Mathematics - 2012-03-08 (page 4 of 5)

Skullpatrol

19:00

@yoda Also, n can not equal 0 and/or k can not equal 1/2.

robjohn

@Jonas: I have been busy offsite, so I have only had a chance to reply to your comment.

Jonas Teuwen

@robjohn No worries. I was just wondering where the $\phi/a$ comes from.

I have fed it to NIntegrate and they give different results.

Skullpatrol

@robjohn Hi Rob

robjohn

@Jonas: I will be away for a bit more and then I will have a chance to look some more

@Skullpatrol Hey there :-)

Jonas Teuwen

Okiedokie!

robjohn

19:02

@JonasTeuwen hmm.. that may be a typo.

@JonasTeuwen fixed. If the answer is not independent of $\phi$, then Mathematica has erred.

Skullpatrol

@JonasTeuwen Okiedokie? is two words: okie = okay, and dokie = do

Jonas Teuwen

Why should it be independent of $\phi$?

Skullpatrol

►

David Wheeler

indeed, including infinity often "breaks" a structure, and one gets "some other kind of structure". another way of saying this is: $\mathbb{R} \cup \{\infty\}$ is no longer a field.

Jonas Teuwen

19:18

@robjohn Strange, if I integrate $$\int_0^{2 \pi} \exp(i \cos(\beta - \phi)) \sin(\beta) \, \textrm{d}\beta$$ with Mathematica it gives me $2 \pi i J_1(1)$. That is about $2.72 i$, but if I fill in $\phi = 0.5$ and integrate numerically I get about $2.42 i$.

David Wheeler

which puts a damper on any arena in which you want to use division in.

robjohn

@JonasTeuwen Have you tried it symbolically in Mathematica?

Jonas Teuwen

I also don't see why it should be independent on $\phi$.

@robjohn Mathematica does not really like that, my computer tries to fly away :-).

robjohn

@JonasTeuwen I don't either. I see that it is an even function in $\phi$, but not independent.

@JonasTeuwen I have to go. I will look when I get back.

@Skullpatrol is advertising Spam, porn?

Jonas Teuwen

@robjohn Mathematica forgets a term $\cos(\phi)$.

t.b.

19:29

@MattN okay, I'm back

Matt N.

Ello : )

t.b.

So, where were we?

Matt N.

I'm still recovering from the pain Tao inflicted on me.

t.b.

Question closed as NARQ by OP?

Asaf Karagila

Heh.

I cast a deletion vote, @tb

t.b.

19:34

@AsafKaragila I prefer to wait at least a few days.

Asaf Karagila

@tb Well, just so you know. :-P

What the...

Jonas Teuwen

@tb What the...?

Asaf Karagila

The front page has real time updates.

t.b.

@MattN So did you get how you can use the characters to pass from a function on $G$ to a function on $\widehat{G}$?

Asaf Karagila

Hmmm. Speaking of Pontryagin I should email my comm. alg. professor...

t.b.

19:39

@AsafKaragila For some reason I don't see those. But luckily, I don't see those bluish favorite things and the community ads either.

David Wheeler

true or false: a group of order 2366 has either a normal 13-sylow subgroup, or a normal 7-sylow subgroup?

Matt N.

@tb I have $\chi_\xi : G \to C$ and I do $\langle \chi_\xi , \cdot \rangle := \int \cdot \chi_\xi (x) dx$ which is a function $\hat{G} \to C$?

Asaf Karagila

@DavidWheeler If the group is Abelian then maybe.

Matt N.

David Wheeler

if the group is abelian, definitely, all subgroups are normal

Asaf Karagila

19:42

Yes, but would it have 7- or 13-sylow subgroups?

t.b.

@MattN Well, $\chi_{\xi}: \mathbb{R} \to S^1$ is a character, so $\chi_{\xi} \in \widehat{\mathbb R}$. Now you need a function $f: \mathbb{R} \to \mathbb{C}$ (at least integrable). Now the Fourier transform gives you a function $\hat{f}: \widehat{\mathbb R} \to \mathbb{C}$

David Wheeler

all groups have p-sylow subgroups if p divides |G|

Asaf Karagila

I know that.

t.b.

sheesh. Sorry for the multiple edits.

David Wheeler

7 divides 2366, as does 169

Matt N.

