Mathematics - 2016-02-10 (page 2 of 3)

1:10 PM

@Idle001 if I say that all squares of positive integers are either divisible by 4 or leave a remainder when divided by 4, I can provide examples for it(and also actually prove it). But on the other hand if I just say that without giving any concrete proof of why that is correct(by proof I here mean a mathematical way to reason why your answer is correct without supporting with a few examples) no one is going to accept your result.

It would be just like saying that I believe P=NP without giving any reason why I believe it to be

Idle001

@Albas every number is either divisible by or not divisible by any other number !

anyways, got downvoted on an already accepted and bounty-granted answer, i feel very sick now

Aayush Agrawal

1:27 PM

I need some advice. I posted a question about my attempt to solve a particular equation up to the point i got stuck. I didn't know there was an error in my computation. I ended up selecting an answer that blatantly went to the solution (Using a different equation), without pointing out the error at all. Then i noticed there's another answer that actually mentions that i made an error at a point and then goes to the solution from there. Should i change the correct answer to that one?

iwriteonbananas

Hello

Semiclassical

@AayushAgrawal the choice of accepted answer is always up to the asker, and in particular you're in no way obliged to stick with one which you now consider inferior to another

you could include a comment to the previous answer explaining why you're making the change, but that's simply a matter of courtesy

Albas

1:56 PM

I have no clue of what you meant there @Idle001

Semiclassical

@Albas as written, that's a pretty weak statement (for any number $b$, every other number is either divisible by it or has some remainder). did you mean "leave a remainder of 1" specifically?

Albas

I mean to say that given the square of an integer it is either divisible by 4 or leaves a remainder of 1. By that if there is a number whose remainder when divided by 4 is 2 or 3 then that number can never be a square@Semiclassical

Semiclassical

i figured as much, yeah. but i think that accidental omission probably explains Idle's comment

Albas

Oh sorry I guess I forgot to write 1 there. My bad@Semiclassical

Semiclassical

no worries.

to engage on that line of thinking in a somewhat different way: suppose someone tells me that the sum of the first $n$ integers is $n(n+1)/2$.

Albas

2:06 PM

Till I dont prove it myself (or read the proof(obviously check if there is any logical error in the proof) somewhere) I cannot believe what he said is true

Semiclassical

i can 'check' that result in a few different ways. first, i can just plug in values of $n$ and see if it succeeds. that is, however, only a way to possibly falsify the statement; it does not justify it.

or i could supply a proof by induction that the formula works. that is logically sound, but arguably not very satisfying because it doesn't explain how one comes up with the formula.

Albas

That is my only issue with induction. You can do it if you have a slight idea of what your end result might be

Semiclassical

third, i could do Gauss's classroom trick and observe that $$1+2+3+\cdots+n=n+(n-1)+\cdots+1 = n\cdot \frac{1+n}{2}$$

and that one, i think, is rather satisfying

since it provides not only a proof of the formula but also gives it a logical interpretation

Albas

Or I could generalize the idea to sum of a series of terms in an arithmetic progression and then use it here

Semiclassical

by that do you mean, generalizing to a different initial value in the progression?

my point being: each of those approaches is a genuine argument, but they're not equal in explanatory power.

Albas

2:15 PM

Given terms $a_1,a_2,...,a_n$ in an arithmetic sequence what is $\sum_{i=1}^n{a_i}$

Semiclassical

right.

i just view that as a linear transformation of the simplest case

Axoren

Can someone give me a sanity check? The rationals are countable, therefore there's an one-to-one mapping function $f$ onto the naturals. Can we select $f$ such that $f^{-1}$ is a continuous mapping onto the rationals? No, right?

Semiclassical

anyways, the first argument establishes only that the formula could be correct. the second establishes that it is indeed correct, and the third explains why such a result is natural

Albas

Hierarchy of proofs

Semiclassical

well, the first one isn't really a proof. but a hierarchy of arguments, yes

one thing the third one doesn't do, on the other hand, is provide some idea of how to generalize this result to squares, cubes, etc.

Albas

2:19 PM

Yes true. I guess for squares you have to use the formula $(n+1)^3-n^3$ somehow if I remember my classes correctly. But I am sure there are 10 more proofs lying here and there

Semiclassical

and while i can pretty easily come up with an argument that any such generalization should end up being some polynomial in $n$ of appropriate degree, coming up with the actual coefficients is harder

Axoren

Rationals are a disconnected set. Nevermind. I thought I was crazy for a second. Just unattentive.

