Mathematics - 2013-05-02 (page 2 of 4)

skullpatrol

12:00 PM

Ooow

Charlie

One does not simply get earlier , it's SP

skullpatrol

OoowooO

Charlie

I'll try to study more...

skullpatrol

@Charlie What is "SP"?

Jayesh Badwaik

@skullpatrol Slow Path.

@Charlie I got a new gravatar.

skullpatrol

12:14 PM

@JayeshBadwaik Thanks for responding :)

Charlie

@skull it's my.city name @jayesh :D

Jayesh Badwaik

Who knew? :P

Charlie

12:33 PM

@skull @jayesh I'm going, bye bye!

skullpatrol

@Charlie later

Dominic Michaelis

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
\coordinate (r) at (5cm,0cm);
\coordinate (h) at (0cm,5cm);
\foreach \x [evaluate=\x as \xeval using 2^\x] in {1,2,...,7}
\pgfmathparse{\xeval-1}\foreach \y in {0,1,...,\pgfmathresult}
\pgfmathsetmacro{\tint}{\x*10}
\pgfmathparse{ifthenelse(mod(\x,2),"blue","green")}
\let\col=\pgfmathresult
     \filldraw[fill=\col!\tint!white,draw=black,very thin]  ($ {\y/\xeval}*(r)+(h)-{1/\xeval}*(h)-{\y/\xeval}*(h) $) rectangle

Charlie

@skullpatrol later

Alexander Jones

Why do I find anything to do with analysis, series or sequences hard? How can I be better at it?

shobon

@AlexanderJones, do you mean like epsilon-delta and epsilon-N proofs?

Dominic Michaelis

12:45 PM

With those explicit explanations what do you expect us to answer? It cries a bit for an answer like "cause you are stupid". What is hard? what have you tried ? ...

Alexander Jones

Well I am a high school student and there is no topics called "Real Analysis" in the syllabus but in the test they always have questions on it

shobon

can you give an example of the sort of thing you found difficult

Alexander Jones

like proving tailor series for e, wallis product, zeta function, etc.

zeta (2)**

Dominic Michaelis

zeta(2) is easy try zeta(3)

shobon

What does proving tailor series mean?

user74918

12:50 PM

hello, might anyone be willing to help me with a bipartite question

Alexander Jones

For example this was a question that was in the exam a few years ago: i.imgur.com/mWl4cSn.png

shobon

@AlexanderJones, if you replace the trig functions with exponentials it becomes a polynomial problem (which is easy)

Alexander Jones

Ok, so I think I really need a lot of practice to see things like that

shobon

no you dont

this a single uniform method that treats all trig identities

Alexander Jones

ok, thanks

CBenni

1:18 PM

Is there any nice theorem about compact subspaces of $\ell^2$? There are plenty about $L^2$, but I really dont know how to apply them in $\ell^2$...

Nimza

@robjohn hi, do you know what the pseudodifferential operator is it $$M^{-1} \frac{\Gamma(z_1+\ldots+z_n)}{B(2,z_1+\ldots,+z_n)} M?$$ $M$ here is the Mellin transform, $B$ is the beta function

robjohn

@Nimza Are you looking for the symbol?

Nimza

@robjohn No, its Mellin symbol is between $M^{-1}$ and $M$, I don't know anything about this operator but I need some its properties, maybe you have encountered it somewhere

robjohn

and isn't the ratio of the Gamma and Beta functions just the ratio of two Gamma functions?

Nimza

aha

robjohn

1:29 PM

@Nimza Can't say that I have sorry

Nimza

:(

tr1n

2:25 PM

The trivial ring is the zero object for the category of rings. Suppose we denote the trivial ring T = {t}. To satisfy the initial object requirement, do we map t to the multiplicative identity of R for any ring R?

Tobias Kildetoft

@tr1n well, we have to

also, it is not a zero object (at least in the definition of zero object I am used to)

tr1n

Ok, I was just verifying.

Tobias Kildetoft

actually, it is not initla. sorry

it is terminal (final) but not initial

unless you are not working with unital rings

tr1n

It's definitely initial.

