Mathematics - 2013-05-30 (page 1 of 2)

12:26 AM

@anon hmm I hope it wasn't me upvoting those. I have been going through and upvoting answers on the unanswered list, but I don't think I have been careless enough with my upvotes to do it to incorrect solutions

Alex Becker

1:25 AM

Anyone around who speaks portugues?

amWhy

@AlexBecker I can sort of make out the question you're referring to, but best if native speaker can translate; I can only rely on my knowledge of Spanish.

Alex Becker

@amWhy Same here.

amWhy

@Gustavo Are you around? @Charlie ?

Tobias Kildetoft

Maybe @PeterTamaroff ?

Alex Becker

@TobiasKildetoft Does he speak Portuguese? I know he speaks Spanish.

Tobias Kildetoft

1:31 AM

@AlexBecker I don't know

I actually thought Spanish and Portuguese were similar enough that a Spanish speaker would be able to read math in Portuguese.

Alex Becker

They're close enough that one can get the gist of it, but not usually close enough to translate.

Tobias Kildetoft

Ok. I thought they were more like Danish/Swedish/Norwegian

amWhy

@TobiasKildetoft Spanish and Portuguese are indeed close, but I would not trust myself to translate on behalf of another person...I can make out enough (given my knowledge of Spanish) to "read" and "understand"...but not "produce" in Portuguese.

Hi, @mixedmath How's the "new atmosphere" with the "change of guards"? I don't mean to ask anything you can't share...just hoping all is going smoothly for you'all, newcomers and "old hats" alike!

mixedmath

1:46 AM

@amWhy Thanks for the concern - actually, I'm at about that age where everyone I know is getting married

and I've been hopping all over, going to weddings

I've barely looked at MSE since the guard changed. I trust robjohn has them all in tow ;)

2

amWhy

@mixedmath I meant "old hats" in the sense of "knowing the ropes" as a mod...

mixedmath

I wonder - @AlexBecker how does the Land of Moderating feel?

as you expected?

amWhy

@mixedmath Yes, marriage of one's friends can change social dynamics...wait until you start getting one invite after another to "baby showers..." :)

Alex Becker

@mixedmath About. I've been cutting back on the moderating the last couple days b/c I had an (overdue!) paper for a humanities class.

Now I have another one...

amWhy

@AlexBecker Oh that's right...you were reading Foucault...how's that going?

mixedmath

1:50 AM

@AlexBecker You're at Chicago, right? Do you ever see Zev?

Alex Becker

@amWhy I didn't actually write anything on Foucault. But now I'm reading Gender Trouble by Judith Butler and have to write about that. Fun.

@mixedmath Yes and no. I don't know many of the grad students.

They hide.

amWhy

@AlexBecker Ahhhh...actually, I'm pretty well versed in Feminist Theory, though not super-acquainted with Judith Butler...

mixedmath

@AlexBecker Yes, we do - in little dark corners to do math where no one can peek in. We're a very secretive cult like that

2

amWhy

@mixedmath When I was a Philosophy grad, I had a fellowship, so didn't have to TA, but the TA's all had offices right next to the boiler-room of an old building, in the basement, no windows...

Alex Becker

@amWhy I got the rundown on it (my mother is a feminist philosopher). I just don't want to actually read it because it is one of the least accessible books I've ever seen. Only less accessible book I've seen is a grad math textbook on quasiconformal mappings whose author must have considered it a sin to prove a result in anything in less than maximum possible generality.

I still shiver when I see weak derivatives.

amWhy

2:18 AM

@AlexBecker Now, if I remember correctly, your wife is/was a double (math, philosophy) major...your mother is a feminist philosopher... you must be somewhat "home" in philosophy?

Alex Becker

@amWhy No, that was my mother. I'm unmarried. (and 18)

Although I won't be 18 for long.

amWhy

@AlexBecker Yes, that make sense...I thought you were young...so I must have misread the post I'm referencing, or else confusing you with someone else...because when I thought I read "my wife"...I had thought you were rather young to be married, particularly being an "academic"...

