Mathematics - 2018-03-03 (page 1 of 2)

Eric Silva

00:32

yo @Balarka you around

Forever Mozart

03:31

yes hohoho

Forever Mozart

04:02

oh my hon

Tanuj

06:24

Guys , if I have to form the differential equation of the family of circles touching x axis at origin,

can I not just simply put $x=y=0$ in $x^2+y^2+2gx+2fy+c=0$ and then differentiate once since I would be left with only $c$ as an arbitrary constant ?

Balarka Sen

07:10

@Eric now I am

Secret

07:28

@Tanuj How will that help. you get c=0

Leaky Nun

09:01

@Secret right, so $x^2+y^2+2gx+2fy=0$, i.e. $2x + 2yy' + 2g + 2fy' = 0$, i.e. $y' = (-g-x)/(y+f)$

Secret

that looks separable

Balarka Sen

@LeakyNun o/

Leaky Nun

hi

Balarka Sen

Learning algebraic topology?

Leaky Nun

yes

Balarka Sen

09:11

Have you gotten to covering spaces yet

Leaky Nun

we just computed the homology groups of a point

so yes

Balarka Sen

I have been thinking about the pathspace formalism in covering spaces a bit

Leaky Nun

what is the pathspace formalism?

Balarka Sen

Remember the construction of covering space of $(X, x_0)$ (a path connected, locally path connected, semilocally simply connected based space) from a subgroup $G \subset \pi_1(X, x_0)$?

Well, yes, you look at space of homotopy classes of paths starting at $x_0$ which are $G$-equivariant.

Leaky Nun

I see

Balarka Sen

09:16

It has an interesting topology

Leaky Nun

hmm?

Balarka Sen

So you construct the universal cover as $\widetilde{X} = \{[\gamma] : \gamma \in X^I, \gamma(0) = x_0\}$, and give it the topology generated by the basis $\mathcal{U}([\gamma], U)$ which consists of paths of the form $\gamma * \eta$ where $\eta$ is a path with $\eta(0) = \gamma(1)$ and stays inside an open set $U$ around $\gamma(1)$.

Under this topology, the projection map $p : \widetilde{X} \to X$ given by $p([\gamma]) = \gamma(1)$ turns out to be a covering map.

Leaky Nun

right

Balarka Sen

And you can prove $\widetilde{X}$ is simply connected: the space is full of noodles starting at $x_0$. Slurp the noodles to contract the space to $x_0$

More or less

Maybe a bit more carefully :)

Leaky Nun

right

and then what about it

Balarka Sen

09:30

Well think about the $G$-action on $\widetilde{X}$ given by composing (homotopy class of) a path with a loop at $x_0$, and then quotient out by it to get the cover of $X$ corresponding to the subgroup $G \leq \pi_1$

I have been thinking about this construction and it vaguely reminds me of the pathspace fibration

Which is the following thing. Suppose $X$ is a (sufficiently nice, just assume CW complex) space, then denote $P(X, x_0)$ to be the space of paths starting at $x_0$ (no homotopy modding here), inheriting the compact-open topology from $X^I$.

Then there is a projection map $P(X, x_0) \to X$ given by sending a path $\gamma$ starting at $x_0$ to $\gamma(1)$.

Slereah

10:29

I'm guessing manifolds with corners only admit diffeomorphisms if you map the corners to corners

Balarka Sen

Yes, diffeomorphism sends corners to corners

Secret

10:53

The stunning geometry behind this surprising equation

►

That's clever, using the inverse square law to make this geometric proof

Leaky Nun

@BalarkaSen can manifolds have corners?

Dattier

Hi, $\sum \limits_{p,q\in\mathbb N, \text{gcd}(p,q)=1} \dfrac{1}{p^2q^2}=?$

This problem is not mine.