Tobias Kildetoft

if the rings don't need to have a unit, you send the element of the trivial ring to 0 (since this has to happen)

if the rings are unital, then there is never a morphism from the trivial ring to any non-trivial ring

tr1n

2:28 PM

I'm working in the category of rings, not necessarily unital rings.

Tobias Kildetoft

ok

then you send the element to 0

tr1n

So I'd imagine I can use the trivial ring.

Tobias Kildetoft

right, in that category, it is a zero object

tr1n

Awesome. Thanks @Tobias

Tobias Kildetoft

for some reason, I always forget that people might be interested in non-unital rings

tr1n

2:30 PM

I'm not sure why you would be, but this question specifically says the category of Rings, so I can't assume otherwise.

Fascinatingly, the categories Sets and Top have the "same" initial and final/terminal objects.

(I find that fascinating, at least)

Tobias Kildetoft

one can also put some nice restrictions to Top to get a zero object

namely by looking at topological spaces with a point

Chris's sister and pals

2:45 PM

Another version $$\int_e^{\pi}\frac{\sqrt{(x-e)(\pi-x)}}{x} \ dx $$

MphLee

How can I improve this question?
http://math.stackexchange.com/questions/379230/how-to-formalize-this-paradox
Some moderators closed it but I still don't have an answer to my question..

Is a kind of paradox..about the truth of a phrase.

skullpatrol

2:59 PM

+1 for effort :)

Dominic Michaelis

it an easy to solve paradoxon

the number 2 appears 3-1 times in the text

@mphlee it is a bit like the barber paradoxon do you know it ?

@MphLee by the way the question was not closed by moderators

Arash Ata Afarin

3:15 PM

hi, can any body explain above pic for me?

how can left pic convert to right pic?

thanks

MphLee

@DominicMichaelis mmh, no i don't know that paradox, but I know the liar paradox, and seems me in some ways related, but I dont know how, anyways where I must ask for reopen my Q?

Dominic Michaelis

nowhere it does already have 3 reopen votes and some guys will look at it

@mphLee the barber paradoxon is, that a barber is a person who shaves every man who doesn't shave himself

MphLee

@ArashAtaAfarin did you do this pics? are two cartesian diagrams?

Arash Ata Afarin

no our teacher did this,but i dont know how he did

it is in our engineering mathematics cource

course

MphLee

@DominicMichaelis ah yeas I know it, the if the barber shaves himself then he can't shave himself, but if doesn't shave himself then he must shaves himself...anyways I dont know the connection (is not clear in my mind)

Dominic Michaelis

3:22 PM

well maybe it is a bit different, but one can solve it by having a female barber ;)

MphLee

xD haha indeed

Arkamis

@MphLee It's a self-referential statement.

But your question is actually not a paradox. It's a pedantic trick of semantics.

MphLee

we can conclude that a barber with such property can't exist in our universe?

Charlie

@skull helloes

Arkamis

@MphLee Essentially, we conclude that the statement is nonsense in the conventional logical schema; it means nothing.

skullpatrol

3:24 PM

@Charlie helloes

Charlie

@skullpatrol :D

skullpatrol

@Charlie :D

Arkamis

@ArashAtaAfarin I can't tell what that diagram means, but maybe you could look into Conformal Mapping.

skullpatrol

Also some context would help >:-)

Dominic Michaelis

3:45 PM

8 upvotes till my first gold badge :D

Arash Ata Afarin

4:08 PM

yes it is about Conformal Mapping

i just want to know how this happened

Arkamis

@ArashAtaAfarin It is a conformal map from a wedge in the upper-half-plane to the upper-right quadrant. Offhand I don't know such a map, but it's probably just $z^k$ or something like that.

anorton

5:02 PM

Do we think this person is the true Jasper? math.stackexchange.com/users/75560/jasper-loy

Lord_Farin

Colleagues.

skullpatrol

Comrade.

amWhy

Interesting: see this "new" user

user19161

Hi everyone.

Chris's sister and pals

Hi

skullpatrol

5:09 PM

Hi

Chris's sister and pals

@Lord_Farin hi:D Have you seen this version? $$\int_e^{\pi}\frac{\sqrt{(x-e)(\pi-x)}}{x} \ dx$$

Lord_Farin

@JasperLoy A good evening (please correct for timezone).

user19161

I have decided to come back to the site, just for fun.