Mariano Suárez-Alvarez

3:00 AM

@AlexBecker it never lasts longer than a year

3

skullpatrol

Unless you were born on Feb 29 :D

DanZimm

4:31 AM

so

today I read on some sequences of functions and came across $$\left( \cos \left( m! \pi x \right) \right)^{2n}$$

and so $$\lim_{m \to \infty} \lim_{n \to \infty} \cos^{2n} \left( m! \pi x \right) = \left\{ \begin{array}{rl} 1 & : x \in \mathbb{Q} \\ 0 & : x \in \mathbb{R} \setminus \mathbb{Q} \end{array} \right.$$

so I was wondering if there's any /weird/ sequences of functions that have a their limit function as a piecewise function

Mariano Suárez-Alvarez

5:39 AM

what do you mean by weird?

skullpatrol

6:02 AM

@MarianoSuárez-Alvarez Could you please continue with what you were saying about $\frac{1}{0}=\infty$ ?

awllower

9:54 AM

http://i.imgur.com/HWhBwXj.jpg?1
Let me congratulate myself, for the second picture...

skullpatrol

congrats

awllower

Thanks!

Theorem

10:48 AM

Hi @robjohn

are you free ? :)

Anyone here who can tell me if any linear map between two modules can be represented in matrix form ? this definitely holds for finite dimensional vector spaces but i don't know if it in general holds for modules .

For infinite dimensional vector space , of course there is no chance

skullpatrol

11:01 AM

2

Q: Injective linear map between modules

If you have an injective linear map between two free modules of equal dimension, is the determinant of the matrix representing the map necessarily nonzero? If not is there an obvious counterexample? (Everything is over a multivariate polynomial ring over a field.) Thanks!

modules

Lord_Farin

12:38 PM

@robjohn Are you here, by any chance?

Or @Mariano, perhaps?

robjohn

yes... what's up?

Lord_Farin

I just had a quick question regarding the "edit tags" button on questions.

I hadn't noticed it before, despite looking for it. This sparked the possibility in my mind that it only appears when reaching 5k rep. Am I mistaken?

Chris's wise sister

Hi all

Lord_Farin

@Chris'swisesister Hello there.

robjohn

@Lord_Farin Only 500

Chris's wise sister

12:44 PM

@Lord_Farin how is it going there? :-)

Lord_Farin

@robjohn Yes, but "when you reach 2k rep, the retag button will disappear". I meant the "edit tags" text that appears when you float your mouse behind the tags currently present.

@Chris'swisesister Quite fine; I'm on a "use a proper title already" spree today.

And over there?

Chris's wise sister

@Lord_Farin I have a crazy question in my mind. I wonder if I might compute math.stackexchange.com/questions/389991/… by Riemann sums.

Lord_Farin

@Chris'swisesister Seems unlikely, because there is no obvious $1/n$ term or power thereof.

robjohn

@Lord_Farin as far as I know, the "edit tags" button appears at 2000. However, I cannot check that myself. I would need to ask an SE dev

Lord_Farin

@robjohn It's not terribly important, of course. But I was just wondering.

Thanks so far, though.

robjohn

12:50 PM

@Lord_Farin find someone who has just over 2000 points and ask them

Chris's wise sister

@Lord_Farin the division of the interval might also follows other rules $d_n=\{1,1/2^n,2/2^n,\cdots,2^n/2^n \}$.

Lord_Farin

@Chris'swisesister True, but you'll have to have that with $n$ the temporary upper bound for your sum (the limiting variable for the partial sums), not with the indexing variable.

robjohn

@awllower: you here?

awllower

Hi!

robjohn

@awllower good day! we have a question...

awllower

1:01 PM

Let me listen then.

robjohn

@awllower when you look at a question, and hover over the end of the tags listing, do you see the option to edit the tags?

awllower

No.

Chris's wise sister

@awllower hi

awllower

@Chris'swisesister Hello!

Chris's wise sister

@Lord_Farin keep in mind that sometimes Riemann sums are well hidden. :-)

robjohn

1:02 PM

awllower

@robjohn I see.
Sorry for mistaking the question.

And then...?

robjohn

@awllower we were just wondering whether you see the yellow "edit tags"

awllower

Oh.
Yes, I see it.

robjohn

@awllower Cool, thanks!

awllower

I guess this question comes from my edit history?

