Mathematics - 2017-06-28 (page 5 of 7)

Sha Vuklia

4:00 PM

yes I did @Akiva

Dattier

Rabbi Akiva : fr.wikipedia.org/wiki/Rabbi_Akiva

Sha Vuklia

but I didn't help me intuition unfortunately

Liad

@SteamyRoot do you agree with my definition of $F $ ?

SteamyRoot

This can never be a complement of $C([0,1])$

Martin Sleziak

@AkivaWeinberger You probably pinged me to let me know that I should have written accepted answer rather than top-voted answer, right?

Liad

4:01 PM

@SteamyRoot what do you mean? maybe i meant completion

Sha Vuklia

So I should be able to show that $\phi(k)$ counts the number of fractions $\dfrac{t}{k}$, for $t\in\{1,\dots,n\}$, where $(t,k)=1$

Semiclassical

There's a big difference between complement and completion...

Akiva Weinberger

@MartinSleziak Oh, I didn't notice that that proof was there

SteamyRoot

@Liad I have no idea what you're trying to do. $C([0,1])$ is already complete.

Liad

@SteamyRoot it is not

Akiva Weinberger

4:03 PM

@ShaVuklia What's the definition of $\phi(k)$

@Dattier Yes that is where my name comes from

Sha Vuklia

there are two equivalent definitions, one counts the numbers relative prime to $k$ up to $k$

Liad

10

Q: Showing that the space $C[0,1]$ with the $L_1$ norm is incomplete

Can anyone think of a relatively easy counter example to remember, which demonstrates that the space $C[0,1]$ with the $L_1$ norm is incomplete? Thanks!

real-analysis

Semiclassical

This is also relevant:

0

Q: Showing that $C[0,1]$ is not complete

Let $C[0,1]$ have the following norm for finite $p$: $$||f||_p=\left(\int_0^1|f(x)|^pdx\right)^{1/p}$$ I want to show that $C[0,1]$ is not complete for this norm. How do I do this? What I know: So I want to find a Cauchy sequence that does not converge in $C[0,1]$. My idea was $$f_n(x) := \...

functional-analysis convergence banach-spaces cauchy-sequences

Sha Vuklia

the other says $\phi(k)=\{\overline a\in\mathbb Z/k\mathbb Z: 1\leq a\leq n, \gcd(a,n)=1 \}$

SteamyRoot

With a non-standard norm... yeah, sure.

Martin Sleziak

4:04 PM

@ShaVuklia If $s/t=a/n$ and $s/t$ is in the basic form then $t=n/\gcd(a,n)$. But I should probably leave this to Akiva, so that you don't have to talk to several people at the same time. (Also it seems that I unintentionally directed you to a different answer from the one you have already been studying.)

Akiva Weinberger

@ShaVuklia Yeah, relative prime numbers to $k$

Liad

im not sure but i think that $L \ ^ 2[0,1] $ is defined to be $C[0,1]$ 's completion (im not sure this is the right word)

SteamyRoot

If you don't mention which norm you're using, I'm assuming you're using the standard norm.

Semiclassical

^

Dattier

Why the symbol of juice is not the dattier (like the just who take is knowledge), but the ...

Akiva Weinberger

4:05 PM

@ShaVuklia So divide the numbers relatively prime to $k$ by $k$

Semiclassical

Completion with respect to what norm? I think that's the key point.

Dattier

@Akiva

Why the symbol of juice is not the dattier (like the just who give is knowledge), but the ...

Liad

induced by $<f,g> = \int fg$

Sha Vuklia

@AkivaWeinberger okay, I guess

your point then is that it doesn't matter that we divide by $k$?

SteamyRoot

That's the $L^2$ norm I mentioned before, then (assuming your functions are real).

Liad

4:07 PM

any way, if i define $F(x) =\sum c_n e_n(x)$ , im sure that $F$'s coefficients are $c_n$ , but , is it well defined ?

Akiva Weinberger

The set of numbers with denominators equal to $k$ is precisely the set of numbers relatively prime to $k$, divided by $k$

Stuff relatively prime to 12: 1,5,7,11

Sha Vuklia

hey but that was my original question

that's what I tried to show

Akiva Weinberger

Stuff with denominators 12: $\frac1{12},\frac5{12},\frac7{12},\frac{11}{12}$

Semiclassical

Some further stuff here: math.stackexchange.com/q/1124176/137524

Sha Vuklia

oh okay that makes sense

oh

i think i see some light

Dattier

4:08 PM

Why the symbol of jewish is not the dattier (like the just who sharing is knowledge, that's in the bible), but the ...

Sha Vuklia

at last

Dattier

@Akiva

Why the symbol of jewish is not the dattier (like the just who sharing is knowledge, that's in the bible), but the ...

Sha Vuklia

okay thanks @Akiva I think I have to reread everything, but I think I'll get it, because the last bit you said made a lot of sense

Semiclassical

why do you keep repeating that

you've said it 4 times now.

