Mathematics - 2018-04-26 (page 1 of 3)

famesyasd

00:23

Any k-regular graph exists by the symmetry, right? unless it is an odd-verticle graph and k is also an odd number which is ruled out by the hand-shaking lemma or something

PVAL-inactive

02:06

The strongest algebraist I've ever had as a teacher en.wikipedia.org/wiki/Kiran_Kedlaya said you just need A&M for Comm. Alg. Pretty sure you just need AM for the question linked at least.

MultivaluedPersonality

Your algebra teacher was was a contestant on jeopardy!? Who is this guy? xD

Secret

sciencedaily.com/releases/2018/04/180423155051.htm

Connection between fluid dynamics and hash functions

Nicholas Roberts

02:37

Yo, given an arbitrary cubic polynomial: $a_3x^3 + a_2x^2 + a_1x + a_0$, how does one convert it to the standard form $x^3 + px + q$?

GFauxPas

i think it's called the Tschirnhauss transformation?

Nicholas Roberts

coefficients in some field

GFauxPas

proofwiki.org/wiki/Tschirnhaus_Transformation here you go

actually proud of myself for how close I was on the spelling :)

PVAL-inactive

x^3+bx^2 is the first two terms of $(x+b^{1/3})^3$.

The rest should follow pretty easily from that.

Nicholas Roberts

So you make a change of variables

Fawad

02:43

I tried writing few more term of integral,is it correcc?

GFauxPas

I dont think you need to make a series like that, you can just iterate once, no?

Fawad

@GFauxPas I am preparing for competitive exams,I think I can save time by making series..., anyhow is that correct for those terms I wrote?

GFauxPas

I have a hard time checking it in my head and i don't feel like bringing out paper, sorry

:(

Fawad

It’s ok 👌

Silent

03:11

Let $\Bbb Q[α,β]$ denote the smallest subring of $\Bbb C$ containing the rational
numbers $\Bbb Q$ and the elements $α =
√ 2$ and $β = √ 3$. Let $γ = α+β$. So, $\Bbb Q[α,β] = \Bbb Q[γ]$, while
Is $\Bbb Z[α,β] \ne \Bbb Z[γ]$. Does this have something to do with fact that while $\Bbb Q[\alpha]$ and $\Bbb Q[\beta]$ are fields, but $\Bbb Z[\alpha]$ and $\Bbb Z[\beta]$ are not.

feynhat

03:34

Hello.

Nicholas Roberts

Yeah, you need to use division to get equality of Q(a,b) and Q(gamma)

In Z, one cannot do this

@Silent

Silent

ok, thank u!

feynhat

@Fawad You need to use do the integration by parts just once. $I_n = \int x^ne^{ax}dx = {x^n}\int e^{ax}dx - \int\big({dx^n\over dx}\big)\big(\int e^{ax}dx\big)dx = {{x^n}e^{ax}\over{a}} - {{n}\over{a}}\int x^{n-1}e^{ax}dx = {{x^n}e^{ax}\over{a}} - {n\over a}I_{n-1}$

feynhat

03:58

@Fawad Were you trying to find a recurrence for the integral or were you trying write the first few terms of the integral explicitly? If its the latter, ignore my post. And yes, your expansion looks correct.

Nicholas Roberts

Does a curve in R^2 have an empty boundary

i suspect it does

Daminark

@Nicholas boundary in the topological space sense? Or boundary of a manifold?

Nicholas Roberts

I think manifold, im integrating a 1 form over a curve gamma and trying to use stokes theorem

Daminark

Yeah not quite, the endpoints are the boundary

Nicholas Roberts

and the one form is exact, so i can move the "d" down to gamma

Daminark

04:06

(So this is fundamental theorem of calc)

Hmm, what's the integral exactly?

