Mathematics - 2014-05-21 (page 2 of 4)

@hb20007 What do you mean by "functional form"? What's wrong with evaluating the function for some values and interpolating the line, i.e. the usual approach?

hb20007

for example 2/(5x+5) can be sketched as a transformation of 1/x. So maybe we can write this in a form that's a transformation of a known function

10:31 AM

in Computer Science, 7 mins ago, by hb20007

do u guys if i ask a very simple math question here? I only get troll replies in the math chatroom

@AwalGarg Good job!

@hb20007 I don't see how that helps plotting a function by hand, at all.

hb20007

I hope the good job is sarcastic

@hb20007 I hope so, too. ;)

@hb20007 He was joking!

@Raphael :D

@robjohn Yeah, that's it!

@hb20007 I don't know what you mean by "functional form", but near $0$, it is like $x/3$ and when $x$ is big, it is approximately $1/x$. Also, its maximum is at $x=\sqrt3$. That should be enough to sketch it.

10:41 AM

@robjohn Indeed.

@robjohn And f(-x)=f(x) as well.

@Sabಠ_ಠ Hey.

hEY @Sawarnik

Anyone familiar with Spivak Calculus?

I can't find problems on Optimization in the book :S

@Sawarnik nope... $f(-x)=-f(x)$. It's odd.

Unfortunately, the joke's only funny if the recipient gets it. A user coming to CS chat with a math question because they felt trolled here is, I should say, troubling.
(I agree, though, that hb20007 started the derailing all by themselves; there's no reason to react to that, though.)

@robjohn Oh yes. Missed the sign.

10:49 AM

I never get troll replies in this room. :O

Can someone please troll me :S

@Sabಠ_ಠ True.

@Raphael it is unfortunate that the only people active in the room at the time were not being very helpful.

@Sabಠ_ಠ Stewart had one section on it I remember.

Plus the problems that Ted Shrifin gave you.

Yeah, Stewart has it. Weird spivak doesn't.

Yup

I got exactly 3 weeks before the exams

You think it's possible to "master" functions, diff, integration?

and the proofs :O

@Sabಠ_ಠ You worry tooo much, I think.

10:55 AM

@hb20007 do you have the ChatJax bookmarklet(s)?

What happened?

@Sawarnik I got 50% in my last test. My section B was nearly flawless and only 1 wrong(structured) Got my mcqs wrong and bam

@AwalGarg See the Computer Science room.

@AwalGarg when?

Is there a CS room? @Sawarnik :O

Gimme the link :O

10:57 AM

@robjohn A while before. I got pinged by some guy called Raphsody... something

@Sawarnik Why? I don't go there

in Computer Science, 34 mins ago, by hb20007

do u guys if i ask a very simple math question here? I only get troll replies in the math chatroom

Awesome :D

I am normally in the Python room haha

@Sawarnik Are you kidding me?

@Raphsody Yeah?

@AwalGarg No.

@AwalGarg And he is Raphael.

No, @Raphael Yeah?

@Sawarnik yes, sorry, just checked it now?

10:59 AM

@AwalGarg This is regarding chat in this room

@AwalGarg He was congratulating (sarcasm?) you.

@Sawarnik oh my goodness. holy shit... this isn't happening.

@hb20007
Math.SE Chat Room
Internet

Subject: Sorry

Sir,
With due respect, I wish to state that I am Awal Garg, a member of the prestigious Stackexchange Community. And, somewhat 1 hour ago, I happened to be irritating you. I hope you will forgive me for my evilness.

With kind regards...

Your Troller

Awal Garg

@Sawarnik Ok, I checked the whole thing now... I didn't know that.

Now I would be rather careful. Who knows, a guy might cry and complain about me in the federal court for me trolling him.

Oh, This damn ecma5script...

@hb20007 What's the question? Also, I see you have a weird surface as an avatar. Is that the cubic surface in $\Bbb P^3$ having maximal number of double points?

Speaking of it, I thought the best we have is Barth's construction with 60 double points at the vertices of the dodecahedron.

@Sawarnik Oh, you are so understanding.

I gtg c ya peeps

11:11 AM

I gtg too, c ya peeps...

@N3buchadnezzar @Chris'ssis @robjohn Here's something I learned from a user of Integral&Series : Riemann Hypothesis is equivalent to stating $$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3} \int_{1/2}^{\infty}\log|\zeta(\sigma + it)|~d\sigma ~dt=\frac{\pi(3-\gamma)}{32}$$

OK, I don't get why $\sigma$ isn't displaying properly.

