Mathematics - 2012-09-08 (page 2 of 4)

6:00 AM

@jay i think using quotes (") instead of apostrophes (') makes the commands expand first, then go. changing to quotes stopped the errors (I still have to see if it worked right though).

Jayesh Badwaik

@Jeff Hmm. I would have thought it should work the other way! The doubles quotes being more stronger than the singe ones.

Jeff

@jay yeah, that seems natural to me, too. the command works perfectly using quotes. apparently quotes means "expand now" and apostrophe means "expand when executed" (or something like that).

Gustavo Bandeira

@JayeshBadwaik Does C has some Symbolic mathematics library?

Jayesh Badwaik

@Jeff the joys of programming :P
@GustavoBandeira I highly doubt it. C is jut not a proper language to do that stuff.

Gustavo Bandeira

Yep.

Jeff

6:07 AM

@jay yeah :D... now i'm trying to see if that's the same in the makefile, so i don't need multiple files

Issac M

If $A,B \in \mathbb{R}$, is the following true: $\{A,B\}^+ = max\{A,B\}$ (notation question)

Jayesh Badwaik

@GustavoBandeira This is there in C++. http://en.wikipedia.org/wiki/SymbolicC%2B%2B
But I am guessing it would be too clumsy.

Jeff

@Jay it seems like that statemen (about quote and apostrophes) is not true in makefiles, just regular bash scripts

Gustavo Bandeira

Hmn, I got curious on what are the libraries of each programming languange

At least a generalization on them: "C has n set of libraries and p set of libraries, Python has some math libraries and some n libraries"

@BenjaLim That guy with the green avatar missed you.

Jayesh Badwaik

@GustavoBandeira there are too many. This is one really good case where ignorance is bliss. Just explore the ones of your interests and be happy!!
But you can know about standard libraries of the language.

For example C on linux has glibc. C++ has stdlib++

@Jeff Awesome. So just call that bash script from makefile. I know ugly ugly but that's the way it is.

Jeff

6:11 AM

@jay it looks like i'll have to do that

:(

@jay how do i display what a variable in a makefile is set to?

Gustavo Bandeira

@JayeshBadwaik I'm just trying to figure out - superficialy - what projects could be developed with each programming languange.

Jayesh Badwaik

@GustavoBandeira ohh okay. that is a good exercise.

@Jeff You might want to use setenv getenv commands inside the makefile

@echo $(FOO)

The command is "@echo $(FOO)"

Also for your question about display

Jeff

@JayeshBadwaik it was the same question :D

now i'm having trouble getting thishost to be set correctly. sheesh

Jayesh Badwaik

See this : stackoverflow.com/a/12111527/796530

Jeff

ok. i have everything working except that I can't set "thishost" to the value I want (in the makefile).

@jay looks like the best way for me to get it working is to add export thishost=$(hostname) to my .bashrc file.

(added "export")

Jayesh Badwaik

6:29 AM

@Jeff Yup. So basically, your bashrc is the bash script you run everytime you start a shell.

Jeff

@jay yeah

Jayesh Badwaik

You can do the same thing by running a bash script at the beginning of your makefile.

that would be cleaner.

Jayesh Badwaik

6:51 AM

Jeff

@JayeshBadwaik probably, but i don't feel like figuring that out now (also, i already emailed the solution to my adviser).

oh, and also, i'm really really tired now. see how sleepy my eyes are: -__-

goodnight.

Jayesh Badwaik

@Jeff Goodnight.

Gustavo Bandeira

7:16 AM

terrytao.wordpress.com/career-advice/…

This is cool

BenjaLim

8:47 AM

@robjohn Hey

are you around?

@JonasTeuwen

Can I ask you something

I have been bashing my brains out in this problem

I thought I worked it out last week

I realised my manipulations were wrong

@JonasTeuwen I am trying to get a uniform bound on $\frac{1}{2} \int_0^x (D_N(t) - 1 )dt$

@JonasTeuwen I reduced the problem to getting a uniform bound on $\frac{1}{2} \int_0^x \sin Nt \cot (t/2) dt$

But the problem is now I am trying to get a bound on this guy, I am trying to say that the integral of this is less than $\int_0^x \sin Nt /t dt$

Nimza

Good day everybody!