Secret

I have no idea how one can exclude composite numbers that are not of the form $(pq)^2$

But I suspect the sum is something of the form $\zeta(2)-(something)$

Slereah

@LeakyNun no, but manifolds with corners can have corners

although it depends what you mean by "corners"

A square is a manifold, certainly

Since it's homeomorphic to a circle

Balarka Sen

@LeakyNun You can generalize manifolds to be locally diffeomorphic to $\Bbb R^n \times \Bbb H^m \times \Bbb H^{\ell}$

Dattier

11:06

@Secret the sum equal to 5/2

Slereah

there are so many generalizations of manifolds

there are manyfold generalizations of manifolds

Balarka Sen

I had to edit that thrice lel

But now it's OK

Secret

hmm that will mean the sum of the complement set is $\frac{3-\pi^2}{6}$

Balarka Sen

But yeah for example the square (with it's interior) is a manifold with corners: the corner points are locally diffeomorphic to $\Bbb H^1 \times \Bbb H^1$

Dattier

@Secret the answer here : forum.prepas.org/viewtopic.php?f=3&t=64924

Balarka Sen

11:08

Notice that manifold with boundaries are special cases of manifold with corners: boundary points are locally diffeomorphic to $\Bbb H^n$

Slereah

it's unfortunate that there isn't a terminology like circle/disk for squares

or polygons in general

Balarka Sen

What's confusing is that manifold with corners are the same thing as manifold with boundaries in the topological category

Slereah

It makes it hard to discuss polygons as manifolds

Balarka Sen

Because $\Bbb H^m \times \Bbb H^\ell$ is homeomorphic to $\Bbb H^{m+\ell}$

For example

Slereah

still wrong!

Secret

11:09

Canwe ever formalise this thing. Isn't that basically category theory:

Balarka Sen

@Slereah? What I wrote is correct.

Slereah

@BalarkaSen I meant the $\mathbb H^($ part :p

Secret

What even is a path in the "space of mathematics" even mean

Balarka Sen

Oh

Slereah

apparently this is a good book for manifolds with corners : books.google.fr/…

But it seems hard to find

Dattier

11:11

@Secret : do you talk with me ?

Secret

the above two messages? No as obvious from the picture

Balarka Sen

Candel-Conlon, Foliations I has a short exposition on manifold with corners

Check it out

Secret

I basically is like talking to no one until someone feeds back

Slereah

amazon.com/…

Dattier

well a solution is here : forum.prepas.org/viewtopic.php?f=3&t=64924

Secret

11:12

So for the above two message, I am talking to thin air and somebody until this superposition is collapsed by somebody responding to it in some form

Slereah

I think it's that one

But it doesn't ship to France

Secret

@Dattier I cannot read french, but my guess is you used some kind of convergence theorem

Slereah

Oh wait

I found it online

The trick is that the authors aren't actually listed as authors

Only the editor

Dattier

$A=\sum\limits_{(p,q) \in \mathbb N^2,\text{pgcd}(p,q)=1} \frac{1}{p^2q^2}$

then $A=(\sum\limits_{p\geq 1} \frac{1}{p^2})^2-A\times (\sum \limits_{n\geq 2}\frac{1}{n^4})$
A.

That give A.

@Secret

Secret

yeah I understand the logic, but how you came up this step is not clear without some number theory, which I am weak at

I mean, I still cannot really comprehend prime numbers

it seems different ways of adding it gives magical results

Dattier

11:17

Sorry, It's not magical, I can explain that, but google translator don't work

first $(\sum \limits_{p\geq 1} \dfrac{1}{p^2})^2=\sum\limits_{p\geq1,q\geq1} \dfrac{1}{p^2q^2}$

ok ?

Secret

yeah by product of summations

Dattier

$A\times (\sum \limits_{n\geq 2} \dfrac{1}{n^2})=\sum\limits_{(p,q)>1}\dfrac{1}{p^2q^2}$ is complamentary

ok ?

Secret

uh this step is not quite clear, the 1/n^2 sum does not contain gcd()=1 terms?

Dattier

$A=\sum\limits_{(p,q) \in \mathbb N^2,\text{pgcd}(p,q)=1} \frac{1}{p^2q^2}$
$E=A\times (\sum \limits_{n\geq 2} \dfrac{1}{n^2}) $

Sorry, I do a mistake

$A\times (\sum \limits_{n\geq 2} \dfrac{1}{n^4})=\sum\limits_{(p,q)>1}\dfrac{1}{p^2q^2}$

power 4 and not 2

n=gcd(p,q)>1

ok ?

@Secret : ok ?

