@Nimza If you know those sums of powers for all $k=1, 2, \dots n$, then you can use the symmetric polynomials to get a polynomial equation of degree $n$ with solutions $x_1$ etc.

But I can't imagine you will get an explicit solution - at least for $n\geq 5$ (Galois)

@OldJohn thanks, ok

(but I might be wrong - I often am!)

@MichaelGreinecker So, my subbasis elements are $p_k^{-1}(O_k)$ with $O_k$ open in $\{0,2\}$, and $k$ a natural number. I can prove that $f^{-1}p_k^{-1}(\{2\})$,$f^{-1}p_k^{-1}(\{0\})$ are open and I'm done. It is trivial $f^{-1}p_k^{-1}(\{0,2\})$ and $f^{-1}p_k^{-1}(\{\})$ are open.

@robjohn How often does it happen that an active user goes away from the site (including removing his account)? I remember ymar and Chandrasekhar not so long ago. (Although I do not know details about the reasons.)

@MartinSleziak I don't know. I have only seen a user deleted three times in the year that I've been here.

That does not seem that bad.

I remember one more guy going away. I think it was after some conflict in the chatroom. But I don't remember the username.

@MartinSleziak No, it's not a great concern, but people gripe when it happens. :-)

@MartinSleziak and I think he came back.

@PeterTamaroff No, that does not work. You would have to show that $f^{-1}(O)$ is open in $\{0,1\}^\omega$ for every $O$ in a subbasis of the topology on $[0,1]$

@PeterTamaroff did Brian Scott leave?

@PeterTamaroff Do you know of the characterization of continuity for first countable spaces in terms of sequences?

@MichaelGreinecker Nope.

@MichaelGreinecker (I know it for metric spaces, though)

@MichaelGreinecker I know that $\Bbb R$ is first countable, also.

@robjohn I think he was less active on main site and he wasn't seen in the chatroom for a long time.

@PeterTamaroff Do you know that $\{0,1\}^\omega$ is metrizable?

(But he posted some answers in the last few days, so he is probably back.)

@MartinSleziak I see that he has not been on chat for 23 days.

But he seems to be on the main site

@MichaelGreinecker Can I metrizize it with a $\max$ metric over the discrete metrics on each $\{0,2\}$?

If $(X_n,d_n)$ is a sequence of metric spaces with $(d_n)$ uniformly bounded, you can metrize the product by $d\big((x_n),(y_n)\big)=\sum_{i=1}^\infty 1/2^n d_n(x_n,y_n)$.

@robjohn He made no post since July 23 until 2 days ago, see here.

@MichaelGreinecker What does it mean for a metric to be uniformly bounded? I mean, I suspect the restriction of $d$ to $[0,1]$ is bounded, but I don't know what "uniformly" means here.

@PeterTamaroff It means there exists B uch that $d_n$ has values in $[-B,B]$ for all n. Uniformly means here simply that $B$ does not depend on n.

@MartinSleziak His usual $200$+/day history was also depressed for that same period

@robjohn It would be pretty remarkable if he capped on a day when he did not post anything :-)

@MichaelGreinecker OK. I'd rather use what I know. I know about subbasis and basis of topologies and how we can use them to characterize continuity, I know the caracterization of continuity in terms of open, closed, and nbhds. I know how the product toplogy is set up, and that it is the weakest topology such taht the projections are continuous.

@MartinSleziak If he did, there should be a badge for that :)

@PeterTamaroff Ok. Show that $f\big((-\infty,a)\big)$ and $f\big((a,\infty)\big)$ is open in the product topology for each $a\in\mathbb{R}$. Do this by showing that each element in these sets has an open neighborhood that still lies in these sets. That trick is that because of the "3^{-n}", what happens from a certain index on doesn't change the value much. I have to leave for 15-20min. I'll be back afterwards.

@MichaelGreinecker You mean $f^{-1}$? Why those, if I'm working on $[0,1]$?

@PeterTamaroff Yes, $f^{-1}$. It actually doesn't matter, you can also take $[0,a)$ and $(a,1]$.

@MichaelGreinecker Why doesn't it matter? AFAIK continuity is characterized in terms of preimages.