Charlie

Hey hey heeeey! Lets stop it, okay? Leave Jaspy alone!

user19161

And yes, I am the real JL, not an imposter.

Lord_Farin

5:10 PM

I just saw something paradoxical happen...

Jasper Loy

Hi everyone.

Charlie

@JasperLoy PROVE

Lord_Farin

@Chris'ssisterandpals Hm, no. Did you manage to solve yesterday's integral?

user19161

By the way, I am using an email address I used before, so when I signed up for the new account, I got all the old notifications, lol.

Chris's sister and pals

@Lord_Farin eventually, yes.

Jasper Loy

5:11 PM

I have decided to come back to the site, just for fun.

Tobias Kildetoft

I am seeing double. Two Jasper Loys

user19161

@Charlie I am only a banana.

user19161

@JasperLoy You are the fake one, you are skullie, lol.

Jasper Loy

And yes, I am the real JL, not an imposter.

Charlie

@JasperLoy prove it

Lord_Farin

5:12 PM

@Chris'ssisterandpals I was so frustrated when I found a way by means of differentiation of a constant.

Jasper Loy

I am only a banana.

Lord_Farin

Only to find I needed to determine a boundary value.

user19161

@Charlie You are MJ. I think I will email you shortly...

Dominic Michaelis

oh jasper welcome back

Jasper Loy

You are the fake one, you are skullie, lol.

Dominic Michaelis

5:13 PM

@jasper BANANANDA

Alexei Averchenko

hi

Lord_Farin

@DominicMichaelis Does pinging even work in this awkward situation? :)

Alexei Averchenko

i have a question about this question: math.stackexchange.com/questions/379163/…

Dominic Michaelis

@Lord_Farin why should it ?

user19161

I suggest the fake JL change his name...

amWhy

5:14 PM

@JasperLoy If it's you, jasper HELLO!!!! if not really you, no one can replace the real Jasper Loy!

2

Alexei Averchenko

It's easy to show that $x^2 + x - 6$ is not invertible in $k[x] / (x^2 + 2x - 3)$

can we conclude the negative answer to the question just from that?

Jasper Loy

I suggest the fake JL change his name...

user19161

@amWhy It is me, I swear.

Lord_Farin

@AlexeiAverchenko I think only if $k = \Bbb C$ (for the eigenvalues will have to satisfy the same equation).

Alexei Averchenko

In general, let $f: k[x] \to A$, where $A$ is an associative $k$-algebra with identity

Jasper Loy

5:15 PM

@amWhy It is me, I swear.

Alexei Averchenko

is it true that if $[a, f(x)] = 0$, then $a \in \operatorname{im}f$?

Dominic Michaelis

@AlexeiAverchenko if you think of the lie bracket the answer is no, take a, f constant and f(0) not a

Alexei Averchenko

what do you mean by constant?

user19161

@jayesh I roam at my favourite shopping centre, lol.

Alexei Averchenko

you mean f(x) = 1? f(x) = 0?

Jayesh Badwaik

5:19 PM

@JasperLoy Cool. :-)

Nice to have you back mate.

skullpatrol

:D

Charlie

I am waiting @jasperloy green

Dominic Michaelis

@AlexeiAverchenko was thinking about [0, 1]

Alexei Averchenko

i don't get it

Dominic Michaelis

you talk about the lie bracked don't you ?

skullpatrol

5:21 PM

@JasperLoy How are you pal?

Alexei Averchenko

yes

the commutator in the associative algebra

actually

amWhy

Will the real @Jasper email me? I would love that, and only the real Jas knows my email address! ;-)

Alexei Averchenko

what if we take a free Lie algebra generated by $x$ and $y$

Dominic Michaelis

ok when a and f(x) commute the commutator is zero

but they are not necessarily the same

Alexei Averchenko

call it $A$

user19161

5:23 PM

Guys, my new email is jasperloy at ymail dot com.

Alexei Averchenko

then consider $B := U(A/([x, y]))$

user19161

And yes, beware of imposters.

Alexei Averchenko

then take $f: k[x] \to B$ given by $x \mapsto x$

Dominic Michaelis

@AlexeiAverchenko the free group is a perfect group isn't it ?