For I remember once I edited a question's tag without using this bottom.

robjohn

1:04 PM

@Lord_Farin there is your answer :-)

Lord_Farin

@Chris'swisesister I know; I was just remarking that in this case, they were hidden well enough to elude me. :)

awllower

Because I didnot know of that function at that time.

Chris's wise sister

@Lord_Farin hehe :)))))

Lord_Farin

@robjohn @awllower Thank you. :)

robjohn

@awllower No, we were just wondering whether that was visible to someone with less than 5000 rep :-)

awllower

1:05 PM

Oh, haha.

robjohn

@awllower thanks for playing guinea pig :-D

awllower

:)

NP

BTW, I have been told that NP problem has been solved: is that true?
I did not find anything related on google?

robjohn

@awllower I've not heard about it. It would probably make some splash.

awllower

I see.
Thanks for replying.
I guess I should stay watching then. :D

Lord_Farin

@awllower You could just solve it yourself. :)

awllower

1:09 PM

If I could, I probably would be too happy to type anyway...
Haha

Because I know almost nothing in this direction. :P

robjohn

@awllower As of May 2, 2013, it was still open.

awllower

Mh, very interesting article.

Chris's wise sister

@Lord_Farin btw, I also wanted to say in the past that I like your attitude since it's always lighty. :-)

Lord_Farin

@Chris'swisesister "Lighty"? What does that mean?

awllower

I surmise that I have received a mistaken source thus.

Chris's wise sister

1:13 PM

@Lord_Farin full of positivity

awllower

@Lord_Farin Like shiny?

Chris's wise sister

:-)

awllower

Light-radioactively. :)

What a word of Chris's choice, haha.

Lord_Farin

@Chris'swisesister Really? Well, thanks. I suppose it's just that I don't usually discuss society and the human stain in these realms.

Chris's wise sister

:D

robjohn

1:19 PM

@Lord_Farin: thanks for the edit. I don't know how I miss the underlined words.

Lord_Farin

@robjohn It happens. :)

@robjohn As a test case, would the questions instigating my inquiry be suitable for merging?

Hello @Tobias.

Tobias Kildetoft

@Lord_Farin Hi

awllower

@TobiasKildetoft Hello! Long time no see!

Lord_Farin

@TobiasKildetoft How are you doing today?

Tobias Kildetoft

good, thanks

writing up my notes so I have them electronically, preparing for going on vacation before returning home to Denmark

JustDanyul

1:43 PM

quick question, in English, when talking about multiple solutions to a equation. Do you say "and" or do you say "or"? I would think its or, but I seen both? whats correct?

Tobias Kildetoft

@JustDanyul depends on the precise formulation

awllower

I think the union is "or", but I prefer saying "and".
One usage is set-theoretic, while the other is the adjunction use.

JustDanyul

Tobias, so it depends on context? For example, the solutions are $x = something$ or $x = something$ , versus ... the equation is true when $x=something$ and when $x=something$

Tobias Kildetoft

@JustDanyul once you write it out with "the solutions are" I would probably also write "and" in the first one

the "or" is mostly for when you write "if the equation holds, then x = .. or x = ..."

JustDanyul

Tobias, ok. thanks :)

Chris's wise sister

2:21 PM

I wonder what is the fastest way to compute $$\lim_{n\to\infty} \sum_{k=1}^{n}\frac{1}{n+2k-1}$$

Lord_Farin

@Chris'swisesister Riemann sum?

Chris's wise sister

@Lord_Farin yeap. And now I wonder what would the most elementary way be ...

(no integration)

Lord_Farin

@Chris'swisesister I think you can make a recursion for the partial sums (with even and odd $n$ separate).

Dominic Michaelis

Hello

Chris's wise sister

Hi

Dominic Michaelis

2:27 PM

Is there a group theoretical equivalent of the cantor bernstein theorem ?

In other words let $G,H$ be groups such that there is an injective group homomorphism of $G$ to $H$, and there is an injective group homomorphism from $H$ to $G$. Can we conclude that there is a group isomorphism from $G$ to $H$ ?

Tobias Kildetoft

@DominicMichaelis what do you mean?