I give more steps

$E=\sum\limits_{(p,q)>1}\dfrac{1}{p^2q^2}=\sum \limits_{n>1} \sum \limits_{(p,q)=n} \dfrac{1}{p^2q^2}$

so $E=\sum \limits_{n>1} \sum \limits_{(p,q)=1} \dfrac{1}{n^4p^2q^2}

Silent

11:33

Let $X_{2n}$ be the group whose presentation is$\langle x,y\,|\,x^n=y^2=1, xy=yx^2\rangle$. From $x=xy^2$, it is seen that $x^3=1$, hence $X_{2n}$ has at most $6$ elements. I have to show that if $n=3k$, then $X_{2n}$ has exactly $6$ elements. I can't see where I am having problem if I assume $x=1$.

Dattier

so $E=A\times(\sum\limits_{n>1}\dfrac{1}{n^4}$

@Secret : ok ?

Secret

I need to digest a bit, it is very hard to picture all the summands without drawing on paper

Dattier

it's a question of set

more multiset question

but you understand the general idea ?

$E=\sum \limits_{n>1} \sum \limits_{(p,q)=1} \dfrac{1}{n^4p^2q^2}$

Secret

hmm... it is obvious that p,q cannot be coprime if they are both multiples of 2, hmmm...

if (p,q)=/=1, it means (pq)^2 can be factored as some n^2(rs)^2 where n is all integers

so $E=\sum\limits_{(p,q)>1}\dfrac{1}{p^2q^2}=\sum \limits_{n>1} \sum \limits_{(p,q)=n} \dfrac{1}{p^2q^2}$

Dattier

no (p,q)=n so p^2q^2=n^4r^2s^2

(r,s)=1

because p=nr and q=ns

Secret

11:46

so (p,q) means gcd of p and q?

Dattier

yes

Secret

ok now I can see how the n bubbles out

Dattier

well you see, so it not magic, but the cooking

lol

@Secret : ok ?

Secret

@Dattier Yeah, I am just not that good with division thus it is hard for me to see how terms bubbles out within numbers

Dattier

but, now it's ok ?

Secret

11:49

It seems fine so far

Dattier

well, you have all ingredient for conclued

$A=\sum\limits_{p,q\in\mathbb N^*}\dfrac{1}{p^2q^2}-E$

Secret

uh, we are summing in the supernaturals now?

$\Bbb{N}^*$ are the supernatural numbers?

Dattier

N^*={1,2,3,....}

all integers without 0

Secret

ok

Dattier

it's a cooking question, not magic question, ok ?

Secret

11:54

don't expect me to learn fast, but I will try my best

Dattier

I am very slow too

All that I know, I give a long time for knowing that

well do you know why : A=5/2 ?

@Secret

Secret

12:14

$A=\sum\limits_{p,q\in\mathbb N^*}\dfrac{1}{p^2q^2}-E \implies (1+(\zeta(4)-1))A = \zeta(2)^2 \implies \frac{\pi^4}{90} A = \frac{\pi^4}{36} \implies A = \frac{5}{2}$

I wonder what does this looks like geometrically...

3B1B showed that $\zeta(2)$ is 4/3 times the total flux received at the origin of a straight line (infinite limit of a circle) with sources at every odd integer. I wonder if similar result also holds for a straight line in $L^4$ so we can reason the same way with $\zeta(4)$

Dattier

12:35

@Secret : Bravo

bye

skullpatrol

@BalarkaSen I just realized that's a squirt-gun for the Janitor :P

Leaky Nun

13:00

@Secret challenge: translate 3b1b's geometric proof to an algebraic proof

(he outlined every step quite clearly I think)

Secret

13:11

@LeakyNun That sounds pretty tedious and uninteresting, cause it is basically applying induction to the use of inverse pythagorus theorem that split the light houses

Leaky Nun

the induction isn't the point

the splitting of the light houses is the interesting part

it's the splitting that makes a constant into an infinite sum

induction is just the formal and tedious and pedantic part

Secret

but he had proved the inverse pythagorus theorem algebraically isn't he in the video, and e.g. for the 8 lighthouses cases, he is applying that 8 times and then showing how the angles are identical?

Leaky Nun

to phrase it differently: what is the equation represented by each stage?

at the stage that has 8 lamps, it represents an equation where the right hand side is the sum of 8 constants

Secret

o..., that I need to think...