@BalarkaSen Interesting.

@BalarkaSen phantom space inserted by the tex parser

Hey Deveno.

Salutations

11:21 AM

@Chris'ssis Isn't it? I would love to see where actually it interacts with RH. In general one can construct such integrals by noting that if all the zeros are in the 1/2 line then you'd only need to take residues at the half line, and nowhere else thus making a closed contour with zeros hidden in the half-strip and construct an real analytic integral with significant effort, but the similar doesn't work here.

Or does it?

Is that the zeta function buried in there?

@DavidWheeler yes.

well then

Let's try again. It looks awful with the phantom space

$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\xi + it)|~d\xi ~dt=\frac{\pi(3-\gamma)}{32}$$

Darn.

Yes!

@BalarkaSen You should probably insert a few spaces in your $\LaTeX$.

11:25 AM

After so many characters (I forget how many exactly) the parser puts in a space of its own

lol

24, I think.

Yeah, Jameson told us.

It's only the first occurrence that is affected, too

$\log|\zeta(\sigma + it)|$

:15637916 ROFL

I would add extra spaces where they wouldn't do any harm

11:27 AM

@KarlKronenfeld $$\log|\zeta(\sigma + it)| ~d\sigma ~dt$$

Aha.

$$ \log \left| \zeta
(\sigma + it)\right|d\sigma dt$$

$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3} \int_{1/2}^{\infty} \log|\zeta(\sigma + it)|~d\sigma ~dt=\frac{\pi(3-\gamma)}{32}$$

What space, @Karl?

What? Who? Where?

like that

11:30 AM

@AwalGarg Ah, I get. You think you are funny, but you really have to work on delivery.

There was this pizza guy tried to be a comic, but his delivery was off...

@Raphael what delivery?

@robjohn Yea; I don't follow the complaint about "trolling". Being unhelpful is not trolling.

@AwalGarg Of the punchline, the joke.

@Raphael I still don't understand. Please try to use plain english...

11:32 AM

Trolling...isn't that like, giving wrong advice, and then asking for payment? I'm not so good with dis internets slang...

Im 12 and what is this

What is what? The eternal tautology....

I missed all the fun, apparently some peeps here drove a math quester to beggar help from compsci...

@DavidWheeler One may consider answering a question "How do I catch a fish?" (on, say, a survival Q&A forum) with "Just buy the frozen stuff at Walmart" to be trolling. A matter of taste, obviously.

Wait isn't that "trawling"?

See, @AwalGarg, that's how you deliver a joke.

11:36 AM

@DavidWheeler is natural at jokes.

Though he is also natural at giving 50-page book-size answers to group theory.

3

@DavidWheeler This might make for a good question to ask for the ethymology of "trolling" on English Language & Usage. There may be a connection!

@BalarkaSen Group theory is a very special kind of humor.

@Raphael ?

I don't want to learn any more English! Please, make it stop!

@Raphael How about category theory? :)

I asked my dad: why can't I marry a transposition? and he told me: son, that's not normal.

David Kirby

11:39 AM

ncf.idallen.com/english.html

Say abscission with precision,
Now: position and transition;
Would it tally with my rhyme
If I mentioned paradigm?

@DavidWheeler This is good.

@BalarkaSen No, its meh like most other math jokes.

@KarlKronenfeld That's like Awal Garg: one never knows whether it's fun(ny) or just a special kind of heroic sadness.

@Sawarnik It's not. You don't know group theory that much yet to understand what it says.

Transpositions are 2-cycles and normality is a property of subgroups.

@BalarkaSen Even if I did, I can think now it will be a very funny meh.

11:42 AM

I am gonna go leave. Bye.

later pal

@Raphael I was playing on the typical remarks that category theory is pure nonsense, e.g. the not so uncommon term "abstract nonsense".

@BalarkaSen Bye.

as opposed to concrete nonsense?

Of course it's meh! It's not like I stayed up all knight searching for the perfect trompe l'oreille!

11:51 AM

Did Euclid put any jokes in the Elements?

or Newton in the Principia?

That's a charming question...it seems possible....

:-)

@skullpatrol Well Fermat certainly put a lot of jokes in his margins..

