1/x) for x>0 and 0 for x<= 0

we want to calclute the taylor poloynomial of degree 10 centered at 0

Sine of the Time

@JackOhara this is not analytic at $x=0$

Jack Ohara

and second part is if this does approximate the function well on [-1,1]

@SineoftheTime how to proceed then?

I am very bad at computation problems

and how do we test if the function is approximated well on that interval

been a while since i took analysis

Sine of the Time

I've seen a similar question on mse a while ago, let me search

Jack Ohara

okay thanks

seems strange to ask for that high degree

Sine of the Time

try to compute the first derivative and generalize

Akiva Weinberger

00:13

0

Q: Create a triangle whose colors are determined by the bitsums of coordinates

Write a program that, for any \$n\$, generates a triangle made of hexagons as shown, \$2^n\$ to a side. The colors are to be determined as follows. We may give the triangle barycentric coordinates so that every hexagon is described by a triple \$(x,y,z)\$ with \$x+y+z=2^n-1\$. (The three corners ...

First time posting on the Code Gold place

Curious to see how it goes

Sine of the Time

@JackOhara math.stackexchange.com/q/2002875/1118406 ,math.stackexchange.com/q/2125155/1118406

Jack Ohara

@SineoftheTime thanks i will check it!

Sine of the Time

@JackOhara :)

D.C. the III

@TedShifrin an issue came up in the quadrilateral problem when I was writing it out:

$f(\theta, \gamma) = \frac{1}{2}(ab \sin(\theta) + cd \sin(\gamma))$

$g(\theta, \gamma) = ab \cos(\theta) - cd \cos(theta) = \frac{a^2+b^2-c^2-d^2}{2}$

$Df(\theta, \gamma) = \lambda gf(\theta, \gamma) \rightarrow \frac{1}{2} [ab \cos(\theta), cd \cos(\gamma)] = \lambda[-ab \sin(\theta), cd \sin(\gamma)]$

Ted Shifrin

@SineoftheTime This is not relevant.

Just compute (carefully) the first and second derivative, Jack. You'll catch on.

D.C. the III

00:26

The only value that satisfies things is $\theta = \gamma = 0$. Kind of messes up the other expression I need to satisfy for inscribing a shape

Jack Ohara

@TedShifrin working on it!

Ted Shifrin

$\tan\theta = -\tan\gamma$, @DCthe

D.C. the III

yes that is the final inequality I got

Ted Shifrin

How are you getting that they're $0$?

$0$ does not remotely satisfy the constraint equation.

D.C. the III

drew the graph of each

Ted Shifrin

00:28

You're being silly.

Think about the meaning of tangent.

How can two angles have opposite tangents?

D.C. the III

ah...my weak geometry and trig comes to haunt again

Sine of the Time

@TedShifrin yes, it was just a remark. it's not relevant since Jack was searching for the polynomial

good night guys:)

D.C. the III

@TedShifrin if one is the inverse of the other.

Ted Shifrin

Yes, the function is very famous because its Taylor polynomial at $0$ is such a colossal failure.

Right, DC, or (since our angles are not negative)?

Jack Ohara

should riemann stieltjes integral give same value of the integral no matter the function alpha we integrating with respect to ?

@TedShifrin

Ted Shifrin

00:36

Why should it not depend on $\alpha$?

Jack Ohara

so it does?

then what are we computing in this case?

i mean shouldnt be the case that we get one value of the integral no matter what function we use?

so if we take this example

f(x) = x^2

and we doing integration from 0 to 1

pick alpha to be 1+x^2 for 0<= x <= 1/2

and 3/2 +x^2 for 1/2 < x <= 1

if we just compute the integral using old riemann we get 1/3

but if we use stieltjes we get 5/8

if I did my math right

@TedShifrin how should we make sense of this?

when we doing stieltjes are we still computing area or some other measure?

Ted Shifrin

Nope, not area at all. In most cases, there is not a geometric interpretation.

What are you doing?

