Mathematics - 2017-05-02 (page 2 of 7)

My bookmarks stopped working too

@blue It is working or it is not working?

Anonymous

It's working for me

Anonymous

9:30 AM

Strange

Astyx

What browser are you using ?

Well, the original host was announced to stop working April 30th

Are you doing anything special?

Anonymous

Chrome

I'm guessing the bookmarks still use the original host or so

And maybe blue still has it cached, whereas the rest of us doesn't

Anonymous

9:30 AM

I'm using the bookmark

When did you last restart your browser?

Anonymous

1 hour back

strange

Astyx

I had the problem 2h back

It seems you have the seed which will be used to bring mathjax back into the world, preserve it!

9:31 AM

Like I said, he probably still has the relevant scripts cached

Anonymous

This (math.ucla.edu/~robjohn/math/mathjax.html) isn't working?

not if he restarted one hour ago

Anonymous

:O

@s.harp Err, why should that imply this? $\partial F = 2(A + B)$ means $A + B$ has order 2; so your thing is $\langle A + B \rangle/\langle 2(A + B) \rangle$ which is $\Bbb Z/2\Bbb Z$

It's working for me too

Anonymous

Yep, I can see Balarka's mathjax

9:32 AM

@s.harp Browser restart doesn't necessarily clean all of the cache

I checked the source code of the bookmark, and it does refer to cdn.mathjax.org a few times, which is the host that's shut down...

Anonymous

Lol, I won't restart my browser then :P

@Balarka the element $[A]$ does not have order $2$ though? Because I mean $[nA]$ is never in the image the derivative

So shouldn't $H_1$ have an element of order infinity?

What does [A] mean? A is not a cycle.

holy shit

yes

:)

9:34 AM

ok

or wait

if you call the point $p$ isnt $\partial A = p-p$?

The complex being two lines $A, B$ based on $p$ and the area $F$ being layed out as $ABAB$?

If you look at the identifications carefully you'll see that the endpoints of the upper $A$ edge are not identified.

(So boundary is actually nontrivial)

yep

thanks, I think I lost half my brain mass this last month^

Okay, I managed to fix my MathJax

You have to (manually) update to version 2.7.1 in the bookmark

Nah, you're good. The picture I like more than that identification story is a disk attached to the circle by a degree 2 map.

This agrees with the identifications if you think of the circle as union of two hemicircles.

$A$, $B$ are precisely the two hemicircles.

@Astyx @s.harp Replace "2.7.0" by "2.7.1" inside the url in the MathJax bookmark. It should work then

3

9:41 AM

Yes, I know that one also, but the square fits the non-oriented surfaces into this whole canonical "oriented closed $2$ manifolds have coverings $\Bbb R^2$ or the poincare disc" story

Well, RP^2 is covered by neither of these.

yeah, but you use the same pictures

namely an n-gon with some boundaries identified

@Steamy I don't see a version number in the bookmark

The bookmark is just a long string inside "Location:"

Somewhere in that string, there's a version number

I found it, I was looking into the wrong bookmark D:

javascript:(function(){if(window.MathJax===undefined){var%20script%20=%20documen‌t.createElement("script");script.type%20=%20"text/javascript";script.src%20=%20"h‌ttps://cdnjs.cloudflare.com/ajax/libs/mathjax/ 2.7.1 /MathJax.js?config=TeX-AMS_HTML";var%20...

9:45 AM

and now it works^nice

Starring that message to help others out

Better ping @robjohn about it too, I guess.

Anonymous

The link on my mathjax url is "https://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS_HTML" rather than "h‌ttps://cdnjs.cloudflare.com/ajax/libs/mathjax/ 2.7.1 /MathJax.js?config=TeX-AMS_HTML"

Anonymous

That explains why my one was/is still working

Oh hey, mine too

9:56 AM

Why are these your links?