Semiclassical

there's a cute 'proof without words' if memory serves

namely, that you can think of $1+4+9+\cdots+n^2$ as a square pyramid built out of cubes

Albas

Yes right.

Semiclassical

i don't remember how that exactly works out, but it's there. but that really only helps for squares

on the other hand, if i recall it's actually pretty straightforward to find formulas for $\sum_{k=1}^n (k)_r$ where $(k)_r= k(k-1)\cdots (k-r+1)$

Albas

2:23 PM

But the interesting part comes when we look at infinite series. Then comes convergence

Semiclassical

depends on what one is interested in, i guess

Albas

And just recently yesterday I did a super beautiful question relating the cantor set and an infinite series

Semiclassical

myself i'm a fan of formal power series, where issues of convergence are pretty much irrelevant unless one wants to take advantage of complex analysis methods

Albas

I dont know about formal power series that much. Can you explain in short @Semiclassical

actually nothing

I just know that they have some kind of ring structure(I may be horribly wrong) associated with them(I dont even know how that comes)

Semiclassical

well, here's why they're useful

suppose i start with a formal power series $A(x) = \sum_{k=0}^\infty a_k x^k$

where $a_k$ is some sequence (of integers, let's say)

Idle001

2:29 PM

@Albas ok its clear now, but how is that linked to my issue ?

Albas

Read the discussion between me and semiclassical @Idle001

Semiclassical

the trick is that i can implement manipulations of the sequence as manipulations of the series

Idle001

now, its raised up to 3 dvts, nice, i have 300 pts if u want to downvote more

@Albas i said it got clearer, but what is the rapport here ?

Albas

@Semiclassical I did not get that statement

Idle001

between division property and my recurrence form

Semiclassical

2:32 PM

I think the objection is that the question asked for an explanation for where the recurrence came from. and while there is some of that in your answer, it's quite thin (compared with the accepted answer, for example)

though I think the example with arithmetic progressions is more comparable to what you were doing, since such a sequence of sums can be represented as a recurrence

@Albas for instance, $(1-x)A(x) = \sum_{k=0}^\infty a_k x^k-\sum_{k=1}^\infty a_{k-1} x^k = a_0 +\sum_{k=1}^\infty (a_k-a_{k-1})x^k$

so it takes the sequence $\{a_0, a_1,\cdots\}$ and replaces it with $\{a_0,a_1-a_0,a_2-a_1,\cdots\}$

Albas

Hmm... Pretty cool. But why formal power series ? Shouldnt this work with any power series?

Semiclassical

well, suppose i have something like $a_k =k!$

those manipulations will still work formally, but now the series converges only for $x=0$.

so thinking about $A(x)$ as something I could plug actual values of $x$ into no longer makes much sense. but as a way of representing manipulations of sequences, the convergence is irrelevant.

Albas

So the question of convergence is out of picture. But why does formal power series gives interesting results?

Semiclassical

well, here's some examples of it in action.

suppose i start with $A(x) = 1+x+x^2+\cdots$. multiplying by $1-x$ then kills of all terms but the first, so $(1-x)A(x) = 1$. thus one has $A(x) = (1-x)^{-1}$, where I can think of the RHS either as an actual function or as a representation of the formal series $1+x+x^2+\cdots$

and since multiplying by $(1-x)$ took the first difference of the sequence $\{a_k\}$, multiplying by $(1-x)^{-1}$ does the inverse of that: $$\{a_0,a_1,a_2\cdots \}\mapsto \{a_0,a_0+a_1,\cdots\}$$

so for example, $$(1-x)^{-2} = (1-x)^{-1}(1+x+x^2+\cdots) = 1+2x+3x^2+\cdots$$

Albas

So I guess due to these abstract properties we get a ring structure on it. Very cool stuff

Semiclassical

2:46 PM

yeah. admittedly, i don't work a lot with the formal properties of such things so much as take advantage of them to count things.

and in truth, it really is nice when your series is actually convergent so that you can use tools from calculus

say, to identify the $n$th term of that series with the $n$th derivative evaluated at $x=0$

Albas

Hmm... I have feeling that the operation on the ring might be a product( Just a feeling since many rings usually have product as their operations). But thats for some other day

Semiclassical

one also sees so-called exponential generating functions (the above are known as ordinary generating functions) with $\displaystyle A(x) = \sum_{k=0}^\infty a_k \dfrac{x^k}{k!}$

Albas

if I let $a_k=1$ for all k that is $e^x$

Semiclassical

right.

