12:01 AM
Back.

@robjohn That problem asked by Chris’s sister has acquired a couple more solutions while you were away.

I like oen’s, though he could have stood to include a few more details.

@BrianM.Scott I knew I had to work fast, because I knew someone knew the arcsin series. Raymond is one to know these things off the top of his head.

@robjohn Which is probably why I don’t run into him very often!

12:10 AM
@BrianM.Scott I know him from sci.math

The name is very faintly familiar, so I may have seen him there years ago when I was still reading it.

@PeterTamaroff: tell me those aren't sock puppets in your avatar....

@robjohn sock puppets?

@PeterTamaroff Sock Puppets is a term used for alternate accounts of one person, there to upvote questions and answers for the main account.

@robjohn Oh, look at that. No, they are plushies. They are $G$ times more evil that sock puppets, where $G$ is Graham's number.

12:18 AM
@PeterTamaroff I've counted higher...
:-D better bragging through bigger step sizes

@robjohn I'm reading about the UCT now. Universal Chord Theorem. It is quite nice.

@PeterTamaroff Is this a musical theorem?

7

Let $f \in C[0,1]$ and $f(0)=f(1)$. How do we prove $\exists a \in [0,1/2]$ such that $f(a)=f(a+1/2)$? In fact, for every positive integer $n$, there is some $a$, such that $f(a) = f(a+\frac{1}{n})$. For any other non-zero real $r$ (i.e not of the form $\frac{1}{n}$), there is a continuous fu...

@peoplepower I wsa just reading that. It seems pretty simple to prove, just look at $f(x+\frac12)-f(x)$

@robjohn The generalization is equally easy, but what I find amazing is that it only holds for reciprocal integers.

12:23 AM
What do you call it when two functions seem to assume the same values as each other within some range, but they don't actually converge to a number/limit.

@peoplepower Yeah, because you cannot break up the interval evenly otherwise
@IssacM what do you mean don't converge to a limit?

@IssacM Huh? I don't understand the "but they don't actually converge to a number/limit." part
@robjohn =)

@IssacM An example perhaps?

Black-Scholes option pricing model :s
I am comparing it to the closed-form pricing model of another option
Over a small interval of inputs they agree almost completely
Then one converges on zero and the other does zigzags

@IssacM you just said, "blah, blah, blardy, blar" :-)

12:26 AM
@IssacM hazarding a guess here, "they are locally convergent" ?

hmm okay
So that's what we say when they marry together over some interval?

way beyond my ken; I will stay out of it

@IssacM Do you mean this?
http://en.wikipedia.org/wiki/Local_convergence
If not, then I don't have a word for it I guess.

Ok thanks :)
Oh robjohn

@robjohn I once had a student who did his senior project on the Black-Scholes model; I felt as if I’d fallen into a black hole.

12:28 AM
@BrianM.Scott: Hey, I can cap if I spend all day answering questions, but how do you get so many acceptances?

Do you know if $\{a,b\}^+$ denotes maximization of $a,b \in \mathbb{R}$, or is it $(a,b)^+$

@BrianM.Scott I'm glad to have good company here :-)

@robjohn I’m not really sure. I like to think that it’s because I write clearly and usually have a pretty good idea of where people are likely to have trouble.

@BrianM.Scott Even though economics is really just math hidden behind a lot of arcane language, it is hidden deeply

@IssacM Naah man. Very very long ago, some 8-9 years I guess, I heard my friend telling me something similar, but I did not take him seriously then and I do not remember now. I have a hunch you are correct, no guarantees though.

12:30 AM
@robjohn So they should award a PhDDD in econ?

The roundbrackets I mean.

@Jayesh ohhh

@BrianM.Scott Deeply Disguised Doctor of Philosophy?

@robjohn Pile it higher and Deep, Deeper, Deepest.

@BrianM.Scott Ah, that sounds good :-)
@IssacM I have seen neither. Sorry. Is it an economics term?