Yours will work for a bit longer

"cdn.mathjax.org" is the domain being phased out

"cdnjs.cloudflare.com" is the new one

yikes

Anonymous

@s.harp Dunno, got it from the Physics SE chatroom

But there's a period where the old one will redirect to the new one

Which is why your bookmark still works (for now)

Anonymous

So it might stop working anytime? :P

9:58 AM

I guess, yes

Anonymous

@SteamyRoot Oh I see

Ok, fixed

Anonymous

javascript:(function(){if(window.MathJax===undefined){var%20script%20=%20documen‌t.createElement("script");script.type%20=%20"text/javascript";script.src%20=%20"h‌ttps://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.1/MathJax.js?config=TeX-

Anonymous

AMS_HTML";var%20config%20=%20%27MathJax.Hub.Config({%27%20+%20%27extensions:%20[‌"tex2jax.js"],%27%20+%20%27tex2jax:%20{%20inlineMath:%20[["$","$"],["\\\\\\\\\\$","\\\\\\\\\\$"]],%‌20displayMath:%20[["$$","$$"],["\\\[","\\\]"]],%20processEscapes:%20true%20},%27%20+%‌20%27jax:%20["input/TeX","output/HTML-CSS"]%27%20+%20%27});%27%20+%20%27MathJax.H‌ub.Startup.onload();%27;if%20(window.opera)%20

Anonymous

{script.innerHTML%20=%20config}%20else%20{script.text%20=%20config}%20document.g‌etElementsByTagName("head")[0].appendChild(script);(doChatJax=function(){window.s‌etTimeout(doChatJax,1000);MathJax.Hub.Queue(["Typeset",MathJax.Hub]);})();}else{M‌athJax.Hub.Queue(["Typeset",MathJax.Hub]);}})();

Anonymous

10:05 AM

Okay, so that's the new one

Anonymous

Phew =P

Anonymous

Done

10:26 AM

Just curious, how is a minimal polynomial defined for an element of a field?

Suppose $\alpha$ is an element of some extension of $\Bbb F$ which is algebraic over $\Bbb F$, consider the morphism $v_\alpha:\Bbb F[x]\to\Bbb F[\alpha]$ that maps $\sum a_ix^i\mapsto\sum a_i\alpha^i$, the minimal polynomial of $\alpha$ is the monic generator of the kernel of $v_\alpha$ as an ideal of $\Bbb F[x]$

(The last bit works because $\Bbb K[x]$ is a PID for $\Bbb K$ a field)

Martin Sleziak

I have not update my bookmarklet yet - I still have https://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS_HTML there. Strangely enough, it still seems to work.

@blue Your code is messing up my ChatJax, it won't start correctly

Anonymous

@LegionMammal978 Just modify your url. Change 2.7.0 to 2.7.1

@blue No, as in, the very code on this page is causing it to fail

Your dollar signs

Anonymous

10:37 AM

@LegionMammal978 No idea. I got the code from robjohn's webpage

Anonymous

It's working for me

...

You shouldn't have posted it here (and outside a <pre> block, nonetheless)

Anonymous

What's a <pre> block?

Little Rookie

Hello all, i need help on question on numerical analysis

https://math.stackexchange.com/questions/2261786/derive-the-approximation-for-integration-using-hermite-interpolating-polynomial

 a monospaced code block
 like this

Anonymous

10:39 AM

@LegionMammal978 How to do so?

@blue Indent it four times:

"    a monospaced code block"
"    like this"

Anonymous

javascript:(function(){if(window.MathJax===undefined)

Anonymous

Like this ^

Anonymous

Okay

Oh, and you can put them in one multiline message that way

Well, that explains my problem

10:48 AM

$\log\lim\limits_{x\to 0} f(x)=\lim\limits_{x\to 0}\log f(x)$ ?

@AlessandroCodenotti Okay, lemme try and interpret this, trying to apply Blömer's denesting algorithm here :p

Hi chat

Anonymous

@Fawad The limit of f(x) should be positive for that to hold true.

