Mathematics - 2016-09-01 (page 1 of 2)

user227867

05:08

I have been reading the recent messages moved to the trash, and also those left in this room.

user227867

'I do math art' is not arrogant. She is a really talented mathematician.

user227867

It seems to me that although she does not attack others, others always attack her, claiming she is arrogant.

user227867

She just feels overjoyed by the results she obtains, and wants to share that joy with us.

user227867

Also, there is no need to move harmless messages to the trash.

user227867

To some of the people who attack her, it is true you cannot solve the problems she can. And she never attacked you.

user227867

05:12

And to those people, maybe it is also true that you will never solve any major problem in your area, because if you are the kind of mathematician who can do that, you will not attack her in this manner.

user227867

Thanks for listening to a banana.

user227867

@amWhy I received your email this morning. Thank you. At least what you wrote in it is a step in the right direction.

I do math art

06:29

@amWhy First, I don't know who you are (even as a user here), but this is less important now. The thing is I don't pretend anything, and seeing your reaction I might ask myself if you worked many hours on my integrals and got nowhere. Then I'm really sorry for your wasted time. And, no, I don't spend any bit of time trying to convince anyone of anything, and especially here where major part of people are interested in other kind of mathematics.

Actually, even if I had the skills of Ramanujan for real, in this chat no one would show more respect than was shown to me in the past, mainly because they don't care or for other reasons that I'm not interested to mention here.

Then I guess here no one of the great figures of mathematics come in, nor people I know for real, and the need of a bragging as an anonimous user, if you use a bit your mind, you immediately realize it has no sense.

@amWhy the bragging comes like that: using a real name, posting all the stuff you published in her, or tons of mathematical stuff created and eventually posted on a known site, and then using all I previously said to do what you said. It's not the case.

But you know what? Not even in that case it should be considered bragging, if my spirit would push others to reaching high performance in mathematics.

@amWhy Why I come here? In general I come not to forget to talk, to find reasons to make breaks from my work, and maybe to inspire others if the case. This is to exist at least a bit in the areas of social activities.

To return a bit to my integral I posted some hours ago. Well, I'm awful for the ones that do not understand my spirit of doing mathematics. How even to ask for a solution without pen and paper here? $$\int_0^1 \left(\frac{\log(1+x)}{1+x}-\frac{\log(1-x)}{1+x}\right) \arctan(x) \textrm{d}x=\frac{\pi^3}{192}+\frac{\log(2)}{2}G$$

Tobias Kildetoft

@JasperLoy You really need to start not only reading the things people say to that user but also the things said by that user. Sure, some of us give him/her a hard time, but it did not come from nothing (and man, does it get tiring to get asked how many problems I have created as if any mathematician cares about that).

I do math art

@TobiasKildetoft I never went in your areas and express opinions about you until you come to me to express opinions that are not fit, justified.

@amWhy and to know for sure I very highly respect people that are dedicated to mathematics for real, and push them selves beyond limits. I know a lot of them and this is what I like to see always!

I challenge mathematics itself, not people. There is much confusion around, or it is not, some just pretend don't understand that I'm not in a competition to anyone.

This is the way to go in mathematics, ready to face anything, with unlimited courage to reach any crazy impossible peak.

Tobias Kildetoft

06:54

I certainly agree that you are not competing with any mathematicians here.

I do math art

@TobiasKildetoft Do you consider yourself a mathematician? Till the end of the year can you provide with a solution to the integral above?

Tobias Kildetoft

Yes

Why would I waste my time trying to solve that integral rather than doing research?

I do math art

@TobiasKildetoft If the students come to you and ask you for a solution what would you do?

(ask for help on MSE?)

Tobias Kildetoft

Tell them to ask someone who cares about such integrals. But why would a student come ask me about such an integral?

I do math art

@TobiasKildetoft Or why integrals even exist in mathematics, right?

Tobias Kildetoft

06:57

Not sure I understand what you mean there

I do math art

@TobiasKildetoft I mean you go too far with your questions, seemly without care about what others might like to do in mathematics.

Tobias Kildetoft

What questions? How can a question go too far?

I do math art

@TobiasKildetoft But why would a student come ask me about such an integral?

Tobias Kildetoft

How is that going too far with anything? I don't mean why a student would ask about such an integral, but why would they ask me about it?

I do math art

@TobiasKildetoft Why not? Aren't you a mathematician (that knows all areas in mathematics) as you just claimed above?