Nicholas Roberts

Hold on a moment

math.stonybrook.edu/~claude/531spring12/531mt12sol.pdf

question 2 on this PDF if you dont mind checking it out. He shows the integral vanishes at the end of the solution but i dont understand

Daminark

No that's not what he's saying here, he's proving the form isn't exact

The idea is this

Let's say $\omega$ is some exact form, so $\omega = df$

Nicholas Roberts

I know, he supposes that it is exact and gets a contra

Ok im listening

Daminark

This would imply that for any closed curve $\gamma$, you have that $\int_{\gamma} \omega = 0$

But he computed the specific form and showed that the integral wasn't zero

Nicholas Roberts

Oh ok, gamma in this case (the question) is a closed curve?

Ya, its a cricle i believe

Daminark

04:10

Yeah

Nicholas Roberts

Gotcha, thats what i didnt make note of. Thank you man.

Daminark

No problem! :D

Nicholas Roberts

So you use stokes to get the integral is 0 right?

over the closed curve Gamma

Daminark

Yup

feynhat

If $(X, d_1)$ and $(Y, d_2)$ are two metric spaces, then, which metric on $X\times Y$ which generates the product topology on $X \times Y$?

Nicholas Roberts

04:13

Product metric

In mathematics, a product metric is a metric on the Cartesian product of finitely many metric spaces ( X 1 , d X 1 ) , … , ( X n , d X n ...

?

Daminark

@feynhat there are a few metrics that will work. I think if you define $d((x,y),(x',y')) = (d_1(x,y)^p + d_2(x',y')^p)^{1/p}$

Lol sniped

Nicholas Roberts

Haha

feynhat

@NicholasRoberts @Daminark Thanks. Suppose I chose this metric: $d((x,y),(x',y')) = d_1(x,y) + d_2(x',y')$, (putting p=1 in Daminark's metric). Then how do I show that this will generate the product topology.

Nicholas Roberts

Hm im not sure, i never quite understood the product topoplogy :/

feynhat

Since product topology is the weak topology generated by the projection mappings (in this case, $\pi_X(x, y) = x$ and $\pi_Y(x, y) = y$), will it suffice to show that these projections are continuous even the in the topology generated by $d$?

Alessandro Codenotti

04:39

@NicholasRoberts a function with a product as codomain is continuous iff all its components are, it is more or less defined with this property in mind

Semiclassical

@Fawad the usual shortcut for these kinds of integrals is to start with $\int e^{ax}\,dx$ and taking n derivatives with respect to $a$.

s. patroller

vignette.wikia.nocookie.net/mlp/images/a/af/…

@Semiclassical^

Semiclassical

Zoidberg Woop Woop Woop!

►

s. patroller

13 hours ago, by Semiclassical

a favor: I need to stay out of here for the next few hours. So if I'm here today, yell at me

Secret

05:11

@Semiclassical * YELLS *

Fawad

05:43

@Semiclassical you mean instead of integrating , dedicating options? I don’t understand what trick you mean/how to use that trick

Zee

Is anyone here ?

s. patroller

wazzup dawg?

Zee

Am not sure

You?

s. patroller

chillin

Zee

Let me ask you this

I feel like this is too clever to work

s. patroller

05:57

gotta run, sorry

Zee

So I wanna prove the max modulus principle for several complex variables , the proof is a generalization of the one variable case but I was thinking to myself if the following would work

Come on man , it’s quick

So you have f(z1,...,zn) so just just fix all the variables except one and apply the one variable max modules principle

Do this to all the variables

Is that a proof or am I doing something silly ?

Oh, f is from Cn to C1

Ight , so it’s like dat??

Silent

Artin's Algebra says'let $g(x)$ be a polynomial in $R[x]$ and let $\alpha$ be an element of $R$, where $R$ is a ring. The remainder of division of $g(x)$ by $x-\alpha$ is $g(\alpha)$. Thus $x-\alpha$ divides $g$ in $R[x]$ iff $g(\alpha)=0$.' In proof, he says: 'This corollary is proved by substituting $x=\alpha$ into the equation $g(x)=(x-\alpha)q(x)+r$ and noting that $r$ is constant.'

I can't understand this proof.

@TobiasKildetoft

Tobias Kildetoft

@Silent Note that the proof is only for the last part about being divisible by $x-\alpha$.

Silent

oh!

philmcole

06:23

Hi, if I have two sets $A,B$ and know $A \cup B=U$ and $A \cap B=\emptyset$, is then $A^c=B$?