Issac M

For $a,b \in \mathbb{R}$, does $\{a,b\}^+$ define maximization or is it $(a,b)^+$?

Jayesh Badwaik

9:13 AM

@BenjaLim Isn't your original integral simply this?
\begin{equation}
\int\limits_{0}^{x} \sum\limits_{k=1}^{n} cos(kt) dt
\end{equation}

If $n$ is finite, you can simply take termwise integration and then add them up which gives
\begin{equation}
\sum\limits_{k=1}^{n} sin(kx)
\end{equation}

Nimza

Is there some algorithm to plot $w = \sqrt{ P(z)}$, where $P(z)$ is some polynomial of $z \in \mathbb{C}$?

Jayesh Badwaik

Calculate the points to a desired accuracy and then use bezier curves?

Nimza

Jayesh, no, only calculate points at some grid on $z$. But correctly :)

Jayesh Badwaik

?

Nimza

I can give a correct example for $w = \sqrt{z}$. Here is an algorithm on Matlab:

phi = pi*(-2*N:2*N)/N;
R = (0:(2/N):3)';

z = R*exp(1i*phi);
w = R.^(1/2)*exp(1i*phi/2);

voila

Is there a way to generalize this on $w= \sqrt{P(z)}$? The easiest case if $P(z) = az+b$

Jayesh Badwaik

9:29 AM

The expression for $w$ will not change w.r.t $P(z)$

Issac M

For $a,b \in \mathbb{R}$, does $\{a,b\}^+$ define maximization or is it $(a,b)^+$?

Jayesh Badwaik

@IssacM Sorry I do not know.

Issac M

Thanks

Jayesh Badwaik

If $P(z) = Re^{i\theta}$ then, $\sqrt{P(z)} = \sqrt{R} e^\frac{i\theta }{2}$

Nimza

yes

but I don't know how many circles on $\theta$ I should take. Angle function will give only angle between $-\pi$ and $\pi$

BenjaLim

9:32 AM

@JayeshBadwaik Yes but how is that uniformly bounded?

Jayesh Badwaik

I guess you are asking of the sequence? Hmm..

Wolfram gives the sum as this
\begin{equation}
\sum\limits_{k=1}^{n} \sin(kx) = \frac{\sin(nx) \sin\left(\frac{n+1}{2} x \right)}{\sin(\frac{x}{2})}
\end{equation}

which can be then simplified to
$$\cot\left(\frac{x}{2}\right) + \frac{sin\frac{3n+1}{2} x}{sin(x/2)}$$

and this is not bounded I suppose.

robjohn

9:57 AM

@BenjaLim Sorry, I was out.

Matt

10:07 AM

Yo bounty hunters, attention please!

Nimza

The easiest question: how to plot $w = \sqrt{z+1}$ ?

ah. okey. I know)

robjohn

@Nimza Plot[Sqrt[z + 1], {z, 0, 5}] (see here)

Nimza

@robjohn no, I plot Riemann surfaces :)

robjohn

10:22 AM

@Nimza Oh :-)

Nimza

I posted an example ($w = \sqrt{z}$) above. I want to generalize such algorithm

Jayesh Badwaik

@Nimza So basically you want to plot $real(z) against $x,y$ where $z=x+iy$ ?

Nimza

@JayeshBadwaik hm, why? There are no problem to do it.

the only problem is ambiguity of function $\sqrt{}$. I don't know any algorithm that constructs and glues sheets correctly for $w = \sqrt{P(z)}$

Jayesh Badwaik

@Nimza ohh.

And i just noticed how ridiculous that italics looks. sorry for that.

Nimza

:))

Jayesh Badwaik

10:33 AM

I forgot a $ is suppose.