19:44

No problem at all : )

Asaf Karagila

I did not try and factorize 2366 to see if 7 or 13 are factors.

I think that Fourier analysis should be taught after some basic course on representation theory.

David Wheeler

the problem is: showing that the 13-sylow subgroup has 14 conjugates, and the 7-sylow subgroup has 169 conjugates are mutually exclusive

t.b.

The function $\hat{f}$ is given by $\chi_{\xi} \mapsto \hat{f}(\chi_{\xi}) = \int_{\mathbb{R}} f(x)\,\overline{\langle x, \chi_{\xi}\rangle}\,dx$.

(since $f$ is integrable and $|\overline{\langle x,\chi_{\xi}\rangle}| = 1$, this is a well-defined complex number by the standard estimate $|\int g| \leq \int |g|$).

Gigili

Whoa, 33.4K

t.b.

@AsafKaragila I don't think representation theory is a necessary prerequisite.

@Gigili 'Ello! Nice to see you!

Gigili

19:49

'Ello @tb.

m. k.

@DavidWheeler: I think you can show that when there are 14 13-sylow subgroups, then there is a normal subgroup of order 13. Dunno if that helps

David Wheeler

what confuses me is the dual interpretation of $\chi_{\xi}$ as a complex number and a function

t.b.

@DavidWheeler I mentioned some time ago that the homomorphisms $\chi: \mathbb{R} \to S^1$ are given by $\chi_{\xi}: x \mapsto e^{i\xi x}$ ,where $\xi$ runs through the real numbers

David Wheeler

to me, it's a blurring of the distinction between a function and its image

Asaf Karagila

@tb It's not, but it makes the course a lot easier to understand.

Matt N.

19:52

@tb Isn't this what I wrote? (Except for forgetting the conjugate bar)

David Wheeler

@mk could you elucidate?

Gigili

'Ello @MattN.

t.b.

@DavidWheeler No, I'm consistently seeing $\chi_{\xi}$ as an element of the dual group $\widehat{\mathbb R} = \operatorname{Hom}{(\mathbb{R},S^1)}$ which just happens to be isomorphic to $\mathbb{R}$

Matt N.

Hi Gigili!

Gigili

'Ello @tb.

Okay, I'll stop now.

David Wheeler

19:54

@tb but you're using it in an inner product

m. k.

@DavidWheeler: Two different Sylow 13-subgroups generate the whole group. If the intersections of sylow 13-subgroups were trivial, then we would have a normal 7-sylow. So suppose they are not. Then there are 13-sylows P_1 and P_2 with intersection of order 13, so the intersection is normal in P_1 and P_2 and thus in the group since P_1 and P_2 generate the group

But again, I dunno if this helps at anything

t.b.

@DavidWheeler No, that's just the usual notation of the evaluation map $\langle \cdot,\cdot\rangle: G \times \widehat{G} \to S^1$. So $\chi(x)$ is written as $\langle x, \chi\rangle$ (which is a complex number).

David Wheeler

@Gigili light of my life! fire of my...oh wait, there's other people here. hello.

m. k.

I'm thinking about it now

t.b.

@Gigili 'Ello for the last time

Gigili

19:56

Ello ello @DavidWheeler.

'Ello @tb.

t.b.

@MattN Well, you wrote "which is a function $\hat{G} \to C$"

David Wheeler

ok, so you're talking about the function $\langle \cdot, \chi_{\xi}\rangle$

t.b.

Exactly.

David Wheeler

which is in the dual space, right

t.b.

It is a function on $G$ (a homomorphism).

Jonas Teuwen

19:58

@robjohn It is a bug in Mathematica.

David Wheeler

well, not just "any function", it's actually linear over $\mathbb{R}$

t.b.

That's why I added the parentheses :)

David Wheeler

my point was, it's more than additive.

t.b.

Just additive. It's the function $\chi_{\xi} : \mathbb{R} \to S^1$ given by $x \mapsto e^{i\xi x}$.

sheesh, homomorphism. I just view $\mathbb{R}$ as its underlying additive group, equipped with its Haar measure (= Lebesgue measure)

David Wheeler

i understand that...it's just that an inner product is bilinear, right? so we get some "extra properties" (we don't have to use them if we don't want to)

Gigili

20:04

Any of you guys knows of a translation website or something like that?