I know the other direction holds but is this direction also true so that it's an equivalence?

Tobias Kildetoft

@philmcole Yes

philmcole

Could you give me a hint how to start to prove this?

I actually want to assume non empty sets if that is important for the proof

I tried $A^c = U \setminus A = (A \cup B) \setminus A = B \setminus A = B \cap A^c$ but that doesn't seem to get me anywhere

Tobias Kildetoft

06:38

@philmcole Write up what $B\setminus A$ is by definition.

philmcole

Do you mean $B \setminus A = B \cap A^c$?

Tobias Kildetoft

no

the full definition as a set

philmcole

ah

$B \setminus A = \{x \in B \mid x \notin A \} $

Tobias Kildetoft

right

philmcole

Now I use that their disjoint union is $U$ and argue, that $B \setminus A = B$?

no wait

Tobias Kildetoft

06:43

no, now you use that their intersection is empty

philmcole

right

okay thanks

Silent

07:16

How to know that $7x+7\equiv 0\pmod n$ if and only if $n\in\{1,7\}$: I thought $7(x+1)\equiv0\pmod n$, but can't see why $n=1$ works

Tobias Kildetoft

@Silent If $n=1$ then everything is $0$ mod $n$

Silent

@TobiasKildetoft thank you!

Secret

07:40

[Capturing the growth of ordinals]

First define a tuple (a,b,c) where a is an operator, b is the argument on the RHS of the operator and c is the number of times to apply it to get to the next term in the sequence

Then we have the following:

0->1->2->3-> ... < $\omega$

The above sequence has a growth of (+,1,1)

$\omega, \omega 2, \omega 3, ... < \omega^2$

The above sequence has a growth of (+,$\omega$,1)

$\omega^2,\omega^3,\omega^4, ... < \omega^{\omega}$

The above sequence has a growth of (+,current term, $\omega$) or ($\times$, $\omega$, 1)

We can easily see from the growth tuple that starting from this stage, it becomes impossible to reach the next term by addition of what we currently have alone

$\omega^{\omega}, \omega^{\omega^{\omega}},{}^{4}\omega, ... < \epsilon_0$

The above sequence has a growth of (+,current term, next term) or ($\times$,current term, current term) or (^,$\omega$,1)

You can now see how fast exponential growth is relative to multiplication, vs multiplication relative to addition

ÍgjøgnumMeg

08:52

It seems I've been typing \Z instead of Z throughout my section on decomposition groups and now I've got loads of subgroups of Galois groups called $\Bbb Z_\mathfrak{P}$

lol

Alessandro Codenotti

\renewcommand{\Z}{whatever you should have typed instead} :P

ÍgjøgnumMeg

hahaha

there are a lot of $\Bbb Z$s throughout the rest of the paper

Alessandro Codenotti

I'm afraid you'll have to correct them by hand then

I like how everyone has \Z \N \Q and \R in their latex files because \mathbb is too long

Tobias Kildetoft

@ÍgjøgnumMeg Do a replace on those places where you have subscripts which you would not have for the integers

Alessandro Codenotti

I'm writing some model theory stuff and I made a \Los command that prints Łoś because Polish letters are hard

I also made a \fip one that prints "finite intersection property" becase I don't like the abbreviation and I don't like having to write the extended name

ÍgjøgnumMeg

09:02

@Tobias Good call!

@Alessandro \Wash

Martin Sleziak

I see you're talking LaTeX. I often use similar thing if I have something that I repeat often. How do you take care of spaceing in such cases?

Tobias Kildetoft

If your editor supports replacing using rexes that might make it even easier

Martin Sleziak

I use something like \newcommand{\fip}{finite interesction property\xspace} and the xspace package.

Alessandro Codenotti

Interesting, I write the commands without leading or trailing spaces and haven't noticed spaceing issues so far

Martin Sleziak

@TobiasKildetoft I definitely prefer the solution using macros for stuff where there is a chance that I might decide to change it completely. If I want to change notation for vectors from $\vec v$ to $\mathbf v$, I only have to change in in one place (in the macro).