D.C. the III

one sec I miswrote it in latex

Jack Ohara

@TedShifrin a little bit more explanations please!

i am a bit lost with this stieltjes thing

D.C. the III

huge mistake of confusing the inverse and reciprocal

Ok I'm not sure of the tidbit I'm missing.

Ted Shifrin

00:52

There was no explanation, Jack. But think about what happens when $\alpha$ has jumps.

@DC Think lines with opposite slopes.

Jack Ohara

@TedShifrin okay thanks!

D.C. the III

only thing that comes to mind with opposite slopes is negative reciprocal

Ted Shifrin

Wrong.

D.C. the III

sigh

It's going to be something so simple and I"m going to hate myself for not knowing it

Rithaniel

Is there anything known about the complement to a proper subring? Like, does there have to exist an ideal of some description that only intersects the subring at 0?

Thorgott

01:08

the zero ideal would do that job

Ted Shifrin

And do subrings have identities?

Thorgott

and if you're asking for a non-zero ideal, the inclusion of any non-field integral domain into its fraction field is a counter-example

(I'm assuming identities)

Ted Shifrin

DC: Draw a line with slope 2 and a line with slope -2.

Rithaniel

@TedShifrin Yeah, that's a requirement. Though, some definitions simply require that they have an element that acts like an identity. So $(4)\subseteq\mathbb{Z}_6$ might be a subring (with 4 acting as the identity)

D.C. the III

I did do that already

I'm looking at them right now

Rithaniel

01:11

@Thorgott Darn, good counterexample

D.C. the III

but nothing jumps out form that notion

Ted Shifrin

Look at the angles.

D.C. the III

was doing that too and setting up their trig identities

w.r.t a right triangle

Ted Shifrin

How are the angles related.

Forget trig identities. Irrelevant.

D.C. the III

I drew a new picture this time with the triangle of slope 2 starting at the origin and the triangle of slope -2 terminating at the origin.

Ted Shifrin

01:21

So the angles measured from the positive $x$-axis are related how?

Rithaniel

Animations can sometimes help: desmos.com/calculator/qh3odeirlv

D.C. the III

so before my computer went on the fritz. I was going to say I can perhaps name those angles $\theta$ and $\gamma$

Ted Shifrin

The two angles sketched in your picture are related how?

D.C. the III

well call the right one $\theta$ and the left one would be $\pi - \theta$, but you want the measurement from the positive x-axis for the second one

Ted Shifrin

your marked angles are equal? But what you just typed is the answer to my question.

D.C. the III

01:29

the marked angles should be equal because I drew w.r.t the same slope, they may not appear like that because of my drawing

Ted Shifrin

Yes, so we are done.

Rithaniel

(Your drawing looks good, not to worry)

D.C. the III

but wouldn't the measurement for the second angle from the positive $x$ axis be $\theta + \pi/2$ ?

Ted Shifrin

Not if it gives $\pi$ when added to $\theta$.

D.C. the III

RIght...........time to whip out a geometry text and do an exercise or two each night....this is getting ridiculous.

Ted Shifrin

01:39

On that note, it's time for my martini.

D.C. the III

Damn...I stressed you like that....

I drove Ted to drinking.

straight or a flavoured martini?

Ted Shifrin

Flavored? Yuck. Dry gin martini, with olive.

D.C. the III

that's something us young folks do. sorry if I offended your purist ways.

Ted Shifrin

I'm OK with mixed drinks, but don't call them martinis.

Rithaniel

Root beer and rum, for me

ペガサス Seiya

02:05

@TedShifrin Your lectures should be called "Ted Talks"

Ted Shifrin

Old news.

ペガサス Seiya

Ted drinks?

Good thing I didn't injure my right arm or no math for almost a month

@D.C.theIII What was the problem here? It's obvious that angles drawn by two slopes of the same magnitude (while one is positive and the other is negative) with respect to the $x$, and even $y$, axis would be the exact same. Am I not understanding the problem right?