@N3buchadnezzar Indeed.

@skullpatrol Euclids jokes would have been quite elementary... Anyway the text is freely translated so you can read it online, or order the books if you want,

12:01 PM

and scour it for jokes rather than mathematical content

lol wut

The thing is: who would know the Greek slang of Euclid's time, and be able to verify it was translated properly?

Possibly greek translators >.<

I figured out something cool, wop wop

Maybe, if they knew "ancient" slang. The meaning of words in the same language change over time: an example in English is "dyke" which transformed from meaning "ditch" to "lesbian".

Or You

12:07 PM

I kinda doubt Euclid put in any jokes, but...hard to say...maybe he did.

You could also study the ancient greek literature...

To see if you can find any jokes from that time.

Let $R$ be a function satisfying $R(x) = R(1/x)/x^2$ for all $x$, and
such that $\int_0^\infty R(x) \mathrm{d}x$ converges. Then
$$
\int_0^\infty (\log x)^{2n-1} R(x) \,\mathrm{d}x = 0
$$
for all $n \in \mathbb{N}$.

I give this a 4/10 on a coolness scale.

What's an example of a function with that property?

I hate integrals. Strange, because I enjoy vector lattices.

$(1+x^2)/(1+x^4)$

Or any rational function on the form $p(x)/q(x)$ where $p$ and $q$ are symmetric polynomials of respectively degree $k-2$ and $k$.

12:13 PM

Your latter example surely bumps the coolness factor up to 4.1

$$
R(x) \equiv \left( \frac{x^2}{x^4 + 2ax^2 + 1} \right)^r \left( 1 + \frac{1}{x^2}\right)
\equiv \frac{x^4 + 3x^3 + 3x^2 + x}{\pi x^7 - x^4-x^3+\pi}
$$
For all $r>1$ and $a \in \mathbb{R}$. Really there are loads :p But who cares.

mm

@DavidWheeler Woo more coolness

12:28 PM

As your coolness is enhanced, mine is diminshed. Cries

I am having a feeling that I really don't understand covering space theory.

sigh

Q: what did one covering space say to the other?

A: do you even lift, bro?

Starred.

Anyway, all hands on deck

I hate or love integrals or both. I exhibit quantum properties :S

So, when you flip a coin, do you wonder how often both sides will land heads up?

12:42 PM

Nop

Because the probability of getting 50% heads and 50% tails increases with number of throws

So if I throw a zillion times, I'm pretty sure number of heads will be approximately equal to number of tail, if not equal.

Personally I believe both cats are dead until I look at the coin.

@Sabಠ_ಠ see you did change your avatar :-)

"Sab"

:P

I changed it a while ago @Skull :)

@BalarkaSen pfft! they used the torus....anyone knows that $\pi_1(S^1 \times S^1) = \pi_1(S^1) \times \pi_1(S^1) \cong \Bbb Z \times \Bbb Z$

@DavidWheeler Yeah, using torus is stupid. The monodromy is just the fundamental group.

The examples I work with are modular surfaces.

But that's the end of my knowledge in algebraic topology.

12:52 PM

The preface to my college thesis started:

We haf da loop, an we haf da group. We loop-de-loop, an group dey loop, to haf da loop group.

6

Hahaha

you can't be sirius.

To demonstrate the plane is almost homeomorphic to the sphere at my orals, I popped a balloon.

sup dawgs

Just illin' wif da homies, yo

dats wat i lyk to hear broseph

oi wanna help me wiff a limit question?

12:57 PM

@Mew the only sensible answer of 'sup' to me is 'limsup'

lim a->infinity of sin(ax)/x

apparanlty it is 2pi*delta(x)

no idea why tho

Sigh.

@Mew In the space of distributions, the distributions $T_a$ given by $$T_a[\varphi] = \int_\mathbb{R} \frac{\sin ax}{x}\varphi(x)\,dx$$ converge to $2\pi\delta$.

salbeira

If I find the directional deriveratives of a function at a certain point by using $$\frac{f(x + h) - f(x)}{h}$$ and then find out that when using the "common" way to derive a function find a way to express the derivation and then find a sequence for which I find another value to be the derivation, is that function total derivatable at that point or not?

@N3buchadnezzar Only 4/10? That's pretty cool.

@AndrewG =)

Finally

1:06 PM

2.36 and 2.37 are also pretty nifty. What book is that from?

