Mathematics

8:00 PM

now pull pieces of the circles up at intersection points appropriately onto the half plane z > 0 to make it the hopf link

Mike Miller

half-space; and you need to pull one up and the other down

else you get something unlinked

Balarka Sen

you can pull one up and the other up too.

Mike Miller

Oh, sure. Misread what you said.

Balarka Sen

robjohn

@Chris'ssis I have not tried it yet.

Balarka Sen

8:02 PM

like this

Mike Miller

Yes, I understand you now.

Balarka Sen

now imagine you have deleted this guy out from R^3. in the positive upper half of R^3, i.e., z > 0, this looks like two handles deleted out.

Mike Miller

Yes.

Balarka Sen

now take a point x_0 in the halfplane R+^3. the noncontractible loops wind around these handles

that's as far as i have right now

still thinking about it

Mike Miller

You're on your way to my two-part van Kampen proof. Make sure to be very careful about basepoints.

Balarka Sen

8:07 PM

i think i'll sleep on it

byes.

user153330

@BalarkaSen bbye

Don Larynx

8:25 PM

Yay! I got my joga bonito hat!

Hi @Pedro

@Studentmath okay, so say we have 5 slots. We have two 1's we want to put into these empty slots. Then we can choose 5*4 ways to put them in, or 20 ways. But three items, we get 5*4*3, so it's not the same

Chris's sis

@robjohn OK

Jasper Loy

@MikeMiller You have a new proof?

Studentmath

Nah. We have (5*4)/(2*1) ways to put them in, or 10 ways. For three it's (5*4*3)/(3*2*1)

10 ways as well

Don Larynx

ohhhh

I see what I did wrong.

Studentmath

Because you don't care which 1 comes first - the 1's are undistinguishable

Mike Miller

8:35 PM

No, @JasperLoy, I meant I could prove his statement using van Kampen twice.

Venus

@user153330 Good luck for the bounty ^^

Don Larynx

Aha! So if we had to arrange "1,2" into 5 slots, it would be 5*4; "1,2,3" would be 5*4*3 @Studentmath

Eureka! My combinatorics level is now $2$.

Studentmath

Yep!

Don Larynx

The order of the objects does not matter.

Studentmath

If they are precisely the same, yep. You can't differ between 101 and 101 (I switched the 1's)

Don Larynx

8:45 PM

It's the same for three, because the slots (empty slots) are also indistinguishable

so rearrange three empty slots in 5 spaces

same as arranging two same items in 5 spaces

Studentmath

Well yeah, you only have two items - filled slot (mark with 1) and empty slot (marked with 0), so the location of the filled slots defines the location of the empty slots.

Chris's sis

$$\int _0^{\pi/2}\int _0^{\pi/2}\cos (x) \cos (y) \csc (x-y) \log \left(\frac{\sin (x)+\cos (x)}{\cos (x) \sec (y) (\sin (y)+\cos (y))}\right) \ dx \ dy$$
$$=\frac{\pi ^3}{16}+\frac{3}{8}\zeta(2)-G-\frac{1}{4} \pi \log (2)$$

Studentmath

Which gives rise to $\binom{n}{k}=\binom{n}{n-k}$

robjohn

@Chris'ssis The answer makes one think it is doable...

Chris's sis

@robjohn Yeah, it only seems like that. Actually, it depends on the tools one uses. My luck is that I created a tool that allows me to compute it quickly.

Jasper Loy

8:53 PM

@Chris'ssis Quickly, not fastly.

Chris's sis

@JasperLoy Is it wrong to say "fastly"?

robjohn

@Chris'ssis fastly is not a word

Jasper Loy

@Chris'ssis Yes. But you could simply say fast there too, though that would be more informal.

Chris's sis

@robjohn Ooooooooo, OK. I'll never use it again.

Daniel Fischer

@robjohn If you have a little time, the first three comments here are now obsolete.

robjohn

8:55 PM

@DanielFischer let me look

Chris's sis

@JasperLoy OK, thanks! :-)

Jasper Loy

I have been feeling very terrible the past few days, but now I feel better again. I hope it stays this way, at least until after I have started studying next Thu.

Don Larynx

terrible about what?

Jasper Loy

Well, just my usual mental problems, which are too long to describe in this chat.

Don Larynx

mental problems suck. brains can be so stupid sometimes.

@Huy: math.stackexchange.com/questions/1076156/…

Khallil Benyattou

9:04 PM

What's the logic behind the step in this substitution when trying to solve an integral?
$$ \sinh(u) = x \implies \cosh(u)\text{ d}u = \text{d}x $$

Don Larynx

@Khallil: Depends on the integral

Khallil Benyattou

$$ \int \dfrac{\text{arsinh}(x) x}{\sqrt{x^2 + 1}} \text{ d}x $$

Does it really matter, @DonLarynx?