David Wheeler

besides google translate?

Gigili

Something that doesn't suck.

t.b.

No, it's just a piece of (maybe unfortunate) notation. Given an abelian group $G$, I consider its dual group $\widehat{G} = \operatorname{Hom}_{\rm cont} {(G,S^1)}$ and I write $\langle x, \chi \rangle = \chi(x)$ for $x \in G$ and $\chi \in \widehat{G}$. So $\langle \cdot, \cdot \rangle: G \times \widehat{G} \to S^1$ is a ($\mathbb{Z}$-bilinear if you want) pairing.

@Gigili I don't. But what languages?

David Wheeler

....parsing....is $\widehat{G}$ isomorphic to its dual?

Gigili

@tb Turkish to English

t.b.

20:09

@DavidWheeler This is just a coincidence in the case of $\mathbb{R}$ or $\mathbb{R}^n$. The dual group of $S^{1}$ is $\mathbb{Z}$ and the dual group of $\mathbb{Z}$ is S^1$.

Gigili

Huh, found it.

t.b.

@Gigili Oh, that could be useful for me. Google translate really sucks when it comes to Turkish.

David Wheeler

turkish...that's a tough one, turkish translators are not very common

Gigili

@tb Exactly, I found this one/; onehourtranslation.com

t.b.

Thanks, bookmarked :)

@MattN still recovering?

Matt N.

20:13

@tb Almost recovered, thank you. I had to deal with the cardboard for the cardboard pick up tomorrow.

t.b.

Oh, given that those happen once a month only...

Matt N.

Yes. Missed it about 3 times.

Asaf Karagila

With the whatboard for the cardwhat pickup??

David Wheeler

so you're not using an "actual inner product", just a kind of "additive tensor"?

t.b.

@MattN well, for that reason I subscribed to the ERZ calendar some time ago. One of the blessings this mobility thing allows... :)

@DavidWheeler exactly.

Matt N.

20:18

@tb Same here : )

I still miss it.

It just doesn't have very high priority in my life.

David Wheeler

so we get an element of G or G-hat, depending on "which one we think is the variable"?

t.b.

No, not quite. If $x \in G$ then we get a homomorphism $\langle x, \cdot\rangle : \widehat{G} \to S^1$, so an element of $\widehat{\widehat{G}}$.

David Wheeler

but isn't the identification of G with $\widehat{\widehat{G}}$ natural?

t.b.

It is and it is an isomorphism by the Pontryagin duality theorem.

But a priori it isn't clear that it is either injective or surjective.

David Wheeler

i am supposing Pontryagin duality is analogous to linear functional duality where you send v to the functional defined by v(f) = f(v)

t.b.

20:24

That's a good way to think about it, but remember that in order to get an isomorphism you need finite dimensionality.

David Wheeler

in the correspondence i just gave, there's no "basis selection" how could dimensionality enter into it?

t.b.

In general it's only injective for vector spaces.

To prove surjectivity you need finite-dimensionality.

Matt N.

@tb True. $\langle \cdot , \chi_\xi \rangle : \hat{G} \to \hat{G}$. Or not, since I'm applying it to $f: G \to C$ and you said $\hat{G} = \operatorname{Hom}{(G, S^1)}$.

David Wheeler

oh, ok, because there could be a lot more "f's" than "v's"

Matt N.

So I almost got it right. Except for a type error.

t.b.

20:27

@MattN no, $\langle \cdot, \chi_{\xi} \rangle: G \to S^1$...

Matt N.

@tb How is that possible? The inner product takes a function as argument.

$G$ is something like $Z_p$

t.b.

Oh, you were talking about your formula. Well, that's a function $L^1(G) \to \mathbb{C}$.

David Wheeler

could i see that integral formula for $\widehat{f}$ again?

t.b.

$\hat{f}(\chi) = \int_{G} f(x)\,\overline{\langle x, \chi\rangle}\,dx$.

Matt N.

$$ \hat{f} (\xi) : = \int f(x) e^{-2 \pi i x \cdot \xi}dx $$

t.b.

20:30

Hey, there, The Chaz!

The Chaz

Howdy!

Ell

hi guys :)

Matt N.

@tb No. That dual pairing takes a function and gives me a function back. So it can't map to $C$.

t.b.

@MattN but you fixed the character.

