@heather remember the following

@0celo7 i am ready to listen =)

Suppose $(a_n)$ is a sequence such that $\sum_{i\ge1}|a_n|<\infty$. Then $\sum_{i\ge 1}a_n$ converges.

The result holds in arbitrary Banach spaces

I just wasted four hours because I forgot it

ouch. well, at least you probably won't forget it again now.

I was trying to use some deep deep theorems to try to prove this shit

Then I eventually rederived what I just told you

and my stupidity clicked

Woah guys

When you think about it...space is expanding

Dear lord I need sleep

Then go to sleep god damnit

Don't make the same mistakes I do

It's 3 AM and I'm browsing the web mindlessly

I can't

Too hot in here

Get out of the furnace then

How?

Kick the door until it gives way

I was just reading about expansion. What a coincidence

@Avantgarde Cosmology is crazy

@SirCumference in a good way

Yep

Weren't you looking for a logo for astronomy a few weeks ago?

@SirCumference

@Avantgarde Yup

Did you get to find anything?

Not yet :/

I see

Listening to Hotline Bling atm

Ew

If you want good Drake listen to "What a Time to Be Alive"

I know

Let me see. I've never heard that

Specifically, "Big Rings, "Diamonds Dancing, "Scholarships," "Jumpman."

This is quite different from the style used in his other popular songs like Hotline Bling

Probably because he collaborated lol. It's nice

@Avantgarde "Sneakin'," No Long Talk," "Gyalchester," "Portland"

Gyalchester is pretty amazing.

On it

make sure whatever you're listening on can bass

Why am I having a hard time finding the songs on youtube? Did he get them removed or something?

Not sure

I use Apple Music

I get all lyric videos where they display the lyrics without sound. Or they put in the beats, without vocals. Weird

@0celo7 Just use iTunes

@SirCumference wot

@Avantgarde

@0celo7 Seriously? Bieber?

This is terrible

@yuggib I am having a really hard time with heat kernels. I'll rant... $\phi_i$ are the eigenfunctions of $\Delta$ and $\mu_i$ the eigenvalues, so $$H(x,y,t)=\sum_{i\ge 1}e^{-\mu_it}\phi_i(x)\phi_i(y).$$ I have uniform convergence on $M\times M\times[a,\infty)$ for any $a>0$. I also have that $$\int \left|\sum_{i=1}^ke^{-\mu_it}\phi_i(x)\nabla\phi_i(y)\right|^2dy$$ converges as $k\to\infty$ uniformly in $(x,t)$. Supposedly we then have that the finite sums converge to $H(x,y,t)$ weakly in

$W^{1,2}$, I think they mean in the $y$ variable, for any $(x,t)$.

It doesn't seem obvious to me.

Maybe one should use Theorem 3 on page 121 of Yosida. Clearly the series is bounded in $W^{1,2}$ because it converges in the $L^2$ part of the norm and the derivative part has a bounded norm.

One can take $C^\infty$ to be a strongly dense subset of the dual. So take $\psi\in C^\infty$, then $$\int\sum_{i=1}^k e^{-\mu_i t}\phi_i(x)\nabla\phi_i(y)\cdot \nabla\psi(y)\,dy.$$

But now one can integrate by parts, noting that the $\phi_i$ are Dirichlet eigenfunctions.

Then one gets something like $$\int \sum_{i\ge 1}e^{-\mu_it}\phi_i(x)\phi_i(y) \Delta \psi(y)\,dy$$ in the limit.

But it's not clear the other term is in $W^{1,2}$, so can one even integrate by parts?

One might be able to use Rellich to show that the kernel is in $W^{1,2}$, but I'm not sure.

@Avantgarde It's a good song

Right, I can use Rellich to get weak convergence of a subsequence in $W^{1,2}$ and strong convergence in $L^2$. But we know the $L^2$ limit, so its weak $W^{1,2}$ limit is $H(x,y,t)$. So it's in $W^{1,2}$ and I can integrate by parts.

Maybe...

It does equal zero on the boundary.

@JohnRennie Apparently he didn't think too much about the Dell

getting a final list put together, trying to compute the price

the god damn tax man is trying to prevent me from having fun

$1482.26

Exactly at max budget

@0celo7 buying the Dell would be the cheap option. If you're willing to spend $1500 then obviously you can do a lot better.

The point about going the Dell route is that you could put together a system for less than half your budget. It wouldn't be anywhere near as fast, but if it was fast enough then it would be fine,

I don't really have a problem dropping that much as long as I'm getting something worth that

Does that make sense?

Ugh, unfortunately I'm going a trip for a week.

@0celo7 Absolutely. If you want to play the latest games at the top resolution then you need the best kit ... and that costs. Shrug :-)

I don't know how this shipping will work out. I suppose I could have everything sent to my brother's house!

I'll call him tomorrow. He'd actually love to build a PC with me

@JohnRennie I earned $0.13 in interest today, I've got plenty of money :P

@0celo7 planning to buy something?

See above

Looks like some laptop or computer. I didn't understand completely. Hence, I asked.