Moreover, I would have to do many regex replacements, since I have many similar macros.

ÍgjøgnumMeg

09:06

I'm a bit of a latex noob, I just do newcommands for things like \mathbb, \mathfrak, \mathcal, \operatorname

Tobias Kildetoft

@MartinSleziak In this case a macro would not have helped, as he had accidently typed the same thing for things that should be different

Martin Sleziak

@AlessandroCodenotti If you have \newcommand{\fip}{finite interection property} then I think that write "since it has the \fip and" will leave not space between the words "finite interection property" and "and".

@TobiasKildetoft I see. I thought you were replying to me, while your response was about ÍgjøgnumMeg's problem.

Or maybe there was problem with \fip, or \fip. and stuff like that.

I do not really recall now what was the problem - it was long time ago when I started to use shortcuts like that. But I definitely remember that there were some problems with spaces after such shortcuts and I had to correct them.

Alessandro Codenotti

@MartinSleziak hm, I'll check the file when I go back to my computer

Martin Sleziak

And I was too lazy to always write \fip{} just for the sake of having space in all places where they were needed.

Alessandro Codenotti

@MartinSleziak I'll probably just steal this if there are problems :P

Martin Sleziak

09:10

Tex.SE: Drawbacks of xspace

Tobias Kildetoft

@MartinSleziak I just checked and yes, there will be a space missing.

Martin Sleziak

From the tag-info:

> {xspace} is a command (provided by the package of the same name) that inserts a space if and only if the space is followed by punctuation. This is one of several techniques to deal with TeX's habit to "eat" spaces after macros without arguments.

@AlessandroCodenotti Probably the thread linked about is worth having a look at. In fact I did not know that there are also some problems.

David Carlisle writes there:

> Conversely with xspace the macro will get the correct space most of the time, but it isn't easy to predict when it will get it wrong, and so it's much harder to learn to enter the markup in a way that is always correct rather than having to always visually check for missing space, which rather negates the purpose of the command.

BTW the LaTeX site (and chatroom) is different from other sites in that you can encouter there the biggest stars in that area on a regular basis.

Tobias Kildetoft

@MartinSleziak Well, not that different from MO then (for the site, not the chat that is).

Martin Sleziak

Although on MO some Fields medalists and other famous mathematicians post on regular basis, it is probably much rarer to meet such person in chat.

Tobias Kildetoft

Yeah, not much activity on the MO chats

Martin Sleziak

09:17

I have to admit that I am not familiar enough with TeX community to be able to judge whether the comparison with Fields medal is appropriate, but I believe I saw there some authors of LaTeX packages, sometimes people who wrote books about (La)TeX and people who are active in TUG and similar organization.

@TobiasKildetoft Yes I remember your post on meta: Specialized chat rooms.

But Homotopy Theory has now close to 100k messages. That's impressive.

In any case, I should leave - I'm teaching in some 10 minutes.

Tobias Kildetoft

Yeah, that is probably the only room that ever really got off the ground

philmcole

10:10

I am kind of confused with relatively open sets in subsets... If I have two arbitrary sets $Y_1,Y_2$ and consider a clopen set $A \subseteq Y_1 \cup Y_2$, what can I then say about $A \cap Y_i$? The answer is it is clopen too, but I don't understand why.

Tobias Kildetoft

@philmcole Well, it is clopen in the subset topology on $Y_i$. It need not be either in the union.

philmcole

no wait I defined $A$ to be clopen in the union.

Tobias Kildetoft

Ahh, right

philmcole

oh you meant the intersection

Tobias Kildetoft

Right

philmcole

10:17

Okay got it. And the reason for the intersection to be clopen is because it is relatively clopen in $Y_i$?

Tobias Kildetoft

no, it is clopen by definition of the subspace topology

philmcole

ok

@TobiasKildetoft Can you explain to me this example of a relatively open subset? We consider the metric space $X = \Bbb R^3$ and the hollow sphere $Y = \mathbb {S}^2=\left \lbrace {v\in \mathbb {R}^3} \mid {\| {v}\| =1}\right \rbrace$ as a subset of $X$. Is now $\left \lbrace {v\in \mathbb {S}^2} \mid {\| {v-e_1}\| <1}\right \rbrace$ open or closed in $Y$?