@LegionMammal978 I don't know what that is but if you have doubts just ask

(There are other ways to define the minimal polynomials, that's the one I'm most familiar with)

@AlessandroCodenotti Okay, so I can see how $v_\alpha$ is defined, but I don't really understand the second part

10:53 AM

Well $f$ has to be a positive function in a neighborhood of $0$, not just the limit.

K

In which case, this is just continuity of $\log$ at $x = 0$ on it's domain of definition

@LegionMammal978 the kernel of that morphism is the set of polynomials with $\alpha$ as a root

And it is also an ideal of $\Bbb F[x]$ (since it's the kernel of a ring morphism)

@AlessandroCodenotti Okay, so how would I find a monic generator from this set?

There are different ways depending on your $\alpha$, we're only interested in its exsistence here

11:01 AM

Hmm

I've gone on a bit of a crash course in this stuff, so I don't know what I'm doing very well :p

Anonymous

@BalarkaSen Uh right, I should have said that

Anonymous

@Fawad Replying with just a K when a couple of people have tried to help you is a bit rude imo.

@AlessandroCodenotti Okay, for an example, how would this work for, say, $\Bbb F=\Bbb Q,\alpha=\sqrt3$?

@blue it's K

Anonymous

@BalarkaSen lol :P

11:05 AM

@blue my apologies

Anonymous

"hmm" and "k" are the two words which make me instantly disinterested in conversation :P (oh well, I know you guys are gonna bug me by repeatedly using them now =P)

@LegionMammal978 you have theorems that help you. If the extension is a field and you can find a monic, irreducible polynomial with $\alpha$ as a root then it's also minimal

Why don't we say $\lim\limits_{x\to0} \sin x=\lim\limits_{x\to0} \tan x =x$ ? It every time correct and makes problems easier

Anonymous

@Fawad That's wrong...The value of the limit you get is 0.

@Fawad You mean $\lim\limits_{x\rightarrow0}\sin x-x=\lim\limits_{x\rightarrow0}\tan x-x=0$?

Anonymous

11:10 AM

Obviously you can take sin(x) approximately equal to x

Anonymous

In the close neighborhood

Anonymous

Similar for tan(x)

Q: Evaluating $\displaystyle\lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)$

14

How to solve this limit $$ \displaystyle\lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right) $$ without using L'hospital's rule?

Anonymous

@Fawad What?

In this problem by taken sin and tan equal will give limit 0

Anonymous

11:12 AM

You need to use your brain. Not everywhere :P

Anonymous

That's an approximation which helps in physics when the functions appear like sin(3x)/sin(5x)

Anonymous

Basically you are taking the first term of the mclaurian series

@blue don't use those words which I don't know,please.

Anonymous

But sometimes the first term isn't enough

Anonymous

@Fawad How will I know what you know and what you don't?

Anonymous

11:14 AM

Oh btw the spelling is McLaurin I think

@blue I usually use hmm as a reply to indicate I am thinking about what I just replied to

@blue NCERT syllabus only

Anonymous

@Fawad I don't read NCERT

Anonymous

Never read

Also it's Maclaurin

Anonymous

11:16 AM

@BalarkaSen I meant it in a different context. When people use hmm as a reply when I have typed over 5 lines to explain them something and they don't even tell a thanks

heh

Anonymous

McLaurin reminds me of.....

Anonymous

Oh! I miss the burgers so much :P

Anonymous

I should go out and have one

where do you live

(are you from WB?)

Anonymous

11:21 AM

Guwahati

ah assam

Anonymous

You live in Calcutta I suppose?