D.C. the III

02:23

"obvious" if one's geometry and trigonometry is strong......now on the other hand if one's skills in those fields are weak.............

Ted I'm working on the question asking to show $A = A^2$ given that $[\proj_V] = A$ directly from the definition you gave which is in terms of $b \in V$ and $b-p \in V^\perp$. I'm having trouble unraveling it

mick

Interesting developments

5

Q: Density of extended Mersenne numbers?

Consider the subset of odd positive integers defined and constructed as follows by these rules : A) $1$ is in the set. B) if $x$ is in the set , then $2x + 1$ is in the set. C) if $x$ and $y$ are in the set then $xy$ is in the set. I call them extended Mersenne numbers because rule A and B alone ...

abstract-algebra number-theory asymptotics sieve-theory mersenne-numbers

Rithaniel

02:58

Some people have natural geometric sense, but, for other people, that's not a thing they ever developed.

Silent

Please let me know how do they get
$\begin{align} \mathbf{a}_x&=\text{cos}\phi\;\mathbf{a}_\rho-\text{sin}\phi\;\mathbf{a}_\phi\\
\mathbf{a}_y&=\text{sin}\phi\;\mathbf{a}_\rho+\text{cos}\phi\;\mathbf{a}_\phi
\end{align}$
[here](https://math.stackexchange.com/q/1445288/94817) ?

Cotton Headed Ninnymuggins

03:35

Does the existence of an inverse imply a bijection?

Ted Shifrin

04:09

@D.C.theIII Think about $Ax$ and $A^2x$.

@CottonHeadedNinnymuggins A bit vague. What do you think?

D.C. the III

I had scribbled some stuff w.r.t to that $A^2x = A(Ax)$

leslie townes

ted: time for more talk about homotopy inverses? :)

Ted Shifrin

I doubt it …

Cotton Headed Ninnymuggins

Yes because the existence of an inverse implies the function is injective and surjective

leslie townes

i do too, i just love callbacks to earlier material

Ted Shifrin

04:11

OK.

leslie townes

themes seem to come in waves, even if it's different people raising different questions.

l'hopital was really "hot" for a while but seems to have cooled off.

Ted Shifrin

There are all sorts of inverses, so it’s important to make context crystal-clear.

Cotton Headed Ninnymuggins

Mapping sets to other sets was the context

Ted Shifrin

I inferred that.

D.C. the III

so if I used $b$, I would've had $Ab = p$ and we know $b - p \in V^\perp$, but I don't think $A(Ab)$ makes sense in this context

Ted Shifrin

04:14

Of course it does.

What is $Ap$?

Cotton Headed Ninnymuggins

Where do you all live?

D.C. the III

Oh if $A$ is projecting onto $V$ and $p \in V$ then $Ap \in V$

Ted Shifrin

Aha. Good. What in $V$?

D.C. the III

I guess it would be onto itself....so on to $p$

Ted Shifrin

Proof?

D.C. the III

04:23

well there is example 3 that used the projection matrix. But this is ebing done without that machinery

Ted Shifrin

Use the defn only.

D.C. the III

Well by defn, this $p$ will have the property that $p - p \in V^\perp$

Ted Shifrin

Yes, but why no other option?

D.C. the III

well it is defined as the unique vector $p$, etc, etc

Ted Shifrin

If $Ap=q$, then …

D.C. the III

04:38

$p-q \in V^\perp$....

Cotton Headed Ninnymuggins

Just multiply everything by $0$. That's what I always do

Ted Shifrin

@D.C.theIII And so …?

D.C. the III

$q = \text{proj}_V(p)$

Ted Shifrin

Why must $q=p$?

D.C. the III

05:01

I still haven't picked it up...

Ted Shifrin

The vector $p-q$ lives where?

D.C. the III

that lives in $V^\perp$

Ted Shifrin

And in …

D.C. the III

and in ???..........hmmm well $p \in V$.. and I just said $Ap = q \in V$, so that would mean this also lives in $V$

I say that because a sum of vectors from a vectors form a vector space will still be a vector in that vector space

Ted Shifrin

Yes. So go on.