12:34 AM
@robjohn Thanks

Surely this has been dealt with here before.

@robjohn My quant finance lecturer put it on the board then rubbed it out because he didn't want to confuse us undergrads
@robjohn Forgot which one it was

@BrianM.Scott I'm sure it has. someone will find the original. Might as well say $(-1)^2=1^2$ so $-1=1$.

@IssacM If he did not want to confuse you, why should you be confused/worried? Its only a notation after all!

I'm doing a 15 pages assignment where I have to do numerous TikZ lattices and it's typographically better without writing Max() or Min() all the time.

12:37 AM
@IssacM I have not taken any econ classes, so I'm not privy to such arcana. Interestingly, there was a grad student Robert Johnson in econ at the same time I was at Princeton for math. He left a lot of overdue charges at the Fine Hall Library and at the dorms as well. I had to deal with all of that.

@IssacM Verbose but a little ugly is always better than confusing and with a proper latex shortcut, of the same effort.

lol

@robjohn
You did the following using binomial theorem I hope?

\int_0^x(1-t^2)^{-1/2}\mathrm{d}t=\sum_{n=0}^\infty\tfrac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\frac{x^{2n+1}}{2n+1}

@JayeshBadwaik The previous series was from the binomial theorem.

12:41 AM
@robjohn Mark Dominus found it; last I looked, it needed one more vote.

@robjohn Damn. What.is.Wrong.With.Me!
Early morning hangover. Need more coffee.

@JayeshBadwaik but you're right, before the integration, the sum was gotten from the binomial theorem. I should edit that in.

@PeterTamaroff HeeeHaawHeeHaaw :P

@JayeshBadwaik NOTE: Chuck Testa does not taxidermize pets.
His name his basicall "Chuck Balls"

12:44 AM
@JayeshBadwaik There, hopefully that is better.

@robjohn yup it is. it was good enough before that too I guess. just did not read the first line properly. but anyway, it is definitely better now.

@robjohn :-)
@PeterTamaroff Hmm. Got that :P

1:09 AM
Is the projective plane usually written as $\mathrm{P}^2$ or $\mathbb{P}^2$?

@robjohn I think the former.
The latter seems to be used for the positive integers, sometimes.

@PeterTamaroff Thanks, I am editing an AG post for MathJax

Yet, it might be just a matter of taste.

I am using $\mathbb{R}^2$ for the plane, so I wanted to make sure.

1 hour later…
2:28 AM
@robjohn You there?

@PeterTamaroff OMFG. That is such a short answer.
@PeterTamaroff That is well written though. Good job.

@JayeshBadwaik I know. I'll try to be less succint next time.

@PeterTamaroff I am now.

@robjohn Spivak has so many exercises. I will never finish them all. Or at least, all of the important ones.

@PeterTamaroff Do you need to do them all?

2:43 AM
@robjohn Not really, just the important ones. Like this one on the Universal Chord Theorem, of those dealing with density arguments and stuff. The relevant and non trivial ones.

3:02 AM

@GustavoBandeira Don't feel special. (You're welcome)

Peter.
I know you love me.
You don't need to hide your love for me, Peter. <3
:D

French Toast Sugar Rush! :-)

@GustavoBandeira HAHAHHAHAH that song is so awesome

Yep.

3:12 AM
@GustavoBandeira Bite dust. Fruit Crush.

Omicron Poranga Shoe Chickenshit¹ Bit Hit!

1 - Chickenshit is a lighter bullshit.

@GustavoBandeira Conversation Food. English Must.
This could go on forever.
@GustavoBandeira What timezone are you in?

GMT -3x³+176x²-56x+8=3
Kidding, It's GMT -3

That was quick. So it must be midnight there?

12:18
And there?

3:19 AM
I was wondering how could you come up with a cubic equation that fast.
It is 0849 here.

Well, I just chose random numbers. =D

Feynman used to choose a random time at the start of the day and would offset his clock by that much time, and then whenever people asked him time, he would give answers like "in 3 hours and 48 minutes, time would be this" and so on.
I thought you had some similar system.