Daniel, can you just prove the limit in it's own right?

i'm not familiar with space of distributions

@AndrewG None

Are you able to prove them ? It is a good exercise =)

The first and second yes, am working on parts 3 and 4 now

Seems like the same basic idea

Yeah

Just a cute little trick

Very much so :)

1:14 PM

@Mew What do you mean with "prove the limit in its own right"? The limit aimed at is a distribution, so we should consider the objects whose limit is taken also as distributions. We could consider everything as tempered distributions, but that wouldn't change much.

The clue is to map $R(x) f(x)$ from $(0,\infty)$ onto $(0,1)$ then
$$
\int_0^{\infty} R(x) f(x) \,\mathrm{d}x
= \int_0^1 R(x) \bigl( f(x) + f(1/x)\bigr) \,\mathrm{d}x
$$
And yeah, thats that.

Yeah

As an example, this integral follows directly from the $(2.36)$, $x \mapsto \tan \theta$
$$
\int_0^{\pi/2} \frac{\mathrm{d}\theta}{(1 + (\tan \theta)^b}
= \int_0^\infty \frac{\mathrm{d}x}{(1+x^2)(1+x^b)}
= \frac{1}{2} \int_0^\infty \frac{\mathrm{d}x}{1 + x^2}
= \frac{\pi}{4}
$$

By the way, I think the constant is wrong. Should be $\pi$ instead of $2\pi$, @Mew.

I'm trying to find a clever solution to a nice equation ...

$$\frac{\{x\}}{x}+\frac{x}{\lfloor x \rfloor}=\frac{3}{2}$$

1:20 PM

@Daniel, how can we prove the set of distributions converge to pi?

@Mew No, they converge to $\pi\delta$, not to $2\pi\delta$. By change of variables, $$\int_\mathbb{R} \frac{\sin (ax)}{x}\varphi(x)\,dx = \int_\mathbb{R} \frac{\sin y}{y} \varphi\left(\tfrac{y}{a}\right)\,dy.$$ Now $\int \frac{\sin y}{y}\,dy = \pi$, so that is $$\pi\varphi(0) + \int_\mathbb{R} \frac{\sin y}{y}\left(\varphi\left(\tfrac{y}{a}\right) - \varphi(0)\right)\,dy.$$ It remains to see that the last integral converges to $0$.

Which, since $\varphi(y/a) \to \varphi(0)$ locally uniformly, is highly credible.

$\operatorname{moderator}$

1:37 PM

thanks daniel

2:13 PM

I've just created a special question ...

Long time no see @AndrewG

A: Spurious space within number

8

The Problem: In a comment to a question or an answer, or on chat, when a string of over 80 characters is entered without whitespace, the pair of characters \unicode{x200C} (zero-width non-joiner) and \unicode{x200B} (zero-width space) is inserted after 80 characters. This is often bad when it h...

Bayern, mia samma mia!

@BalarkaSen I fixed your comment

How are things going in chat, what have I missed?

2:21 PM

@Chris'ssis this is why you need a space every 80 characters

@robjohn Yeah, it's OK.

Haindling Bayern des samma mia!

@Chris'ssis What kind of formulas/mathematics are you up to?

Hey @Mike

Take a look at this

@MatsGranvik Seen this?

3 hours ago, by Balarka Sen

@N3buchadnezzar @Chris'ssis @robjohn Here's something I learned from a user of Integral&Series : Riemann Hypothesis is equivalent to stating $$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3} \int_{1/2}^{\infty}\log|\zeta(\sigma + it)|~d\sigma ~dt=\frac{\pi(3-\gamma)}{32}$$

@N3buchadnezzar No I have not. I will have a look at it.

That statement with integrals looks very interesting.

2:32 PM

Let's consider the family of roots $ R_{n}(x):\left \{\displaystyle \frac{ \{ x \}}{ x } + \frac{ x}{ \lfloor x \rfloor}=\frac{1}{n}, \space n\in \mathbb{N} \right \}$.
Test for convergence $$\sum_{\displaystyle R_{n}(x) \in \mathbb{Q}} $$

@MatsGranvik see above

$$\bigcup\limits_{n \in \mathbb{n}} \left[ \frac{1}{n} , 1 - \frac{1}{n}\right) $$

Is this subset of $\mathbb{R}$ half open, closed, or open with the standard topology defined on $\mathbb{R}$?