Nimza

Ok. I'll try nearest neighbors now

BenjaLim

@robjohn hi

robjohn

1:06 PM

@BenjaLim Hey there...

Jayesh Badwaik

1:23 PM

@JohnJunior Do you even have to ask?

Well I am going for dinner now. Talk to you later.

John Junior

later

robjohn

1:54 PM

Cool. I was just fixing up a question posed in an answer to someone else's question and extended it to show that $$ \frac1{\sqrt{4n+1}}\le\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\le\frac1{\sqrt{3n+1}} $$

The asymptotic limit is $\dfrac1{\sqrt{\pi n}}$

Jonas Teuwen

2:26 PM

@robjohn Nice :-).

Peter Tamaroff

2:44 PM

Guys, I have a load of work to do on Spivaks Ch.7. Expect me. I'm not anonymous.

@JonasTeuwen Joney.

Jonas Teuwen

@PeterTamaroff Honey.

@robjohn Would you know when the weak* topology is first countable? If and only if it is metrizable I presume... that sucks.

Jayesh Badwaik

@JonasTeuwen what is weak* ?

as opposed to weak?

Jonas Teuwen

Standard usage.

Peter Tamaroff

@JayeshBadwaik Well, that one managed to be a star.

Jayesh Badwaik

@PeterTamaroff I thought you had a chapter to complete.

:P

Matt

3:18 PM

@JohnSenior I didn't mean cope mathematically though, many others have helped me lots, too. Yet none of them is special to me. : ) But never mind.

@JonasTeuwen Hi Bro.

Haz teh hungre. See you all later.

robjohn

3:48 PM

@JonasTeuwen I'm not sure, off hand.

I haven't seen Ilya on chat in a few days.

robjohn

4:18 PM

For $k\ge1$,
$$
\begin{align}
12k^2-8k^2-n+1&\le12k^3-8k^2\\
(2k-1)^2(3k+1)&\le(2k)^2(3k-2)\\
\left(\frac{2k-1}{2k}\right)^2&\le\frac{3k-2}{3k+1}\\
\frac{2k-1}{2k}&\le\sqrt{\frac{3k-2}{3k+1}}\tag{1}
\end{align}
$$
and for $k\ge0$,
$$
\begin{align}
16k^3-12k^2&\le16k^3-12k^2+1\\
(4k-3)4k^2&\le(2k-1)^2(4k+1)\\
\frac{4k-3}{4k+1}&\le\left(\frac{2k-1}{2k}\right)^2\\
\sqrt{\frac{4k-3}{4k+1}}&\le\frac{2k-1}{2k}\tag{2}
\end{align}
$$
Taking products of $(1)$ and $(2)$ gives
$$
\prod_{k=1}^n\sqrt{\frac{4k-3}{4k+1}}

Jonas Teuwen

@robjohn Ilya works hard.

robjohn

@JonasTeuwen but it doesn't mean he is not missed.

Jonas Teuwen

@robjohn Yes, but it means he has a good reason to not be here! :-).

@robjohn But you still have meeee!

I'm a pretty bad excuse for that right? 8-).

robjohn

@JonasTeuwen I miss everyone who is gone for a while.

Jonas Teuwen

4:34 PM

@robjohn Yes, me too :-(. But I know where to find Ilya but I do not want to disturb him in deep concentration :-).

robjohn

@JonasTeuwen I was just noticing that Ilya had flagged an answer recently

Peter Tamaroff

4:55 PM

@JonasTeuwen Duuuuuuuude.

Jonas Teuwen

What?

Peter Tamaroff

@JonasTeuwen Would you care to help me with something? Or maybe @rob ?

Suppose $f$ is such that $f(x+y)=f(x)+f(y)$. Then prove there exists a $c$ such that $f(x)=cx$ for all rational $x$.

Jonas Teuwen

Try it for natural $c$ first.

robjohn

@PeterTamaroff It's not true unless $f$ is continuous at at least one point, sorry, i missed the rational

Jonas Teuwen

And that.