Matt N.

@tb No. Then I wouldn't get a function!

t.b.

20:31

(so you evaluate the function at the character).

David Wheeler

is G pre-supposed to be a subgroup of the multiplicative group of $\mathbb{C}$?

Gigili

'Ello @TheChaz.

Asaf Karagila

I have come up with an interesting characterizations of trivial topological spaces!
$X$ has only one point if and only if the trivial topology on $X$ is completely metrizable.

The Chaz

Hi

Gigili

'Ello @Ell.

Ell

20:32

has anyone got time to answer this question super-quick. I would ask on the main site but I don't actually know if its maths :S

The Chaz

@Gigili aka "LOL"

Ell

what is an inversion in this context: "The file contains all the 100,000 integers between 1 and 100,000 (including both) in some random order( no integer is repeated).

Your task is to find the number of inversions in the file given (every row has a single integer between 1 and 100,000)."

t.b.

@TheChaz no, laughing out @ loud

Ell

I do apologise for not opening a question, but i genuinely dont know if it means in maths or a palindrome or what :S

The Chaz

Parity of a permutation

@Ell

??

Right, t.b. my bad!

Ell

20:33

@TheChaz an inversion :s what is an inversion?

if its not maths I apologise, send me away

David Wheeler

in that context, i am guessing "an inversion" is an order-reversal

The Chaz

inversion

David is right. It has to do with how many bigger numbers precede a number, positionally

Ell

@TheChaz thank you! Wikipedia has many defenitions of "inversion" in maths

t.b.

@MattN Given a function $f: G \to \mathbb{C}$ (from $L^1(G)$, say) you get a function $\hat{f}: \widehat{G} \to \mathbb{C}$ (in $L^\infty(\widehat{G})$, say. But you fixed your character $\chi$, so what you wrote is the map $f \mapsto \hat{f}(\chi) \in \mathbb{C}$.

The Chaz

:)

David Wheeler

20:35

can G be any abelian group?

Matt N.

Sometimes it's a topological group.

In the book it's finite abelian groups. No mention of topological.

t.b.

@DavidWheeler make it locally compact, but if you just think of an abstract group, take the discrete topology.

David Wheeler

the reason i am wondering is because you are integrating over G, implying you have a topological structure

Matt N.

You don't need a topology to integrate, I think you need a measure space.

t.b.

@DavidWheeler if it's just the abstract group, the integration becomes summation (counting measure)

David Wheeler

20:37

well you do need some idea of "+"

Matt N.

That's where the measure maps to: R or C.

Besides: what has + to do with a topology? You mean a group structure?

David Wheeler

if you have a discrete space, you get "sums" right?

Matt N.

No you get "sums" if you have a group structure with a +.

David Wheeler

isn't that what an abelian group is?

Matt N.

Exactly.

No topology involved there.

t.b.

20:39

Okay, the setting is this: given a locally compact group, there's a unique (up to a positive scalar) Radon measure on $G$. If $G$ happens to be discrete, it's counting measure. If $G$ happens to be $\mathbb{R}$ then it's Lebesgue measure. If $G$ happens to be the circle then it's the usual angular measure.

David Wheeler

so, if G is finite, we get some sort of finite sum

t.b.

Exactly.

David Wheeler

i don't know too much about Lebesque measure, but i assume it returns |b-a| on the interval [a,b]

t.b.

Yes, it's the usual "volume" in $\mathbb{R}^n$.

The Chaz

^THAT, I can recall :)

t.b.

20:42

So "length" in the one-dimensional case, "area" in the two dimensional case, etc.

@TheChaz I'm flabbergasted....

David Wheeler

i assume Lebesgue measure is used because it can "measure more sets" than the normal forms of integral used

t.b.

...and proud of you.

The Chaz

Oh man. Thanks to some heavy-hitters, including one in this room and Tao's MT notes :)

t.b.

@DavidWheeler well, it's basically the convenient setting to treat integration. Riemann theory is a little painful to carry out Fourier analysis.

David Wheeler

i am guessing a measure makes m(AUB) = m(A) + m(B) true when A and B are disjoint

there's probably some other technical conditions required, too

t.b.

20:45

Yes, but more than that. If you have a sequence of disjoint sets you want $\sum_n m(A_n) = m\left(\bigcup A_n\right)$. It is this property that allows you to give handy criteria to interchange summation/limits and integration.