Alexei Averchenko

then $[x, y] = 0$, yet clearly $y$ is not a polynomial expression of $x$!

@DominicMichaelis What the hell are you talking about? :)

skullpatrol

5:25 PM

4 mins ago, by skullpatrol

@JasperLoy How are you pal?

@JasperLoy

Dominic Michaelis

oh misread something

Tobias Kildetoft

@DominicMichaelis no, a free group is not perfect

and this is a free Lie algebra, not a free group

user19161

@skullpatrol Same, which means bad, lol. Now that reply confirms it is the real JL, lol.

skullpatrol

@JasperLoy I was just having some fun with you pal, nice to see ya :)

Tobias Kildetoft

@AlexeiAverchenko Are you asking if $a$ is in the image of $f$ given that it commutes with the image of $f$?

Alexei Averchenko

5:28 PM

yes

and i've just provided a counter-example :(

Tobias Kildetoft

yeah, you need a lot of extra assumptions to get something like that

since you can always just hit something very small in a commutative algebra

Alexei Averchenko

does it hold when we constrain ourselves to matrices?

Tobias Kildetoft

@AlexeiAverchenko yes, since those are CSA

Alexei Averchenko

no, probably not

Tobias Kildetoft

ohh, probably not, no

I was thinking that $k$ is in the image and the center is also $k$

but of course something could commute with the image without commuting with everything

Alexei Averchenko

5:31 PM

Tobias, I was thinking about this question: math.stackexchange.com/questions/379163/…

it's trivial to show that $x^2 + x - 6$ is not invertible in $k[x] / (x^2 + 2x - 3)$

but this doesn't account for the possibility that this matrix has an inverse for which there is no polynomial expression :(

maybe i'm shooting too far?

i mean, an inverse is not just some commuting element

user19161

I will make it a point not to downvote on my new account.

Chris's sister and pals

I downvoted meta.math.stackexchange.com/questions/291/… this morning I think, and the reason in my mind was exactly related to what happens right now. Someone enters the site and spreads all kind of stuff using your name.

Charlie

@jasper $\huge \text {JASPEEEEER!}$

2

@JasperLoy you lost so much drama... But I'm happy you're back. Very happy.

Lord_Farin

@JasperLoy Then what to do with blatantly wrong answers?

user19161

@Lord_Farin Comment.

Lord_Farin

5:48 PM

@JasperLoy Well it's your call, of course. I don't see any harm in downvoting. Given that my up/down ratio is always floating around 10:1, I don't see the need to alter my practice. What reason do you have not to downvote?

(On a completely unrelated note, I just crossed the psychological barrier of 4k rep. Yay :).)

skullpatrol

I believe we should be allowed a range of values from -3 to 3.

Lord_Farin

@skullpatrol You mean one person can assign each of those scores to a question?

skullpatrol

Yes, you can vote: -3, -2, -1, +1, +2, +3.

IMO

Lord_Farin

@skullpatrol There has been psychological research. A vast majority of votes will reside at the extremes, rendering the distinction virtually useless.

Perhaps it could work if each of the categories had different vote limits.

E.g. you gain one +/-3 vote per day, three +/-2 votes, and the +/-1 are as currently implemented.

Alexei Averchenko

how do you know if you're gonna see a +3-worthy question/answer today?

should you conserve your +3 or give it away to that very nice answer?

Charlie

5:55 PM

Or instead of numbers, two buttons: good/bad

Alexei Averchenko

you'll probably never conserve

Lord_Farin

@Alexei That's why I said you gain one, not you have one.

Alexei Averchenko

oh

Lord_Farin

I.e. you could save them.

user19161

It is strange, my network and chat profiles have the same ID as before.

skullpatrol

5:58 PM

Are you still at the same IP address?

Lord_Farin

@skullpatrol Perhaps related: this.

user19161

@skullpatrol Hmm, maybe, I don't know.

Lord_Farin

I'll have to leave. Maybe I'll be back later today, I'm not sure yet.

Bye all. (Oh and nice to finally encounter the elusive @JasperLoy :).)