Lord_Farin

@DominicMichaelis "If $G$ is iso to a subgroup of $H$, and vice versa, then $G \simeq H$?"

Chris's wise sister

@Lord_Farin I think the first point to notice is that $$S= lim_{n\to\infty} \sum_{k=1}^{n}\frac{1}{n+2k-1}= \lim_{n\to\infty} \sum_{k=1}^{n}\frac{1}{n+2k}= \lim_{n\to\infty} \frac{H_{3n}-H_{n}}{2}$$

Dominic Michaelis

@Lord_Farin jupp

Tobias Kildetoft

@DominicMichaelis only for finite groups

Dominic Michaelis

2:29 PM

Do you have a counterexample for non finite ?

Tobias Kildetoft

for example, the free group of any finite rank is a subgroup of the free group of any other finite rank (ranks at least 2)

also actually up to countable rank

Lord_Farin

@Chris'swisesister Assuming $n$ is even?

Meh, -10 due to user removal.

Chris's wise sister

@Lord_Farin I don't think you need to assume n is even.

Lord_Farin

@Chris'swisesister Then you're employing nontrivial trickery in the last equality.

Dominic Michaelis

Oh I need to calculate that later. This should even work for: we have a groups $G,H$ and and injective homomorphism from $G$ to $H$ and a surjective homomorphism from $G$ to $H$, then $G$ doesn't need to be isomorphic to $H$

Tobias Kildetoft

2:34 PM

@DominicMichaelis that one I am not so sure of

since the example I gave will not work

Chris's wise sister

@Lord_Farin then, using the fact that $$\lim_{n\to\infty} \left(H_n-\log(n)\right)=\gamma$$ we're done.

Lord_Farin

@Chris'swisesister Yes, that last limit is easy peasy.

Dominic Michaelis

really? oh i am really bad with free groups

Tobias Kildetoft

@DominicMichaelis if the quotient of a free group is free, then it has at most as large a rank as the original one

(this just isn't the case for subgroups)

Chris's wise sister

@Lord_Farin here is another cute integral $$\int \frac{x^{2000}}{x^{2668}+1} \ dx$$ It was given on a high school math contest

Lord_Farin

2:37 PM

@Tobias Could you explain how $F(3)$ can arise as a subgroup of $F(2)$?

Tobias Kildetoft

@Lord_Farin take the element $xy$, $yz$ and $zx$ (I think)

Dominic Michaelis

@Lord_Farin I don't see that yet either

Tobias Kildetoft

or something along those lines

maybe easier is to note that the commutator subgroup is free (being a subgroup) and of countably infinite rank, so it certainly contains subgroups of any finite rank

Lord_Farin

@TobiasKildetoft Thanks, much clearer now.

Tobias Kildetoft

free groups are very counter intuitive, since they behave so much differently from free abelian groups

Mariano Suárez-Alvarez

2:41 PM

they are only counterintuitive if your intuition is wrong

and if you are basing your intuition about free groups upon your knowledge of abelian free groups, it most surely is wrong :-)

Tobias Kildetoft

@MarianoSuárez-Alvarez true

Mariano Suárez-Alvarez

intuition is a great thing. It can like one thing one day and the next day the exact opposite, without any qualms

Sometimes I've believed as many as six impossible things before breakfast, as she said.

Dominic Michaelis

@Lord_Farin -112 points because a user was removed ...

Lord_Farin

@DominicMichaelis X(

jerrysciencemath

@Lord_Farin A topological argument can explain why a free group of rank 2 can contain a free group of any rank as a subgroup.

Dominic Michaelis

2:48 PM

It happened several times that i have lost 10 or 5 points but never that much O_o

@jerrysciencemath Tell me :)

jerrysciencemath

Look at the covering space of $S^1\wee S^1$

Lord_Farin

\wedge

jerrysciencemath

Sorry it should be \vee

one point union of two circles

Lord_Farin

@jerrysciencemath You can edit it. :)

jerrysciencemath

The universal covering space of it is a tree with every vertex has degree 4

Lord_Farin

2:52 PM

@jerrysciencemath Yes, I know that.

jerrysciencemath

Given n circles, you can glue the i-th circle with the (i+1)-th circle at a point

the result is a chain of circles

This space is a covering space of $S^1\wee S^1$!