Albas

Where do you use them @Semiclassical I suspect you are a physics graduate student

Semiclassical

2:50 PM

and more generally, if $[x^k]A(x)=a_k$ (the $k$th term) then $[x^k]A'(x) = a_{k+1}$

yep

i don't use them a lot in my own research, i should admit (though they have come up once or twice)

but i've always been fond of them.

Albas

Your research is in?

Semiclassical

theoretical condensed matter physics. though for me that basically boils down to "eigenvalue problems"

especially asymptotic methods for such, which in quantum mechanics (if you think of Planck's constant as being a 'small' parameter for a given system) amounts to semiclassical approximations of the system.

Albas

Hmm... linear algebra. I plan to do it after my real analysis course. I myself have only done linear algebra till polynomials and determinants(including linear transformations).

Semiclassical

right. though the usual ones i do involve sparse structured matrices that can be arbitrarily large

Albas

Well since I am in high school I do not know much of the physics you are talking about@Semiclassical. One physics concept that really shocks me is entropy. But that is just I guess a property of a macrostate and cannot be used in microstates

But I do know plancks constant

Semiclassical

2:57 PM

well, that is connected with all that stat mech stuff

so if you end up doing any of that kind've thing you'll see generating functions, albeit in a rather disguised way

Albas

That is cool

Semiclassical

yep.

i wouldn't say that that generating function logic is at all the heart of the matter, but it does make understanding certain calculations a bit simpler

Albas

Do you also use topology @Semiclassical in your research

Semiclassical

depends on what you mean by topology

point set topology, not really

algebraic topology, definitely

Albas

I know algebraic topology only till fundamental groups of a few spaces

Semiclassical

3:00 PM

technically, what i do is all in the realm of 'topological insulators' and 'topological superconductors'

though i usually just boil it down to specific toy models

Albas

superconductors and topology? they seem totally unrelated to me

Semiclassical

well, it's not so strange once one understands it

a state of a physical system can be thought of as a map from one space (the physical sites of a system, say) to another (the possible values we can associate with each such site)

so, for example, I can think of an electric field as telling me, at each point in space, what vector to associate with it

and then you can ask questions about the nature of said mapping, and try to classify them topologically

topological insulators/superconductors represent one line of development of that idea

so at the level of analogy, at least, it's not so strange. but that doesn't make the details easy.

Albas

when you say "try to classify them topologically" what do you mean

Semiclassical

well, suppose i require that my electric field be continuous

in that case, i can just as well envision it as a set of field lines by following the 'flow' of those vectors

and one thing i know about electric field lines is that they can only begin/end at positive/negative charges

Akiva Weinberger

So, like, because of the Hairy Ball Theorem, if I have an electrical field on a metal ball, I know that there must be a point with zero electricity. Or something. Right?

I'm probably completely wrong, as I don't know anything about this stuff.

Semiclassical

3:11 PM

well, the physics doesn't work for that. the electric field on a conductor is perpendicular at its surface

Akiva Weinberger

Oh

Semiclassical

but that basic idea is the right one, and would work directly in the context of spin systems for example

for the electric field case, a consequence of said begin/end logic is that if i draw a sphere around a given portion of space and 'count' the field lines passing through the surface, i can figure out how much charge is inside

Akiva Weinberger

I'm imagining some variant of the winding number

But with spheres rather than loops

Semiclassical

you're not wrong. with winding number, you're considering mappings of the circle $S^1$ into a given space

for example, i can map it into the real plane, and classify that map according to how it winds around the origin

with the electric field case, one is instead considering mappings of the unit sphere $S^2$ into the relevant space

Akiva Weinberger

Right

Semiclassical

3:17 PM

(the space being, in this case, $\mathbb{R}^3$ minus all the points where the charges sit)

Akiva Weinberger

And if two maps of $S^1$ are homotopic, they have the same winding number

Semiclassical

wellllll

Akiva Weinberger

And I guess the same idea works for $S^2$ in $\Bbb R^3\setminus\{0\}$

Homotopic might be the wrong word.