Nope.
I should make some time of transformation with time, but I can't think of something cool.

We can design an equation which has two complex roots and a real root equal to 1.
Such that all coefficients are integers
and the constant term is 1.

What's the condition for a equation having two complex roots?

3:24 AM
There is a big discriminant third order I guess. I don't rem the exact formula.

I must be careful...
There's one guy on MSE that ALWAYS say: "Take care when saying root on Australia!"

Why?
You should be careful when you are a root in UNIX.

Yep.
0

Don't use "root" in Australia. It has a very different meaning.

It's him! xD
Gerry Myerson

Duuude.
That is why they drink so much real beer. I guess root beer would not be exactly welcome there.

I'm doing something... XD

3:32 AM
:P

Australian Mathematics Axiom: For the root of $N=1$ we have a fap. For the root of $N = 2$, we have a couple. For the root of $N = 3$ we have a threesome. For the root of every $N > 3$ we have an orgy. $\square$ — Gustavo Bandeira 1 min ago
2
=D
I guess this dude post this thing about root on every root question.

@GustavoBandeira :P
By the way the real root of
$x^3 - (1+c)x^2 + (t+c)x - tc$ is c and other roots are complex if $t>\frac{1}{4}$
Chose a random value of $t$ everyday and spook people.

@GustavoBandeira You are geting so flagged!

Why?!
This is racism, only because I'm brazilian. =/

3:50 AM
@GustavoBandeira so we can say that there is no unique alphabetical representation of root for N>3? Wow Abel

XD
Jayesh
This is a topic that is still no available at my level but, it seems that Abel and Galois proved that polynomials of degree > 5 couldn't be solved by ordinary algebraic operations.
So, what operation is used then?
I don't remember what degree is exactly, if it's 5 or 6.

4:21 AM
degree $\geq5$, not >

@MarianoSuárez-Alvarez Yep.

what they showed is that theere is no general formula for the root of a polynomial of degree $\geq5$ using roots
much later, it was shown one can use elliptic functions to solve quintics
(by Klein, for example, and others)
the general case... I don't know

And what about the other degrees?
Oh, I typed before you typing your last message.
So, we can only solve polynomials of the 5th degree?

much of analytic number theory consists in finding analytic functions whose values at specific complex numbers generate specific number fields
just like the values of square roots generate quadratic number fields
no
the statement is that the general polynomial is not solvable
but there exist solvable polynomails

Got it.

4:24 AM
for example, we know the roots of $(x-1)^{23456}$.

What did they used to get there?

Galois came up with a precise criterion which allows us to decide if a polynomial (with, say, integer coefs) can be solvable with roots
it is an algorithm, so given such a polynomial we can in principle decide if it is solvable by roots —but the algorithms become impracticable as soon as the degree is larger that 15 or so
@GustavoBandeira who used what to get where? :-)

@MarianoSuárez-Alvarez On the $(x-1)^{23456}$ polynomial.

hm, think a bit: you can also figure out all the 23456 roots of that polynomial!

@MarianoSuárez-Alvarez Why they become impracticable? Not enough computational power?

4:29 AM
that is the definition of impracticable :-)

Yesterday i was reading on a kind of Lie Algebra
The E8

more technically: the complexity of the algorithms (in the technical sense of complexity theory —see wikipedia) is very huge

Which was made completely on computers.

what they did was compute some polynomials attached to that Lie algebra
the so called Kazhdan-Lusztig polynomials

From what I understand about Algebra, it seems to deal with a generalization on the structures of operations - I guess.

4:31 AM
that does not mean much,really :-)

Why?

"a generalization on the structures of operations" is absolutely devoid of content :-)

Operations between elements?

it is rather difficult to explain what the K-L polynomials are without entering into technical details, because they are very technical gadgets
in any case, a Lie algebra is usually studied through its «representations», and the K-L polynomials encode (in a rather complicated way) very important information about those representations

I shouldn't expect the existence of a K-L Polynomials for dummies, should I?