@robjohn Wait... are your sure that is also needed for rational $c$?

Oh, his $x$ is rational.

Peter Tamaroff

4:58 PM

@JonasTeuwen Yes, first I have to prove it for integers. Actually, the scheme is like this

robjohn

@JonasTeuwen Yeah, I just added to my previous comment that I missed the rational part.

Peter Tamaroff

First, prove $$f(\sum x_i)=\sum f(x_i)$$

Jonas Teuwen

That is like yeah duh. So skip that.

Peter Tamaroff

@JonasTeuwen Hehhee yes.

Well, the second part is to prove the formula for integer $n$, then for $1/n$; $n$ integer, and then for rationals.

@robjohn You're so irrationaly minded.

Jonas Teuwen

Yes, take $n$ times $x$, then you get $f(n x) = n f(x)$.

Take $x = \frac1n$, so you get $f(1) = n f(n^{-1})$ for all $n$, so this means that...

Then go on. Bye!

Peter Tamaroff

5:17 PM

There is a crucial moment in a proof when one has to change the pen for the pencil.

user19161

@PeterTamaroff I always use the pencil for own work. Better than wasting ink.

Peter Tamaroff

@WillHunting Ink forces you to be correct. I like it.

user19161

@PeterTamaroff You will find that your thinking will change soon, because you will meet many problems whose solutions you only get after trying 9000 different things.

Peter Tamaroff

@WillHunting OK. So, I alrady have this is true for all rationals of the form $1/n$

With $c=f(1)$

user19161

Ink is only for work to be handed up for me, and also no need to use correction fluid, a waste of time, fluid and bad for health.

user19161

5:21 PM

@PeterTamaroff Then the rest is trivial.

Peter Tamaroff

@WillHunting FUUUUUUUU

I can't see it.

Let me think.-

user19161

@PeterTamaroff Hint: Use f(x+y)=f(x)+f(y).

Peter Tamaroff

@WillHunting I think I got it.

@WillHunting $$f\left( {\frac{1}{m}} \right) + f\left( {\frac{1}{n}} \right) = f\left( {\frac{{m + n}}{{mn}}} \right)$$ ???

user19161

@PeterTamaroff You are making it too complicated. You have proven it for all 1/n. Now you need to prove it for all rationals. But rationals are of the form m/n. Can you split up m/n into copies of 1/n?

Peter Tamaroff

@WillHunting Yes.

Done

=)

Jonas Teuwen

5:26 PM

Man.

What?

user19161

Let's all have beer.

Peter Tamaroff

@WillHunting I have a party tonight, I'll have vodka.

user19161

@PeterTamaroff Very good. Also, remember to stay safe, hehe.

Peter Tamaroff

@WillHunting From what? Face-numbness?

user19161

@PeterTamaroff Er, from making babies.

2

Peter Tamaroff

5:29 PM

@WillHunting LAWL

user19161

@PeterTamaroff Always bring protection just in case!

2

Peter Tamaroff

@WillHunting I have a bigger problem than that. Suppose now that $f$ additionally satisfies $f(x)f(y)=f(xy)$, and that $f$ is not identically zero. Then prove that $f(1)=1$ (easy piecy) that $f(x)=x$ for rational $x$, that $f(x)>0$ if $x>0$, that $f(x)>f(y)$ if $x>y$, that $f(x)=x$ for all $x$, knowing that given any $x,y\in \Bbb R$, there is a rational such that $x<r<y$

Peter Tamaroff

5:44 PM

@rob I have already proven $f(x)=x$ for all rationals. Now I need to prove that $f$ is positive for positive $x$; that $f$ is increasing and then that $f(x)$ is the dientity on the reals. Why do I need the previous inequalities to work the last thing out?

Matt

Is there any way to filter comments w.r.t. votes?

robjohn

6:42 PM

@PeterTamaroff you mean $f(x) = cx$ for all rational $c$? Oh, wait you're on a new problem...