(which you need constantly when calculating)

David Wheeler

i think you left out an "m" somewhere

Matt N.

@tb Ouch. Ok. You asked me how to turn $f: G \to C$ into a $g : \hat{G} \to C$. And I ballsed up the sides in the pairing. If I do $\langle f, \cdot \rangle$ I get a thing taking a character and mapping it to a complex number. Sorry I don't know what took me so long.

t.b.

@DavidWheeler right. better?

@MattN Now we're talking :)

David Wheeler

so Riemann theory doesn't let us move the "lims" where we want very easily, hmm?

Matt N.

@tb The upshot is: double check what I reply to your questions before posting it.

t.b.

20:50

Exactly. The only thing you can establish easily is that you can interchange limits and integration provided you have uniform convergence. Moreover, if you have less, you normally need to prove that the limit of sequence is still measurable (read (improperly) integrable if you don't know what that is). This is automatic in measure theory.

David Wheeler

stupid question: does measure let us create a metric?

robjohn

@JonasTeuwen okay. I wonder if the people in the Mathematica group know the easiest way to report such a bug.

t.b.

@DavidWheeler metric where?

Jonas Teuwen

@robjohn I have filled in a bug report.

David Wheeler

on G

t.b.

20:52

Well, if $G$ allows a metric giving its topology, the integral allows you to get an invariant metric. That is $d(x,y) = d(gx,gy)$ for all $x,y,g \in G$.

robjohn

@JonasTeuwen cool!

David Wheeler

is that the same as saying the topology on G is metrizable?

t.b.

@MattN yeah, sorry, I was distracted by having too many conversations at the same time. David asking good questions and the joyous surprise of seeing Gigili again.

And The Chaz entering the room for the first time in a looooong time.

@DavidWheeler Yes, if a group is metrizable it always admits an invariant metric. If you have a locally compact group you can get that invariant metric using the measure.

David Wheeler

cool, so locally compact groups are "well-behaved" then

t.b.

Yes, I see the local compactness condition as analogous to the finite dimensionality condition in linear algebra.

Some things become easier, but finite dimensionality is not needed, other things are just not true without it. (Same holds when using local compactness in this sentence)

The Chaz

20:57

@tb I actually came to rant, but this isn't a good time. You're juggling multiple lessons!

David Wheeler

the way i've always thought of "locally compact" is: we can study neighborhoods of a point, rather than the whole space

t.b.

@TheChaz Rant away! Please.

The Chaz

Let me start a load of laundry at least.

Matt N.

@tb Sure : )

Gigili

Did you know this usage of Wolfram|alpha?

Matt N.

20:59

No.

Gigili

howtogeek.com/106925/10-amazing-uses-for-wolfram-alpha

David Wheeler

@Gigili that's comparing "oranges and apples"

Gigili

Especially the "Am I Drunk?" part ..

Matt N.

Ah.

t.b.

@TheChaz oh, no! my teaching lessons has stopped a rant from the Chaz!

David Wheeler

21:03

@tb i, too, am over-joyed to see Gigili, but i don't want HER to know that, because that puts me at a strategical disadvantage. oh, shoot.

The Chaz

Well there are rules in the chatroom... this is not 'Nam

;)

t.b.

You're wrong Walter

2

The Chaz

In essence, my issue is with how petty many of the personal interactions are on MSE, and how petty I have become.

Matt N.

So if the dude asks me to talk about the duality $\mathbb{Z} \leftrightarrow \mathbb{R}/ \mathbb{Z}$ then he means $k \mapsto e^{2 \pi i k x_0}$, I suppose.

David Wheeler

@TheChaz what is it you wish to rant about (under the theory that a meta-rant may be within the rules whereas the rant itself may not)

t.b.

21:06

@MattN You need to flesh out the facts that a homomorphism $\mathbb{Z} \to S^1 = \mathbb{R/Z}$ is given by an element of $\mathbb{R/Z}$ (fairly clear) and that a homomorphism $S^1 \to S^1$ is given by $z \mapsto z^n$ with $z \in \mathbb{Z}$.

The Chaz

(@David see above ??)
Recent meta articles about acceptance rates, voting patterns, snarky comments, and moderators/top-tier members quitting the site have left a sour taste in my mouth.

