The h Bar

user223506

12:03 AM

@AccidentalFourierTransform maybe... who knows? who cares? Little point if efforts are wasted here (as you stated)

smaude

1:01 AM

Who wants to check out this question? physics.stackexchange.com/questions/307146/…

heather

1:18 AM

it's quiet today.

Bernardo Meurer

@heather Sh

heather

::yells in the library::

Ramanujan

@heather hello

2

heather

@Ramanujan, hello

Ramanujan

I have 168 points but still it is showing to me

heather

1:23 AM

hmm...

one moment

i don't know, to be honest

@Kenshin, what do you think of this?

@Ramanujan, kenshin would be the one to ask, as he knows more about the source code for the site.

my apologies that I cannot help you more. I approved your meta question, as well.

ZeroTheHero

@GPhys: L&L definitely not a favorite as a textbook. The Russian style is very dense and Landau had such intuition that too many statements are presented as obvious.

dmckee

2:46 AM

@AccidentalFourierTransform I do believe that moderator super-powers can fix that. My fingers have their own opinions about how things should be spelled, and I hadn't noticed.

Bernardo Meurer

@dmckee Do you know how to do induction proofs?

dmckee

@BernardoMeurer Easy ones, probably.

You have to show two things:

(A) That what you want to prove is true in some limiting special case.

(B) That is case (n) is true then case (n+1) must also be true.

Bernardo Meurer

dmckee

with the direction leading from the special case to all other cases.

Bernardo Meurer

"Prove using induction that $a_n \in ]0,6[$ for any $n\in \mathbb N_1$"

So I must prove for n=1 which is trivial

dmckee

2:52 AM

So, try to show that $a_{n+1}$ must be positive and less than $a_n$ as long as $a_n$ is positive. Then use the fact that $a_1 = 6$ to set the upper bound. Maybe?

Yeah. Looks like that might work.

@BernardoMeurer $n=1$ is the starting point and, as you observe is particularly easy in this case.

Bernardo Meurer

I just wrote $0<\frac{a_n}{3}(1-e^{-a_n^2})\le 6$

But then I'm not sure how to conduct the proof so to speak

I see why it's true

But I don't know how to "say it"

dmckee

Might be clearer in the form $0 < (1 - e^{-a_n^2})/3 < 1$.

Bernardo Meurer

Hm? How'd you do that?

dmckee

@BernardoMeurer I looked and saw that $a_{n+1} = a_n (\text{some factor})$ and thought 'Ah! We want some factor to lie between 0 and 1, so I'll write it that way.'

Sir Cumference

Hm, question

Has any SE site used public domain images for its design, or have they all made their designs by scratch?

dmckee

2:58 AM

You 'll have to argue that you know that $a_n$ is positive to insure that $(1 - e^\{\dots\})$ is positive but smaller than unity.

Bernardo Meurer

Hm

dmckee

@SirCumference They have a dungeon full of tormented designers. I presume they do them all.

Sir Cumference

@dmckee Welp, I feel like one

It's a huge pain to create a ton of vector images from scratch

dmckee

@SirCumference But you're torturing yourself.

Sir Cumference

And even worse to make them look good

dmckee

3:00 AM

That's different.

Sir Cumference

@dmckee I know, but it's a project that I started, and thus need to see through

dmckee

@SirCumference Yes. Yes it is. That's why you capture and enslave a designer.

Bernardo Meurer

@dmckee Okay, I get the first part of that, but why can you state that the factor has to be between 0 and 1? This is as if you had stated that $a_n=6$ and then divided by it

Kenshin

@heather more reputation is required to send private messages

I haven't updated the text with the new reputation figure

dmckee

Well. You can start with a case where the value is known to be 6. So $e^\{\dots\}$ is less than 1 (because $\dots$ is negative), and thus $(1-e^\dots)$ is less than one.

That makes $a_2$ smaller than six but still positive.

Kenshin

3:02 AM

I think it is 200 rep I will check and update soon

dmckee

Wash. Rinse. Repeat.

Kenshin

@heather yes 200 reputation is required

2

this was increased to stop users from creating multiple accounts and upvoting to get to the required reputation

this can still be done but the effort to do so has increase4d

Bernardo Meurer

@dmckee What a boring thing to do :/

Sir Cumference

@Kenshin I assume you're talking about qandaexchange

Bernardo Meurer

Thanks for the help!

Kenshin

3:04 AM

@SirCumference that's right

dmckee

Induction proofs are like the drug pushers dream product.