4:34 AM
not really :-)
if you want to get there

pick a good book on rep.. theory

But this ain't no problem, I'm going to a math bachelor.

do not try to run too fast: you'll trip
3

Yep.
These KL polynomials have something to do with metamathematics?

4:36 AM
not at all
do not try to guess, there is no point in it

Yep. I guess that I'm unable to do so without the shoulders of giants.
I don't "guess", I'm sure.

no, that is ot wha I am getting at
it takes a cconsiderable amount of time to get to the point where you can pick up the meaning of things like KL polynomials

What you mean?

simply because you need to know a lot of things
be patient
learn
study

And hard work
K-L Polynomials seems to be something amazing.

4:39 AM
even if you sat down a week reading up all definitions needed to make sense of what a KL polynomial is
you'd find them pretty meaningless, because you would nt be able to appreciate why on earth one is interested in them

I believe mathematicians solve problems in a sort of abstract world. Someone would be interested on it because of the building of this abstract world.

let alone, find them amazing :-)
but that is just generic line which does not explain anything
people are interested in such things because they allow them to understand specific problems

Yes.
You're studying them now?

What are you studying now?

4:42 AM
I work on homological algebra
currently, studying a certain class of algebras which are of interest in non-commutative geometry

Should I expect a 4dummies version?

no
there are no 4 dummies versions of anything
you are in all likelihood not a dummy
so it would be pointless

Strange, it's an algebra for studying other algebras.

hm
that again does not mean much :-)
"homological algebra" is a subject
the algebras which belong to my class of algebras are actual algebras, that is, vector spaces with an associative product

But it isn't an Algebra?

4:45 AM
I have no idea what you mean by "an Algebra"

I guess I haven't it either.
Ok, I erased the suposition that I know what's Algebra.

one way you can tell when you are starting to run too fast and begining to risk tripping is when you start using words whose meaning you do not know
3

Yes.

rest assured that the number of things you know will increase with time (assuming you keep up the hard work, of course)

Yep.
Once I tried to learn what a monoid is.
It was in vain

4:49 AM
you talk about yourself as if you really thought you are a dummy
don't do that

The only thing I have in mind is "It's a set and an operation" - Deeply meaningless.

well, that is a monoid
definitions by themselves carry very little information

@MarianoSuárez-Alvarez I believe it's a temporary state. I am now, because I know little about math. But I'll leave this state soon.

well, there is always more one math one does not now that math one does know because there is a loooooot of math

Somedays ago I posted a quote.
Let me get it.
Here:
Very few people realize the enormous bulk of contemporay mathematics. Probably it would be easier to learn all the languanges of the world than to master all mathematics at present known.
The languanges could, I imagine, be learnt at a lifetime; mathematics certainly could not. Nor is the subject static. Every year new discoveries are published. In 1951 merely to print brief summaries of a year's mathematical publications required nearly 900 pages of print. In january alone, the summaries had to deal with 451 new books and articles.
The publications here mentioned dealt with new topics; they were no restatements of existing knowledge, or very few of the were. To keep pace with the growth of mathematics, one would have to read about fifteen papers a day, most of them packed with technical details and of considerable length. No one dreams of attempting this task.
@MarianoSuárez-Alvarez btw, thanks for the advice.

5:01 AM
np :-)

@robjohn where would one know the RGB hex for woody brown (browny wood or whatever) color?
And, hello to you folks...

@KannappanSampath hi. What would you like to do? I mean do you have the image of the color in mind? Or do you have the name in mind?

@JayeshBadwaik I'd like to use that for the handle of an axe... :)
(Axe that Pedro is fond of...)

@KannappanSampath Yo!

Hello Gustavo.

5:14 AM
I want to write about something...
But I don't remeber what.
Anon.