Semiclassical

not quite. reason being is that, for some manifolds, the homotopy group can be non-abelian

Akiva Weinberger

…right.

Semiclassical

3:18 PM

on the other hand, the first homology group is always abelian

and in fact the first homology group is the abelianization of the first homotopy group

Akiva Weinberger

Oh, so thats what that is

Semiclassical

so i usually associate winding numbers with homology rather than homotopy

Akiva Weinberger

Got it

Semiclassical

(though given that physicists like to use non-commuting 'grassmann numbers' in certain places, i'm not as insistent on that as i might otherwise be)

Mike Miller

@Semiclassical I don't see what the complaint is or where you're talking about abelianity

Akiva Weinberger

3:21 PM

Is the second homotopy group of $\Bbb R^3\setminus\{0\}$ commutative?

Mike Miller

all higher homotopy groups are abelian

Semiclassical

i have in mind the first homotopy group of a sphere with three punctures

Akiva Weinberger

Which is the same as a plane with two punctures

Mike Miller

Sure, but what does that have to do with "And if they're homotopic the winding numbers are the same"

were you just worried about the lack of a notion of winding number?

Akiva Weinberger

By letting one puncture be the North Pole and projecting

Semiclassical

3:23 PM

i think i was misreading the earlier statement, actually. what i was getting at was that "two maps being homotopic is not equivalent to them having the same winding numbers."

Akiva Weinberger

Are they equivalent with the plane with the origin missing? (The winding number being around the origin.)

Semiclassical

a sphere with $n$ punctures is equivalent to a plane with $n-1$ punctures, yeah

Mike Miller

no objections, your honor

Semiclassical

hah

bottom line being: "Two maps having the same winding numbers is a necessary but not sufficient condition of them being homotopic."

Akiva Weinberger

Can you give me a link where I could find the variant of the winding number that deals with mappings of $S^2$? I don't know how to define the winding number of a sphere around a point.

Semiclassical

3:27 PM

well, i wouldn't call it 'winding number' in that case

Akiva Weinberger

Right, but I don't know what to call it.

Semiclassical

right. 'topological charge' is one word i've heard for it

in which case, you can define it using Gauss's law

i suppose the story goes as: suppose $F:S^2\to \mathbb{R}^3-\{0\}$ is a continuous map.

Mike Miller

I mean, maps from $S^2$ to what?

Semiclassical

well, i'm envisioning this as a system with some number of charges at the origin

but hrm. something's not clicking in my head.

robjohn

@Idle001 Sorry, I was not around when you pinged. I am familiar with recurrences.

Mike Miller

3:33 PM

that was aimed at @Akiva

Semiclassical

ahh

Mike Miller

This is also the issue when one says winding number; that is meaningful in subsets of the plane

Akiva Weinberger

@MikeMiller $\Bbb R^3\setminus\{0\}$

Hey, this looks like what I'm looking for (almost):

en.m.wikipedia.org/wiki/Degree_of_a_continuous_mapping

Mike Miller

ok

Semiclassical

@MikeMiller well, i'd consider it to also be useful in things like $\mathbb{R}^3$ with the $z$-axis removed

Akiva Weinberger

3:35 PM

Specifically, if $F:S^2\to\Bbb R^3\setminus\{0\}$ is any mapping, and $\pi:\Bbb R^3\setminus\{0\}\to S^2$ projects out from the origin…

robjohn

I am surprised at the number of upvotes my answer to this question has gotten, but completely baffled at the number that the most upvoted answer has gotten. Although it looks plausible, no formula is given, just the first few terms, but no proof is given.

Semiclassical

though i guess that's kind've a silly example (since that's just $\mathbb{R}^2\setminus\{0\}\times \mathbb{R}$)

Akiva Weinberger

Then the "winding number" of $F$ (the 3D version of a "loop") would be the degree of $\pi F$.

robjohn

It must be a hot question or something. Yes, I see it is. That is why there are so many upvotes.

Semiclassical

the way i'd define topological charge for mappings of $S^2$ is pretty much just physics

Mike Miller

3:38 PM

sure.