Sir Cumference

See, a rep requirement would make PMing viable here

Bernardo Meurer

@dmckee Lol

Sir Cumference

No help vampires or noobs

dmckee

The first one gives you a high. The second one a little boost. The third one you barely notice, but by then you're hooked and can't quit.

4

Bernardo Meurer

3:07 AM

Gives me low grades that's what it does

koolman

Hii @Kenshin

Kenshin

hi @koolman

koolman

@Kenshin After a long time to come online

Secret

3:53 AM

ugh, I suck at integrating curved trajectories. I need to think carefully on what happens for each second in this system

It is true I assumed the rods are rigid, but with a tangential acceleration, this cannot be circular motion

Secret

4:23 AM

o wait nvm

rob

5:19 AM

3

Bragging time: from now on, when I say something like "I'm speaking with my moderator's hat on," you'll know to take me literally

DanielSank

5:31 AM

O

M

G

I should have run for mod.

Bernardo Meurer

Damn you mods

rob

I also got SE and Physics.SE shirts and stickers

Bernardo Meurer

wtf man

rob

I am like the coolest person on my entire couch right now

Bernardo Meurer

How come I get nothing

I'm the only reason this site functions!

DanielSank

5:32 AM

@rob lkaooiuwoiy98y4329873498yu3 t4p9nt8yne p98yt 0oe9tugy 89yu erpoiuh ert

rob

@DanielSank Translation?

DanielSank

I'll send you Google stickers in exchange for PSE stickers.

Bernardo Meurer

@rob Let me see the stickers!

@rob You don't want to know what he told you

That's russian btw

DanielSank

@rob Oh, yes, in English that's "I envy your stickers, good sir".

rob

@DanielSank Somehow I suspect this translation lacks nuance

DanielSank

5:33 AM

@BernardoMeurer нет, этот не русский.

@rob Well, a translator takes certain liberties, you know.

Bernardo Meurer

@DanielSank If it's not ASCII I ain't reading it

(extended ASCII counts)

DanielSank

rekt

So @rob, how is being a mod?

Champagne and caviar as promised?

Secret

Hey guys, I have a one year old question which I just recently (get through my procrastination and) update it with new modelling details. I'd like to know whether the question looks clear to you for the section after it states "Update"

rob

@DanielSank It's mostly not remarkable.

DanielSank

@Secret Never, ever put an "update" section in a question.

Just edit it.

Nobody cares about the chronological development of the question.

rob

5:38 AM

The flag queue stays pretty empty unless something strange develops over a few days.

Lately our users have been mostly nice to each other.

DanielSank

@rob Yeah, I noticed the tone around here has improved remarkably.

Secret

Well, for that, the issue is Mark have answered the old version of the question, thus I want to briefly acknowledge that in the update so that future users will not see his answer as off. Maybe there's a better way to do so?

rob

I no longer feel bad about letting the close review queue build up since I don't like casting super-close votes

DanielSank

There are noticeably fewer antagonistic comments.

@Secret Leave a comment under that answer.

@rob That's one of the reasons I didn't want to be a mod. It seems like it discourages doing reviews.

rob

@DanielSank Lately at least, when I've peeked into the review queue, the close votes I've cast have been fourth or fifth. I guess the queue is showing me the questions that are nearest to closing before the questions. So that's not terribly different either.

DanielSank

5:42 AM

oh that's interesting!

rob

I think that we're lucky here to have a fairly large community of useful high-rep people.

DanielSank

The fraction of high-rep/total-users is probably high here.

That would make an interesting chart.

rob

@DanielSank That's the kind of thing that I'd like to be able to extract from the SEDE, but not enough to actually wrap my head around the database structure.

Secret

6:01 AM

Ok, advices taken. Be warned. While I do try to treat this seriously at the newtonian level, the question does cover a rather taboo topic in physics. As a disclaimer, my expected answer is a mechanism explaining how the device can fail, although I also consider the slim possibility of a surprise

0

Q: Can energy be extracted indefinitely from a mass dipole/mass negative mass pair in perpetual motion?

Some days ago I have read this short piece of article by Hammond which caused me to revisit the theoretical properties of negative mass. It is known that a pair consists of ordinary mass and negative mass of equal magnitude, known as a mass dipole in some sources, can accelerate indefinitely by ...

My major issue in the modelling is it is not clear how entropy is calculated in such system as the total energy of the driving component is zero

user228700

Hi, everyone.

Secret

finally feel a bit alive by not procrastinating

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