@Abcd quoi?

@aminliverpool Think about the example of the line $x = 0$ in $\matbb{R}^2$. Every subspace that complements that line passes through the affine line $x = 1$, and for each point on that affine line there is a corresponding complementary subspace.

The same idea works in general.

(In fact this is an important example in topology and geometry, because this is how you put charts on projective space.)

why doesn't the typesetting display? anyone got troubleshooting ideas?

You need chatjax!

@aminliverpool Can you identify the one dimensional subspaces in $R^2$ that complement the subspace $x = 0$?

@aminliverpool Right. And each one of those will pass through the line $x = 1$, correct?

@Abcd it isn't correct

So, the assignment that takes a complementing line $L$ to the intersection $L \cap \{x = 1\}$ sets up a function. You can prove this is a bijection.

And that bijection is your identification with the affine line $x = 1$, so that's where the "affine structure" comes from.

[Chemistry]Finally submitted that very bulky error report to the technical centre to have a look. Hopefully they will finally get that stubborn molecule to converge

After that, I might need to start reading up how to do monte carlo on these things...

Monte Carlo is something I only understand in the broad but not technical sense

otherwise, I might just try to see if I can write a python script to make 360 input files one for each molecule orientation in order to scan the potential energy surface

Does this sound clunky "Recall that if $H$ is a group of order $17$ then $H \cong \mathbb{Z}/17\mathbb{Z}$ (i.e. groups of order $17$ have only one isomorphism class) and that if $K$ is a group of order $4$ then $K \cong \mathbb{Z}/4\mathbb{Z}$ or $K \cong (\mathbb{Z}/2\mathbb{Z})^2$ (i.e. groups of order $4$ have two isomorphism classes)."? To someone with experience in algebra and English language! lol

Why is it that if $f:Y\to Z, g:X\to Y $ are given with $f $ injective, then the image of $f\circ g $ is the same as that of $g $? I.e. $fg(X)= g(x) $

because the problem of using the inbuilt scan option of Gaussian 16 is that if any of the calculation fails because of the energy not converging, it does not do the remaining scans

@Richard of course the image isn't the same

myweb.liu.edu/~nmatsuna/gamess/input/GLOBOP.html for the monte carlo stuff I mentioned earlier

@ÍgjøgnumMeg well I understand it :p

I would add "either" after the second "then"

Have you ever run into the sign problem in Monte Carlo?

@LeakyNun Indeed it's understandable, but the phrase "groups of order $4$ have two isomorphism classes" seems a bit weird to me!

@aminliverpool Can you prove that it is a bijection? (You should first prove that it is well defined, i.e. that each line intersects x = 1 exactly once. To prove surjectivity, it will be easier to define an inverse -- to each point on $x = 1$, assign a complementary line passing through that point.)

@LeakyNun I'm sorry, I forgot that $g$ is surjective

@ÍgjøgnumMeg "there are only two groups of order $4$ up to isomorphism"

@Richard they still aren't the same

@LeakyNun Yeah I suppose that is better

Not sure, I actually have not done a monte carlo calculation before, though I am aware of its working principles somewhat

@ÍgjøgnumMeg or "exactly" in lieu of "only"

@LeakyNun Really? Let me check what my professor wrote, this even is an exam correction

Mmkay. The sign problem shows up when you want to do problems with fermions vs bosons

ah, the (anti)symmetry of indistinguishable particles?

@aminliverpool You do the same thing. If your subspace is defined by the linear equation $f = 0$, then take $f = 1$. Each complementary line to $f = 0$ passes through $f = 1$ exactly once.

Right

en.m.wikipedia.org/wiki/Numerical_sign_problem

@aminliverpool All hyperplanes can be defined by such an equation.

hmm, afaik, the teams I work with have not encountered that problem yet, probably because we are modelling classical molecular dynamics

I discovered this today, is this true:

@aminliverpool No this is true for any vector space over a field

@LeakyNun Please tell how to solve it

your approach is right, i.e. divide into right limit and left limit

but your calculations are wrong

@aminliverpool To obtain the equation, note that $B / A$ is a one dimensional space, so isomorphic to $k$. So the quotient map $B \to B /A$ followed by the identification gives a linear functional with kernel exactly A.

@LeakyNun They aren't

@Abcd who said they aren't?

@AlexKChen hmm?

$$ \lim_{n \rightarrow \infty } \frac {\sum_{i = 1}^{n} {\frac{\sigma (i)}{i}}}{n} = \frac{\pi^2}{6}$$ [?]