In particular, there they do claim $L^2(\Omega)$ as the completion of $C^0(\Omega)$ with respect to the $L^2$ norm.

Akiva Weinberger

@Dattier What does "dattier" mean

Dattier

4:12 PM

@Akiva : Why the symbol of jewish is not the dattier like (the righteous who shares his knowledge), but the

dattier :
date-palm

Why Akiva don't choose this symbol

for his community ?

but choose the ...

Just win baby

ask here^

:-)

Dattier

Yes, a fruitless tree, why ?

Just win baby

ask on that site

they would know

Dattier

I ask @Akiva

Liad

why does this equation legal: $<\sum c_n e_n(x) , c_m> = \sum c_n < e_n, e_m> $ ? it is defined as an integral so don't i need uniform convergence or something for this to equation to hold?

Semiclassical

4:16 PM

And you've asked him half a dozen times now.

Balarka Sen

I like this tree

(From the cover page of Rolfsen)

nice

That's terrifying.

yes, I think too

@Dattier Why choose the Star of David, you mean?

Semiclassical

4:18 PM

The horned sphere, I mean.

Sha Vuklia

Thanks @Akiva and @Martin I had to reread everything you said, and I understand everything now, and I think I've practised well with ggd's and the like

Akiva Weinberger

Is the date tree a reference to the "tzadik katamar" verse (at the end of Psalm ninety-something)?

Dattier

The start of David is present in the first, in the flag of Morocco (a muslim country)

@Akiva

Akiva Weinberger

I don't remember the origin of the Star of David, actually. I vaguely remember something about a shield

Balarka Sen

@AkivaWeinberger "Star of David" brings this to (my obviously ignorant) mind:

Akiva Weinberger

4:20 PM

@Dattier Yeah, it's used a lot in Muslim art since that's very geometric. Well, not as much nowadays.

Dattier

google.fr/…

Akiva Weinberger

@BalarkaSen …Why?

Semiclassical

Wikipedia: "The use of the hexagram in a Jewish context as a possibly meaningful symbol may occur as early as the 11th century, in the decoration of the carpet page of the famous Tanakh manuscript, the Leningrad Codex dated 1008."

Balarka Sen

Cover of David Bowie's Blackstar yo.

Semiclassical

It's definitely a symbol whose origins are lost to history, though.

Krijn

4:22 PM

What are the vertical components of a divisor on a surface?

Akiva Weinberger

@BalarkaSen Oh, that's really clever

Dattier

Why @Akiva choose, for the symbol of jewish a fruitless tree ?

Balarka Sen

Maybe that cover really is inspired from the Jewish symbol you two are talking about, I dunno.

Akiva Weinberger

@Dattier What tree?

@BalarkaSen The Jewish symbol is six-sided

It's a hexagram

See my profile pic

Semiclassical

I think he's referring to Justwinbaby's linking to Mi Yodeya above

Akiva Weinberger

4:24 PM

(which I might just change to a picture of myself at some point)

Balarka Sen

Ah, I see, @Akiva.

Akiva Weinberger

@Semiclassical Well, the Bible calls itself the "Tree of Life" (etz chayim), so maybe the logo for the Stack Exchange site Mi Yodea is a reference to that

("Mi Yodea" literally translates to "Who Knows")

Krijn

I always forget that the Mathematics part of this chat room is more grandeur than reality.

Akiva Weinberger

I didn't start this fire

(Song reference, for the non-English speakers)

Dattier

@Akiva The tree live (symbolised by the menorah)

Semiclassical

4:27 PM

@Astyx I think I see your argument: If $\det(M+xJ)$ is a linear function of $x$, then it's enough to compute it at two different values of $x$. But for $x=-a,-b$ the determinant can be found immediately.

Akiva Weinberger

I don't think the menorah is meant to symbolize the tree of life

At least, I've never heard that before

Balarka Sen

@Krijn I dunno the answer to your question, so there's that.

Dattier

Ask you master

@Akiva

Akiva Weinberger

In any case, Krijn's right; this is off-topic for a mathematics room.

Semiclassical

What's not immediately obvious to me is why it's a linear function of $x$.

Akiva Weinberger

4:28 PM

@Dattier …You mean "Rabbi"?

Krijn

@AkivaWeinberger That's not really a problem, I think

Only when it obstructs actual mathematics, which in this case is not the case

Dattier

yes, but in cabbalah the name is master, no ?

Akiva Weinberger

I don't know any cabbalah.

Krijn

Anyway, I'm done with mathematics for today, on to KFC

Akiva Weinberger

(But I doubt it?)

Dattier

4:30 PM

Well, what's the tree symbol of jewish for you (@Akiva)

Il 's fun you study the matematics, and you say you don't know the cabbalah

Akiva Weinberger

I don't understand the question

@Dattier I've really never been interested in it.

Math is not numerology

Dattier

For you, If you must choose a tree for the symbol of jewish, what 's the tree you choose

Akiva Weinberger

Uh, olive, maybe?