For n=3 and 4, the examples occur in Page 58 of Hatcher's algebraic topology.

Lord_Farin

I'm having trouble identifying the covering map.

Mariano Suárez-Alvarez

weeee

Lord_Farin

Since the middle circles will have more than one link to another circle.

jerrysciencemath

A way to find the covering map is to look at the graph of this chain of circles

Then every vertex has degree 4 too

You put a label of A or B on the edges

Give all the edges orientation

Such that at every vertex, there are exactly 2 edges labeled A, 2 edges labeled B

And for a same label, the two edges have different orientations

Lord_Farin

3:01 PM

@jerrysciencemath Ah I see now; the middle circles are split into two arcs, each of which surject onto a full $S^1$. Thanks, very nice!

jerrysciencemath

@Lord_Farin Exactly what I mean~

Lord_Farin

I'll have to go now, see you around!

jerrysciencemath

Bye bye!

Dominic Michaelis

@Lord_Farin Bye

Charlie

3:25 PM

@DominicMichaelis hhi dominichen, wie gehts?

Dominic Michaelis

@Charlie mpfh not really better

Charlie

@DominicMichaelis ooh...

Chris's wise sister

@DominicMichaelis when I'm in a bad mood I create things (just to forget things that upset me)

Charlie

@Chris'swisesister hi

Chris's wise sister

@Charlie Hello

Charlie

3:31 PM

@Chris'swisesister wassup?

Chris's wise sister

@Charlie creating here some integrals that contain nested integrals. How about you? :-)

Charlie

@Chris'swisesister not much

Chris's wise sister

@Charlie and I'm a bit worried as regards one of my former teachers.

Every time I send him a problem that puts him in trouble he doesn't talk to me a few days. The good part is that after a while things go back to normal.

Julian Kuelshammer

3:49 PM

hi

Charlie

@Chris'swisesister hehege once I annoyed so much a fermilab doctor...he stopped replying me xD

Chris's wise sister

@JulianKuelshammer HI

@Charlie hehe. It happens sometimes. :-)

Charlie

@Chris'swisesister ;)

Charlie

4:16 PM

How are you @ian ?

Ian Mateus

@Charlie quite fine! And you? What have you been doing?

Charlie

@IanMateus I'm fine, doing the usual :)

Ian Mateus

@Charlie you do mathematics, huh? Which semester? Is the course going well?

Charlie

@IanMateus fifth semester, everything going right and you?

Ian Mateus

@Charlie under some stress :S I haven't been sleeping well since two weeks ago

Chris's wise sister

4:26 PM

55 points are lost (removed user)

Julian Kuelshammer

I was lucky, lost only 10 points

Chris's wise sister

By the way, I wonder if it's a known user ...

Dominic Michaelis

@Chris'swisesister 112

Chris's wise sister

@DominicMichaelis oh, sorry

Charlie

@IanMateus oh no!

Chris's wise sister

4:33 PM

1

Q: Losing votes despite being over the 200 rep limit

I was a little surprised to have found that I lost 132 rep due to some user that was deleted (!). What I do not understand, however, is that this has happened on a day that I am (so far) 17 votes beyond the 20 of the daily limit, and yet I still lost the rep. How does this work?

support reputation deletion deleted-accounts

Ian Mateus

@Charlie what are you studying now?

Παρθ Κοχλι

I lost 5 points

WiseStrawberry

Also stressing for exams :(

Charlie

@IanMateus number theory, set theory, topology

@WiseStrawberry :(

WiseStrawberry

4:54 PM

well its in 3 weeks time. But I NEED to get them. Its mostly about graph theory, paramater estimation (stats), and linear programming.

@Charlie :(

Charlie

@WiseStrawberry I hate exams :-/ relax, Wisey, it's almost over

WiseStrawberry

argh. I love my course though.

However I learn everything with every exercise I make, but asking help on everything on stack feels... wrong.

Charlie

@WiseStrawberry yes, yes, I have no doubt :)

WiseStrawberry

I love the insight everybody has.