Peter Tamaroff

@robjohn No,no. I have alraedy proven that. I have an extra condition now, that is $f(x)f(y)=f(xy)$

1 hour ago, by Peter Tamaroff

@WillHunting I have a bigger problem than that. Suppose now that $f$ additionally satisfies $f(x)f(y)=f(xy)$, and that $f$ is not identically zero. Then prove that $f(1)=1$ (easy piecy) that $f(x)=x$ for rational $x$, that $f(x)>0$ if $x>0$, that $f(x)>f(y)$ if $x>y$, that $f(x)=x$ for all $x$, knowing that given any $x,y\in \Bbb R$, there is a rational such that $x<r<y$

robjohn

yeah, I was away and just noticed that. I was just going from my inbox and not looking back :-)

@PeterTamaroff So you are now assuming that $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$?

Sorry, I have to go again. BBL.

Peter Tamaroff

@robjohn Oh noes.

Steenrod

I played the part of a wisecrack so well tday

Some users seem to frequent this place. I can hardly see new ones.

@BrianM.Scott, hello

I was wondering what become of Reid Barton(I am more interested in finding his papers).Any idea where I could at least have a look(I do not claim to understand his papers)?

Brian M. Scott

6:57 PM

Hullo. Don’t expect much: I’m not really awake yet!

Steenrod

@BrianM.Scott, got it at last.I am not sure I understand it.So, no need to take th trouble.Though i am still curious about Reid does nowadays.

Brian M. Scott

@Steenrod No idea, I’m afraid; I don’t even recognize the name.

Steenrod

@BrianM.Scott,en.wikipedia.org/wiki/Reid_W._Barton He is a legend in Olympiad folklore.

Brian M. Scott

He seems to be active on MathOverflow.

Steenrod

He was active

Brian M. Scott

7:12 PM

He was apparently there today.

Steenrod

@BrianM.Scott, good day and bye.

Peter Tamaroff

Hello

@BrianM.Scott Hello

Brian M. Scott

I’m still here.

Gustavo Bandeira

@Gustavo: I mean that once we've fixed the definition of a polynomial, we don't get to consider a function as being a polynomial or not. It either is or it isn't. — Qiaochu Yuan 29 mins ago

Peter Tamaroff

@BrianM.Scott Good!

Gustavo Bandeira

7:17 PM

What does Qiaochu wants to tell me with this?

Peter Tamaroff

@BrianM.Scott I'm working on $f(x+y)=f(x)+f(y)$

2 hours ago, by Peter Tamaroff

@WillHunting I have a bigger problem than that. Suppose now that $f$ additionally satisfies $f(x)f(y)=f(xy)$, and that $f$ is not identically zero. Then prove that $f(1)=1$ (easy piecy) that $f(x)=x$ for rational $x$, that $f(x)>0$ if $x>0$, that $f(x)>f(y)$ if $x>y$, that $f(x)=x$ for all $x$, knowing that given any $x,y\in \Bbb R$, there is a rational such that $x<r<y$

2 hours ago, by Peter Tamaroff

@rob I have already proven $f(x)=x$ for all rationals. Now I need to prove that $f$ is positive for positive $x$; that $f$ is increasing and then that $f(x)$ is the dientity on the reals. Why do I need the previous inequalities to work the last thing out?

Gustavo Bandeira

I'm seeing it kinda as: Once you learn what a car is, you don't need to think if it's a car or if it isn't, It either is or isn't

Brian M. Scott

@GustavoBandeira You wrote that Barbeau ‘considers’ two functions to be polynomials. Qiaochu was pointing out that consders is the wrong word: they are polynomials. Considers implies that it’s a matter of opinion, and it isn’t.

Gustavo Bandeira

@BrianM.Scott Oh, ok. I didn't think "consider" would express a matter of opinion.

John Senior

Good evening

Brian M. Scott

7:21 PM

@PeterTamaroff Which inequalities are you calling the previous inequalities?

@JohnSenior Evening!