I just wanted to understand how they went from the LHS of the implication to the RHS when using the LHS as a substitution, @DonLarynx.

Don Larynx

@Khallil: Consider the integral $\int_0^1 x dx$.

Khallil Benyattou

I'm lost.

It's just one half.

Don Larynx

Well cosh is the derivative of sinh

Daniel Fischer

9:08 PM

@KhallilBenyattou Plain old substitution. If we substitute $x = g(u)$, then we have $dx = g'(u)\,du$.

Khallil Benyattou

I'm more used to seeing $x = g(u) \implies \frac{\text{d}x}{\text{d}u} = g'(u)$, @DanielFischer.
Are the two forms equivalent? I'm really uneasy with 'multiplying' (if that's even what I'm doing) by $\text{d}u$.

Huy

@DonLarynx: You could have asked before posting.

Don Larynx

I did, @Huy.

You ignored it, hence I posted.

Huy

@DonLarynx: I never received a question from you in this chat.

Daniel Fischer

@KhallilBenyattou If you think in differential forms, the two are really equivalent. Before that, they're only functionally equivalent. But remember the substitution rule $$\int_{g(a)}^{g(b)} f(x)\,dx = \int_a^b f(g(u))\cdot g'(u)\,du$$ for bijective differentiable $g$ to see that they are indeed functionally equivalent. "Replace $x$ with $g(u)$ and $dx$ with $g'(u)\,du$".

evinda

9:21 PM

Hello @DanielFischer!!! I want to write an algorithm that deletes any element of a heap $H$ in time $O(\log m)$. Could you give me a hint how we could do this?

Huy

@DonLarynx: That's not even the differential equation I wrote down, afaik.

Studentmath

@Evinda is it max or min heap, or something else?

Don Larynx

@Huy: chat.stackexchange.com/transcript/message/19188358#19188358

Huy

@DonLarynx: I only saw a message containing "@Huy" and thus had no idea what you wanted.

Don Larynx

Does this question make sense?? sosmath.com/CBB/viewtopic.php?t=24832

Oh, sorry, @Huy.

Huy

9:25 PM

And the ODE you wrote in your post is a different from the one I posted in the chat a few days ago.

evinda

@Studentmath I assume that a min heap is meant...

Don Larynx

@Huy is the same ODE as the one here math.stackexchange.com/questions/1076156/…

Huy

@DonLarynx: The ODE I posted a few days ago is a different one than you posted on MSE.

Don Larynx

@Huy: Click the arrow.

Huy

?

Don Larynx

9:27 PM

The arrow, next to "@Huy"

Huy

And?

Don Larynx

Compare that ODE to the one on MSE

Huy

The ODE I posted a few days ago was a different one.

Don Larynx

@Huy: I have a typo.

OK!

evinda

Nice hat, @MikeMiller :D

Don Larynx

9:29 PM

The typo has been dealt with and the ODEs are the same now

Suppose we have three empty slots. We can put two $1$'s in these slots. How many ways are there? Clearly, $3*2 = 6$. However, the $1$'s are identical! So really, there's only (dividing by the number of identical $1$'s = $2$) $3$ ways to arrange these $1$'s into these three empty slots.

Are the $1$'s necessarily next to each other?

No, they're not.

i.e. 110, 101, 011

Khallil Benyattou

9:47 PM

Sorry for the late reply, @DanielFischer. I haven't actually seen that rule before.
Do you know of any books in which I can read more about it and prove the substitution rule?
(I feel like proving it would help my understanding a bit more.)

Don Larynx

@KhallilBenyattou This is called u-substitution

Jasper Loy

@DonLarynx What if I use another letter? What if I don't like u?

Mike Miller

I don't like him either, @Jasper.

Jasper Loy

@MikeMiller Of course, you are referring to the letter and not to Don, lol.

Daniel Fischer

@KhallilBenyattou Any introductory book on analysis or calculus, whichever applies to where you are, dealing with integration should have and prove the substitution rule. At least for piecewise continuous $f$. Then it's a matter of applying the fundamental theorem of calculus with the chain rule.

Jasper Loy

9:52 PM

@DanielFischer Now you make it sound so simple.

evinda

What does it mean that the points are $\mathbb{Z}-$linearly dependent?

Daniel Fischer

@JasperLoy For piecewise continuous $f$, it is simple. It only becomes non-simple if you go and prove it for all integrable $f$.

Don Larynx

@evinda one of the points is a multiple of the other

in terms of integers

i.e. (2,4) Z-L.D. of (1,2) or (-1, -2)

but not (-1, 2)

evinda

@DonLarynx So is it for example $P_1(x,y)=\lambda \cdot P_2(x,y), \lambda \in \mathbb{Z}$ where $P_1, P_2$ are points?