@PeterTamaroff It doesn't matter because the range of $f$ lies in $[0,1]$. So $f^{-1}\big((a,\infty)\big)=f^{-1}\big((a,1]\big)$. Gotta go now.

@MichaelGreinecker Oh, I meant about $f^{-1}$ vs $f$. Thought you were talking about that. =P OK.

Maybe somebody knows an explicit solution of $x_1 + x_2 + x_3 + x_4 = a_1, \ldots, x_1 x_2 x_3 x_4 = a_4$ ?

@Nimza What do the dots mean?

I can't see the pattern.

@PeterTamaroff standard symmetric polynomials

@Nimza they would be the roots of the polynomial $x^4-a_1x^3+a_2x^2-a_3x-a_4 = 0$ and then you use the formula for solution of a quartic, I believe

???

@OldJohn ah, yes, Vieta's formulas. thanks

@PeterTamaroff yeah)

I think that knowing the values of the symmetric functions is equivalent to knowing the polynomial - if there were a quick and easy way to do it, there would be a quick and easy way to get a solution of an $n$th degree polynomial, and we know that is not true, unfortunately

@OldJohn Hehheeh :)

@PeterTamaroff I spent some hours going down that route a couple of decades ago, before convincing myself it was a lost cause :)

when I was teaching symmetric functions to school kids

@OldJohn Hehe. Cool.

Hi!

@JonasTeuwen welcome back

ouch my brain

@OldJohn Hi!

doesn't want to think about anything that makes @anon's head hurt :)

@OldJohn On the other hand, Vieta is quite nice for numerics...

@J.M. Oh yes - I tend to forget about the numbers-side-of-things :)

Nystatin yuck.

@robjohn That guy has actually been deleted twice, for reasons I don't fully understand.

Deleted...twice?

@OldJohn Hah. I talked about it here, and it happens to be one of my favorite algorithms... :)

@JonasTeuwen Yep. He was deleted, restored, and now deleted again.

Heh.

@JonasTeuwen It's the only thing that works on those particular mushrooms, see...

The only?

A specialist, if you will.

I hope it works then! 8-). Perfect.

The nuclear bomb did not work well enough. Pretty well though.

@J.M. Very neat!!

@JonasTeuwen Yes, it's a messy affair. At least they are now using sniper rifles instead.

@J.M. Heh 8-).

@J.M. was I right in my assessment earlier of the setup if we are not looking for numerical solutions?

@J.M. I was at the doctor again, and I was like: "what is this...?" she is like: "I'm afraid to say it but it is again..."

"Oh good, so it is only one thing!"

@OldJohn Oh yes, Abel and Galois pretty much shot the possibility of doing anything meaningful on the symbolic front.

@JasperLoy I know at least one more user here who got reinstated after being deleted.

@JasperLoy Yes, maybe. I'm not privy to those methods.

But... but... special functions for my polynomials?

"Thrush is rare if a person's CD4+ cell count is above 500" da fuq.

@JasperLoy To solve higher order degree polynomials.

@JonasTeuwen Theta functions are a bit beyond the methods of Abel and Galois. ;)

@J.M. 8-). But lovely.

(Wow, you really went through that paper by Umemura?)

@JM Yes.

@JasperLoy They're special functions, somewhat more complicated than trigonometric functions.

Only slightly.

But the principle is the same. To use the cubic formula as an example, you're pretty much forced to use trigonometric functions if 1. your roots are all real, and 2. you don't want complex numbers in the representation of your solutions.

@JasperLoy Oh, okay. It's actually a common question that gets asked whenever I bring them up, see... :D

@JasperLoy Are you an undergraduate? Or graduate?

@JM Symmetries ftw?

@JonasTeuwen Oh, certainly. :D

@JasperLoy Okay.

How long does a "differentiation" cost (blood test)? Nobody told me.

@JonasTeuwen Is that about your tongue?

@PeterTamaroff Yes.

@JonasTeuwen Haven't checked recently, but they were pricey the last time around...

...but maybe things are better now.

@JonasTeuwen What happened?

Oh, not the price, but the amount of time! :-). (insurance pays).

@PeterTamaroff Not much.

@JonasTeuwen Oh, that. Shouldn't take more than three days.

@JonasTeuwen I donated blood last year. Bloody (and) long time.