Peter Tamaroff

@BrianM.Scott $f(x)>0\iff x>0$ and $f(x)>f(y)\iff x>y$

The scond follows from the first right, by putting $f(x-y)>0\iff x-y>0$

Brian M. Scott

Yes. And you need them to show that $f(x)=x$ for irrational $x$.

Peter Tamaroff

@BrianM.Scott OK. But that is supposed to be proven forall $x$

That is, $f(x)>0$ for all $x$.

I only have my formula is true for rationals.

That is, $f(x)=x$ for rational $x$.

Brian M. Scott

What exactly are your hypotheses on $f$. It’s additive and multiplicative and not identically $0$; anything else?

Peter Tamaroff

@BrianM.Scott Yes, that is all.

Brian M. Scott

7:31 PM

@PeterTamaroff Okay; I’ll have to think about it for a bit.

Peter Tamaroff

@BrianM.Scott OK. I'll have to go "work" for a while. Maybe I'll just post it as a question

Brian M. Scott

Hang on a minute.

Suppose that $x>0$. Then $x=y^2$ for some $y$. But $f(x)=f(y^2)=f(y)^2\ge 0$.

That’s not quite there, since we still have $\ge$ instead of $>$, but it’s a start.

John Senior

@PeterTamaroff What are you aiming to prove?

(ignore me - I think I found it)

Brian M. Scott

@JohnSenior That if $f:\Bbb R\to\Bbb R$ satisfies $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ for all $x,y\in\Bbb R$ and is not identically $0$, it’s the identity function.

Peter Tamaroff

@BrianM.Scott But if $x^2>0$ whenever $x\neq 0$; yes? Maybe we need to consider the fact that $f\not \equiv 0$.

John Senior

7:37 PM

so presumably the multiplicative property is going to allow us to rule out the nasty discontinuous functions that satisfy the rest of the conditions

John Junior

Hi all :)

Brian M. Scott

@PeterTamaroff If $x\ne0$ and $f(x)=0$, then $f(xy)=0$ for all $y$, so $f\equiv0$. This gives you $f(x)>0$ for $x>0$. And since $x+(-x)=0$, it gives you $f(x)<0$ for $x<0$.

Peter Tamaroff

@BrianM.Scott Come again?

Are we still using the $x=y^2$ hypothesis?

Brian M. Scott

I’d already shown that if $x>0$, then $f(x)\ge 0$. But $f(x)$ can’t be $0$, because if it were, I could write any real number in the form $xy$ for some $y$ and get $f(xy)=f(x)f(y)=0$, maiking $f$ identically $0$.

Peter Tamaroff

@BrianM.Scott Yes, I see now =D

Brian M. Scott

7:41 PM

Thus, $x>0$ implies $f(x)>0$. And then I can use $x+(-x)=0$ to get $f(x)<0$ when $x<0$.

Peter Tamaroff

@BrianM.Scott Yes. Now $f(x)>f(y)$ when $x>y$ follows from that, right?

I have 3 minutes to go =P

Brian M. Scott

Yes: if $x>y$, then $f(x)=f(y)+f(x-y)>f(y)$.

John Senior

@PeterTamaroff plenty of time - almost finished :)

Brian M. Scott

Now that you know that $f$ is strictly monotone increasing, you can easily use the density of the rationals in the reals to ‘trap’ each irrational and get that $f(x)=x$.

John Senior

Is the result now established for all rationals

Brian M. Scott

7:44 PM

@JohnSenior Yes, Peter already did that.

John Senior

Great!

Peter Tamaroff

@BrianM.Scott Enlight me, but just a little!

When I return, I'll try and solve it.

Brian M. Scott

@PeterTamaroff About the trapping?

Jaakko Seppälä

Is it better to learn C or Matlab if I want a job related to mathematics?

Peter Tamaroff

@BrianM.Scott Yes. I know that for any $x<y$ there is a rational such that $x<r<y$. I have formally proven that!

I'm off, peoples of the math.

John Senior

7:47 PM

Bye Peter

John Junior

@JaakkoSeppälä Do you know BASIC?