D.C. the III

05:17

So $p-q \in V$ and $p-q \in V^\perp$..........

is the only vector that could be in $V$ and $V^\perp$ the $0$ vector?

granted I should know this considering all the linear algebra I've done...

and recollecting....yes this is the case.

leslie townes

👀

D.C. the III

so then $p - q = 0$

Ted Shifrin

That does it.

Cotton Headed Ninnymuggins

If my linear algebra isn't rusty, the only vector in both a subspace and its compliment is the $0$ vector

D.C. the III

and as such $Ab = A(Ab)$

Ted Shifrin

05:22

Not compliment. Not complement, either.

D.C. the III

I respect the level of deduction and unravelling needed. I suspect $A = A^t$ is going to be just as fun

Cotton Headed Ninnymuggins

Orthogonal compliment? Is that the term?

Ted Shifrin

I believe there is a hint.

Right. Orthogonal complEment.

Cotton Headed Ninnymuggins

Oops haha.

You know what else I remember?

$\langle Ax,y \rangle=\langle x,A^Ty \rangle$

leslie townes

dc3 not to fry your circuits but ya oughta remember how to prove that V intersect V^perp is {0} from more fundamental axioms. (hint: compute the length of something in V intersect V^perp in terms of the dot/inner product)

Ted Shifrin

05:26

@CottonHeadedNinnymuggins Super important. It’s Ted’s favorite formula.

Cotton Headed Ninnymuggins

I know. I used it on my final too

What are yall's opinions on Jimi Hendrix?

I think he's the 2nd coolest person that ever lived. Right behind Ted Shifrin

Ted Shifrin

What was your name before Cotton .. ?

D.C. the III

@leslietownes I think I did it in a previous section of the linear algebra book I'm doing. The idea sounds familiar

Cotton Headed Ninnymuggins

My name before what? My name is a reference to the movie "Elf"

Call me Johnny

if you want

Ted Shifrin

Oh, I assumed you were in this chat with a different name before.

copper.hat

05:36

What about donald?

Ted Shifrin

Bad assumption, I guess.

D.C. the III

I recognize the avatar and it not having such a long name beside it

Funny I can't find in my linear algebra exercises showing that $V \cap V^\perp = {0}$. I would've thought it would be in the inner product section of the text

Ted Shifrin

Leslie suggested that you consider $v\cdot v$.

copper.hat

or $\langle v , v \rangle$. what is that $\cdot$ thing?

leslie townes

i recommended both of those things.

copper.hat

05:41

speechless

Cotton Headed Ninnymuggins

@copper.hat huh?

@TedShifrin How come? Profile picture or name?

copper.hat

just trolling

Cotton Headed Ninnymuggins

ahh lol

D.C. the III

if we say $v = x + y$ with $x \in V$ and $y \in V^\perp$ then $\langle v, v \rangle = \langle x + y, x + y \rangle = \dots = \|x\|^2 + \|y\|^2$

copper.hat

yep.

pythagoras

D.C. the III

05:57

what about it here applies?

leslie townes

applies to what? you're the one who brought it up

sheesh

D.C. the III

Lol....I was trying to show the claim you suggested I do ......😪

copper.hat

the square of the 'diagonal' $x+y$ equals the sum of the squares of the sides $x,y$.

Kumar Shuvam

The equation "x² + y²- 6x+ 8y + 25 = 0" represents... ??Can anybody tell me what this 2nd degree curve represents..according to my text I know that a 2nd degree curve represents a circle when coefficients of x²= y² and coefficient of xy term = 0 and it seems like both the conditions are fulfilled here..but I also know that a general 2nd degree curve represents a pair of straight lines if determinant is equal to zero i.e. abc+2fgh-af²_bg²_ch²=0

What am I missing 😕 here

leslie townes

dc: i was thinking more in terms of the likely definition of V^perp , which was something of a gamble as i don't know what source you are using

copper.hat

06:03

i suspect there are not many points that lie on that 'curve'.