Ah, perfect. So they know already.

(should've asked "how long does it take?" ;))

@JasperLoy Same here. It was just OK.

Sentences like: "well, if that doesn't show anything, well do a bone marrow puncture!" make me scared.

Yes.

@JonasTeuwen It's normal. I'd be scared shitless too if it came to having to sample my marrow...

@JasperLoy Well, the acute ones.

@JasperLoy Good. Do you love your potty?

Perfect.

@JasperLoy Let $X_n=\{0,2\}$ with the discrete topology, and set the product space $X=\{0,2\}^\omega$. I'm trying to prove that $X \to [0,1]$ defined by $$f(x)=\sum_{n=1}^\infty \frac{x(n)}{3^n}$$ is continuous. Now, I proved it is one one. Suppose $O$ us open in $[0,1]$. We must show $f^{-1}(O)\subset \mathfrak I_\pi$ where $\mathfrak I_\pi$ is the product topology on $X$.

Since $f$ is one one, $f^{-1}(O)\subset \mathfrak I_\pi$ means $O \subset f( \mathfrak I_\pi)$. Do we want to show $f$ is an open map?

@mixedmath ?

hmm?

@mixedmath Could I borrow some of your time?

sure - what's up?

@mixedmath See up there.

Forget about the open map thing.

@PeterTamaroff I'm back.

Driving to the beach. Be back when we're settled there.

@JasperLoy Thanks.We have a lot of relaxation and food in our schedule.

@MichaelGreinecker I'm trying to solve this.

@JasperLoy What secrets?

:P

@PeterTamaroff Is it true that if you show that the inverse image of a each element of a basis for $[0,1]$ is open, then the map is continuous?

that seems easier, maybe, to me

and perhaps doable

@mixedmath It is one of the characterizations of continuity I have, yes.

@mixedmath I'm thinking as follows: Let $(a,b)$ be a basis element.

@JasperLoy Hmm. Well, tell me sometime. I have to go now. Be back later.

Let $a=0.a_1a_2a_3\dots$ and $b=0.b_1b_2b_3\dots$

Bleh. No, too compicated

oh - well then

@mixedmath I mean, the preimage will be functions $x$ for which $x(n)=0\text{or} 2$. The fact that each $\{0,2\}$ has the discrete topology gives us a lot of freedom from where to chose a suitable open set.

@PeterTamaroff Show that for each $x=(x_n)\in\{0,1\}^\omega$, and each $\epsilon>0$, there is an open neighborhood $U$ of $x$ such that $f(y)\in(f(x)-\epsilon,f(x)+\epsilon)$. You can explicitely give the neighborhood $U$ in this case.

@MichaelGreinecker

oh, that's true

what Michael said, that is

@MichaelGreinecker What do you call an open neighborhood?

@JasperLoy But all open sets are neighborhoods of their points... why give them that name?

@MichaelGreinecker What is $y$ there?

@PeterTamaroff Mendelson actually defines neighborhoods so that they may not be open. And: yes, $y\in U$

@MichaelGreinecker What about something like $$\bigcap_{n=1}^k p_n^{-1}(\{x(n)\})$$, where I choose $k$ for the desired accuracy?

@PeterTamaroff Yes, that's the idea.

@MichaelGreinecker =D

@JM "amphotericin B-resistant chronic mucocutaneous candidiasis" is hell of a long name for lots of shrooms!.

Apparently it is not such a bad thing. Good, whatever. I need beer.

@JonasTeuwen Dude. Beer has shrooms! Problem solved. Quit drinking.

That's quite some leftish radical statement eh.

@JonasTeuwen Candidia is yeast, not shrooms. Both fungi, though.

It contains yeasts, but not the ones that cause infections in humans (unless you're basically dead).

@HenningMakholm We layman call fungi shrooms.

@HenningMakholm I know, it is a yeast which is a fungus which amically call "shroom".

Too much of any of those in your body is bad anyway, so whatever 8-).

It is basically impossible to make beer using the candida genus :-(.

Well, maybe that is actually good.

I thought shrooms were only those fungi that develop discrete fruiting bodies with a stem and a cap.

@HenningMakholm Yes, but back in the day, we joked about having "a mushroom farm in your mouth"... :D

@HenningMakholm Yes, that's the proper botanical definition.