@LeakyNun Ok. So have I written opposite stuff? Means have I written LHL calculation in RHL and vice versa?

@Abcd no

I conjectured it while working on a HMMT (havard Mathematics matheamtical tournament) problem.

$\sigma$ is...?

The sum of positive divisors equal to or less than the number itself.

@Secret gotcha. It mostly comes up when doing stuff with fermions in field theory

@Abcd none of the answers is right

O **** number theory and combinitorics, Leaky might be a better person to check that sum

@LeakyNun Please tell what's wrong then

@Abcd Did you try to understand the approach I gave you?

@Abcd well you showed none of your steps so I can't really tell you which step is wrong

@aminliverpool Yes. And the action on $f = 1$, is by adding vectors in $f = 0$.

@AlexKChen I can attempt to calculate the sum using program

@ÍgjøgnumMeg It's complicated

@LeakyNun I did exactly that.

#include <stdio.h>
#include <stdint.h>
#define N 20000 //List lenght

int t[N];
int p[N]; //storage for sigma(n)


void prime(void){
    int i=0, j = 3, n = 1;
    t[0] = 2;
    for (j = 3; n<N; ++j){
        for(i = 0; (i<N)&&(t[i]!= 0); ++i){
            if ((j % t[i]) == 0){
                break;
            }
            else if (j < t[i]*t[i]){
                t[n++] = j;
                break;
            }

        }
    }

}
int sahore(int a){
    int i, j, k, l, m;
    j = a;
    k = 0; l= 0 ; m = 1;

Ahah, so show that the average abundancy index of all integers is zeta(2)

(I know just enough to know what those words mean)

Eh that's true ??

@LeakyNun Alright. For LHL : -3/(0-2) I think I am facing problem in evaluating. Please tell me how to do it

I don't even know what zeta(2) means, I just guessed it.

@AlexKChen no, that's a rephrase of your conjecture

@AlexKChen zeta(2) = 1/1^2 + 1/2^2 + 1/3^2 + ... = pi^2/6

The exercise is the following: Let $X $ be such that there exists an injective $f_1:\mathbb{N}\to X $ and such that, calling $Y=X\setminus f(\mathbb{N}) $, we have an injective $f:\mathbb{N}\to Y$. We are asked to construct $h: X\to Y $ and $k:Y \to X$ such that Im (h)=Im (k)=$\mathbb{N}$. My professor starts by saying that the injectivity of $f_1$ implies the existence of a surjective $g: X\to \mathbb{N}$, so far so good. He then says that $h:=fg:X\to Y $ satisfies Im(h)=Im(g)=$\mathbb{N} $

Yeah, thats all

@Abcd wrong

@LeakyNun edited

I suspect the thing to do is look at the Lambert series of divisor-sigma

@Abcd still wrong

@LeakyNun I mean help me evaluate

@Semiclassical Heh I have no idea what your're talking.

@Abcd I still can't see any step

-3/(0-2) is wrong

@Abcd Do you know how to write $f\left( \frac{1}{n} \right)$?

@LeakyNun That's what I wanna know. What are the steps? How to solve?

@ÍgjøgnumMeg No

@ÍgjøgnumMeg this is not necessary

@Abcd just substitute $x=\frac1n$

@LeakyNun I've posted the exercise

@AlexKChen en.m.wikipedia.org/wiki/Lambert_series

@Richard As you see I'm doing 1000 tasks at once

@LeakyNun It's somewhat more rigorous and I prefer it that way.

@LeakyNun Okay

@ÍgjøgnumMeg no it isn't more rigorous

@aminliverpool The vector spaces is $B$. Suppose that $B$ is the zero set of the linear equation $f$. Then $B$ acts on the set of solutions to $f = 1$. This is because, $f(b + x) = f(x)$, so if $f(x) = 1$, then $f(b + x) = 1$.

@LeakyNun ?

Okay, it is either $\sqrt{e}$ or $\frac{\pi^2}{6}$; but for very large I can see it oscillating (but you can show by trivial arguement that it never exceeds $2$ ), I think it's for C's handling of int.

@Abcd tell me the definition of $|x|$

@LeakyNun Sure, sorry, I'd be grateful if you could look at it in a while

@AlexKChen use python

@LeakyNun Alright well it's preferable to $x \to 0^+$ and $x\to 0^-$ for me...

@AlexKChen See also the identities in here: en.m.wikipedia.org/wiki/Euler_function

@ÍgjøgnumMeg I don't see the difference between $x\to0^+$ and $n\to\infty$

@LeakyNun Absolute value of x which = -(x) if x is negative and x if x is positive

in terms of rigor

@Abcd yes

Then??

can you use that to compute one step of LHL?