Tobias Kildetoft

07:01

Why would being a mathematician mean that I was the right one to ask about some complicated integral when those are not my area of research?

I mean, what would you do if a student came to you to ask what the dimension of the first Steinberg module for $SL_2$ is in characteristic $3$?

I do math art

@TobiasKildetoft Here is the point. Because of your poor knowledge (sorry I couldn't skip that part) it seems tou you a complicated integral. No, it's not. In your class you should make sure that all students can do such integrals without no problem, without no difficulty.

@TobiasKildetoft Please go in your school and ask all the students there of they can do this integral.

Then do a meeting with all professors there and CHANGE SOMETHING!, if they fail to do it!

It's unacceptable to say you did integrals and you cannot do it!

Please reconsider this part of mathematics, the perception is completely wrong. Chane, improve something, and later you'll thank me!

Tobias Kildetoft

I have no idea why anyone would benefit from being able to do that integral (this is the sort of thing you never seem willing to say anything about)

2

I do math art

@TobiasKildetoft Do you admit you should organize a meeting with all professors there if the students fail to solve such an integral?

Tobias Kildetoft

No, I disagree rather strongly in fact

Quite frankly, if the students were taught to solve such integrals I would have to question whether their time could not be spend better learning some abstract material instead.

but again, this is because I have never seen any background for why such integrals are important

An old man in the sea.

Hello all, could anyone help me with this question? http://math.stackexchange.com/questions/1910156/lateral-limits-of-an-endpoint-of-the-interval

Thanks

syzygy

09:07

I fully agree with @TobiasKildetoft and @amWhy

Tobias Kildetoft

@syzygy agree about what?

Balarka Sen

10:38

Good god, this chat has turned into a public fightplace.

Should save some popcorns to munch at an appropriate time.

user227867

There is no fighting, only peaceful talking.

user227867

@BalarkaSen How is your health? It seems you often fall sick.

Balarka Sen

I see lots of heated discussion.

It's much better now.

user227867

@BalarkaSen Good, don't fall sick like Ramanujan.

Tobias Kildetoft

11:06

@BalarkaSen Remember I said yesterday I was writing a grant application and a job application?

Today I got offered a 1-year postdoc position, so I am no longer writing those.

amWhy

@TobiasKildetoft Congratulations!

Tobias Kildetoft

@amWhy thanks

Balarka Sen

@TobiasKildetoft Oh that's great.

Tobias Kildetoft

@BalarkaSen Yeah, it really is

Balarka Sen

I am happy for you. Congrats.

So, what's new in terms of math?

Tobias Kildetoft

11:16

Looking to apply some recent work with Mazorchuk and two others to something we studied when I first came here.

We hope to be able to do some further special cases using that recent work

I do math art

$$\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{\Gamma (k)^2 \Gamma (n) }{\Gamma (2 k+n)}((\psi ^{(0)}(n)-\psi ^{(0)}(2 k+n)) (\psi ^{(0)}(k)-\psi ^{(0)}(2 k+n))-\psi ^{(1)}(2 k+n))$$

Balarka Sen

Nice. What's your research about, in a nutshell?

Tobias Kildetoft

This part is about the restriction of projective functors on the principal block of the BGG category $\mathcal{O}$ to parabolic subcategories and how these decompose.

Balarka Sen

What's the BGG category?

Tobias Kildetoft

A suitable category of modules for a Lie algebra which contains enough to be interesting but not so many as to be impossible to understand

Balarka Sen

11:19

Hm.

Tobias Kildetoft

We described these restrictions for the special linear Lie algebra previously, and we hope to be able to describe them for all types when the parabolic has a special form using some new results

I do math art

$$\sum _{n=1}^{\infty } \frac{\psi (n) \psi \left(n+\frac{1}{2}\right) \psi \left(n+\frac{1}{3}\right)\cdots\psi \left(n+\frac{1}{m}\right)}{n \left(n+\frac{1}{2}\right) \left(n+\frac{1}{3}\right)\cdots \left(n+\frac{1}{m}\right)}, \ m\ge2$$

How do we finish this one?

Balarka Sen

What's the motivation behind doing this, @Tobias? A question out of curiosity from a person who doesn't know much of algebra.