David Wheeler

every integer determines a rotational symmetry of the circle, yep

t.b.

@TheChaz who quit the site?

The Chaz

Oh that wasn't so recent.

Bill getting suspended, Robin (?) quitting, and PLC almost quitting, if memory serves

t.b.

Don't forget Akhil

The Chaz

21:09

(et. al!) ... so I have talked myself out of a meta question to the effect of "Can't we all just get along?" a few times in the past week.

Of course an absence of conflict is impossible, but people are really losing touch with the human element

David Wheeler

my feeling is this: any time you assign a way of assigning integers to individuals, people almost invariably equate the natural ordering of the integers with some form of social ordering

Matt N.

@tb I think the second homomorphism should've been $S^1 \to \mathbb{Z}$. Thanks, I didn't realise there were maps involved. : /

The Chaz

(Which goes back to my first rant about how we berated new users for poorly worded questions)

t.b.

I haven't followed the drama around Bill. I think the suspension was quite harsh (and meanwhile we had Math Gems as a substitute), but on the other hand, I think that the more visible persons should held to higher standards for obvious reasons and his comments were a bit ... well ... unhelpful(?)/cryptic(?)/condescending(?)

Matt N.

Unnecessary? Unnecessarily aggressive?

The Chaz

21:11

Then my rant is directed at him, not just about him!

Of course, all of this might sound strange coming from the village idiot (who also happens to have a low integer assigned to him, @David)

David Wheeler

i have observed myself that well-educated mathematicians tend to be a bit haughty towards normal peeps

The Chaz

Matt, that's exactly what is wrong with this site

People should keep their mouths shut (or fingers still) about others.

Matt N.

Of course I'll have to delete this. : )

The Chaz

What's more important is that these things are not said! sigh

David Wheeler

in this chat-room alone (ignoring all else for purposes of discussion) there is substantial talk about badges, capping and vote solicitation

The Chaz

21:14

^

t.b.

@MattN Well, the duality is $\widehat{\mathbb{Z}} \cong S^1$ and $\widehat{S^1} \cong \mathbb{Z}$, that's why I wrote what I wrote and I don't think there's a typo. The map is given by evaluation $x \mapsto \langle x, \cdot \rangle$ which is a map $G \to \widehat{\widehat{G}}$ (this works exactly the same way as the inclusion of $X$ into $X^{\ast\ast}$ for vector spaces)

Matt N.

I give up.

t.b.

DON'T

Asaf Karagila

Giving up? That's my thing. You can't have it.

David Wheeler

i think i see it: an element of the dual of $S^1$ has to send 1 to 1, and if it's a rotation, must therefore be an integer multiple of a "full circle" rotation

t.b.

21:18

Right.

But you still need to argue that a homomorphism has the form it has. This is basically the fact that the functional equation $f(x+y) = f(x) f(y)$, $f(0) = 1$ has the exponential function as unique (continuous) solution.

The Chaz

I'll conclude the rant with a few personal policy/perspective changes: I duped myself into caring about rep. While I would like the privileges associated with higher rep, I will work to lessen the impact that fickle voting patterns have on me. This will probably result in me:
a) asking more questions. It used to bother me that 45 minutes of thought and Latexification would get me maybe one vote and one hint. No longer!

2) answering more questions. If I think that my solution or advice is worthwhile to the OP and/or community, I will face the threat of downvote!

Gigili

@TheChaz In short, the whole system sucks. I agree.

David Wheeler

@TheChaz i still have trouble "finding questions" i'd like to answer....any suggestions?

The Chaz

Uhh... I'd say "the way I have responded to the system sucks". @Gigili

@DavidWheeler As t.b. will tell you, I'm always on the lookout for "low hanging fruit"!

David Wheeler

@t.b. is that enough? don't we need some sort of co-domain restriction?

The Chaz

21:23

But seriously, I am most comfortable with lower-level questions, of which we have relatively few, and more conceptual/big-picture questions.

Matt N.

@tb Ok, it's not a typo since the range is always $S^1$ of those homos. And that map is something like the evaluation map, I assume. I quit reading futile attempts before I got as far. Sorry : /

@tb Now you've done my work for me :,(

t.b.

@DavidWheeler Yes you do. But using the logarithm you can quickly reduce to what I said (speaking only of functions $\mathbb{R} \to\mathbb{R}$.