Charlie

@WiseStrawberry yes, it's nice

@pourjour hi

pourjour

5:01 PM

is true that $\dfrac{1}{2}mg^2=a$ as a is the acceleration and m is the mass and g is the constant od gravity field

@Charlie hi

@Charlie any news

Charlie

@pourjour no news, today is holiday

pourjour

@Charlie good

Παρθ Κοχλι

I wonder why people rarely upvote on Super User and Stack Overflow.

Charlie

Hmm

Charlie

5:35 PM

@wisestrawberry whats the origin of your name?

robjohn

5:57 PM

@Chris'swisesister That should be $\frac12\log(3)$ if I am not mistaken.

Chris's wise sister

@robjohn that's correct.

robjohn

@Chris'swisesister my scrollback was on that problem, so I haven't looked to see if anyone else answered. Sorry if that was a duplicate.

@Chris'swisesister was that supposed to be a definite or indefinite integral?

Chris's wise sister

@robjohn At that point above I was wondering what was the most elementary way to compute it.

@robjohn it's an indefinite integral

I'll be off for a few minutes. There is a storm here.

J. M.

"BRB, my house is being blown..."

robjohn

@J.M. do you need a condominium?

J. M.

6:06 PM

@robjohn Nah, I don't live in one. Thankfully, the place I'm in has stood up to not a few storms.

Joe Stavitsky

howdy folks

how can I calculate h here en.wikipedia.org/wiki/Circular_segment given only s and c?

J. M.

@JoeStavitsky You can compute $R$ from $s$, no?

Joe Stavitsky

@J.M. how, please?

J. M.

@JoeStavitsky Did you look at the second formula in that page?

Joe Stavitsky

@J.M. o duh ty

WiseStrawberry

6:16 PM

@Charlie, well I am a wise Strawberry of course.

robjohn

@J.M.: That looks like a Borg-Hilbert Cube

J. M.

@robjohn You could say that, yes. :D This was even more difficult to render than the previous iteration.

(Damn exponential growth...)

Joe Stavitsky

@J.M.: but, wait... what if I dont have s either?

robjohn

@J.M. I need to write up my algorithm and proof of my n-dimensional hilbert curve generator. It gives $\Lambda^{1/n}$ smooth curves (Hölder-1/n continuity)

J. M.

@robjohn Ooh, cool! I'll wait for it...

@JoeStavitsky Then your problem's underdetermined.

Joe Stavitsky

6:24 PM

so from s anc c alone it is imposible to get R or H?

robjohn

@JoeStavitsky you should be able to get everything from s and c

$\frac{c}{s}=\frac{\sin(\theta/2)}{\theta/2}=\mathrm{sinc}(\theta/2)$

Joe Stavitsky

@robjohn: what's that last one?

and how does that help pls?

robjohn

Sinc function

In mathematics and engineering, the sinc function, denoted by sinc(x), has two slightly different definitions. In mathematics, the historical unnormalized sinc function is defined by :\mathrm{sinc}(x) = \frac{\sin(x)}{x}.\,\! In digital signal processing and information theory, the normalized sinc function is commonly defined by :\mathrm{sinc}(x) = \frac{\sin(\pi x)}{\pi x}.\,\! The normalization causes the definite integral of the function over the real numbers to equal 1 (whereas the same integral of the unnormalized sinc function has a value of π). As a further useful property, a...

J. M.

@JoeStavitsky Now you are confusing me. First you say you have $s$, then you say you don't. What gives?

Joe Stavitsky

@J.M., sorry, typo

robjohn

6:35 PM

@JoeStavitsky The sinc function is 1-1 on $[0,\pi]$, so you can get $\theta/2$ from $c/s$

Joe Stavitsky

@robjohn: OK, that works, ty

robjohn

@JoeStavitsky It is just not a function whose inverse is on your run-of-the-mill calculator.

Joe Stavitsky

@robjohn: I'm in a timecrunch, so a little stupid atm :)

J. M.

...so numerics are needed. You know Newton-Raphson?

Joe Stavitsky

@J.M.: for me? Are you sayin robjohn's method is wrong?

robjohn

6:39 PM

@J.M. that is how I would solve it

J. M.

@JoeStavitsky No, I was continuing with what rob's saying.