Anonymous

(Kolkata)

Right. I suppose you have a tiny bit idea about what's a popular snack in Bengal then

i hate burger/pizzas or other junk foods. moori's are fine

Anonymous

I suppose there are many popular snacks in Bengal :P

11:23 AM

i guess it's not the trend nowadays

Anonymous

Oh even I like moori with chanachur :)

Anonymous

But plain moori is too much for me

Isn't it mur muri?

ofc with chanachur. otherwise it tastes like garbage

Anonymous

hehe ^

Anonymous

11:24 AM

yeah

Anonymous

wth is mur muri =P

in The h Bar, 22 mins ago, by Slereah

For branched manifolds $K$ there's a collection of subsets $\{ U_i \}$ such that $\bigcup \text{Int}(U_i) = K$

read from there

Little Rookie

12:06 PM

Need help here math.stackexchange.com/questions/2261826/… =/

1:05 PM

How to do this question:
tan 38 - cot 22
Evaluate,
in terms of cosec and sec
Can someone tell me the identities to be used?
I will do it myself then.

write in terms of sin and cos

sin 38/cos 38 - cos 22/sin 22

is it solved now @Abcd

1:19 PM

@BAYMAX I had already done that :(

I am stuck at this point:

ok

that will lead to

sin 38sin22 - cos 38 cos 22/cos 38 sin 22

yes

What do I have to do next?

??

now there is an identity

1:21 PM

WHICH??

The numerator becomes -cos(38+22)

by using

cos(A+B) = cos A cos B -Sin A sin B

Oh! Tysm !

List of trigonometric identities

In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables where both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles. They are distinct from triangle identities, which are identities potentially involving angles but also involving side lengths or other lengths of a triangle. These identities are useful whenever expressions involving trigonometric functions need to be simplified. An important application is the integration of non...

Angle sum and difference identities

np

2:10 PM

What is the difference between -+ and +- (I mean their symbols with - above + and + over - respectively)

I mean the difference between: ∓ and ±

Is there any difference or are they same?

no

there is a difference

arctic tern

If both are used in the same expression, it's likely implied one is the opposite of the other (whichever sign is chosen for the one).

Leaky Nun

For example $a^3 \pm b^3 = (a \pm b) (a^2 \mp ab + b^2)$

which encodes two separate identities:

1. $a^3 + b^3 = (a + b) (a^2 - ab + b^2)$

2. $a^3 - b^3 = (a - b) (a^2 + ab + b^2)$

$5 \mp 3 = |3 \mp 5|$

Leaky Nun

per what @arctictern says

@BAYMAX what?

2:14 PM

@LeakyNun Understood. Thanks :)

Leaky Nun

@Abcd no problem

Like 5 - 3 = 2 in Lhs

arctic tern

2+2=4

and also |3-5| = 2 in Rhs

ok bye

[Random] Consider a finite bag containing infinite number of gas molecules such that the concentration of gas in the bag is given by $\Bbb{Z}/n$ where $n\in \Bbb{Z}$. Initially, the concentration of gas in the bag is $\Bbb{Z}$. Now make a hole to leak out the gas, then as the gas fill in in the surrounding space, its concentration will go down indefinitely as $n$ increase without bound, but always less than $\aleph_0$

So, a world where physical infinity can exists has cool things like concentration that is independent of the amount of stuff

Q: Recovering the Lie derivative of a covector from Cartan's formula on 1-forms

2:27 PM

This looks like a nice question for one of our differential-geometry people:

2

The coordinate expression of the Lie derivative of a covector field, $w$ with respect to a vector field $X$ is given by: $$ \mathcal{L}_{X}w = \left(X^{\alpha}\frac{\partial w_{\mu}}{\partial x^{\alpha}} + w_{\alpha}\frac{\partial X^{\alpha}}{\partial x^{\mu}}\right)\mathrm{d}x^{\mu} $$ But sin...

differential-geometry exterior-algebra

2:49 PM

How to do this question:

Mike Miller

I am so tired

Sleep? Or drink (even more) coffee?

@Abcd Suppose $s_n$ is the sum of $n$ terms in an arithmetic progression. How does $s_n$ differ from $s_{n+1}$ and from $s_{n-1}$?

@Semiclassical by the difference between them ?

Right.

2:52 PM

So?