I do math art

There is a post of mine that annoys me a lot

(just a bit, to find it)

Also this one

2

Q: Another way of doing integration

What's your option for calculating this integral? No full solution is necessary, it's optional as usual. Calculate $$\int_0^1 \frac{2 \zeta (3)\log ^3(1-x) \text{Li}_2(1-x) }{x}-\frac{2 \zeta (3) \log ^2(1-x) \text{Li}_3(1-x)}{x}+\frac{ \log (x) \log ^5(1-x)\text{Li}_2(1-x)}{x}+\frac{\log ^4(1-...

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

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

Tobias Kildetoft

@BalarkaSen Hard to describe shortly. The functors (without restricting them to parabolic subcategories) have been used extensively for the past 30 or so years to better understand representations of semisimple Lie algebras. And the parabolic subcategories have some relation to some geometrical stuff that I don't really know. So it was natural to ask for their restrictions (since they happen to preserve the parabolic subcategories).

Also, I know that a similar result for maximal parabolics in type $A$ (i.e. a special case of the work I did with Mazorchuk) was used for some categorification results related to Khovanov homology.

I do math art

11:26

No closed form so far.

Balarka Sen

@TobiasKildetoft Interesting.

I do math art

Found it!

0

Q: A double integral with origins in the area of multiple gamma function

What tools would you recommend me to use for a fast evaluation? No full solution is required, but your thoughts on it only, largely speaking. $$\small\int_0^1 \int_0^1\frac{1}{(1-x) (1-y) \log (x) \log (y)}\biggl(\frac{6 (x+y-x y-1)+\log (x) (6 x y-(2 x y+x+y+2) \log (x)-6)}{6 \log ^2(x)}+\frac{...

calculus real-analysis integration definite-integrals

$$\small\int_0^1 \int_0^1\frac{1}{(1-x) (1-y) \log (x) \log (y)}\biggl(\frac{6 (x+y-x y-1)+\log (x) (6 x y-(2 x y+x+y+2) \log (x)-6)}{6 \log ^2(x)}+\frac{x+y-x y-1}{\log ^2(y)}+\frac{x y-1}{\log (y)}+\frac{1-x y}{\log (x)+\log (y)}\biggr)\textrm{d}x \textrm{d} y$$

No answer from March! Maybe it comes in a few months.

Time to research!

BBL

user227867

12:05

@Idomathart Good luck!

I do math art

@JasperLoy Thanks. If I don't do it, then who will do it? Actually preparing a new article for publishing.

Tobias Kildetoft

Almost certainly nobody will.

user227867

LOL. We need to laugh more, then this room will be healthy.

I do math art

@JasperLoy then ask @TobiasKildetoft (also @BalarkaSen and @amWhy) learn and say here better jokes.

user227867

@TobiasKildetoft I am asked by @Idomathart to tell you to say better jokes.

I do math art

12:15

I would like to laugh more. :-)

Maybe after finishing some more stuff here.

BBL

Huy

13:59

@BalarkaSen: given c = 7cm, $\gamma = 60^\circ$ and the radius of the inscribed circle of a triangle ABC is 2cm, any idea how to construct the triangle? I'm having trouble making use of the inscribed circle

Balarka Sen

@Huy Radius of incircle has an explicit formula.

Huy

??

I know, but that seems way too difficult

Balarka Sen

Let me try to recall it.

Huy

should again be a very elementary construction

Balarka Sen

hmm ok

off the top of my head, no idea. maybe someone else can help; I am not going to think too much about it, sorry.

Huy

14:05

np

Balarka Sen

Here's an idea in any case. Take your line $AB$. Draw a ray from $A$ with angle 60 degrees.

The center of the incircle has to lie in the bisector of the angle $BAD$ ($AD$ the ray)

Huy

@BalarkaSen: the 60 degrees angle is the one disjoint from AB

Balarka Sen

oh crap

Huy

yup, otherwise it would be much much easier

its easy to draw the arc "over" AB such that any point on the arc together with AB has angle 60°, and obviously the midpoint of the incircle must be on the line parallel to AB with distance 2, but I need one more condition for C or for the incircle to be done

Balarka Sen

Oh well. Maybe ask on the main.

Juan Fran

14:19

@Idomathart of course no one will do it.

I do math art

14:36

@JuanFran First you have to be able to do it. There were many contradicting me during the time, actually no one succeeded to win because all was only about subjective opinions, and nothing about mathematics.

amWhy

Ahhh...I've just now smelled the foul scent of grandiosity (delusions of grandeur) from one who fails to realize that some challenges don't, and shouldn't, involve mathematics.