@MattN No, I have given an outline of what you have to do and think about. There's nothing wrong about that, I think.

Gigili

@TheChaz Right, you can also say it implicitly.

Skullpatrol

@robjohn "is advertising Spam, porn?" I'm not sure what you mean, oh Great Mean Square, by this question? I said advertising porn is spam, in my opinion. How does one advertise spam (other than the semi-eatable kind)?

David Wheeler

@tb what i meant is, there are lots of functions $f:\mathbb{R} \to \mathbb{R}$ satisfying f(x+y) = f(x)f(y), f(0) = 1, but only one has the property f'(0) = 1, as well.

Matt N.

21:26

@tb Well the painful part would've been to figure out what he meant when he wrote that. And you solved that puzzle for me.

Anyway, I don't like crowded places. So I'll see you tomorrow, I suppose. Or are you going to disappear again?

t.b.

@DavidWheeler Right sorry. Anyway, there's essentially one and that's the exponential function.

David Wheeler

or: perhaps more relevant to the situation at hand: only one of those same functions on $\mathbb{C}$ has the circle as co-domain

t.b.

@MattN No, I'll be around at least until Sunday.

(I think I left plenty enough to think about and work out)

Matt N.

Nice : ) Nice to have you back. (til Sunday)

@tb Yes, don't worry.

t.b.

@TheChaz I hope you're not on the verge of heading out, it's gonna take me a moment to react.

David Wheeler

21:30

@MattN oh, dear...i've driven you away again...i was just fascinated by what you were talking about, it seems like a very cool subject

Matt N.

Bye folks.

t.b.

Bye Matt, have a nice evening!

The Chaz

I will be here for a while. No reaction is necessary - I've had my catharsis !

(But I will stay close to the computer @t.b.)

Asaf Karagila

@TheChaz Why do you need a catheter?

The Chaz

Lots of buildup.

Matt N.

21:31

@DavidWheeler What do you mean again? I told you: I didn't mind your "personal attack" at all. Seriously. And today I'm leaving because there is a bunch of people in here that rub me the wrong way so I'm grumpy and hence leaving.

Asaf Karagila

For crying out loud. I can't find my pen!

Matt N.

@tb And you! And thanks for today.

Asaf Karagila

How can I start working without my pen??

Skullpatrol

@Gigili Hi

Matt N.

Pen...cil?

Asaf Karagila

21:32

No. I want my pen.

David Wheeler

@MattN i'm sorry, my brand of humor takes a bit of getting used to. no worries, mate. :)

Matt N.

No offense taken.

Asaf Karagila

I think al-Qaeda took my pen. Oh! There it is!!!

t.b.

What is upsetting me more and more with the whole rep-thing is that the whole system works towards having a rather low level. It is hard to resist the temptation and not answer the low hanging fruit and leave them to others who are also eager to contribute. Moreover, slick and unexpected solutions instead of going the hard but more helpful/insightful way are way overrated.

David Wheeler

@AsafKaragila therefore: al-Qaeda must be IN YOUR ROOM. look out!

Asaf Karagila

21:38

@DavidWheeler This is my room.

Are you saying that one of the users here is associated with al-Qaeda?

That's quite an offense you drive there, well done.

David Wheeler

it logically follows

Skullpatrol

@Gigili Have you seen this clip explaining infinity yet? youtube.com/…

Asaf Karagila

Only if you're Spock.

The Chaz

@tb Ok... just one more point. Let me copy it from my activity history...

David Wheeler

@Skullpatrol that's actually an excerpt from a much longer show Horizon:To Infinity And Beyond (i think you can find the whole show on youtube)

Gigili

21:42

@DavidWheeler My favorite quote.

The Chaz

The conditions listed when you hover over the voting arrows are
Research effort,
clarity, and
usefulness,
none of which are strongly associated with highly up(down)-voted questions.

Skullpatrol

@DavidWheeler Really? Thanks

Gigili

@Skullpatrol Hi. I'm not interested myself but there was a guy talking about types of infinity and I thought he was wrong.

The Chaz

I could start a new account, find an old (obscure) unknown problem or diophantine equation, and get 15 upvotes in an hour.