D.C. the III

@leslietownes V^perp as in the usual sense of x in V and y in V^perp would mean $\langle x, y \rangle = 0$

copper.hat

@KumarShuvam consider the point $(3,-4)$.

leslie townes

dc: so v in V and v in V^perp means... ?

kumar: sounds like you have a lot of general tools, but maybe not an exhaustive list? the solution set of an equation like that might become more recognizable if it were rewritten. the algebraic tool here would be to separately complete the squares in x and y and move the number stuff over to the right hand side

Kumar Shuvam

@copper.hat basically I wanna know what this curve will represent a circle, line or a point

copper.hat

do as @leslietownes says. complete the square and see what is left over.

want to

my son drove the 2h to santa cruz by himself today. a bit nervr wracking for me.

Kumar Shuvam

06:07

this is what i got (x-3)²+ (y+4)²=0

D.C. the III

@leslietownes errr......thier inner product is $0$?

@copper.hat how old is he?

copper.hat

19

D.C. the III

I see....first time doing a long trek then

copper.hat

on his own without parental reminders to slow down and leave a few mm between you & care in front etc

Kumar Shuvam

@copper.hat ah I recently turned 19....

copper.hat

06:08

:-) its been a while since i turned 19

D.C. the III

He had the fear of his parents banishing him from driving until he's 30 to keep him in line...

copper.hat

:-) we could drive by 14 or so

things were more relaxed back then

my daughter on the other hand abhors driving

Kumar Shuvam

@copper.hat so it represents a point right?

copper.hat

$(x-3)^2 + (y+4)^2 = 0$ has exactly one solution.

(in the reals)

Kumar Shuvam

Yes got it :)

copper.hat

06:12

when my daughter comes home, she takes the google bus back to work.

i wish my company had a google bus for me

leslie townes

@D.C.theIII yeah. V intersect V^perp = {0} is just one consequence of the definition of V^perp (and the hypothesis in an inner product space that the only thing with length zero is zero)

D.C. the III

perks of working for an evil conglomerate

copper.hat

do no evil

D.C. the III

@leslietownes to prove that was wonky. Was I right in thinking of "taking" one $v \in V$ and the other $v \in V^\perp$ since I'm allowed to say such since $v \in V \cap V^\perp$?

leslie townes

dc: yes

copper.hat

06:18

good night folks. i need my ugly sleep.

D.C. the III

excellent. Well you West Coast boys kept me up, it is bed time now.

copper.hat

i feel like there should be a song about west coast boys

D.C. the III

Thanks for the ayuda and I will bother you all again tomorrow with some spectral theorem questions

@copper.hat oh there is plenty..........can I interest you in a listen of a few hits from the Good Dr?

copper.hat

:-)

leslie townes

@copper.hat you mean, like, westside connection "westside slaughterhouse"?

Magnus Alexander

10:06

Hello everyone! Why $W=\{(x_1, ..., x_n) \ | \ \sum x_i=0\}$ is the same as $W = span(e_1-e_2, e_1 - e_3, ..., e_1 - e_n)$? I mean, why those things are equivalent.

Sine of the Time

@MagnusAlexander try to see what happen in $\Bbb{R}^3$

Magnus Alexander

It's weird, but for some reason it doesn't even work. I took $(0,0,1), (0, 1, 0), (1,1,1)$. So we're looking for $span((0,-1, 1), (-1,-1,0))$, right?

Sine of the Time

10:22

@MagnusAlexander I think you're looking for $\text{span}((1,-1,0),(1,0,-1))$

Magnus Alexander

How did you get those?

Sine of the Time

@MagnusAlexander $e_1-e_2$ and $e_1-e_3$ no?

Magnus Alexander

@SineoftheTime No, this gets us vectors that I've wrote, or else I'm just going nuts xd

Sine of the Time

@MagnusAlexander are these your vectors: $e_1=(1,0,0) , e_2=(0,1,0), e_3=(0,0,1)$?