ÍgjøgnumMeg

If $K[X]$ is a polynomial ring over a field and the quotient $K[X]/(m(X))$ by the minimal polynomial $m(X)$ of some $\alpha$ gives you a $K$-vector space, does the same construction with $K$ replaced by some ring $R$ give you a free $R$-module?

Dunno if this is a silly question

Balarka Sen

13:33

woo i got the +50 bounty

right before the grace period ended

skullpatrol

cool

Secret

13:51

[Drawsomething]

Blind contour drawing

Blind contour drawing is a drawing exercise, where an artist draws the contour of a subject without looking at the paper. The artistic technique was introduced by Kimon Nicolaïdes in The Natural Way to Draw, and it is further popularized by Betty Edwards as "pure contour drawing" in The New Drawing on the Right Side of the Brain. == Technique == The student fixes their eyes on the outline of the model or object, then tracks the edge of the object with his or her eyes, while simultaneously drawing the contour very slowly, in a steady, continuous line without lifting the pencil or looking at the...

Slereah

What's the generic word for... objects that are homeomorphic to polytopes

Polytopes that aren't necessarily flat

I guess it might just be a complex

Secret

won't that just be any n+1 manifold with corners that bounds an n volume?

Abcd

14:07

Let $T_r$ and $S_r$ be the rth term and sum up to the rth term of a series. If for an odd number n. $S_n= n$ and $T_n= \dfrac{T_{n-1}}{n^2}$, then $T_m$ ($m$ being even) is?

$S_{n+2}= n+2$

$S_n= n$

$S_{n+2}- S_n = 2$

Thus every number at even positions is 2.

Can someone please tell me whats wrong with my method?

(Answer is in terms of $m$, I don't get why the answer is just not $2$)

Secret

> If for an odd number n. Sn=n

For all n or just one n?

Abcd

Its not given in the question, even I am wondering about that.

Secret

I suggest to expand the recurrence relation for $T_n$ and see what you get when you sum that all up, there should be no constants I think

Abcd

Maybe its just for one odd number. Let me retry.

Secret

Also the difference $S_{n+2}-S_n$ does not really give you the value of the even positions, it only gives you how much consecutive odd terms differ

Abcd

14:15

I don't see how.

$S_{n+2}= a_1+a_2+...a_{n+2}$

$S_n = a_1 + a_2 +a_ 3.....a_n$

So $S_{n+2}- S_{n}= a_{n+1}$ (= even position)

Secret

if you are really summing odd terms the index would be $S_{2n+1}$ not $n+2$

Abcd

@Secret In the question its given that $n$ is odd.

so n+2 = odd

Secret

The you should have:

$S_{n+2} = a_3 + a_5+a_7+\cdots$

Abcd

No.

11 mins ago, by Abcd

Let $T_r$ and $S_r$ be the rth term and sum up to the rth term of a series. If for an odd number n. $S_n= n$ and $T_n= \dfrac{T_{n-1}}{n^2}$, then $T_m$ ($m$ being even) is?

sum up to (i.e. including every)

Secret

$S_{n+2} = a_1 + a_2 + a_3 + \cdots + a_n + a_{n+1} + a_{n+2}$
$S_{n} = a_1 + a_2 + a_3 + \cdots + a_n$
$S_{n+2} - S_{n} = a_{n+1} + a_{n+2}$

and n odd

Abcd

14:22

Oh, okay!

This was my mistake then

Thanks.

Balarka Sen

Hey @anakhronizein

0celo7

@BalarkaSen morning

Balarka Sen

evening

0celo7

@BalarkaSen Consider Banach spaces $X,Y$

Suppose I have a map $F:X\to Y$ such that for each $x'\in X$, uhh

the map $X\to Y$ that takes $x$ to $dF_x(x')$ is continuous

Is $F$ $C^1$?

Like, I don't know how to show that $dF:X\to L(X,Y)$ is continuous

But I can show that $dF_x(x')$ is continuous in $x$ for each $x'$

Is this good enough?

Dattier

14:39

A problem of computer : Find the integer $a$ is verify $3^a<2^{2^{2018}}<3^{a+1}$

anakhronizein

Hola.