I do math art

@amWhy Maybe that scent comes from the sentence you just typed above.

amWhy

So long...the stench has grown larger.

I do math art

@amWhy I come in the room to talk about math, not other stuff. Of course, sometimes it happens I talk with Jasper about other things but this doesn't involve any dispute, and long replies that lead nowwhere.

@amWhy by the way, who are you to come to me and tell me your opinions out of the blue? Did I ever ask you anything? Or do you want to teach me some math I like, maybe some integrals? If it's about it, you're my invitee.

Please share some opinions on $$\int_0^1 \left(\frac{\log(1+x)}{1+x}-\frac{\log(1-x)}{1+x}\right) \arctan(x) \textrm{d}x=\frac{\pi^3}{192}+\frac{\log(2)}{2}G$$

0celo7

@Idomathart math art? You do geometry?

I do math art

14:44

@0celo7 I did in the past, but I suppose everybody here did.

0celo7

Differential geometry

I do math art

@0celo7 No, I didn't study yet.

Balarka Sen

brings out some of the popcorns and starts munching

I do math art

@BalarkaSen I'm busy now, don't expect more things to be spread around.

amWhy

@BalarkaSen lol...hilarious...!!!

0celo7

14:46

@Idomathart huh?

I do math art

Guys, I have a question for you: aren't you busy with some mathematical research? I don't know, I feel there is too much energy inside of you that wasn't consumed yet.

amWhy

Sincerely, @BalarkaSen, given your earlier comment regarding popcorn, your answer is so spot on. That's not me being sarcastic.

0celo7

While I disagree with JuanFran and others' attitude, your attitude is really...bad too

I do math art

For example, now I'm really tired after many hours of research to answer back to any nonsense spread on channel.

Balarka Sen

@amWhy Sure, sure.

I do math art

14:48

@0celo7 Which part? Defined the precise point where I have a bad attitude and use rational arguments.

0celo7

you say you've made fundamental contributions to math, IIRC

@Idomathart FYI, people thinking you're annoying has nothing to do with rational arguments.

I do math art

@0celo7 I said that I did contributions to my mathematical area, not more, no.

0celo7

@Idomathart which is?

amWhy

@Idomathart If you're now tired, go away and go to sleep, but if you can't sleep, practice your English...

I do math art

@amWhy This is all you could say back after you talked so much nonsense?

0celo7

14:51

@Idomathart AFAIK you haven't said anything about any papers you've published, conferences you've talked at, etc.

Balarka Sen

I need more popcorn.

0celo7

If you're so great, why won't you share some of your accolades with us?

I do math art

@0celo7 Because beyond the decriptions I saw here, I really don't like to brag at all. It's that simple.

0celo7

hahaha

@BalarkaSen I need a drink, not popcorn

amWhy

@Idomathart Oh, I have much more I can say to you, but I'm not going to waste energy with you; you are far too unstable. <ignores Idomath>

I do math art

14:53

@amWhy You only said nonsense.

@0celo7 who said anything about greatness?

Balarka Sen

@0celo7 Bring a glass for me too, if it's soft drinks.

0celo7

@Idomathart If someone asks you for papers, etc. it's not bragging

@BalarkaSen Bourbon straight up.

Not even good Bourbon

I do math art

@0celo7 First of all, although I initially defended you here when some kicked you, I should have not defended you, you lack the respect. You can't even imagine to have a decent conversation with me in these conditions.

amWhy

@BalarkaSen I'll join the spectators here and munch on popcorn; 'll bring some wine!

Balarka Sen

Still too strong; I prefer soda.

0celo7

14:56

@Idomathart I am having a decent conversation with you.

I'm simply asking for papers you've published, etc.

If you don't have any, say so. But to say you don't like to brag is simply BS.

4

I do math art

@0celo7 I'm here just anonymous, I don't need anyone here appreciation, I can live without that.

0celo7

@Idomathart I will give you my email and you can send me some. I don't tell anyone who you are.

I do math art

My satisfaction comes only from my mathematics not from the beautiful words that could come to me from some users.

Only mathematics.

0celo7

Then you must understand why some people don't think you're a real mathematician. You're unwilling to share your "great" results.

4

I do math art

@0celo7 I never said that I'm a mathematician, I only said that at most I would be glad to be considered an artist in mathematics.