@LeakyNun I can't undrstand which value to substitute. -1?

@Abcd you don't need a value to substitute

you know that $x$ is negative when you say $x \to 0^-$

yes

@LeakyNun Perhaps rigour was the wrong word for me to use but it's more illustrative of the direction from which one approaches $0$, in my opinion :)

so can you substitute the definition?

In particular, see the formula that shows up just before the "special values" section @AlexKChen

@LeakyNun I got LHL = 1

@Abcd how?

I told you to compute one step...

-3x/ (-x-2x)

@Abcd why -3x?

x is negative

that's why @LeakyNun

x is negative so 3x = -3x?

@LeakyNun Oh :(

\I got LHL = 3 and RHL = 1 @LeakyNun

just as a reminder to people : if you think that your conversation is getting lost in the shuffle, you can create a new room to speak in

@Richard if $h:X\to Y$ then how can Im(h) be $\Bbb N$?

@aminliverpool Indeed that is true. Again, look at the case of x = 0 in R^2 to see this vividly. The general argument can be done as I have outlined (and in some other ways, that may feel more invariant.)

@Abcd good

Ok, thanks

@LeakyNun In other questions I just have to substitute right? (cept in indeterminate cases)

@Abcd depends on the content of "other questions"

@LeakyNun See an example:

@LeakyNun I have to factorise, then substitute right (in such questions)?

@Abcd yes

@LeakyNun One more thing. I understood that why we can eliminate indeterminacy using apparent fallacy

@Abcd because it is equal point-by-point to a continuous function except for the point at which we are taking limit

@LeakyNun Because we are using tends to and not the actual value so we divide and get 1 instead of 0 by 0

Let $f(x) = \dfrac{x^2-4}{x-2}$ and $g(x) = x+2$.

Is my understanding correct @LeakyNun?

we see that $f(x)=g(x)$ except for $x=2$.

@Abcd yes

Ok. Thanks.

@LeakyNun Since $f:\mathbb{N}\to Y $ is injective (where $Y= X\setminus f_1(\mathbb{N})$), don't we know that the cardinality of $Y $ is at least as large as that of $\mathbb{N}$, hence the plausibility of Im(h) being N?

@Richard it has to do with the codomain of $h$

which is $Y$

and may not contain any natural numbers

@aminliverpool They're the same dimension, so there is no contradiction. Also, it really is the same. Some people prefer your description though. To see the connection, think about complementary lines as graphs of a map $A / B \to B$ (inside of $A / B \times B$, which we identify with $A$ by picking a fixed complementing line - in the running example of the plane, take $y = 0$), and note that such a graph is determined by where it intersects $f = 1$.

@aminliverpool I cannot stress enough how helpful it is to draw this in the plane.

@LeakyNun On the other hand, I am not allowed to do that kind of factorization to a non limited function, right?

@Abcd I would say so

@LeakyNun Oh, you're saying that $Y$ for example might have the cardinality of $\Bbb R $, but still might contain any natural numbers

@Richard might not.

@LeakyNun You mean "yes" in plain language

@Abcd I mean "it depends on the context" in plain language.

@LeakyNun Might not, yeah, typo

@Richard yes, that's what I'm saying.

@LeakyNun Please give an example to prove that it depends on the context

maybe your professor meant $g \circ f$.

@aminliverpool Good luck puzzling this out. Feel free to message me, I just may not get back to you for a while since I need to go. The ideas in this example end up being important for an important class of interesting spaces (Grassmannians), so it's worth struggling with it for a while.

@Abcd $\displaystyle \frac{\mathrm d}{\mathrm dx} \frac 1 {x^2} = \frac {(0)(x^2) - (1) (2x)} {(x^2)^2} = \frac {-2x} {x^4} \color{red}= \frac {-2}{x^3}$

@Abcd don't bother dealing with that level of rigor at high school.

@LeakyNun You can't do that

in fact, don't bother with that rigor even in university except in real analysis.

@LeakyNun Alright, I think too much, unfortunately

@Abcd indeed

@LeakyNun Well, he did write $h=fg: X\to Y $

@Richard could you by any chance upload the question asked and the response your professor gave?

If he intended to write $gf $ then the whole approach is kinda meaningless because $f $ goes from $\Bbb N $ to $Y= X\setminus f_1(\Bbb N)$ and $g$ from $X $ to $\Bbb N $, hence $gf$ goes from N to itself, being of no use to answer the question

@LeakyNun You mean posting a question on the main site?