Magnus Alexander

@SineoftheTime No, the third one, just for fun, I took (1,1,1)

Sine of the Time

10:30

@MagnusAlexander that's not $e_3$

the text specify $e_1-e_3$

Magnus Alexander

@SineoftheTime $e_1= (0,0,1), e_3=(1,1,1) \rightarrow e_1-e_3 = (-1, -1, 0)$

Sine of the Time

@MagnusAlexander this is correct, but it's not what the text of the problem is referring to

Magnus Alexander

@SineoftheTime It said $span(e_1-e_2, e1-e_3)$. In our case that is $span((0, -1, 1), (-1, -1, 0))$

Sine of the Time

@MagnusAlexander $e_1$ is not $(0,0,1)$ and $e_3$ is not $(1,1,1)$

Magnus Alexander

@SineoftheTime But I set them to be that way. They are linearly independent and they form a base of $\mathbb{R}^3$.

Sine of the Time

10:36

@MagnusAlexander it doesn't matter how you see them, you must follow the definition

Magnus Alexander

@SineoftheTime It seems to me that I'm following it. I took a base of $\mathbb{R}^3$ to be $e_1 = (0, 0, 1), e_2 = (0, 1, 0), e_3 = (1, 1, 1)$. Then we are taking a span with subtractions...

Sine of the Time

@MagnusAlexander the definiton of $e_i$ is $e_i=(0,\dots,0,1,0,\dots,0)$ where $1$ is in the $i$ position

Magnus Alexander

@SineoftheTime I thought I could take any base of $\mathbb{R}^3$.

Sine of the Time

@MagnusAlexander see the definition in the book you're using

Magnus Alexander

10:51

@SineoftheTime It is not specified in the book, since this should work for all vector spaces, and, I guess, all bases...

Sine of the Time

@MagnusAlexander en.wikipedia.org/wiki/Basis_(linear_algebra) look at the examples

Magnus Alexander

@SineoftheTime I see what you mean now, I thought I could take any base. Sorry for the inconvenience, and thanks for the help.

Sine of the Time

@MagnusAlexander no problem, in general look at the definition and notation, since they can change from a book to another

Sine of the Time

11:07

hi @ペガサスSeiya. how are you today?

ペガサス Seiya

@SineoftheTime Pretty tired

Sine of the Time

@ペガサスSeiya is your injury hurting?

ペガサス Seiya

@SineoftheTime not unless I move too much. Its not that bad

Sine of the Time

hope you'll recover soon

PNDas

13:32

For any set $X$, is it possible to give a metric on $X$ which would make it incomplete or complete?

This may be useful: math.stackexchange.com/a/3789234/612798

Xander Henderson

14:18

@PNDas A complete metric is easy: just put the trivial metric on the space. You can do that for any set.

For an incomplete metric, you need an infinite number of points, since you need at least one Cauchy sequence which fails to have a limit. But you can probably just pick out some countable sequence, map it to $(1/n)_{n\in\mathbb{N}}$, then do something silly like putting the trivial metric on everything that's left.

This is unlikely to be compatible with any other structure you might want on your set, but if all you are concerned with is the metric structure, something like this should work.

Mats Granvik

Does the consective ratios of higher order derivatives of a function give the analytic continuation of that function somehow?

$s = (2 + 14\,i)$
$$\lim_{n \rightarrow \infty}
\left(
\left[
1-
\left(
\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta(\tfrac{k}{n}+s)}
\Bigg/
\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta(\tfrac{k}{n}+s+\tfrac{1}{n})}
\right)
\right]^{-1}
+\frac1n + s
\right)=\text{1:st non-trivial Riemann zeta zero }$$

where $Re(s)>1$, $n>0$ and $k>0$.

and therefore there the analytic continuation in the zeta function is not needed. So does this mean that the limit somehow uses/involves analytic continuation?