@KannappanSampath

@anon
The Virtuoso Pianist (Le Piano virtuose) by Charles-Louis Hanon, is a compilation of sixty exercises meant to train the pianist in speed, precision, agility, and strength of all of the fingers and flexibility in the wrists. First published in Boulogne, in 1873, The Virtuoso Pianist is Hanon's most well-known work, and is still widely used by piano instructors and pupils. However, the applicability of these nineteenth-century exercises has been questioned by some piano instructors today. Overview The exercises address common problems which could hamper the performance abilities of a stud...

eh?

Everytime I see your name, I remeber this.
Hanon was the author.

I see

5:16 AM
Hanon -> Homological Anon.

@KannappanSampath you could have just used some desktop application with a color chooser.

I thought it was an alien!!!!

@GustavoBandeira It was Donald Knuth Duck.

5:22 AM
Newby feeling invades me . :\

Why?

Please don't laugh at what I have (trying to learn inkscape):

@KannappanSampath No one is laughing here my friend. For record, I cannot myself use inkscape. And the axe is very nice I will say.

@JayeshBadwaik Just a request but you can still laugh. ;-)

@KannappanSampath I'm laughinh because It reminds me of the pic Peter Trollaroff posted some days ago.

5:30 AM
@GustavoBandeira Yes, axe was becoming popular here!

Your pic instantly reminded me of my music making.