Joe Stavitsky

J.M.: No, I'm afraid I don't know that one. Another option I was given was plug in r from the first into the second and solve for h.

robjohn

6:52 PM

@JoeStavitsky what class is this for?

Joe Stavitsky

@robjohn, this is a practical problem of estimating beam deflection from change of length of beam

robjohn

@JoeStavitsky so the angles are very small?

Joe Stavitsky

Hookes law (physics) gives change in length of steel beam from pressure, but safety rating compares vertical deflection to overall length

@robjohn, on the contrary, nearly 180

robjohn

180° of deflection?

bending back on itself? Hooke's law is an approximation that will probably be of no use there

Joe Stavitsky

no, the deflection is from the horiontal, thus s and c aer hopefully very close

the closer theta is to 180 the better

@robjohn i was thinking, I could choose any d ( and therefore r)

robjohn

7:05 PM

@JoeStavitsky when $\theta$ in the diagram is near $180^\circ$, $c$ is near $2R$ and $s$ is near $\pi R$

Joe Stavitsky

@robjohn, yes youre right, sorry. But like I said, an arbitrary selection of r would give accurate h,no? the trig ratio would still hold

robjohn

@JoeStavitsky in the digram, yes. However, extreme bending may not follow Hooke's Law

Joe Stavitsky

@robjohn: safety limit is h is 1/260 c. I dont think thats extreme :)

Peter Tamaroff

7:56 PM

@CBenni You haven't changed your answer yet...

CBenni

@PeterTamaroff in what way should I edit it? Not use the word big-o notation?

Peter Tamaroff

@CBenni You have addressed something that has nothing to do with what the OP is asking.

CBenni

I did

in the last line

Peter Tamaroff

You're talking about little-$o$ and $\lim_{x\to 0}$.

CBenni

that should basically answer the OPs question

the $\lim_{x\to 0}$ is indeed fail

Peter Tamaroff

7:59 PM

@CBenni No, again, you're talking about little $o$ and the OP has big $O$.

CBenni

that wasnt the question back when I answered

the OP was with little o

some editor changed that

I dont agree with that, the question is legitimate for little o aswell

Peter Tamaroff

@CBenni I you look into the linked webpage, you'll see it has only big O.

CBenni

well whatever

i edited it now

Peter Tamaroff

@CBenni But what you wrote is wrong...

$f(x)=\mathcal O(g(x))$ means that there exists $C$ and $x_0$ such that $|f(x)|\leq Cg(x)$ when $x\geq x_0$.

CBenni

My answer was correct in the beginning

after the question changed, my answer was invalidated.

now its basically the same as everyone elses

I cannot delete it sadly

I never thought it would earn so much hate

Peter Tamaroff

8:09 PM

@CBenni Hate? Nobody hates anything here.

I was preoccupied because the OP accepted a wrong answer, that is all.

CBenni

well it was correct back when I wrote it

still got downvoted

Chris's wise sister

8:41 PM

For instance, the guy here math.stackexchange.com/questions/402141/… is not a genius but a super genius (I know him, he doesn't know me) and if you look at the points he got so far one would say is a poor student. I think one shouldn't care too much about these points. I've never needed some people around me to tell me I'm good or not at doing some things.

Here: math.stackexchange.com/questions/402141/…

By the way, his last question seems interesting.

Peter Tamaroff

Patern recognition is so popular...

Chris's wise sister

Yes, it is.

J. M.

@PeterTamaroff Certainly, the OP has every right to be arbitrary and capricious with his acceptance, in the same way you have the right to be arbitrary and capricious with your downvotes. ;)

Peter Tamaroff

@J.M. Hey! I am not arbitrary and capricious with my downvotes!

=D

J. M.

@PeterTamaroff Good to know. ;) I'm just saying you could choose to be.

Peter Tamaroff

8:50 PM

@J.M. Muahahahaha.

J. M.

@PeterTamaroff Remember: rationalizing vote counts is a sure path to insanity...

Peter Tamaroff

@J.M. Yep. Instant madness.

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$Mathematics$ Mathematics