Semiclassical

put a unit charge at the origin (with field $\mathbf{E}(\mathbf{r})=\hat{r}/r^2$). then for any $f$ mapping $S^2$ to $\mathbb{R}^3\setminus\{0\}$, i can compute the flux of $\mathbf{E}$ through $f(S^2)$

and then this will be (up to a factor of $4\pi$?) the topological charge of $f$

Mike Miller

@robjohn: I was going to call you picky but I realized I actually have no idea what his series is supposed to be.

robjohn

@Semiclassical Do you mean $\hat{r}/r^2$ (inverse linear) or $\hat{r}/r^3$ (inverse square)

Semiclassical

inverse square. but $\hat{r}$ is a unit vector here

robjohn

@MikeMiller infinite sum of irrationals that is rational.

Akiva Weinberger

3:41 PM

@MikeMiller Who's, RemcoGerlich's?

robjohn

@Semiclassical Oh, okay, I read $\hat{r}$ as a position vector.

Semiclassical

yeah, i reserve $\mathbf{r}$ for that

Akiva Weinberger

(Infinite sum of linearly independent irrationals.)

Semiclassical

thou admittedly in actual calculations i prefer $\mathbf{r}/r^3$

Idle001

@robjohn what do u think about my last two answers which are succintly delete-voted

if u confirm they are wrong , i would delete em geez

Mike Miller

3:43 PM

@robjohn: I mean the terms.

robjohn

@Idle001 let me look

Akiva Weinberger

@Idle001 If they were deleted, we can't see what they were…

Semiclassical

"wrong" and "useful" are not the same thing when it comes to voting.

robjohn

@MikeMiller you have to look at the picture to see the pattern. It is hard to work up a formula so working up a proof that they are rationally independent is very hard.

Idle001

@AkivaWeinberger not deleted but voted to be deleted

Semiclassical

3:45 PM

i don't really agree with downvote = vote to delete

robjohn

@MikeMiller His first answer was definitely rationally dependent, yet he got an enormous number of upvotes before it was fixed.

Akiva Weinberger

This? math.stackexchange.com/questions/1648088/…

Semiclassical

to me downvote can just mean "needs improvement"

Idle001

yes and its predecessor

Semiclassical

obviously if a lot of people downvote an answer, there's a likelihood that it'll be deleted on the basis of it being low-quality

but i don't consider the two one and the same

Idle001

3:46 PM

i m waiting confirmation of robjohn

Akiva Weinberger

What are $a_n$, $b_n$ and $c_n$?

Idle001

recurrence formulas indeed :/

Semiclassical

Recurrence formulas for what, though?

Akiva Weinberger

You haven't explained what they mean.

Your section labeled "explanation" doesn't explain very well.

Idle001

20 lines explanation isnt sufficient ?

yes, maybe my weak english

its open for edit though

Akiva Weinberger

3:50 PM

Wait, never mind

It was mostly because my browser wasn't showing all of it

Sorry

Idle001

u use phone ?

Akiva Weinberger

I opened it on the app now (I'm on my phone)

Yeah

Idle001

usually people face visuality problems on phone

Akiva Weinberger

What does "right-ended" mean

Idle001

there is a little option in the bottom-right entitled mobile, for this case

it means xxxxxxxxA , where A is the right-end

Akiva Weinberger

3:53 PM

So, $a_n$ is the number of sequences of length $n$…

$b_n$ is the number of sequences of length $n$ ending with $0$…

Idle001

it all sums up to $a$ yes

exactly

Mike Miller

@Semiclassical: It's worth noting that all that stuff you said is really just an instantiation of degree. Let $M$ be a closed smooth $n$-manifold, $f: M \to \Bbb R^{n+1} - 0$ a smooth map. Project to the unit sphere. Take degree of this map; that's the winding number around 0.

Akiva Weinberger

And $c$ is what?

Idle001

and $c$ is 0000...01

Mike Miller

My phone likes to try to correct instantiaton to instanton.

Akiva Weinberger

3:54 PM

@MikeMiller Ah, like I said, right?

@Idle001 What do you mean?