@Sid the ultimate gaming machine! :-)

sounds fun..

sounds interesting

I really do not understand the heat equation

sup Secret

@0celo7 Get a ROCCAT keyboard. That is for gamers only.

@Sid I have a Corsair

That works too..

Hi, everybody.

Hey Daniel

@Avantgarde hi

Can someone help me with this question?

@Abcd What's the problem here?

@Avantgarde I can't solve, it's difficult.

@Abcd It's just using standard equations that related time, distance, speeds and acceleration

@Avantgarde Nope. I tried solving using those equations, I couldn't.

@Avantgarde How do I solve it? Any hints?

Just equivalent ways of arriving at the same answer. Check your calculations again

@Abcd analyse the motion in two parts.

@JohnRennie Yes. I was doing that. First upward then total downward, but couldn't arrive at any answer.

First consider the upward part of the motion. You know the initial velocity $u$, the final velocity $v=0$ and you know the acceleration $g$, so you can calculate the time using $$s = ut + \tfrac{1}{2}at^2 $$

@JohnRennie We don't know the $s$ for upward motion.

$$ v^2 = u^2 + 2as $$

Actually the time is just calculated using $$v = u + at $$

Also, the sign is important. Take upwards as +ve and downwards as -ve

@Sid Yes.

displacement is h.

Solve that using the equations of motion

(considering there is no resistive force by air)

for total downward path, take the height as $h+$ height above tower. The only way you could've gone wrong is by taking just $h$

I think the answer is B. (note- I am too lazy to pick up a pen and paper to do it. I have solved mentally. It might be wrong)

Can it be solved using displacement =h only?

Simply using the equation: $S = ut - \frac{1}{2}gt^2$

I think so.

@Abcd the problem is you don't know how high it goes on the upward part of the motion

Actually, you don't need to know that ...

@JohnRennie I am just unable to solve it. Why can't it be done using displacement only?

Take the first part to be the ball moving from your height up then down again back to your height.

Why can't it be done using displacement only?

The initial velocity is $+v$ and the final velocity is $-v$, so the time for this part is given by: $$ 2v = gt $$ Is that OK so far?

If you throw a ball up with some velocity $v$ then (ignoring air resistance) when it gets back to you it must be moving at the same speed but in the opposite direction.

Ah, since you deleted your comment I guess you've got this now ...

@JohnRennie yes

@JohnRennie v=gt

not 2v

Right?

foliations.org/surveys/FoliationLectNotes_Milnor.pdf

that's one ugly ass paper

@Abcd If you use $v=gt$ that gives you the time for the ball to leave your hand and come to a stop at the highest part of its motion. Then it takes the same time to fall back to your hand again. That's why I've used $2v=gt$

and then, you have to find its velocity at that height, then, at the ground, and then, find the time. Add both the times..

Sounds tedious.

@Abcd: would it help if I drew a diagram?

Hi :-)

How is life in Chennai / driving / food / everything else this morning?

Still scary?

A three point turn?

Oh, @Kaumudi.H You in Chennai?

That's one of the standard things you get taught (and tested on) in the UK :-)

That and reversing round a corner

I visited Chennai... I think a decade back.

I don't even remember much about what sightseeing I did apart from the fact that the food was very spicy

Mmm :-)

FYI, I'm not sure if I will be able to make it to tomorrow's (today's?) chat session. I'll try though.

I don't believe we have anything in particular to chat about though.

@JohnRennie I know the diagram. I just can't solve it.

@DavidZ I was wondering if we need a canonical What do we expect from a question or answer on the PSE post.

We've just had two users kick off because what they were posting is at odds with what we expect.

@Kaumudi.H pretty good. It's cooled down in the UK and now it's about 18C, which I consider ideal.

@JohnRennie That would certainly be worth discussing

Cool enough that I can go racing around Chester on my pushbike without dying of heat exhaustion :-)

@DavidZ there may already be one lurking in the Meta somewhere - I haven't looked. I think it would be nice to have a definitive post on the subject we can point users to.

@Kaumudi.H I deliberately cycle fast for the exercise.

You'll get used to Canada. It's surprising how quickly the body adapts to a change in environment.

@JohnRennie I obtained: $gt^2 + \frac{1}{2} gt^2 + h = 0$ . SHould I substitute $v= -gt$

@Kaumudi.H Though in winter the liquid nitrogen condensing on the pavements can make it a bit slippery :-)

@JohnRennie I can't think of one offhand, but there are a lot of meta posts.

@Abcd see:

@JohnRennie There's a problem with signs there.

For simplicity I've assumed everything is positive i.e. $v$ is the modulus of the velocity and $g$ is the modulus of the acceleration. Solving the equations I've written for $t$ will give you the correct result.

@JohnRennie Please see: clay6.com/qa/52505/…

Can you please tell why he took only h instead of 2h in the second equation below the answer.

since he has multiplied the whole equation by 2.

@Abcd That looks like a mistake to me ...

He has got the correct answer (A) but I think there is a mistake in his working

Oh, it was A.

@Abcd that's why you should never trust mental mathematics. See, I made a mistake in the signs. Always write things up in a paper while solving

@Sid there's an easy way to see it must be (A). Set $h = 0$ in which case $t = 2v/g$. That rules out (C) and (D). Then set $v=0$ in which case $t=\sqrt{2h/g}$. That rules out (B). Only $A$ is left.