Gustavo Bandeira

@JasperLoy I knew the other Jasper wasn't you. He wasn't saying "lol" at every message in the chat.

Charlie

@lord_f bye

@jasper we're both green, it's a sign

robjohn

6:10 PM

@JasperLoy so, you're back, I see...

anorton

No one can stay away from this site for long... the addiction is too strong.

4

Charlie

@anorton yes

robjohn

@JasperLoy should have taken my advice, but then you wouldn't know if you would have been able to stay away if you had deleted your account...

user19161

@robjohn Nah, I don't regret deleting my old account at all!

user19161

And I really messed up my old accounts, it is good to start afresh...

robjohn

6:13 PM

@JasperLoy now you know that resistance is futile...

user19161

@robjohn I now no longer need to keep my rep at a multiple of 5, and I will keep my downvotes at 0, lol.

Charlie

@skull are you there?

@Charlie Yes.

@skullpatrol :)

@Charlie :)

6:19 PM

@skullpatrol ;O

@Charlie :-|

@skull :?

@Charlie :-\

:D

user19161

So caveman is shobon, hahahaha.

robjohn

6:29 PM

@JasperLoy how do you figure that?

user19161

@robjohn Erm, someone just told me, but it might not be true, and you don't need to tell me as a mod. =)

robjohn

@JasperLoy I wouldn't tell you as a mod.

skullpatrol

>8(

Peter Tamaroff

@JasperLoy Dafaq?

skullpatrol

@robjohn I find this question interesting, do you?

user19161

6:36 PM

@PeterTamaroff My dear Pedro!

Peter Tamaroff

@JasperLoy Is it really you?

user19161

@PeterTamaroff Yes, it is.

Peter Tamaroff

@JasperLoy Hmmm....

robjohn

@skullpatrol If you say "E equals m c squared", then that is one way to interpret it. Luckily, Einstein wrote it down $E=mc^2$

shobon

thats frustrating

first question I could answer in ages, so could lots of other people

user74918

6:40 PM

actually that is not how Einstein wrote it down

user19161

@shobon Hi caveman!

skullpatrol

@robjohn There is an interesting comment section discussion about writing math equations for the blind...

robjohn

@skullpatrol Braille iPads!

shobon

@JasperLoy, who told you

user74918

fourmilab.ch/etexts/einstein/E_mc2/www for the curious

user19161

6:41 PM

@shobon Anyway, I hope we can be friends and not enemies...

shobon

@JasperLoy maybe if you answer my question instead of ignoring it

user19161

@shobon I just guessed from reading the transcript...

shobon

this is clearly a lie

you earlier said "someone just told me"

who told you?

Peter Tamaroff

@shobon I did.

somaye

Hi

skullpatrol

6:45 PM

Hi

somaye

how are you?

where is charlie?

skullpatrol

Fine thanks. How are you?
Dunno.

user19161

@PeterTamaroff To verify that it is really me, I asked you if you are gay in one of my emails, lol.

Peter Tamaroff

@JasperLoy Oh, yes you did. Silly pants.

somaye

fine too

thanks

user19161

6:47 PM

@PeterTamaroff I also once thought anon was gay.

Arkamis

@JasperLoy Who haven't you thought was gay?

shobon

@JasperLoy, I dont think we will get along since you lie directly to me and act so weird ignoring my questions

user19161

@Arkamis Almost nobody.

user19161

@shobon Well, I think we will get along just fine.

shobon

@JasperLoy, if you are seriously about "I hope we can be friends" then treat me with some more respect

user19161

6:48 PM

@shobon Nobody ever disrespected you. But you should stop calling everyone rude.

shobon

@JasperLoy, you are just being an obnoxious bully

skullpatrol

Etiquette

Civility is required at all times; rudeness will not be tolerated.

Be nice.
Treat others with the same respect you’d want them to treat you. We’re all here to learn together. Be tolerant of others who may not know everything you know. Bring your sense of humor.

user19161

@shobon See, that's what I mean. You are calling people names again...

shobon

ignored, im not talking to you anymore. i gave you a chance

Arkamis

@JasperLoy What is your accuracy rate with your gay-classifier?

skullpatrol

6:50 PM

(-:

Arkamis

Let's get an ROC curve drawn up.

user74918