Secret

$a \ln 3 < 2^{2018} \ln 2 < (a+1) \ln 3 \implies a < 2^{2018} \frac{\ln 2}{\ln 3} < a+1$

Dattier

Well, you must have a very good approximation of ln(2) and ln(3)

how calculate that ?

I don't choose this path

but I don't know if it give the solution

Secret

I guess we can check what integers bound $\frac{\ln 2}{\ln 3}$ by making use of exp, but I need to think about how to set up that inequality

because $2^{2018} < 2^{2018} c$ where $c > 0$

so $a$ is bounded below by $2^{2018}$

Dattier

yes

Secret

14:46

The cool thing about exp and ln is that they are order preserving injections, thus they will not affect inequalilties

we might be able to use that to our advantage

Dattier

Maybe, I don't have choose this path

anakhronizein

@BalarkaSen overleaf.com/read/tnbjjgqnrcnx

Hopefully linking it as an overleaf doc works.

anakhronizein

15:12

I have to go, but I will back at a later date. So read on your own time.

Secret

So the interval $(a,a+1)$ must include the interval $[2^{2017},2^{2018}]$

Dattier

a=E(ln(2)/ln(3)*2^{2018})

E the integer part

0.5<ln(2)/ln(3)<0.7

Secret

I don't know whether you can evaluate E without some kind of recursive algorithm, though

We can repeat the process to arbitrarily shrink the bound for $\frac{\ln 2}{\ln 3}$, but that is not very illuminating. So what's your other way?

actually wait a sec...

I don't think there is a pair of consecutive integers that bounds $2^{x}$ and $2^{x+1}$

so I must have done something wrong...

ah I knew what's going on... I cannot bound $2^{2018}\frac{\ln 2}{\ln 3}$ outward, i must bound it inward otherwise I will overestimate the bound

Dattier

15:30

what do you think about the calculus of 2^i/3^j ?

Secret

what do you mean, its derivative?

Dattier

you can build an algorithm with this idea

which calculate $a$

do you see ?

Secret

Newton's method?

Dattier

No, but maybe it 's working

Abcd

@Secret You specialise in which branch of chemistry?

Secret

15:40

Organometallic and computational

Abcd

Okay.

Secret

what is your question?

skullpatrol

what do you compute?

Abcd

@Secret I was going to link these bounty questions to you: 1 , 2, if you had said "organic/ inorganic or both"

Secret

in order to do organometallic, one must have adequate understanding of both inorganic and organic

so let me check

Abcd

15:45

@skullpatrol See Martin's answers on Chemistry.SE, you'll understand what a computational chemist does.

Secret

https://chemistry.stackexchange.com/questions/15620/what-is-bents-rule
I found it amazing that there are so many rules that I don't know even exist. UNSW must be really good at explaining most things using MO theory...

Abcd

@Secret What is UNSW?

Bent's rule is quite popular on Chem.SE

Secret

my uni back in undergraduate

Abcd

Okay.

@Secret Do you think Soumik Das' answer is valid? Those are all his (undergraduate) opinions therefore I am somewhat doubtful.

HRJ

15:59

Hey People. Computer Engineer here. I have spent about two weeks trying to understand the results of this paper (free access).

I am not able to verify the result in chapter 7 (example deconvolution). If I take the computed X and convolve with the given A, I don't get the expected B. Could someone help me understand?

nbro

Hi!

I have a question about notation.

Which angle are we exactly referring to in the following picture:

?

Secret

abcd: Analysing...

nbro

The problem of notation is its rigidity. It helps to achieve rigor and conciseness when presenting formulas or concepts, but, if ambiguous or incorrect, it can be more painful than a discursive explanation of the formula.

Secret

Now analysing Soumik Das' answer

Abcd

@Secret For fluorine too I effect is stronger than M effect

Therefore fluorine is also a deactivating group

But fluorine is the weakest deactivating group.

So $I \approx M$

Semiclassical

16:13

Usually, it’ll be the angle between the rays $\overrightarrow{p_k p_{k-1}}$ and $\overrightarrow{p_k q_i}$

Abcd

But you cant say $M>I$

Semiclassical

Fixed

Abcd

Source: Advanced Organic Chemistry By March @Secret

i'll edit my question(after a while) to mention that.

nbro

@Semiclassical Why do you call them rays and not segments?

Secret

btw, I got the positions of my $\delta^-$ wrong, let me fix that...