5:47 AM

6:46 AM
@KannappanSampath Three different shades of $\color{#503000}{\Huge\text{ Wood Brown}}\color{#784800}{\Huge\text{ Wood Brown}}$$\color{#A06000}{\Huge\text{ Wood Brown}} 7:10 AM What's the meaning of max(deg p, deg q)? Is it an instruction to chose the biggest value? maximum of degree of p and degree of q 7:24 AM ty 1 hour later… 8:24 AM @robjohn Thank you, Dear Rob. @KannappanSampath you already found a brown it seems @robjohn But, perhaps the second of yours would suit, no? @KannappanSampath They all work, depending on the type of wood. Whatever suits your needs Well, yes. :-) 9:07 AM "A brown"..ie? I need a brownie. Get one for me, Sir. @JonasTeuwen Good morning @JohnSenior Morning. 9:58 AM @JonasTeuwen anything in your brownie? @robjohn Uh... other than the standard ingredients? @JonasTeuwen I know some people who used to put some non-standard ingredients into their brownies. @robjohn Yes. I am boring. I eat boring brownies. They still are lovely. @JonasTeuwen Chocolate is a nice thing. Non-standard... you mean like small nails and other hard objects? "It has a surprise!" 10:01 AM Yes. Sir Cleese. @JonasTeuwen "Spring Surprise" 10:18 AM how much cpu power would I need to run mathematica, so its not slow? 20-30 GFlops @JayeshBadwaik you know of a laptop that can do that? @MaoYiyi I guess Corei5 does that. Corei7 goes to 80 GFlops. I am not serious by the way. I mean the numbers are real, about the processors at least. I am joking about the mathematica stuff. But I think you should be okay with Core i5. i have tried using mathematica before but it was so slow, so I want to try using it again but this time I don't want to wait for a responce. What laptop do you have? 10:27 AM and old dell, its like 5 years old @MaoYiyi Hmm, then it is possible that mathematica might be slow. I have a dell which turned 4 three days ago. However, I have a huge L2 cache a good FSB and a good graphics processor among other things, due to which it still performs really well. what is fsb? Front Side Bus That and good RAM is also important. so is there a real different between i5 and i7? Core i5 has two cores Core i7 has four cores Even if mathematica runs on a single core. The other applications you are using can run on other cores, and hence you can have better performance. 10:41 AM oh. can you have matheamtica be on more than one core? I don't know. Mathematica claims to use OpenCL, so I would not be surprised if it does though, since OpenCL is a hybrid structure for use of both GPU and multiple CPU cores as computing resources. It advertises only GPU computing though. However, if it uses openCL then extending the capabilities to multiple cores is not very difficult. Apart from that, I guess, it can still use multiple cores to handle different threads, so may be it is possible. Core i7 also has some precompiled caching. so when I buy the new laptop look for large fsb, ram and l2 chase? Not all applications benefit from parallelization... You need to transfer data between cores (= slow). Also, I wonder why you would need a kickass one, what are you going to do? I would look for something with decent driver support 8-). If need more computing -> send to cluster. @JonasTeuwen precompiled caching? @MaoYiyi Jonas is right, unless you need to do some heavy-lifting, you should be good with i5 I guess. 10:58 AM thanks, just wanted to it not to be painfully slow. 2 hours later… 12:55 PM$$7 \pi -\text{Log}\left[\frac{7}{2} e^{-7 \pi /2}+\frac{5}{2} e^{-5 \pi /2}+\frac{3}{2} e^{-3 \pi /2}+e^{5 \pi /2}+2 \pi \right] = 14.13472514154629716253329494571302508888...$$Zeta function FTW!! :-) 1:13 PM @JayeshBadwaik Things like the branch predictor. Heya =) @JonasTeuwen They are in a lot of CPU's. Core i7 does has an improved method of prediction which uses two target buffers instead of a single one thus reducing the incorrect prediction, but the technology has been there for a long time. Interestingly :-) http://cs.stackexchange.com/questions/73/which-kind-of-branch-prediction-is-more-important/76#76 I am trying to show that the function$$f(a) = \lim_{h \to 0} \frac{a^h - 1}{h} $$equals \log a. Is it enough to show that the function satisfies f(ab)=f(a)+f(b) and f(1)=0 ? 1:29 PM @JayeshBadwaik I know, that is what I mean. @N3buchadnezzar Does that uniquely determine the logarithm for you? @JonasTeuwen Okies :-) @N3buchadnezzar As in: how do you define the logarithm? @JonasTeuwen I know there are various ways to define the logarithm, but the simplest I guess is to use the integral definition @N3buchadnezzar That is important. @N3buchadnezzar But then you must show that the solution to your functional thingie is unique.$$\log a = \int_1^a \frac{\mathrm{d}x}{x}$$@JonasTeuwen Yes, and I was unsure about the f(1)=0, I do not think the restrictions are strong enough.. =( Bleh 1:33 PM @N3buchadnezzar So, what about trying to compute f(a) - \log a? And bound it. Maybe it requires rephrasing the definition. My motivation is to either express e as a sum or as a limit Asumption: There exists a function f(x) = a^x, where a is some real number, such that f'(x)=f(x) Now I want to use the definition of the derivative to show this, it leads me to Hmm.$$ \lim_{h \to 0} \frac{a^{x+h}-a^x}{h} = a^x $$You use the existence theorem? And then you call that e. Or rather, after uniqueness.$$ \lim_{h \to 0} \frac{a^h - 1}{h} = 1$$1:37 PM So you want to show that for small h that$$a^h - 1 \sim h \int_1^a \frac{\text{d}x}{x}?$$So I need to find a number "a", that satisfies that equation. And I do not want to plug in some formula for e and show that it fits. I want to derive either a closed familiar sum or limit. First you have to show it exists bro. Otherwise your manipulations are only formally true. @JonasTeuwen Yeah! Let us do what I always do when I don't know what to do: split. \int_1^a = \sum \int_1^{1 + a 2^{-m}}. Mm.. Boring. That a is retard, let us make it 2. So,$$2^h - 1 \sim h \int_1^2 \frac{\text{d}x}{x}.$$And$$\int_1^2 = \sum_m \int_{1 + 2^{- m - 1}}^{1 + 2^{-m}}.$$1:41 PM So now let us take given$\epsilon > 0$an$M$such that$2^{-M} < \epsilon$. Well, you catch my drift. I need to get some beer. I voted to reopen this. @N3buchadnezzar You can plug-in the binomial theorem for$(1+(a-1))^h\$ and then show that the result tends to the series of log?
@Matt How is it too localized?