Brian M. Scott

Let $x$ be irrational. For each $n\in\Bbb N$ pick rationals $p_n\in(x-2^{-n},x)$ and $q_n\in(x,x+2^{-n})$. Then $f(p_n)<f(x)<f(q_n)$. From there you should be able to finish it off.

Jaakko Seppälä

@JohnJunior Yes. And Java too.

John Junior

@JaakkoSeppälä How about FORTRAN?

Jaakko Seppälä

@JohnJunior Nope.

John Junior

7:51 PM

Fortran

Fortran (previously FORTRAN) is a general-purpose, imperative programming language that is especially suited to numeric computation and scientific computing. Originally developed by IBM at their campus in south San Jose, California in the 1950s for scientific and engineering applications, Fortran came to dominate this area of programming early on and has been in continual use for over half a century in computationally intensive areas such as numerical weather prediction, finite element analysis, computational fluid dynamics, computational physics and computational chemistry. It is one of ...

Jonas Teuwen

Anyone know a bit of measure theory icm with their convergence theorems and nets?

John Senior

@JonasTeuwen don't think my rusty measure theory is likely to help - sorry :((((

Jonas Teuwen

@JohnSenior Just wondering if Fatou's lemma works for nets if every integral makes sense :-).

Actually, life is quite okay when you only talk to people you can shut off!

John Junior

@BrianM.Scott Would you be interested in a question relating Math to English?

John Senior

@JonasTeuwen you have found an on/off switch for people? - I want one!

Jonas Teuwen

8:05 PM

@JohnSenior Only online.

John Senior

@JonasTeuwen drat :)

robjohn

@PeterTamaroff back :-)

John Junior

Attention is sort of an on/off switch...

@robjohn Would you be interested in a question relating Math to English?

Here.

Brian M. Scott

@JohnJunior My only real comment is that I’ve not heard it called an equal sign.

robjohn

@JohnJunior Cool Elf's answer looks good.

Jaakko Seppälä

8:16 PM

@JohnJunior Hmm. There's only two open Fortran positions in Finland but both requires a Ph.D education so I won't get them.

John Junior

Thanks for having a look. @BrianM.Scott @robjohn :)

robjohn

@BrianM.Scott you've never heard = being called and "equal sign"? I've heard "equal sign" and "equals sign". The "s" after the "equal" gets hidden by the "s" at the beginning of the "sign"

@JaakkoSeppälä I would think that the PhD of anyone who still programs in Fortran would be so old as to be almost obsolete.

Brian M. Scott

@robjohn I’ve heard equals sign pronounced rapidly and carelessly enough that the two sibilants blend, but I interpret that as equals sign, not as equal sign. The latter has a hiatus that the former lacks even when run together.

John Junior

Does anybody call it the "sign for is."?

robjohn

@BrianM.Scott If you enunciate "equals sign" clearly enough. Usually I say "equal sign" although I guess that could mean "the same sign".

Brian M. Scott

8:25 PM

@robjohn For me there’s a very clear distinction between the two even in rapid speech. Even if the /z/ of equals loses its voicing, the resulting /s/ is longer in equals sign than in equal sign. Moreover, the latter has a slight break between the /l/ and the /s/.

Jonas Teuwen

Some people claim I sometimes speak so fast they cannot follow 8-).

Matt

Saturday night does not appear to be the best time to ask questions on SE.

I can't think of anything better to do on a Saturday night than sitting at home on a comfy sofa and thinking about stuff though.

@BrianM.Scott I got your ping and I will read your helpful comment regarding that teddy answer. But right now I have lots of homotopy stuff in my cache and I'd rather not swap that right now.

Thanks a lot for the comment though.

Brian M. Scott

@Matt That’s fine; it’ll keep.

Matt

I hope it will : )

@BrianM.Scott Are you "back"?

I thought you'd left chat.

Brian M. Scott

@Matt For a while I was too busy answering questions and cycling to spend any time here.

Matt

8:38 PM

: )