PNDas

15:46

Is it bad to star someone's message? When should I star a message? I put a star on Xander's message because then it'll be easy to find it.

robjohn

16:50

@PNDas that’s a fine use for the stars. If it’s a comment that you think many people will find helpful, then a star is appropriate. Sometimes people use stars to support a comment that they really agree with.

ペガサス Seiya

It has become an upvote

robjohn

You can look back at all the starred comments and see what people have done in the past.

Peter

@PNDas I did not yet find out the logic when a star is considered as OK and when it is considered as a "like". One of many things that are unlucky handled on this site.

ペガサス Seiya

17:07

True

Sine of the Time

18:38

hi @TedShifrin

can you help me with a problem?

Ted Shifrin

What sort of problem?

Sine of the Time

computing a fundamental group :(

Ted Shifrin

What is your issue?

Shinrin-Yoku

i am in chat

Ted Shifrin

@Shinrin-Yoku Congratulations!

Shinrin-Yoku

18:42

lol thanks

Sine of the Time

We have on $\Bbb{R}$ the following base for the topology: $B=\{\emptyset, (a,a+1)_{a\in \Bbb{Z}}\}$ and we have to find $\pi_1(X,1/2)$

$X$ is $X=(\Bbb{R},U_B)$

Shinrin-Yoku

@TedShifrin. one thing that has been irritating me to no extent is the defintion of arc length

compared to the definition of a riemann integral

Ted Shifrin

Darn, @Sine. Why does your class love these idiotic topologies?

Too vague a comment, @Shin. What is irritating you?

Shinrin-Yoku

the thing is to calculate riemann integral we take upper and lower sums to "squeeze" out the area

shintuku

!!!

i am not involved here

warning to all:

there is no involvement of my part in this

i don't want trouble with anyone

Ted Shifrin

18:45

Oops. When two names have the same first four letters, you get pinged.

Shinrin-Yoku

but in arclength there is no squeezing @TedShifrin

Sine of the Time

@TedShifrin Don't ask me :'(. It was in a past exam and we never did it such example (i.e. fundamental groups of spaces with a specific basepoint)

Shinrin-Yoku

you just take rectifications

Node.JS

What is a pre-order that is not reflexive nor irreflexive?

Sine of the Time

@shintuku don't speak until your lawyer arrives

Shinrin-Yoku

18:47

which is particulary problematic for me

Ted Shifrin

@Shinrin You are still taking the sup. It is not defined as an integral. You have to prove that the length is an integral in some situations.

Shinrin-Yoku

I know taking sup, but why not take inf from some angle to squeeze out the length, you only have a lower approximation not an upper one by taking sup

Ted Shifrin

@Sine I don't find these questions interesting as actual geometric topology questions. But approach like yesterday. Understand what functions into the space are continuous.

@Shinrin If arclength bothers you, wait until you get to surface area. Then you can't approach it with piecewise-linear approximations at all.

Shinrin-Yoku

I am not talking about integral def of arclength at all rather comparing defintions of area and length

Sine of the Time

@TedShifrin me neither. thank you for the advice

Node.JS

18:50

Can someone please help with this question: math.stackexchange.com/questions/4638649/…

Shinrin-Yoku

If its so ok just to take lower limits why care about upper riemann sum at all @TedShifrin?

Ted Shifrin

The point is that the area/volume definition is working in the top dimension of the space. Lengths are not (nor surface area, etc., for lower-dimensional subspaces).

You know better than that, Shinrin. That's a silly question.

Shinrin-Yoku

Then why dont we take upper limit for arcs?

and consider the length from the upper arc length and the lower arc length...

if equal then good and arclength is defined

Ted Shifrin

So you're totally ignoring the explanation I gave you.

But go ahead and investigate this. How are you going to get upper arclengths? Tell me.

Shinrin-Yoku

its a good point for things like circles you can but i guess there are weird curves out there

Ted Shifrin

18:55

You don't need a weird curve. How do you even define a "circumscribed" polygonal approximation?

Shinrin-Yoku

yes good point, by considering tangents on convex curves maybe...