Abcd

16:20

(You can also use [google draw] (which provides hexagon)(chrome.google.com/webstore/detail/google-drawings/…) for drawing better structures.)

Secret

Translation of my thought process:

Abcd

@Secret Are you sure about the H bond order? I think it should be F>>Cl>Br>I

Secret

Well, what I remember is F is much stronger to the order of similar to O and N, and Cl is the 4th guy the follows that but a bit weaker (not rmb how much weaker, but Br and I has negligible H bonding

Abcd

Okay

Secret

There are 3 major effects at play here:
1. The mesomeric effect/resonance effect (M/R) [e- shifts due to p orbital overlap]
2. The inductive effect (I) [e- shift due to electronegativity]
3. Intermolecular hydrogen bonding

There orderings, are respectively:
1. F > Cl > Br > I
2. F > Cl > Br > I
3. F >> Cl >(>?) Br >(>>) I

So the overall acidity is the relative strengths of these factors. Now:

Using the diagram up there (which is basically I am so lazy thus I smush the resonance diagrams altogether by using the deltas), we can see that I effect and M effect competes with each other. Therefore, a molecule is less acidic if:

(+M effect) + (H bonding) > (I effect)

Now let's start with F:

uh sorry, I brainfarted

2

let me try again

For F:

F has strong I, M and H. Since it is the least acidic, we thus conclude that M+H > I, but we don't have enough information to say how is I vs M, thus it is possible that for F, M > I, that you will need further check to know

Cl has strong I and M, but H is a lot weaker, thus it is a bit more acidic

Br has weak I and M, and negligible H, thus it is more acidic than Cl. Same applies for I

Now since I of Br > I, we have that Br is more acidic than I

@Abcd While you find I > M for all halogens, the fact that F is most acidic just show relative to the Cl case, how big the contribution of the H bonding is

and so, Soumik Das answer is a more detailed version of mine and is ok

Abcd

16:46

@Secret Right, got it , thanks. I think I should award the bounty to him...

Secret

do you found the I effect of F is also stronger than the M effect?

cause that's the only thing that is not correct about his answer if this is the case (everything else is fine)

Abcd

34 mins ago, by Abcd

But fluorine is the weakest deactivating group.

35 mins ago, by Abcd

So $I \approx M$

Secret

ah gotcha

Abcd

Fluorine is deactivating because I effect is stronger than M effect.

Secret

but its deactiviting ability is also the weakest (i.e. |I-M| is smallest), I see

Abcd

16:49

Yup.

Secret

I don;t recall on top of my head how to compare |I-M| though

and now, to check martin's answer

and see how that translates to bent's rule...

Abcd

@Secret Martin's which answer?

bent's rule question is answered by Ron

Secret

https://chemistry.stackexchange.com/questions/34842/why-is-bcl3-a-monomer-whereas-alcl3-exists-as-a-dimer/64600

I followed that link from the comments

pardon my occupation sickness lol...

> To make this straight: There are no hyperbonds in (XCl3)2;X={B,Al}; this includes three-centre-two-electron bonds, and three-centre-four-electron bonds. And deeper insight to those will be offered on another day.

This is different from what I am being taught in my undergrad

Abcd

I see.

Secret

17:07

Martin's an excellent work on the analysis, but it is clearly too high level since the description involve quantum effects and it really does not explain why the angle is so other than stronger back bonding of the boron monomer. Now to check Ron's answer

@abcd I see there were two links, the first one is the one I meant. Here is the important bit for your question: "So after all we have two kinds of bonds in the dimers four terminal X−Cl and four bridging X−μCl bonds. Therefore the most accurate description is with formal charges (also the simplest)." So, if it is correct to describe the dimer with formal charges, then the deviation in bond angles is in perfect compliance with Bent's rule. — Stian Yttervik Oct 17 '17 at 15:49

hmm...

Tanuj

@Secret got any hints about the monkey and the banana ?

Secret

@Abcd Ah I see it now. Ron's answer explain how Bent's rule works. Meanwhile Martin's calculations have shown that the Al2Cl6 (in the gas phase) bonding is basically described by formal charge, and how ionic it is, which means we expect Al to be quite positive. Since the Cl-Al-Cl angle is much smaller than tetrahedral, it agrees with the bents rule prediction. James Gaidis's answer address the issue of Al2Cl6 in the solid phase

So the short answer is, yes you can apply bents rule, but the long answer will need to unite Martin's, Ron's and Jame's answer into one

Abcd

@Secret Please elaborate. I didn't get you.