(check well and see that no where there is such a statement)

An artist in mathematics finds no meaning in things like bragging, such a thing cannot even be conceived.

Anyway.

I have stuff to do.

BBL

0celo7

15:01

@Idomathart An artist wants to share his art with the world. You're no artist.

I do math art

@0celo7 You're not the world, you only represent now the way a bad conversation can look like.

0celo7

@Idomathart I don't claim to be the world, I'm simply trying to explain why some users are frustrated with you.

I have nothing against you.

Albas

@Huy if you do not mind me asking what is c and gamma in the question about the incircle you had written above?

I do math art

@0celo7 hahaha, I saw enough to understand that. It's your problem what you do, but if I were you I would need more data before expressing such opinions.

@0celo7 Did you try my integral?

$$\int_0^1 \left(\frac{\log(1+x)}{1+x}-\frac{\log(1-x)}{1+x}\right) \arctan(x) \textrm{d}x=\frac{\pi^3}{192}+\frac{\log(2)}{2}G$$

@0celo7 no need to calculate it, just asking.

0celo7

No. Did you try my problem? Show that the Riemannian holonomy group satisfies $\mathrm{Hol}_{(p,q)}(M_1\times M_2,g_1\oplus g_2)\cong \mathrm{Hol}_p(M_1,g_1)\times\mathrm{Hol}_q(M_2,g_2)$.

(if anyone present knows how to do this, ping me :) )

I do math art

15:13

@0celo7 I write it now down and try it when my knowledge will permit this (I write it down carefully).

Secret

@Idomathart What is the significance of this integral? Is it a physically important integral, or describing some kind of dynamics, its graph is a type of geometric art, use in mathematical proof or somehting else?

I do math art

@Secret It must be a physically important integral to calculate it?

Secret

not necessary

I do math art

@Secret It might be.

Secret

but the shape of this integral seemed to suggest there might be something interesting in it

because it kinda reminds of Li(x) related stuff for some reason

I do math art

15:18

@Secret I can tell you this integral is very important for calculating some particular cases of a class of infinite series related to polygamma function.

Secret

Is G the polygamma function?

I do math art

@Secret Catalan constant

Secret

ok, makes sense

I do math art

@0celo7 It would have been cool to see how you would like to start there. Maybe you have a nice precious idea. Don't worry if you fail.

It happens, in general some stuff still bothers me, so I need to learn more before getting the desired results, and this happens to everyone.

0celo7

I have a pretty good idea of how the proof goes, but it's local coordinate heavy and not interesting at all.

I want something nice and slick tbh

Secret

15:24

$$\frac{2}{\log(2)}\int_0^1 \left(\frac{\log(1+x)}{1+x}-\frac{\log(1-x)}{1+x}\right) \arctan(x) \textrm{d}x-\frac{\pi^3}{192}=G$$

Hmm...

0celo7

Oops.

I do math art

@Secret It's not really correct how you put things there.

Secret

I am just rearranging the given result, because it makes me think of something

However I forgot the big brackets

I do math art

@Secret I know, but you didn't arrange it properly. :-)

(note how you multiplied both sided by $2/\log(2)$)

Secret

Yup, I moved the pi term to the left, and then multiply the whole equation by $2/\log(2)$.

We knew the equation holds thus one can rearrange terms like an algebraic equation

My gut feeling when looking at the rearragned equation is as follows:

If there exists another nice closed form for this expression, $\int_0^1 \left(\frac{\log(1+x)}{1+x}-\frac{\log(1-x)}{1+x}\right) \arctan(x) \textrm{d}x-\frac{\pi^3}{192}$, then we can learn more about what number set G is in

> Is Catalan's constant irrational? Is it transcendental?

I do math art

15:32

Catalan's constant

In mathematics, Catalan's constant G, which occasionally appears in estimates in combinatorics, is defined by G = β ( 2 ) = ∑ n = 0 ∞ ( − 1 ) n ( 2 n + ...

It is not known whether G is irrational ...

Secret

The stuff in the integrand $ \left(\frac{\log(1+x)}{1+x}-\frac{\log(1-x)}{1+x}\right) $ suggest of Li(x) related functions

by other than that, I have no idea, I am not good at integrals at all because of how the integral in general is not even closed on $C(\mathbb{R})$

I do math art

@Secret I agree. However, if using the integration by part things become ugly and complicated with using the dilogarithm function, and we don't want this happens.