I know that an open set in $Y$ takes the form $O \cap Y$ for some open set $O$ in $X$

Tobias Kildetoft

Right, so if you replace the sphere with $\mathbb{R}^3$ in the definition of that subset, what do you get?

philmcole

I'd say the set $A$ is then open?

Let's call this set $A = \left \lbrace {v\in \mathbb {S}^2} \mid {\| {v-e_1}\| <1}\right \rbrace$

I think $A$ is open in $\Bbb R^3$ because there is a $\lt$ sign and not a $\le$

DarkRunner

10:36

Hi

I'm dealing with the problem, "A sphere is inscribed in a regular tetrahedron. If the length of the altitude of the tetrahedron is 36, what is the length of a radius of the sphere?"

My idea is to project the 4 incenters of each face of the reg. tet., an their intersection would be the center of the circle. But, I have no idea how to get started

philmcole

If I consider $\left \lbrace {v\in \mathbb R^3} \mid {\| {v-e_1}\| <1}\right \rbrace$. This is a sphere around $(1,0,0)$, so this is not even a subset of $\Bbb S^2$ anymore, so we can't talk about it wrt $\Bbb S^2$?!

But I think its open in $\Bbb R^3$

Secret

10:58

(More details on ordinal growth in SBA chat room)

user548331

11:16

Hi stack exchange do I need to download something for the latex to work in these rooms?

Silent

What does this mean:(this is from Artin's Algebra): Let $\Phi: \Bbb R[x,y]\to\Bbb R[t]$ be the homomorphism that is the identity on the real numbers, and that sends $x\to t^2, y\to t^3$. Then it sends $g(x,y)\to g(t^2,t^3)$.

@AkivaWeinberger

Akiva Weinberger

$g$ there is a two-variable polynomial

It's essentially an arbitrary element of $\Bbb R[x,y]$

So it's $\sum_{m,n}a_{m,n}x^my^n$ where finitely many $a_{m,n}$ are nonzero

Silent

11:31

ok. What does ' the homomorphism that is the identity on the real numbers' mean?

on, got it! coefficients don't change under this homomorphism, only variables change, right?

Akiva Weinberger

11:51

Yeah

Homomorphism means $g(x+y)=g(x)+g(y)$, $g(xy)=g(x)g(y)$, and $g(1)=1$; identity on the real numbers means $g(r )=r$ for $r\in\Bbb R$.

Alessandro Codenotti

12:05

@user548331 See here

user548331

thankyou

@AlessandroCodenotti it is rendering on that page you directed me to but not on here\

Alessandro Codenotti

You have to add it to the bookmarks in your browser and click on it while on this page

user548331

ah beautiful that's better thankyou so much

philmcole

Hi, in the proof that for continuous functions on metric spaces the pre image of an open set is again open, why can we say for $A,B$ both open sets is $A \subseteq f^{-1}(B)$ equivalent to $f(A) \subseteq B$?

Or does this hold generally for any function (not necessarily continuous)?

Silent

12:22

@AkivaWeinberger thank u so much

Alessandro Codenotti

$f(f^{-1}(B))\subseteq B$ for any function, so from $A\subseteq f^{-1}(B)$, by applying $f$ on both sides, you get $f(A)\subseteq f(f^{-1}(B))\subseteq B$

philmcole

okay thanks!

Silent

Please someone suggest where should i begin my complex analysis study. I started with Ahlfors, but it became unbearable. I also tried Gamelin, but it mentions Riemann geometry in ch 1, which i have never learned. (I have basic understandin of real analysis upto ch 6 (integration) from Baby Rudin).

s. patroller

why not just finish rudin?

Hi, I'm your new meteorologist and a former software developer. Hey, when we say 12pm, does that mean the hour from 12pm to 1pm, or the hour centered on 12pm? Or is it a snapshot at 12:00 exactly? Because our 24-hour forecast has midnight at both ends, and I'm worried we have an off-by-one error.