Idle001

00000001000 is $b$

Semiclassical

i forget what the word instantiation means

Idle001

00000001 is $c$

Secret

Some algebra construction fun:
Consider the set $$A=\left\{\begin{pmatrix}r \\ \theta\end{pmatrix}:r,\theta\in \mathbb{R}\right\}$$
with binary operator $\cdot$ to form a $\langle$insert name$\rangle$

$$(A,\cdot)$$

Define
$$\begin{pmatrix}r_1 \\ \theta_1\end{pmatrix}\cdot \begin{pmatrix}r_2 \\ \theta_2\end{pmatrix}=\begin{pmatrix}r_1r_2 \\ \theta_1+\theta_2\end{pmatrix}$$
where $+$ and multiplication is as defined in $\mathbb{R}$
Then the identity is
$$\mathbf{1}=\begin{pmatrix}1 \\ 0\end{pmatrix}$$ since $$\begin{pmatrix}1 \\ 0\end{pmatrix}\cdot \begin{pmatrix}r \\ \theta\end{pmatrix}=\b…

Idle001

0000011 is $a$

00000012 is forbidden

Semiclassical

3:56 PM

more to the point, i don't actually remember what the degree of a map is

Akiva Weinberger

@Idle001 Wait

Idle001

if i understand the core of the question well

Akiva Weinberger

Why is $c_2$ equal to $6$? What are the six sequences of length $2$ that are counted?

Mike Miller

special appearance of

Akiva Weinberger

($c_2$ is listed as $6$ in the last example)

Semiclassical

3:57 PM

ahh

Idle001

lemme check again the text is foggy

Semiclassical

and looking at the $S^n \to S^n$ bit of Wiki's page on the degree of continuous maps, i think i see the point

Mike Miller

take a regular value in the codomain; it's locally a diffeomorphism near the points in the inverse image. count the points, with sign depending on whether that local diffeo is orientation preserving or reversing

I'm an artist

@robjohn hey. The series I showed you yesterday cannot be computed in that form. After doing the calculations I fixed a mistake and add another series to it to make it have sense.

Mike Miller

of course $M$ needs to be orientable to do this.

Idle001

4:00 PM

$c_2$=01xx there is 6 possibilities for xx to be (00,01,02,10,11,12) @AkivaWeinberger

Akiva Weinberger

Ohhh

See, that wasn't clear at all in your answer

Idle001

i skipped detailed explanation thinking people would decipher it from first look-at, it is simple for people who are usual with this stuff

Semiclassical

@MikeMiller that's basically what's in the differential topology part of the Wiki page

Akiva Weinberger

@Idle001 Don't assume that people are familiar with your specific approach to a problem.

Semiclassical

which concludes by linking it up with de Rham's theorem and therefore presumably implies the construction I gave for $n=2$

Akiva Weinberger

4:03 PM

@Idle001 Maybe later I'll try to edit your answer and make it more clear (or, if it's deleted by then, I'll write it up as a separate answer and give you credit)

Semiclassical

with mine having been constructed so that $\int_Y \omega =1$

Idle001

@AkivaWeinberger thank you anyways, no need to credit there exist another answer which i fail to understand

thank u @robjohn for ur comments

Mike Miller

@Semiclassical: My only complaint about yours is that you let $f$ be merely continuous

to pull back a 2-form you'd better be at least $C^1$

I'm an artist

@robjohn I gained a profound view on such series, I see a lot of connections that I didn't see before with other important results I got during the time.

Semiclassical

@MikeMiller that's fair

the other thing one could complain about is that taking $\hat{r}/r^2$ as one's field gives $\int_Y \omega = 4\pi$ not $1$ :)

but w/e

Mike Miller

4:20 PM

eh, like you said that's inconsequential; renormalize

Semiclassical

yeah

Idle001

wats going on here why m i serial-dvted ?

robjohn

@MikeMiller I computed the first two terms and have shown that they are not rationally independent. However, the upvotes keep rolling in.

Balarka Sen

4:35 PM

@Huy if $H$ is transverse to $\beta$, $H^{-1}(\beta)$ is a 1-subfold. Y shape isn't a 1-fold

iwriteonbananas

4:53 PM

Hey @BalarkaSen and @MikeMiller

Balarka Sen

Hi @iwriteonbananas.

iwriteonbananas

That notification scared me. Had volume up as high as possible, without knowing that.

Balarka Sen

:P

iwriteonbananas

I'm gonna start studying homotopy theory soon. You too?