@Secret I appreciate you at least do efforts and find this stuff interesting. The integral is amazing and the journey to the solution will offer you much satisfaction.

No need to hurry with that, build clever strategies to put it down.

Many don't like integrals because they simply seem impossible, extremely hard to do, but things improve after a period of time, like some years.

Just much patience is needed. No need to do all kind of integrals over night. This would be impossible I think. Just patience and much hard work.

If one fails to do it, one can return to it after gaining more experience. It happens to me all the time.

@amWhy can you write $$\log^7(1-x)$$ in terms of a (multiple) series containing the generalized harmonic number?

@amWhy eventually I would also like to learn something from you (if we talked to each other so much - at least to have some gain from it)

@0celo7 what do you think?

Or how do you write $$\arctan(x)\operatorname{arctanh(x)}$$ in terms of an elegenat seris?

Then we go further a bit

How about the version

$$(\arctan(x)\operatorname{arctanh(x)})^3$$?

The last one looks crazy, indeed. @BalarkaSen, can you confirm that?

Look the game I do now!

I put together $\log(x), \log(1-x), \log(1+x), 1/(1+x), 1/1(1-x), \arctan(x), \operatorname{arctanh(x)},$, I take some at randomand find their series.

Balarka Sen

15:52

No idea what's going on.

I do math art

How about finding the series of

0celo7

@Idomathart I don't really care tbh

I do math art

$$\frac{\log(x)\log^3(1-x)\arctan(x)}{1-x}$$?

It's taken at random. Can we write it elegantly in terms of an infinite series? How?

Just look at it, it looks awesome!

Forget all!

How would you write $$$$

$$\frac{\log(1-x)}{1+x}$$

in terms of an infinite series?

Secret

Have you tried expanding it in terms of taylor series and then rearrange until some pattern pop up?

I do math art

@Secret I use different tools I created.

No Taylor, no Cauchy product, only my tools.

Secret

15:59

Elaborate briefly, is it some kind of identity or make use of some theorems?

I do math art

@Secret I make use of the theorems I created, but it might be too much to call it a theorem (maybe).

Secret

can you show me one example and how to motivate, extend them or somehting similar?

as every tool has a nice intuitive example to demostrate its ability

and that's how we learn how to use new tools

I do math art

@0celo7 the thing you might like to accept after the meeting with me is that you have a huge lot to learn. And don't focuse on persons. You say you don't care, but I don't know it this is a good attitude.

0celo7

I never said I was amazing at integrals

it's not something I've ever wanted to do, either

I do math art

@Secret The tool I use can be used in many situations, I cannot tell it here with all users present, not now. Sorry.

Secret

16:03

...

Albas

@Idomathart don't you get tired of writing so much?

I do math art

@Albas I actually get tired. :-) I was just preparing to leave. ;)

@0celo7 That's fair.

@Secret Can you get an interesting series for $$\frac{\log(1-x)}{1+x}$$?

@Secret let me know if you do. There is a very beautiful way to derived such a series and the series itself is awesome, definitely.

I have to leave now for some hours.

Don't miss me much, I'll be back later.

bbl

Secret

I am not a good series person, the best I can is $\left(\frac{1-x+x}{ln(1-x)}\right)^{-1}=\left(\frac{1-x}{ln(1-x)}+\frac{x}{ln(1‌-x)}\right)^{-1}=((\frac{d}{dx}Li(x))^{-1}+(\text{too lazy to write the taylor series of} \frac{x}{ln(1-x)})$

@Idomathart Let's suppose it is because of the fear of ideas being stolen, then you should get the papers published so that it is secured and thus others can build upon your ideas, cause this is how discovery works in collaborations

typo: forgot the -1 power on the last bracket

0celo7

@Idomathart Aha, maybe you know? Is $\sqrt{|y-z|}\le \sqrt y+\sqrt z$ when $y,z>0$?

Yes, it is.

Wonderful.

Secret

(To math chat) I have a tendency to use the do nothing technique, otherwise known as cheat, trick and other names

x=x+y-y, 1=x/x

It makes me wonder, can we abstract this technique so that we can study the properties of the do nothing technique itself to find out where it will and will not work...?

(NB ...? are not really questions, but more of thought dumps)

I do math art

16:28

@0celo7 I'm one step away from the house. Is that a question? Consider the cases $y$ higher or smaller than $z$, and you're done.