Khallil Benyattou

Ah, gotcha. I've only done the introductory study of sequences, series and completeness for Analysis, so I'll get to it eventually.
Thanks for your reply, @DanielFischer! ^_^

Don Larynx

9:58 PM

@Evinda yes

and $x, y \in \Bbb{R}$

evinda

@DonLarynx Nice.. :)

@DonLarynx Thanks :)

@DonLarynx Could you maybe take a look at an exercise in algebraic geometry?

Don Larynx

I am unable to help right now

evinda

@DonLarynx Ok, no problem :)

Don Larynx

I am working on a combination with repetitons problem

Huy

@DonLarynx: I posted a solution.

evinda

10:00 PM

Aha...

Random Variable

@DanielFischer Indeed the two integrals are related, as a couple answers to the question show. But the OP specifically asked if an evaluation using contour integration was possible. And the answer I posted doesn't involve the other integral.

Don Larynx

Yay! I will take a look at it later.

Huy

@DonLarynx: It doesn't answer your question though, I think. But it's a solution to the ODE.

@Venus: How was your date?

Daniel Fischer

@RandomVariable Yes. I just wanted to indicate that it wasn't totally absurd to close as duplicate, since different integral limits alone don't guarantee it's not a duplicate.

I should maybe have also mentioned the request for contour integration, but you didn't mention that either, @RandomVariable, so that makes two of us who forgot.

Jasper Loy

@Huy I have a date every day. My date is now 23 Dec.

Random Variable

10:12 PM

@DanielFischer Should I edit my post on meta even though the question has already been reopened?

Daniel Fischer

@RandomVariable I don't think that's necessary. If it gets re-closed as duplicate, mention it in the new post.

Don Larynx

►

this just happened to me.

Daniel Fischer

@RandomVariable And by the way, nice job.

Random Variable

@DanielFischer Thanks.

Alexander Gruber

10:34 PM

@Behaviour Ahh, somebody finally got me. What's the next best 0-downvote post?

Behaviour

@AlexanderGruber The one about $\sum 1/n^2$. Here's the full list. Change type to 2 to get answers instead of questions.

Don Larynx

Hi @Studentmath, are you here?

Alexander Gruber

@Behaviour Nice. The answers are doing better. Good to see Arturo still on top.

Jasper Loy

@AlexanderGruber Is Arturo ever coming back?

Alexander Gruber

@JasperLoy Doubt it. I met him at a conference once, though. His research is going good.

Studentmath

10:42 PM

@Don yep

Don Larynx

@Studentmath math.stackexchange.com/questions/1078150/…

Studentmath

Both answers provided are correct, how can I be of use though?

Don Larynx

You were too late :p

evinda

10:58 PM

Could you take a look at this: math.stackexchange.com/questions/1078104/… ?

N3buchadnezzar

Is $[f(x) - f(y)] /(x - y)$ anything special?

It almost looks like the mean value theorem?

Kaj Hansen

@N3buchadnezzar, it is. You can find at least one point between $x$ and $y$ with that value as the derivative.

With the appropriate hypothesis, of course.

N3buchadnezzar

11:14 PM

@KajHansen Ok.I thought it was a plus and not minus

Kaj Hansen

Nope, that's how it appears in the MVT.

N3buchadnezzar

I am trying to find the solutions to
$$
\frac{f(x) - f(y)}{x - y} = \frac{f(x+y)}{x+y}
$$

Kaj Hansen

Do you mean to have $f(x) + f(y)$ in the numerator on the RHS?

Or correct as is?

N3buchadnezzar

@KajHansen It is correct. Otherwise the function would be a constant right?

Kaj Hansen

Probably, yeah.

Mike Miller

11:17 PM

Mm, I'm tied in hats again. Unless one of the top 3 forgets a timed hat, I'll be remaining this way until I get Treasure Hunter.

N3buchadnezzar

@MikeMiller Hats of for you

Behaviour

@MikeMiller That's a good strategy: keep a hat up your sleeve, so they think they're safe.

Mike Miller

Perhaps I should stop pursuing my academia Treasure Hunter and only get it on MSE (where I'll get it the day before the event ends).

Behaviour

Is that Electorate that you are shooting for?

Mike Miller

On Academia, yes. On MSE, I've yet to get Steward, and I've done 759 close reviews.

Of course, if I wanted to annoy everyone, I could just make 400-odd edits for Copy Editor.

Behaviour

11:24 PM

That'd be risky: things can go wrong with Steward.

Mike Miller

@Ted's here just in time to see us talk about hats.

Behaviour

One can speed up Electorate by downvoting some closed posts between 0 UTC and 3 UTC, so that they become eligible for roomba. Roomba runs daily at 3 UTC, and you get the votes back.

Transcript for

$Mathematics$ Mathematics