More precisely, there should be some common value such that $s_{n+1}-s_n=s_{n}-s_{n-1}$.

yes.

The key point is that that's not possibly true for one of those answers.

Mike Miller

@SteamyRoot I will do both

I did it using this vague method right now: Formula for sum of AP ->
sum = n(a1 + an)/2

Thus,

second term must have "n"

SInce option D doesnt have "n" as second term, therefore it cannot be the sum

Did I use the right method?

2:54 PM

Yeah, that also works.

You can also note that the first three answers only differ by a constant multiple.

I can't understand that :/

n^2+n differs from 2n^2+2n by a factor of 2.

And 3/2*n^2+3/2*n differs from n^2+n by a factor of 3/2.

Right..

Point is, suppose that n^2+n is the sum of some arithmetic progression.

ohkay

2:57 PM

Then I could double all the terms of that progression. That would still be an arithmetic progression, but its value would be doubled---i.e. 2n^2+2n.

Oh.

And similarly for 3/2*n^2+3/2*n.

Understood.

Thanks.

The reverse is also true: If one of those weren't the sum of an arithmetic progression, the others couldn't be eiither.

So if A were false than so would B,C. But that's absurd, since there's only one right answer.

So it has to be D.

robjohn

@LegionMammal978 which stuff on the page was causing problems? I didn't see any problem in the transcript.

3:10 PM

@abcd Actually, taking another look at that problem, I find myself annoyed. The only way for one of those is to be right is if you consider only arithmetic progressions in the mold of 1,2,3,4,...

Ignore that. I am being nonsensical. Your method is right and the answer is right.

Daminark

Hey

robjohn

@LegionMammal978 If it was the ChatJax code pasted on the page, it would only mess that code up. You should just go to the ChatJax installation page.

Sorry typo: $\Bbb{Z}/n$ should be $n\Bbb{Z}$.
From that we observe the following interesting property: The bijective map $x\mapsto nx$ for integers $n>1$ on $\Bbb{Z}\to \Bbb{Z}$ actually maps $\Bbb{Z}$ to a proper subset of $\Bbb{Z}$ with the same cardinality

Rosie F

@LegionMammal978 or $$\lim_{x\to 0}\dfrac{\sin x}x=\lim_{x\to 0}\dfrac{\tan x}x=1$$?

So very loosely speaking, unlike in finite sets, where the map defined by multiplication by an integer on all elements in the set generally change it into a disjoint set of the same cardinality, the same map can end up mapping an infinite set to a proper subset of itself with the same cardinality

e.g. 1,2,3,4,5... under the map y=2x gives 2,4,6,8,10... which is a proper subset of the former

3:17 PM

oh now i'm grumpy. a question i'd given an answer to recently appears to have been deleted/withdrawn.

Actually no, typo, consider the following finite set 1,2,3,4, after x 2, gives 2,4,6,8, neither sets are subsets of each other, but they have nontrivial intersection 2,4

So... in the infinite world, things can get "smaller" by doubling it

depends how you define "smaller"

Given two sets of the same cardinality, if one set is a proper subset of the other, then the set that is being contained is "smaller" naively speaking

I am not sure if there is a formal definition other than saying one is a proper subset of the other

That's formal enough

But that definition of "smaller" isn't particularly interesting IMO

So basically, once we entered the infinite world, there are two different ways to talk about the "size": of things, one is its cardinality, and the other is $\subset$. In the finite world, the two coincide

3:25 PM

Intuitively, you find that $2\mathbb{Z}$ is "smaller" than $\mathbb{Z}$ because it "only has half the number of elements"

Exactly

Similarly you'd claim that $2\mathbb{Z}+1$ is also half of $\mathbb{Z}$

But this definition does not allow any way of comparing $2\mathbb{Z}$ and $2\mathbb{Z}+1$

yup, because their intersection is empty, hence not contained within each other. That scenario also happens in finite sets

@Semiclassical Okay. Thanks.