Semiclassical

16:55

hi chat

0celo7

@Semiclassical hi

Semiclassical

17:22

just posted a new question. the title is a bit of a pious fraud, but i like it nonetheless :)

a more honest way to pose it, i suppose, is to ask for conditions on $A,B$ such that $OACB$ and $OA'C'B'$ are both parallelograms where $A',B',C'$ are inverses of $A,B,C$ w/r/t some circle through $O$

user227867

18:00

@Semiclassical Interesting you type w/r/t and not wrt or w.r.t.

Semiclassical

i tend to default to that, yeah

not sure why

0celo7

most people do wrt.

user227867

The British often omit full stops and the Americans often include periods in their abbreviations.

0celo7

the star wall is a PoS

Semiclassical

i don't like doing w.r.t. for some reason

Balarka Sen

18:02

@Semiclassical Getting physics done right now.

Semiclassical

@balarka nice

on that note, it looks like i'll be TAing introductory electromagnetism this semester

Balarka Sen

cool

0celo7

how is that cool

0% of the people in that class want to be there

Semiclassical

regardless of whether thats true, i definitely prefer TAing second semester intro physics to first semester

Balarka Sen

0% of the people in this chat like you

0celo7

18:04

@BalarkaSen such a lie

Semiclassical

...

I like it better mostly because using circuits and such is more interesting than pushing carts

And, heck, they get to use Helmholtz coils :)

Balarka Sen

@Semiclassical The statics problems are confuzzling

Semiclassical

do you have one in mind?

Balarka Sen

Yeah, let me write it down

Semiclassical

usually what people get confused about is the condition for rotational equilibrium

and how to choose the rotational axis for that conveniently

Balarka Sen

18:10

I have a homogeneous cylinder of diameter 8cm on an inclined (not friction-less) plane making an angle of 30 degrees with the ground. What's the maximal height of the cylinder so that it won't topple?

what I have done is this

user227867

I hear physics in this chat.

Balarka Sen

Since the problem asks me to use friction explicitly, let's use it. It acts on the bottom of the cylinder: it's $F = \mu mg\cos(30)$. The force it's own weight gives it acts through the center of mass: that's $F = mg \sin(30)$.

I wanted to do a moment argument, but $\mu$ isn't given to me, so I have no clue.

Semiclassical

You probably don't need it, then.

Balarka Sen

also I am not using the diameter information in any way I see it

Semiclassical

So just leave the friction force as some unknown $f$.

Balarka Sen

18:14

Alright.

What next?

Semiclassical

Well, you want the system to be stable if the cylinder is small enough. So you need conditions for equillibrium

There should be 3 of those: 2 for balance of forces (translational equillibrium), and 1 for balance of torques (rotational)

For the balance of forces, you need to pick some coordinate system. probably picking the $x$-axis to be parallel to the plane is convenient here.

Balarka Sen

Hm ok

Semiclassical

from that you get that the normal force $F_n$ balances a component of the weight force, and the friction force $f$ balances the other. And since you know the angle of the ramp, you can write both of these components in terms of the weight $mg.$

Agawa001

@Secret since you mentioned it, exclusive property of an idea to the owner is a critical abused topic in this site

Balarka Sen

@Semiclassical hm

Semiclassical

18:19

If that's not clear, play around with the free body diagram.

Once you work that out, you've got everything specified in terms of weight. But things aren't entirely constrained yet---for that, you need to enforce rotational stability.

Balarka Sen

well, when you say balance, you mean $f$ is > than the planar component of the weight force, yes?

Semiclassical

if it was greater, then the net force would be nonzero :)

In other words, if I'm standing on the ground then what magnitude of force is the ground exerting on me?

If it's less than my weight, i'd be going through the floor. If it was more, I'd be accelerating upwards. So it has to equal it.

Agawa001

as long i saw later answers than mine, replicating my though in different tems, not payin me a tiny credit/reference or even an innuendo

Balarka Sen

OK. I had been reasoning something wrong here, I just realized. I reasoned that $f$ > than the planar component by myself standing on the ground. Then my weight has no horizontal component, yet static friction is $\mu mg$.

But that's not right. If I am just standing, there's no such thing as friction.

I have to move forward, and then there will be a thing called friction which will try to stop me.

Semiclassical

there's zero friction needed to not move horizontally, yeah

Balarka Sen

18:25

Well, duh. That was silly.

So, yes, $f = mg \sin(30)$.