@Kaumudi.H I'm free for the next 20 minutes ...

@Kaumudi.H though my coffee is getting low ...

@Kaumudi.H Hello?

@Kaumudi.H there now ...

@Sid ok, thanks.

Next PSE priviledge is moderator tools

rubs hands

@HritikNarayan: nice answer to the Rindler question. I was holding off answering because I suspected it might just be a homework question.

@Slereah Next privilege has to do with tag wiki edits :(

I had to create a few tags

Because of my weird questions

@JohnRennie I pretty much held off till the last moment but decided to go for it anyway

@HritikNarayan in any case I don't mind homework questions if they're about a reasonably high level topic like relativity.

I only get annoyed when they're basic questions that could be Googled if the OP could be bothered.

Black Holes have an associated Hawking Radiation because they have an event horizon, and it's similar with the Unruh effect and the Rindler horizon, if I'm not mistaken. The earth doesn't give out hawking radiation because there's no horizon as such. ($r_s<R_E$). Why is a horizon an important element for thermal radiation?

@JohnRennie Those hurt the most, yes

The horizon affects the way the radiation propagates to infinity

@JohnRennie Yeah I did read your answer! I agree that it might be almost impossible to explain at a layman's level. I've been self learning GR so I think I'll learn the true explanation someday

@HritikNarayan If I'm honest I don't understand it myself.

I kind of see it, but not on anything approaching a rigorous level.

Here's a super random question inspired from last night dream: Suppose we have a absorption spectra of some complicated molecular species, I wonder if it will be useful if in addition to examining it visually, to also convert it to audio form and examining it acoustically?

@Kaumudi.H I find the idea of performing a three-point turn in a large Indian city to be a potential source of endless nightmares.

@DawoodibnKareem :-)

I guess the point is that you can do it not that you should do it :-)

I guess even large cities have quiet roads.

Sometimes engineered to be that way, like the insane one-way system of central Toronto.

"Our notation and conventions are those of Birrell and Davies. In particular, the metric has the signature (+- . . .-). "

Reeee

@Slereah dianajuvenille.files.wordpress.com/2012/08/captain_haddock.jpg

@BernardoMeurer: did you know King Gizzard and the Lizard Wizard played at last weekend's Glastonbury Festival? I can get the video of it if you want.

@Kaumudi.H Well that sounds like a whole lot less fun.

bbc.com/news/technology-36517340?SThisFB

Hurts the human in an unpredictable way? Sounds easy. if(rand()%2) //KILL THE HUMAN

What does AI got to do there? :P

^I concur this is not a conundrum with AI at all, saying "The robot makes a decision that I as a creator cannot predict" would equally apply to just a pseudo-random generator.

if (x^n + y^n + z^n = 0 has solution for n > 2) release_poison_gaz();

@Slereah That only kills Fermat (and puts Andrew Wiles to a coma?)

Are non-Fermat humans immune to poison gas?

Also Fermat's dead, so I guess his hypothesis was false

This paper defines the "past light cone" on a spacetime

$C^-(p)$

Oh wait, it's for null curves

So it's not quite as standard as $I^-$

(Although most books that define it it's still $E^-$)

What a hoax

Umm, x = y = z = 0?

#rekt

I know a guy called Gaz. I guess he was released.

I think has solution should be has non-trivial solution

de minimis non curat lex

xyz \neq 0 is the extra condition, yes

(1, -1, 0)

x, y, z must be positive integers

I just wrote down the condition...

No, they need not be positive integers.

they should be real numbers

No, never!

They are integers.

Just not necessarily positive.

nvm, n > 2

x^n + y^n = z^n has infinitely many solutions for all n if you allow x, y, z to be real

yea, I considered 3i, 4i, etc. to be integers. Silly me.

WHen we say that the particle is accelerating vertically upward with 4 m/s^2 ... Is it after taking gravitational acceleration into consideration or not?

after taking it into consideration

Thanks.

you don't have to worry about gravitational forces

because the particle is accelerating upwards with 4m/s; that implies that the force is large enough to overcome the gravitational force and cause an acceleration of 4m/s in the upward direction.

Why can negative acceleration not cause increase in magnitude of velocity?

It can if the initial velocity is zero or if you give enough time.

If velocity and acceleration have opposite signs, the object is slowing down.

True or false?

@Yashas ok

true

How?

@Yashas How?

because the acceleration and velocity are parallel (or anti-parallel) vectors

they are on the same line

$\vec{a} = \frac{d\vec{v}}{dt}$

as they are on the same line, you can write it as

$-|a| = \frac{d|v|}{dt}$

$-|a|$ because the acceleration and velocity are in opposite direction

you can write it as $a = \frac{dv}{dt}$

"because the acceleration and velocity are parallel (or anti-parallel) vectors" must be "because the acceleration and velocity are anti-parallel vectors"

if (x^n - 1 = 0 has solution for n > 2) create_black_hole();

It's hard to find a definition of horizons independant of specific spacetimes

Apparently the original definition of horizons is from Rindler in 56