@MichaelGreinecker So how would you put it formally? Let $x\in \{0,2\}^\omega$ and $\epsilon >0$ be given. Let $B(f(x),\epsilon)\subset [0,1]$ be an open ball about $f(x)$, with radius $\epsilon$. We show there is some $k$ such that$ f^{-1}(B)\subset P_k$, where $P_k=\bigcap_{i=1}^k p_n^{-1}\left(\{x(n)\}\right)$ is a nbhd of $x$.

@J.M. Wait... what? Were you toddlers. Who had this? 8-).

Basically, "mushroom" was being used in a loose, humorous sense.

@JonasTeuwen Well, we saw this really nasty case of thrush a long time ago...

@J.M. Oh right. "Come and see this guys!"

(We weren't above morbid humor then, see.)

"This site is currently in read only mode, we'll return with full functionality soon."

Hmm...

@J.M. Hmm...?

@PeterTamaroff yes, that is a good way to write it down.

Site's locked down; I wonder how long it'll take...

In Danish the lay catch-all term for fungi is "sponges".

@J.M. whenever I see something like that, I always wonder if it was me that broke it :)

@HenningMakholm Hmm, okay. I see it's a matter of culture... :D

@OldJohn What did you do? ;-)

@MichaelGreinecker nothing that I know about - I am not a secret hacker

@OldJohn "Damn, what did I press now?!"

@J.M. yep :)

but - if it broken for everyone, I guess I am innocent

@J.M. Wikipedia claims that "fungus" is derived from a Greek word meaning "sponge"... on the other hand "mycology" comes from a Greek word meaning "mushroom".

I did not know that piece of etymology about "sponge".

Though with the second bit, that's also why the names of those fungal groups end in "-mycetes".

Shrooms.... SHROOMS.

I'm pretty sure also that greek myco- and myxo- are akin. Myxo- means "slime", and is the origin of English "mucus".

And mucilage, etc.

@MarkDominus A sticky subject...

Oh hey, @Mark, since you're here... would you mind if I ask you a somewhat personal question?

@OldJohn your avatar is interesting. Who is that?

Davide's answer is quite nice, but I have to wonder why it has that many votes...

@MarkDominus Wiktionary says they are cognate, but that "mykos" began meaning "mushroom" rather than "slime" somewhere on the way from PIE to ancient Greek.

@Tim Just some cartoon character I copied from the web - and modified a bit - just thought it looked suitably old :)

@J.M. Because it is pretty easy to understand and reasonably hard to come up with.

I suppose.

@JonasTeuwen I suppose that, too. :)

Still, 25 would not be much of a surprise, but 40+... wow.

@J.M. Remember... batman?

Oh, that is easier. People love Batman, see. :D

@HenningMakholm Thanks.

@J.M. Sure, ask.

Everybody, stop.

Mars time

@MarkDominus Okay: are you still getting royalties from your book to this very day?

@PeterTamaroff Nice!

@MarkDominus Dude, you're in Wikipedia!

@J.M. Yes.

@MarkDominus Okay, that was it. Thanks. :)

@PeterTamaroff Yes, that only means that I am not clearly less important than Jigglypuff.

@J.M. Sure!

As personal questions go, that was really not very intrusive.

I issue a grateful tweet every six months when my royalty check comes.

It's the happiest day of the year.

@MarkDominus Okay. The last guy I asked that question squirmed a bit, see...

eh, why is m.SE locked?

I bet that was because it never earned out the advance. :)

@ZhenLin Maintenance stuff, apparently.

hmmm

Some strange maintenance where users can't even log in...

@ZhenLin Maintenance.

@ZhenLin Conspiracy team, assemble.

I paid them to lock it with my brilliant pseudocompactness question right at the top.

So that…

Are there methods to solve Diophantine equations of degree 2 in general?

Hmm, why did I have them do that?

@MarkDominus You were beaten by "What do we mean by an elegant proof?"

That's #1 on the active list, but mine is #1 on the newest list.

@JaakkoSeppälä That's a good question.

And we're back on track.

And as soon as we say that, the site is on, atleast for me.

@JaakkoSeppälä The "general" makes it tough...