5

Akiva Weinberger

12:43

I appreciate

"What is it doing?" "It's raining."

philmcole

lol the math guy would make the weather forecast much more interesting to listen to at least

Silent

@s.patroller Oh! till then no complex analysis

?

s. patroller

Sure, why not?

the second book will cover it

Semiclassical

@Silent I liked Stewart and Hall

Silent

thank u!

s. patroller

12:55

Welcome back @Semiclassical :P

philmcole

should we still yell at you Semi?

s. patroller

*Tall

Semiclassical

@s.patroller bah, I knew it was that but I still wrote it wrong

s. patroller

np

*typed :P

Silent

13:16

Artin says: since $\sqrt[3]2$ is irrational, and root of $x^3-2$, we see that $x^3-2$ is not product $f=gh$ of two non-constant polynomials with rational coefficients. I can't see why we can't factor.

philmcole

There is this result which says $a$ is a zero of a polynomial $f(t)$ iff $(t-a)$ is a factor of $f$. Since there is no rational zero there is no factor $(t-a)$ with $a$ rational.

Silent

@philmcole Thank u very much. So, there is no divisor polynomial of degree 2, because if there were $g$ with degree 2 with $f=gh$ for some polynomial h, then since $f$ has degree 2, $h$ would have degree 1, impossible as you showed. Is this reasoning correct?

philmcole

there is a divisor polynomial of degree 2. By the above you can factor $f(t)=(t-a)q(t)$ where $q(t)$ doesn't factor anymore and has degree 2 (since $(t-a)$ has degree 1 and $f$ has degree 3).

But if I look at it I don't know if my theorem helps you though..

Silent

13:34

@philmcole it definitely helps! you let me know that divisor polynomial can't have degree one, and if there were divisor polynomial of degree 2 then there was a divisor polynomial of degree 1, so it can't have degree 2 polynomial divisor either. thank u

philmcole

just constant polynomials have degree 1 so non-constant polynomials allow degree 1 and 2 right?

Astyx

14:13

Don't constant polynomials have degree 0 ?

GFauxPas

14:29

So I used the term "boundedish" when asking my professor a question and he used the term himself later that class and it made me happy

philmcole

@Astyx ups right I was confused myself

that makes more sense

Waiting

15:37

@Secret

An attractive bounty for you, math.stackexchange.com/questions/2749560/…

vzn

@Secret way cool fascinating thx for sharing luv bitcoin, fluid dynamics + have found similar connections in number theory/ collatz maybe post to Physics chat also o_O vzn1.wordpress.com/code/collatz-conjecture-experiments

GFauxPas

that xkcd in the starred posts is so good

r9m

@Waiting Hello! How are you?

Waiting

@r9m Hey! I just entered the room to see what's going on. You?:-)

r9m

15:53

@Waiting I am doing okay .. as usual with a problem bugging me for a while that's all :)

any news of the awesome 'book'?

Waiting

@r9m I have a 300 bounty for a nice question involving the calculation of a particular case of a slightly modified form of the Faddeeva function by means of contour integration exclusively.

r9m

@Waiting lemme check! ..

Waiting

It's almost crazy to see no one knows how to deal with it by contour integration. Did you ever try it?

@r9m Perfect.

@r9m Lately I'm thinking of the butterfly wings contour (roughly). I'll tell you the details if I can make it working.

r9m

@Waiting cool! :)

Twink

Let $p$ be a prime number and $a$ an integer and $n$ a nonnegative integer. If $p | a^n$ then $p | a$? Is that true?

Semiclassical

15:58

Is that much like the pochhammer contour?

Waiting

To some extent.

Semiclassical

hmm

I like the pochhammer contour, so that's on my attention

Waiting

Pochhammer contour

In mathematics, the Pochhammer contour, introduced by Camille Jordan (1887) and Leo Pochhammer (1890), is a contour in the complex plane with two points removed, used for contour integration. If A and B are loops around the two points, both starting at some fixed point P, then the Pochhammer contour is the commutator ABA−1B−1, where the superscript −1 denotes a path taken in the opposite direction. With the two points taken as 0 and 1, the fixed basepoint P being on the real axis between them, an example is the path that starts at P, encircles the point 1 in the counter-clockwise direction and...