Ted Shifrin

What does tangent even mean for a curve?

Shinrin-Yoku

differentiable convex curve

Ted Shifrin

So we're no longer defining arclength of a (say) continuous curve.

Shinrin-Yoku

yes good point. very good point. I guess thats all thats there to it then, it can be done for some curves but not alll

Ted Shifrin

19:00

If you want to play with differentiable convex curves, then write down a definition and prove that you'll get the right answer. But this is surely not a reasonable approach in general.

But you should eventually learn about things like Hausdorff measures for non-top-dimensional subsets.

Shinrin-Yoku

yes thanks for the help @TedShifrin I plan to learn about fractals and hausdorff dimension in the future

anyway have a nice day all in chat. see you later

Ted Shifrin

See ya.

D.C. the III

in showing $A = A^t$, in your hint about writing $x$ and $y$ as the sum of vectors in $V$ and $V^\perp$, do you mean write $x$ and $y$ individually or $v = x + y$ with $x \in V$ and $y \in V^\perp$?

Ted Shifrin

19:18

Your question doesn't make sense. Aren't we considering $Ax\cdot y$?

D.C. the III

Yes which as itself equals $y^tAx$

Ted Shifrin

Whatever. We're trying to prove it equals what?

D.C. the III

trying to prove it equals $x \cdot Ay$

Obliv

Anyone know some good sites that you can practice your knowledge of an intro to diff eq course

Ted Shifrin

Right, DC (although you still need one more step to be done).

Obliv

19:24

ooh I didn't know khan academy had diff eq, gonna check that out

D.C. the III

So I haven't used the definition of projection yet. Let's call $Ax = p$.

so $x - p \in V^\perp$

Ted Shifrin

Which is why I told you to write $x = u+v$ with ....

D.C. the III

yea that's what I was asking initially. Maybe I phrased it wrong. So what I have written down I used $z_1, w_1 \in V$ and $z_2, w_2 \in V^\perp$ such that $ x = z_1 + z_2$ and $y = w_1 + w_2$.

Ted Shifrin

OK.

So now what are $Ax\cdot y$ and $x\cdot Ay$?

D.C. the III

writing those out now.

$Ax \cdot y = Az_1 \cdot w_1 + Az_2 \cdot w_1$ and $x \cdot Ay = z_1 \cdot Aw_1 + z_2 \cdot Aw_2$

Which reduces to $Ax \cdot y = Az_1 \cdot w_1$ and $x \cdot Ay = z_1 \cdot Aw_1$

Ted Shifrin

19:41

But you're not done "reducing."

D.C. the III

I'm rewriting the transpose of one rigt now

It is the only thing that would fly, but I don't feel fully on board with it: $z_1 \cdot Aw_1 = w_1^t A^tz_1 = A^tz_1 \cdot w_1$

I guess I feel it doesn't fly because I'm thinking of $A$ as not necessarily being a square matrix

Ted Shifrin

Well, of course it's square. But you're not thinking.

D.C. the III

I did something wrong there?

Ted Shifrin

I didn't even look at it.

D.C. the III

Well I concluded that the only way that $Ax \cdot y = x \cdot Ay$ is if $Az_1 \cdot w_1 = A^tz_1 \cdot w_1$.

Ted Shifrin

19:52

Stop and think. How are the $z_i$, $w_i$ related to $A$?

D.C. the III

Well $Az_1 = z_1$ and same for $w_1$

Ted Shifrin

Aha.

D.C. the III

oh...very nice. So using that relationship I now have equality.

so as a result $Ax \cdot y = x \cdot Ay$, now manipulating $x \cdot Ay$ I do eventually get to $A^tx \cdot y$

Ted Shifrin

But make sure you justify why you can conclude $A=A^\top$. (This was probably an exercise way back in chapter 1.)

D.C. the III

To be able to do this, the middle step was having to establish equality of the two objects

copper.hat

19:59

@leslietownes was thinking something more melodious...

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$Mathematics$ Mathematics