Balarka Sen

@iwriteonbananas I have to get calculus out of my way first.

When are you starting studying homotopy theory?

iwriteonbananas

4:57 PM

Ok, sure. shouldn't take you any longer than a week or two

Balarka Sen

Yes, I am going to devour Ted's book.

I'm an artist

Thinking to offer a bounty for this one

Balarka Sen

Exams are over too.

I'm an artist

$$\int_0^1 \frac{ \log (x) \log ^5(1-x)\text{Li}_2(1-x)}{x}+\frac{\log ^4(1-x)(\text{Li}_2(1-x){})^2 }{x}-\frac{ \log (x) \log ^4(1-x)\text{Li}_3(1-x)}{x}$$ $$-\frac{2 \log ^3(1-x)\text{Li}_2(1-x) \text{Li}_3(1-x) }{x}+\frac{\log ^2(1-x)(\text{Li}_3(1-x){})^2 }{x} \textrm{d} x$$

@BalarkaSen ^^^ great for calculus practice

Balarka Sen

No, it's not.

iwriteonbananas

4:59 PM

I wanna start as soon as possible with homotopy theory. Need to learn a shitton for upcoming seminar presentations which I committed to.

Balarka Sen

Still, a tentative date?

iwriteonbananas

Pretty sure I'm on a great way to a burnout

@BalarkaSen The presentations are at the beginning of July

I'm an artist

@Semiclassical ^^^

Balarka Sen

@iwriteonbananas Oh, that's a long way from now.

So I'll have time to catch up with you :D

OK, I am going to study some calculus. Let's start with Ted's exams.

iwriteonbananas

@BalarkaSen Did he give you some old exams?

Balarka Sen

5:05 PM

Yes, lots. And a lot of homework too.

Get started then!

Hi @iwriteonbananas.

Hey Mike

5:47 PM

@robjohn love the solution provided by CS guy, it s distinguishably upvote-worthy

Akiva Weinberger

@Idle001 Yeah, that's what I thought, too. It gets upvotes for creativity/novelty.

Idle001

as a cs graded i see this answer is candidate to be accepted

robjohn

@Idle001 The one that is not rationally independent?

Idle001

@robjohn oh, i missed that part, there is a linear rational relation indeed

robjohn

@Idle001 math works because things are proven. That answer was missing its proof.

Akiva Weinberger

6:03 PM

@robjohn I had upvoted the first version of it before you showed that it didn't work. Theoretically, I could rescind my upvote, but I decided against it.

robjohn

@AkivaWeinberger I am not urging people to change their votes, I was just amazed at the influx of votes, not only for my answer, but the flood of votes for an answer which is incomplete and ultimately was not rationally independent.

I have to go for a while. BBML

Balarka Sen

What is the answer we are talking about?

Akiva Weinberger

130

A: Can an infinite sum of irrational numbers be rational?

EDIT: Pardon me, but it has been shown in the comments by robjohn and Michael that these are not linearly independent. Indeed:$$91a_1-10a_2=10$$ — Akiva Weinberger Think of a series of real numbers with decimal expansions like 0.1100110000110000001100000000110000000000110000000... 0.001...

Balarka Sen

People probably just missed the Q-linear independence bit.

Akiva Weinberger

@BalarkaSen No, he originally had a different series, but when it was pointed out that they were linearly dependent, he changed the series to this one.

Balarka Sen

6:16 PM

Ah, OK.

PVAL

7:03 PM

@Semiclassical wow that banded method helps alot. It did in 30 seconds what it couldn't do in 12 hours.

I'd like to show that (3(signature)+2(dimension of the matrix) is always positive. This seems to increase pretty rapidly.

It did the first 30 in a few seconds. Well the first 50 seems like a nice round number. Seems like it will take a little time though.

(First 50 examples took about 5 minutes)

Ted Shifrin

7:21 PM

@PVAL: Glad to know you're getting advice from an expert ;)

Huy

hey @Ted, how's life ?

PVAL

Ya his optimizations did in about 15 seconds what mine couldn't do in 12 hours.

Ted Shifrin

I expected some numerical methods tweaking were required.

Hi @Huy

mr eyeglasses

hi @TedShifrin

Ted Shifrin

Hi @mreyeglasses

Balarka Sen

7:23 PM

Hi @TedShifrin.