@Richard no, I mean uploading an image of the question

@LeakyNun Ah, unfortunately I don't have a file of it

But what I first wrote is almost precisely the text of the question, apart from the names of the objects

I could rewrite it word by word, maybe I made a mistake in changin the names?

where is the question now?

is it a hard copy?

I copied it after the exam, and I have the handwritten correction, but they're not in English

@Richard which language is it in?

Italian

ok, upload it

Ok, but can I do it on my mobile? I see no button for it

maybe you can write it here in Italian...

Sure, I asked because you asked for an image

In this question L Hospital rule is being violated @LeakyNun ^

@Abcd why?

@LeakyNun See:

@LeakyNun I am getting answer in indeterminate form again so L' Hospital rule is being disobeyed

@Abcd Check your subtraction of those fractions

@SimplyBeautifulArt I am getting zero in the denominator so that's not a problem

No no, that's not what I'm saying. You ought to check your algebra, that is, the first step.

@SimplyBeautifulArt My differentiation is also correct

Not your differentiation, your algebra.

@SimplyBeautifulArt What is wrong with my algebra?

$$\frac1{x-2}-\frac{2(2x-3)}{x^3-3x^2+2}\ne \frac{(x^3-3x^2+2x)-2(2x-3)}{(x-2)(x^3-3x^2+2x)}$$

WHAT!?

That's how we subtract fractions

You just forgot a part

@Abcd it's a careless mistake

@LeakyNun Which mistake?

@SteamyRoot Which part?

The step I've pointed out.

@Abcd try to justify that step

Note that:$$\frac{2(2x-3)}{x^3-3x^2+2x}= \frac{2(2x-3)(x-2)}{(x-2)(x^3-3x^2+2x)}$$

@LeakyNun Man but that's how we subtract fractions.

@Abcd think again

^

$\frac ab-\frac cd=\frac{ad-bc}{bd}$

@LeakyNun Man I used a multiple in the denominator

@AlexKChen Oh my god that's amazing

@Richard scrivi?

Ted Shifrin is "Hit Friends"? xD

@SimplyBeautifulArt Isn't that correct?

@AkivaWeinberger then should I call you a viewing breaker?

@Abcd yes, but it's not what you did

I think Ted's, Semiclassical's, and Balarka's things are the only good ones, to be honest

Oh yes! I forgot to multiply by (x-2)

@LeakyNun Sì ma aspetta qualche minuto, devo risolvere una cosa

@Abcd bingo

"Blank Areas" sounds so cool

@Richard ok

@SimplyBeautifulArt But still, denominator is creating the problem not numerator

If you have a different numerator, it might also vanish

It should be fine after you fix it

ok

@SteamyRoot Your name is an anagram of "Oyster Moat" and "Meaty Torso"

Yup

I use Meaty Torso as a username at some other place

Also Tasty Romeo :P

Also my real name

Didn't know the Oyster Moat one, though, thanks. I may end up using it somewhere :D

is your real name also an anagram

I like TastyRomeo a lot

Well, SteamyRoot is an anagram of my real name :P

Yea

I don't think my name will pop up on any anagram solvers, though

that's what I was asking

You're Dutch?

Belgian, but close enough :P

I'm going to just assume your actual name is Tasty Romeo and move on with my life.

Fair enough

I've used that one as a nickname too, so I'll probably respond out of habit if you call me that

Meaty Torso is cool also

@SimplyBeautifulArt Heh, funny

@SimplyBeautifulArt why is it invalid?

oh, because the limit doesn't exist

Assumption that $L$ exists in the first place

or more intuitively, $L=0$ isn't the only solution of $L=2L$

Lol

if we include infinity

Alternatively, allowing $L$ to lie in the extended reals, it's lack of recognition that infinity solves $L=2L$ as well

I am slow at typing

:-P define number @user685272

(EDIT, extended reals, not extended integers. Thinking about Typhon's thing too much, perhaps.)

Well, if you don't want to go all "extend the reals with infinity", you could also rephrase the mistake as "we assumed the limit was real in the first place"

@SimplyBeautifulArt $\Bbb C$, nothing more, nothing less :P

But but

:P

The Riemann Sphere is so much nicer than just $\mathbb{C}$ :(

@SimplyBeautifulArt a number counts things :P

@user685272 we have infinite counting numbers

We have infinitely many counting numbers...

Well, I'm off. Cya later

cya pal

@user685272 I meant we had tranfinite numbers...

:-)

:-D

@Richard this room is getting full of Italians

@AlessandroCodenotti duo itliani