Semiclassical

yeah, that's a fun picture

Waiting

@r9m r9m It's long time done. Now I only need to wait. :-)

r9m

16:03

cool! :)

Semiclassical

The fun bit of pochhammer in the complex plane is that, if you've got a function with poles at the highlighted points and analytic everywhere else, then integrating around that contour always gives zero

Waiting

@r9m What are you doing lately?

Preparing for PhD? :P

r9m

@Semiclassical it may do well as a good interview question .. to see if one can relate the homology and homotopy versions of cauchy residue theorem ..

@Waiting aye aye captain :)

Semiclassical

Well, depends on what you're interviewing for

for a math position, maybe

r9m

yas .. I meant for a math program :)

Semiclassical

16:06

For anything else, you'd probably see it posed in terms of "how do you hang a picture on two nails so that removing one nail makes the picture fall"

r9m

nice!! :D

Balarka Sen

Hey @r9m

r9m

@BalarkaSen hey man! How are you?

Waiting

@Semiclassical good point

Semiclassical

(which in homology terms is of course the same as what you were getting at: that you want a curve which is not null-homotopic but is null-homologous)

an interesting variation on that is to do setups with more than two nails

Balarka Sen

16:08

@Semiclassical Only in the context of $S^1 \vee S^1$.

Secret

I like how the Age of Geometry (2018) makes integrals have a geometric context such as contour integrals

Semiclassical

@BalarkaSen hmm?

Balarka Sen

@r9m So-so. Going to take the admissions soon.

r9m

@BalarkaSen awesome!! :D

Balarka Sen

Eh, opposite of that really.

Secret

16:09

still I am trying to be better at contour integration. Perhaps those complex plotters might come in handy to gain more intuitions...

r9m

@BalarkaSen I was only expressing how that made me feel :)

Twink

@BalarkaSen why does 2 have to divide $n+1$?

Balarka Sen

@Semiclassical Well, the nail-and-picture formulation and the nullhomotopic vs nullhomologous formulation only relates in the context of the wedge of two circles.

Semiclassical

eh. doubly-punctured plane

is enough for me

Twink

@BalarkaSen .

Balarka Sen

16:11

That's the same as the wedge of two circles upto homotopy equivalence.

Semiclassical

yeah, fair enough

and when I write stuff like $\alpha\beta\alpha^{-1}\beta^{-1}$ I do admittedly have that setting in mind

Balarka Sen

I was commenting on the "of course". It's not "of course" the same formulations.

Just happens to be the same in the specific space.

Semiclassical

Well, in the context of the problem it is.

Wasn't trying to insist beyond that.

Balarka Sen

Gotcha.

@Twink $2$ divides $n^a = (n+1)^b$. If $2$ divides $(n+1)^b$, it divides $n+1$ (why?)

Semiclassical

On that note, I still want to get you to do stuff on magnetic helicity :)

Twink

16:13

yes that's my question, why?

Balarka Sen

@r9m Hah, from my perspective it's quite scary

Twink

is this by Euclid's lemma?

Balarka Sen

@Twink You answered your own question :)

Twink

but Euclid's lemma is not that general

Semiclassical

@BalarkaSen it's awesome in rather the same way that an avalanche is.

Twink

16:14

it's just for squares

Balarka Sen

@Semiclassical Hahah I really do want to study that after my exams are over

Twink

if $p|a^2$ then $p|a$

Balarka Sen

@Twink Euclid's lemma says $p$ divides $ab$ iff $p$ divides $a$ or $p$ divides $b$.

$2$ divides $(n+1)^b = (n+1)(n+1)^{b-1}$, so ...

Twink

what if it divides $(n+1)^{b-1}$?

Balarka Sen

Do the same argument :)

Twink

16:16

do I need to use induction

:(

Semiclassical

yeah, but it's easy in this case.

Balarka Sen

To be entirely rigorous, yes. The point is Euclid's lemma in general says if $p$ divides $\prod a_i$ iff $p$ divides $a_i$ for some $i$.