Secret

17:27

wait a sec..., I think I focused on the wrong angle...

Secret

17:38

@Tanuj without further info, not really, cause the pulley like answer that is given will mean the banana had to move down as the monkey climb up, and I cannot figure out any real life setting of a pulley system where two heavy objects can be on one side and the rope not fasten to some support such as the branch without slipping off

Perhaps Johnrennie will have better ideas

Lembik

can anyone help me with notation. What does 1[{i}] mean in section 3.2.1 of arxiv.org/pdf/1107.4466.pdf

Balarka Sen

@AlessandroCodenotti I wrote a bunch of jackshit in the endspace room if you want to look at it

Simon Sirak

17:54

Hi!

DarkRunner

How can you conclude $(35)^{ 2222 }+(14)^{ 5555 }\quad (mod\quad 7)\equiv 0$?

I mean I get that 35 and 14 are multiples of 7, and thus they "become" 0, but why does this work?

if someone can he;p that would be great

Astyx

You can prove that if $a\equiv b \pmod c$ then $a^n\equiv b^n \pmod c$

Also if $a\equiv b$ and $c\equiv d$ then $a+c \equiv b+d$

William Oliver

I am really struggling to get this question across. I am hoping someone can help me make it more formal. Basically, when dealing with co and contravariant vector components under a certain basis for several co and contravariant vectors, its clear that any function of quantities of the form $v_iw^i$ where $v_i$ are components of a covariant vector and $w^i$ are the components of a contravariant vector is invariant under a change of basis (invoking Einstein summation convention here).

My question is, is the converse true? That is, is any invariant function of vector coordinates under a certain basis a function of quantities of the above form?

Eric Silva

18:10

@Balarka I needed a sanity check cause I think I was being stupid last night. What is $H^1_c(\mathbb{R}^2 -\{0\})$

Astyx

So since $35 \equiv 0$ and $14 \equiv 0$, you get $35^{2222}+14^{5555} \equiv 0^{2222} + 0^{5555} \equiv 0$

William Oliver

This is my attempt at formalizing it, but it feels weird and overly complicated, gyazo.com/1f8a86469d2431c3cdaaef831f0f83bd and the "product of all possible projection maps" is vague.

Astyx

Yo Dami

Mike Miller

@Eric Kunneth says R

You get something from the radial direction and nothing from the circular direction

Eric Silva

hmmk

Mike Miller

18:21

The circle bounds a noncompact submanifold, so if you have a closed compactly supported form Stokes says if must integrate to 0 on the circle

Eric Silva

right

Semiclassical

What does the $c$ in $H_c^1$ stand for? @EricSilva

Eric Silva

18:36

Compact support

Joe Shmo

is there a formula for a pullback of a product of two functions?

Abcd

18:53

0

Q: $f(x)=a\sin(x)^2+2b\sin(x)\cos(x)+c\cos(x)^2.$ Find $\max(f), x\in [0,2\pi]$.

For $a,b,c\in \mathbb R $ and $f :[0, 2\pi ] \to \mathbb R$ defined with: $f(x)=a\sin(x)^2+2b\sin(x)\cos(x)+c\cos(x)^2.$ Find $\max(f), x\in [0,2\pi]$.

the OP writes $\sin x^2$

Does he/she mean $\sin^2x$? or $\sin (x^2)$?

Skortya

Can somebody help me remember the specifics of this decomposition of a vector using the inner product: $v=\sum_i (v,v_i)e_i$, does the basis have to be orthonormal or not?

user193319

0

Q: Finest Locally Convex Topology on a Group Ring

Let $G$ the free group on $n$ generators, and let $\Bbb{C}G$ denote the complex free group ring (which is the same the group algebra, since $G$ is a discrete group). Endow $\Bbb{C}G$ with the finest locally convex topology. Is much known about such a topology on $\Bbb{C}G$; are there any study re...

general-topology locally-convex-spaces group-rings

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$Mathematics$ Mathematics