David Wheeler

one of the people interviewed on the show is Max Tegmark...famous for his hypothesis that our universe might actually be a mathematical structure

Asaf Karagila

21:43

@TheChaz It is associated with top voted answers methinks.

t.b.

@AsafKaragila you're kidding.

The Chaz

(top = ??? 50 + ?)

Not batman. Not dy/dx

Not "The unilateral..." oh wait...

Skullpatrol

@Gigili What is your favorite quote?

t.b.

Not Hilbert's hotel, not $W$ not the decimal expansion of $\pi$.

Gigili

@Skullpatrol "To infinity and beyond"

The Chaz

21:44

But anywho.
So some new guy throws up an integral that immediately gets 6 slick/advanced answers and all of a sudden that is the model for asking questions.

Asaf Karagila

@tb The Batman answer; Qiaochu's piano answer; Arturo's dx/dy answer...

The Chaz

@Asaf - I misread the above as "questions" (not "answers")

Skullpatrol

@Gigili Yes. "To Infinity And Beyond " does sound like an oxymoron.

Gigili

Right, I was excited when @DavidWheeler said it.

The Chaz

But the only criterion for answers is "usefulness"

2

t.b.

21:47

@AsafKaragila I don't think either one of them compares to this or this in terms of effort, insight, usefulness and clarity.

Just two examples that came to mind immediately

David Wheeler

another interesting look at the pandora's box cantor opened is "Dangerous Knowledge", also can be found on youtube

t.b.

@TheChaz entertainment?

The Chaz

W

Asaf Karagila

@tb Of course that the top ranked are not actually the best. They are not badly written though, and I don't think we should feel bad that those are the top answers. The top questions, though... that's a whole other story.

t.b.

I think anything beyond 20 votes is a combination of luck, easiness and popularity

2

Asaf Karagila

21:50

Take for example my answer on AC/AD.

The Chaz

W

David Wheeler

when "completed infinity" was accepted as a well-defined concept, one starts of think of: infinite numbers of "completed infinities"...if we package these a certain way, we get a new, bigger infinity

anon

MSE turned into twitter

t.b.

@AsafKaragila Yes, those are all great answers and we shouldn't feel bad about them.

Asaf Karagila

I've worked really hard on that answer, and revised it a lot. The OP insisted that advanced topics in mathematics can be explained intuitively.

I'm still getting furious when I read the comments there.

t.b.

21:52

I understand that. It is an excellent answer...

The Chaz

link?

Skullpatrol

@DavidWheeler Bigger infinity sounds like an oxymoron

Asaf Karagila

link

Or that question which sparked my thesis, with the automorphisms of vector spaces.

That should have been my top rated answer. Not a drunken rave about Russell's paradox.

t.b.

Anyway, we got distracted. At the end of the day I think the system levels itself out. There are very few, say, 10k+ people who don't deserve to be in that club. For slightly different reasons in each case.

David Wheeler

it is, if cardinalities are finite (our usual experience of "sizes"). but some "infinities" are the same size: the usual example is the number of natural numbers, and the number of even natural numbers.

Gigili

21:56

@AsafKaragila "So yeah, this was pretty unhelpful, unfortunately, though I appreciate the attempt." Hugh.

Asaf Karagila

Hmmm... we are nearing the point where the first page of all-time rep. ranking are 20k'ers, nice.

Gigili

I'm wondering how a silly question like that got many upvotes.

David Wheeler

@AsafKaragila i always pick "the left sock" and you can't prove i don't.

Jonas Teuwen

@robjohn mathematica.stackexchange.com/questions/2740/…

And why do people downvote my question? 8-).

t.b.

@Gigili It takes a whole s-load of balls to write something like that...

Asaf Karagila

21:59

@DavidWheeler I can. Simply because the sock is left only after choosing it to be the left sock, your argument is therefore circular.

Jonas Teuwen

Assholes.

The Chaz

@tb I do appreciate how the system favors answerers by making upvotes to answers worth more rep. Then when I have earned a sufficiently high rep, it will be obvious that I have really earned it :)

See, here's Jonas with a perfect example of what I started the rant with.

Skullpatrol

@DavidWheeler To begin with, I don't really agree with talking about "a number of numbers" because it seems circular... secondly big, bigger, and biggest can not be applied to "infinity," in my opinion.

Transcript for

$Mathematics$ Mathematics