Now, how many people were twitching in the time it was down... :D

@MarkDominus How important are royalties for authors? Financially and not sentimentally?

that was the question in a Finnish math forum.

@JayeshBadwaik I suppose it depends on the author and the book.

For me the royalties on the book are a nice twice-yearly bonus, but nothing like enough to live on. For many books, perhaps the majority of technical books, there are no royalties.

Donald Knuth made quite a lot of money on royalties, enough to live very comfortably, retire happily, and outfit his home with a custom-built pipe organ.

I also get royalties on Computer Science and Perl Programming: Best of the Perl Journal (vol. 1), which amount to US$5–10 per quarter, and I believe I get a lot more from that book than any of the other contributors.

@MarkDominus Reason I am asking is when I was doing my senior thesis project, many of the books I had to refer to, each costed almost half the monthly salary of my thesis-mate's dad. And my thesis-mate's dad has the average salary of an Indian.

Those books probably sell very few copies, mostly to university libraries.

Academics do not write academic books for the money. They write them for reputation and to disseminate their ideas.

Yeah, our libraries are also not very good. Even though my college was one in the list of "National Importance"

Yes, the specialty books are priced a bit on the high side most of the time. "Why" is a somewhat sticky matter...

We are in a time of transition. Now that publishing and distribution via internet is so cheap and easy, a lot of academic writing is being published that way rather than by traditional publishing houses. This will only increase.

Huh, we just lost a bunch of users.

@ZhenLin ?

@JayeshBadwaik A bunch of users in this room went off.

For some reason.

As with any "time of transition", a number have to be dragged kicking and screaming... ;)

@J.M. ;-)

@MichaelGreinecker Are you there?

@JasperLoy Yes, but not five or so people all of a sudden... at least, in this room.

One person in a math forum asked to find all integer satisfying x^2 + x = y^2 + y + z^2 + z. How can I determine the solutions?

@JaakkoSeppälä Sounds like you should ask on main...

what is main?

ah, ok.

Methinks we can do it by congruences. If x is even, then so are y and z

What we do in here is trivial.

perhaps completing the square

@anon integers are asked, and completing the square gives a $(x+ 1/2)^{2}$ term which is not desirable I guess

multiply lhs=rhs by 2

or maybe 4

thing is coefficient of $x^2$ and $x$ will be the same, but yeah, now you have to prove the indentity in only odd numbers I guess. @anon brilliant!

@JasperLoy and trivia pursuit ;-)

@JaakkoSeppälä This seems to be the same as finding triangular numbers which can be expressed as sum of two triangular numbers.

yup

Kind of like Pythagorean equation with triangular numbers instead of squares.

This should be known....

it's $u^2+1=v^2+w^2$ after a variable change ($u,v,w$ odd), which should also be known

remainder with respect to 6 of squares of odd numbers leaves either 1 or 3 as remainder. Hence $u$ must be a multiple of $3$ and one of the $v$ or $w$ must be a multiple of $3$ and the other should not be, but should be odd. hence either $6k+1$ or $6k-1$

@JaakkoSeppälä Seen this?

According to this all solutions of $x^2+y^2=z^2+w^2$ are known. but only statement is given there.

nice

Nice find. Although it does not give all solutions.

It seems to be an interesting problem.

hi @KannappanSampath

@BillD I had requested that my account be nuked. I came in just to ask if anything formal is expected of me for the process to reach completion.

But why?

@JasperLoy Yes, did that, except possibly editing my "About Me" (to read "please delete me").

@JasperLoy But, I'd like an answer from @BillD too

me too ;-(

@JasperLoy Well, I'll be around with the TeX.SX people.

Not sure I like internet anymore.

or three or four ...

blah.

Just remember count negative people on the negative side of the number line :-D

What did I miss?

oh, that's sad. I thought mass panic might have erupted when the site went read-only

@Kannappan: Hey - I haven't seen you around for a while

@mixedmath Please see here.

oh - I did miss something

so what is it that you want?

it's a big thing, and a mistake would be very bad

I'd like to delete the account.

@mods Please tell me if you guys can delete an user?

@JasperLoy Yes, but AFAIK, it's irreversible

@KannappanSampath Yes, a mod can delete an account

I don't know if they've changed, but I've deleted accounts by request on Phil.SE before