Semiclassical

right. as a check, if the angle goes to zero then indeed $f=0$.

and the normal force $N=mg\cos 30^\circ$ for similar reasons.

Balarka Sen

Sure, sure.

The rest is easy.

Semiclassical

pretty much, yeah. the only thing that could trip things up is the choice of rotational axis

Balarka Sen

Take the normal to the surface of the plane passing through the center of mass, no?

Semiclassical

that's one option, yeah. in that case the weight won't provide a torque.

another option is to pick a point on the plane itself, say the midpoint of the cylinder's base. in that case the friction force won't present a torque on that axis

the nice thing about the latter option is that it makes it clear that the 'location' of the friction force on the cylinder won't matter.

It can't actually matter what axis you choose, to be clear; the result is the same regardless. But sometimes one is simpler than another.

Balarka Sen

18:31

I am unsure what your axis is, can you explain a bit better?

Semiclassical

Sure. In your free-body diagram, the cylinder is basically just a rectangle. The axis you suggested runs through the center of that rectangle.

Balarka Sen

Right

Semiclassical

But you could equally well have it run through a corner of that rectangle---say, one of the corners touching the plane.

Balarka Sen

Ah

Semiclassical

And my suggestion is to put it in between those two bottom corners

Balarka Sen

18:34

I see

Semiclassical

As I said, it can't actually matter.

Balarka Sen

That's not too different, yes

Semiclassical

It's a matter of convenience, not of principle.

you also need to know the moment of inertia of a cylinder, which I forget. but i imagine you've seen that.

Balarka Sen

Well, I haven't, actually. I computed the moment of the friction force on the bottom of the cylinder, but I can't figure out what to equate it with to state equilibrium.

I was going to ask you that.

Semiclassical

ahh

Balarka Sen

18:38

I actually don't know what moment of inertia is, really. It's after the chapter I am working on.

Semiclassical

Wait, I'm being silly.

You'd need to know what the moment of inertia is if you wanted to see something turning

When the net torque is zero, you don't need to know what the moment of inertia is

Balarka Sen

Toppling doesn't count as turning, I suppose?

Semiclassical

it would, but you don't want it to topple :)

Balarka Sen

That's true.

Semiclassical

What i'm getting at is basically just: If $\vec{a}=0$, then $m\vec{a}=0$

So if the net force is zero, then it doesn't directly matter what the mass is. (it still enters in through the weight, but not through $m\vec{a}$)

Balarka Sen

18:42

Agreed

So, how do I do this?

Semiclassical

Well, let's agree to use your axis through the center of mass.

Balarka Sen

Mhm

Semiclassical

Hmmm

Well, there's definitely no torque from the weight force (which acts at the axis point)

by contrast, there will definitely be a torque from friction. that needs to be counter-balanced by something else, and the only other force present is the normal force.

user227867

@Idomathart The current state of this room is very unhealthy. Ignore the negativity and keep doing great things!

Balarka Sen

@Semiclassical Hmm

But that's orthogonal to the friction force, so not sure if it'd affect the torque

I do math art

18:46

@JasperLoy Yeah, that's because I'm hard to understand. It's not bragging, just natural joy, excitement to the top, but it's still natural.

Semiclassical

Well, it had better. Otherwise there's nothing to balance the torque from friction.

user227867

@Idomathart You know you always have a supporter in me.

Semiclassical

The issue is that the cylinder has a base and therefore the forces are applied over that surface.

0celo7

@Idomathart @JasperLoy Marry already.

I do math art

@JasperLoy hehe, yeah.

Balarka Sen

18:47

The normal acts through the center of mass though

Semiclassical

Eh, no it doesn't. It acts at the point of contact

user227867

@0celo7 Your username reminds me Oneiric Ocelot.

Semiclassical

(which in this case is a surface of contact)

Balarka Sen

@Semiclassical That's a good point.

Semiclassical

there's still an issue, though. suppose that the normal force is equally distributed over the bottom of the cylinder

in that case, there'd be just as much normal force acting at the lower part of the base as the upper. so those two torques would cancel out.

which seems to imply that the normal force can't be equally distributed if it's to provide a counter-torque.

Balarka Sen

18:52

hm

Semiclassical

Hmm. It really is that normal force that's the tricky bit here.

Balarka Sen

I have to leave for a bit. I'll think about it meanwhile. Thanks for the help, by the way.

Semiclassical

np

Transcript for

$Mathematics$ Mathematics