Twink

no it's not what it says

Euclid's lemma says $p$ divides $ab$ iff $p$ divides $a$ or $p$ divides $b$.

Balarka Sen

That's the punchline formulation, and nobody spells out the generalization I wrote because it's a "trivial" corollary

Semiclassical

I'd say it's not Euclid's lemma in that case, but it's an obvious generalization/application of it

Twink

16:19

it just has to be proven by induction

Semiclassical

You start with the two-factor case and bootstrap to the full case

Balarka Sen

Induction on the number of factors, correct.

@Semiclassical Also, that's accurate, I'd say

Semiclassical

Mostly I just like the excuse to use the word 'bootstrap'

r9m

@BalarkaSen all I can say is that if the interviewers start from a normal frame frame of mind .. they will be shocked soon enough :P

Twink

and if use the corollary of Euclid's lemma that says that if $p|a^2$ then $p|a$, and I want to prove that if $p|a^n$ then $p|a$, I need to use induction on $n$ right?

Balarka Sen

16:21

@r9m If I get to the interviews that is!!

r9m

@BalarkaSen (-_-) okay modest guy! :-)

@Semiclassical can you help me with a frechet derivative stuff?

Semiclassical

no

Astyx

(rude)

r9m

okay :)

Semiclassical

both because I should be doing other stuff and b/c I just don't know that stuff so well

Parth Kohli

16:24

@BalarkaSen Good luck, man.

Balarka Sen

@r9m Here's a more constructive question instead of panic-channeling: How did you approach the CMI paper? Was it more with a psychological goal of absolutely nailing down every problem, or writing down observations which may or may not lead to a full solution?

@ParthKohli Hey, long time. Thanks a lot!

How's life?

Parth Kohli

@BalarkaSen I'm going to be writing the CMI paper as well (even though I wouldn't be joining -- I'm already done with a year of college).

Balarka Sen

Ah nice.

Twink

¬¬

Parth Kohli

I'm not even sure why I'm doing this... it's possibly because I committed a few computational errors the last time in the first section, and I need to check this thing off my bucket list.

Did you write JEE Main?

Balarka Sen

16:29

Nope.

Parth Kohli

Is CMI your final bet then?

Balarka Sen

I'm taking both CMI and ISI, let's see what happens.

Semiclassical

16:42

@BalarkaSen if you do get to a math interview, the helicity thing might be a good example since you can link it with both topology and physics

Balarka Sen

Lol

Semiclassical

though I guess that includes some assumptions as to how such interviews work, which I'm probably wrong about

Balarka Sen

I think it's more along the lines of, they'll ask people to solve pretty concrete down-to-earth problems

I imagine there's a chance at being able to demonstrate what kind of math one likes, though.

Waiting

@BalarkaSen Wish you much luck at the interview in the sense nobody will ask you to calculate a tricky integral. ;)

Balarka Sen

Hahahah god I hope not

Parth Kohli

16:48

The ISI interview is very elementary, as far as I know. :|

r9m

@BalarkaSen writing down observations which displays you are on the right track to the solution will definitely add positive score .. as far as I have heard each problem is broken (unless it's something utterly trivial) has a number of essential steps. Guessing or proving each essential step adds credit. Out of the box solutions are greatly appreciated.

Balarka Sen

@ParthKohli The ISI exam in general is quite elementary, it's the time constraint of 2 hours that fucks with me.

@r9m Got it. Thanks!

r9m

@BalarkaSen I believe .. for ISI they expect you to nail shit down! :P

Balarka Sen

Yeah I imagine so.

Semiclassical

@BalarkaSen well, keep in mind how I motivated the helicity stuff: thinking about how the Biot-Savart law together with Ampere's law gives you Gauss's winding number formula

Balarka Sen

16:57

I liked that idea.

Semiclassical

So if Ampere's law comes up you could potentially use that as a jumping off point

Again, though, it depends a lot on the context of the interview

Balarka Sen

Oh it won't. It's just a honest to god math test.

Semiclassical

right

this would be more like if you were wanting to do a grad school program tbh

Balarka Sen

Or @Astyx's admission exams into the Ecole Woahs

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$Mathematics$ Mathematics