Ted Shifrin

and hi @Balarka

Balarka Sen

I am revising the previous chapters a bit. I.e., working through the earlier exams.

Ted Shifrin

well, my exams are very low level by comparison with the material

Balarka Sen

Oh, and the school exams are over. :)

Ted Shifrin

I presume you have not been kicked out of school, @Balarka

Balarka Sen

7:26 PM

Well, I'll know by May. That's when the results come out.

Ted Shifrin

Wow ... surely it doesn't take them 3 months to grade them.

oh, if it's nationalized exams, maybe it does

Balarka Sen

Well, madhyamik doesn't work like US high school graduation.

Ted Shifrin

I hope you managed not to fail your maths exams.

Balarka Sen

Oh, no, I expect 100.

Ted Shifrin

Good.

Balarka Sen

7:29 PM

There were very vague word problems (what the heck does a retailer and a manufacturer do? Is that even math?) but there were a lot of optionals.

Ted Shifrin

It's math for the "real world," yes

Balarka Sen

So I survived, lack of general common sense notwithstanding.

Ted Shifrin

The manufacturer sells to the retailer, who in turn sells to the customer.

Balarka Sen

Huh.

Ted Shifrin

And everyone wants to maximize his profit.

Balarka Sen

7:31 PM

That makes sense.

Huy

you need to play some Sim City or Cities Skylines

Ted Shifrin

Huh?

Huy

that's where I learnt about how the real world works

Balarka Sen

lol

Ted Shifrin

Donald Trump just insults everyone and the world explodes ... That's the real world.

Huy

7:32 PM

that certainly happens in Sim City

Balarka Sen

There was a very stupid problem: Milk contains 89% of water (really?). Given a sample of Milk containing 90% of water, what is the excess of water in 22 litres of such a sample?

I mean, the answer is clearly (90-89)x22/100, but what really is mind-boggling is that they gave 5 points for this problem. Confuzzling!

I think it was a printing mistake done by the editor of the question papers. The "excess" should have appeared before the 90% instead of where it is.

Huy

so just 22/100

Balarka Sen

Yes, @Huy.

Huy

wow I'm so good

I should do this for a living

Balarka Sen

We know.

Huy

7:36 PM

I think 89% sounds about right

I remember it slightly less

but 80+

Balarka Sen

who cares, I am allergic to milk.

speaking of, I think I am allergic to everything I can think of.

Ted Shifrin

We're allergic to you, though, so it works out.

Balarka Sen

My kind of allergy is contagious so, tautologically, everyone should be allergic to me.

I'm an artist

@RandomVariable did you try this integral?$$\int_0^1 \frac{ \log (x) \log ^5(1-x)\text{Li}_2(1-x)}{x}+\frac{\log ^4(1-x)(\text{Li}_2(1-x){})^2 }{x}-\frac{ \log (x) \log ^4(1-x)\text{Li}_3(1-x)}{x}$$ $$-\frac{2 \log ^3(1-x)\text{Li}_2(1-x) \text{Li}_3(1-x) }{x}+\frac{\log ^2(1-x)(\text{Li}_3(1-x){})^2 }{x} \textrm{d} x$$

Balarka Sen

@TedShifrin Anything you'd recommend I should do from your book in chapter 2+3+4 as revision? I just finished your exam #2.

Hippalectryon

7:48 PM

Hi :D

Huy

@BalarkaSen: there are pills if you're lactose intolerant

Idle001

@Hippalectryon hey le pegase

Huy

very practical

Hippalectryon

@Idle001 o/

Balarka Sen

@Huy I am everything intolerant.

Huy

7:49 PM

I can see that

you should run for president of the USA

Balarka Sen

that's the last thing USA would want: an Indian president.

PVAL

@MikeMiller I was told the following showed $\Sigma(2,3,5)#-\Sigma(2,3,5)$ embeds in $\Bbb R^4$. Take an embedding of a punctured $\Sigma(2,3,5)$ and delete a neighborhood of the punctured to get a smooth 3-manifold with boundary. Take a regular neighborhood.of this manifold is $(Sigma(2,3,5)-D^3 )\times I$ and take the boundary. Does this work?

Balarka Sen

Hi @AndrewThompson.