I'd say it is open but....

@N3buchadnezzar But what?

Not sure how to argue that it is equivalent to $(0,1)$

@r9m hey, did you finish the candy limits I showed you yesterday? ;-)

@FernandoMartin What's the justification behind this category?

Why are we looking at it?

wow, I take that bacl

@FernandoMartin That was incredible

2:50 PM

@N3buchadnezzar For $n\in \{0,1,2\}$, $\left[\frac{1}{n},1-\frac{1}{n}\right)$ is empty, therefore a subset of $(0,1)$. For $n \geqslant 3$, $0 < \frac{1}{n} < 1 - \frac{1}{n} < 1$, so the union is contained in $(0,1)$. Pick $x\in (0,1)$ and find an $n$ so that $\frac{1}{n} < x < 1-\frac{1}{n}$.

0?

You mean 1 and 2 right?

@N3buchadnezzar ??

$0\in\mathbb{N}$

no ? :p (yes, it is a matter of taste)

I still don't understand why 0 isn't the initial object

@Mike

@FernandoMartin What are the morphisms? The isometries?

2:57 PM

@FernandoMartin Because $L^1$ is, clearly

@Daniel: It's somewhat complicated, it's explained in the paper

Just opened it. Reading.

they're "unit" preserving contractive maps, preserving some sort of mean on the space

it's really quite good

Ahh I'm stupid

obviously the 0 object of the space 0 maps to 0

2:59 PM

@FernandoMartin You just saw that $0$ can't have unit-preserving maps?

so there's no map from 0 to a banach space $V$ if $V$'s $u$ isn't 0

I type too slowly.

Yes. I think too slowly :)

@FernandoMartin damnit

I got back here right after you guys said that

3:11 PM

@Chris'ssis I don't think there will be any solutions.

@Chris'ssis Am I right?

@Sabಠ_ಠ Hi. Did you see Madras Cafe?

hI

I'm actually buffering it as we speak

coincidence much :P

G.T.R

Is there any geometric interpretation of this math.stackexchange.com/questions/801157/… ?

@Sawarnik No. There are solutions ...

@Chris'ssis Oo. Then for example what are the solutions?

@G.T.R sure... just a minute...

3:21 PM

@Sabಠ_ಠ Means? :/

@Sawarnik That's the part you need to find on your own. ;)

(if you like it)

I mean I was just about to watch it @Sawarnik

@Sabಠ_ಠ Oh. Then go and watch. No disturbing in that time! And tell me when you have seen.

:)

This is probably the only day I'll be able to watch a movie

after that exams willcome :O:

:D

@Chris'ssis I took x/floor(x)=y, then y-1/y=2, which gives y=1+-sqrt2, which has no solution in x? :O

3:26 PM

@Sawarnik Do you refer to this one? $$\frac{\{x\}}{x}+\frac{x}{\lfloor x \rfloor}=\frac{3}{2}$$

@Chris'ssis Yes.

@G.T.R this is not it :-)

@Sawarnik That one has $3$ solutions.

@Chris'ssis But where did I go wrong? :O

@Sawarnik There you have both fractional part function and floor function. Did you consider them both?

sdf

3:28 PM

If you have a rational function whose zero set is dense in the image then is it identically zero?

G.T.R

@robjohn I think it's the usual mean value theorem interpretation. The result stated by the user is quite different. Correct me if I'm wrong

Studentmath

How can you even deal with such a thing @Chris @Sawarnik

@G.T.R That's what I was thinking.

@Chris'ssis Yes. $\frac{\{x\}}{x}+\frac{x}{\lfloor x \rfloor}=\frac{x-\lfloor x \rfloor}{x}+\frac{x}{\lfloor x \rfloor}=1+\frac{x}{\lfloor x \rfloor}-\frac{\lfloor x \rfloor}{x}$. Is this wrong?

@Studentmath Do you refer to this one? $$\frac{\{x\}}{x}+\frac{x}{\lfloor x \rfloor}=\frac{3}{2}$$

@G.T.R let me check again...

3:31 PM

@Sawarnik It seems OK.

Studentmath

@Chris'ssis Yes, but I just saw what Sawarnik did, so figured out.

@Chris'ssis Then if $y=\frac{x}{\lfloor x \rfloor}$, $y-\frac1{y}=\frac12$, which means $y=1\pm \sqrt2$?