Dattier

The cabbalah is the science of the name

Akiva Weinberger

…of God

Chris Nguyen

4:32 PM

Help plz on question 18:

mualphatheta.org/useruploads/files/…

Dattier

No

of God but of djinn

Akiva Weinberger

What the hell is a djinn

Dattier

How catch a djinn

Akiva Weinberger

Isn't that an Islamic thing?

Semiclassical

This is getting decidedly esoteric.

Balarka Sen

4:33 PM

I agree.

Akiva Weinberger

I said olive, by the way, because it's on the emblem of the State of Israel

Dattier

no, no, Salomon (ps) known who catch the djinn

Akiva Weinberger

@Semiclassical Well, I have no idea what he's talking about

@Dattier I have no idea what you're talking about

Semiclassical

And, frankly, it's a bit odd to wander into a chat room and start interrogating one specific person as to a belief system they may or may not have.

Dattier

So, there are forum where I can ask that ?

Semiclassical

4:34 PM

...the one which was linked earlier.

that's why it was linked.

Akiva Weinberger

Mi Yodea

Semiclassical

I mean, I have no idea if they'll know what you're talking about either.

Akiva Weinberger

It's another site on Stack Exchange

Dattier

Thanks

Akiva Weinberger

So, uh.

Anyone have a cool math thing?

Semiclassical

4:36 PM

This is a neat looking problem on this month's AMM:

Balarka Sen

I also don't see why one's interest to mathematics has anything to do with they knowing or not knowing Cabbalah. Even if there is any mathematical content to it, one is feel free to not read it. That's such a weird thing to say.

But yeah whatever

Semiclassical

For any $x\in(0,1)$ show that $$\prod_{n=1}^\infty(1-x^n) \geq \exp\left(\frac{1}{2}-\frac{1}{2(1-x)^2}\right)$$

Akiva Weinberger

Oh, god

Balarka Sen

Huh

Akiva Weinberger

I bet $e^x\ge1+x$ enters this

Semiclassical

4:37 PM

probably.

Liad

how can i show that $\sum_{n = 1}^N cos(2 \pi n x) $ is bounded for $x \notin \Bbb Z$ ?

Semiclassical

My immediate thought is to start by taking log of both sides.

Akiva Weinberger

Wait, the exponential is on the wrong side. Hm.

Liad

for each $N \in \Bbb N$

Akiva Weinberger

@Liad That can actually be summed explicitly.

Do you know $\cos(x)=\frac{e^{ix}+e^{-ix}}2$?

Balarka Sen

4:38 PM

Maybe use $e^{-x} \leq \frac{1}{1-x}$, by inverting both sides?

Semiclassical

So that'd become $\displaystyle \sum_{n=1}^\infty \log(1-x^n)\geq \frac{1}{2}-\frac{1}{2(1-x)^2)}$

Akiva Weinberger

Using that, the above turns into a geometric progression. @Liad

Balarka Sen

And use expansion of $\frac{1}{1-x^n}$, something something.

Akiva Weinberger

Alternatively, you can find formulas for the sum online, or from Wolfram Alpha :P

Liad

@AkivaWeinberger i did not know that expression

Akiva Weinberger

4:39 PM

Oh

Do you know $e^{ix}=\cos(x)+i\sin(x)$?

Liad

well im trying to show that $\sum cos(2 \pi n x) /log n $ converge so maybe there is a simpler way

@AkivaWeinberger yea

Chris Nguyen

Help plz on question 18:

mualphatheta.org/useruploads/files/…

Akiva Weinberger

@Liad What a coincidence! I had essentially the same problem a few days ago

I had no idea how to do it. Ted helped me.

But you seem to be on the right track; $1/\log n$ is decreasing and $\sum\cos(2\pi nx)$ is bounded

Liad

@AkivaWeinberger great. now is your turn :P btw, where is Ted?

Akiva Weinberger

and it can be shown that if $\sum a_n$ is bounded and $b_n$ is decreasing to $0$, then $\sum a_nb_n$ converges.

Liad

4:43 PM

Dirichlet test

Akiva Weinberger

(I think I have a sort of geometric proof interpreting those as areas)

Liad

Dirichlet's test

In mathematics, Dirichlet's test is a method of testing for the convergence of a series. It is named after its author Peter Gustav Lejeune Dirichlet, and was published posthumously in the Journal de Mathématiques Pures et Appliquées in 1862. == Statement == The test states that if { a n } {\displaystyle \{a_{n}\}} is a sequence of real numbers and { b n } {\displaystyle \{b_...

yea, so why does $\sum cos(2\pi n x) $ is bounded? and btw @SteamyRoot hinted me of this way earlier :P

Akiva Weinberger

I already knew that $\sum\cos(nx)$ is bounded (for $x\ne0$) and $\sum\sin(nx)$ is bounded (everywhere)

(Well, for $\sum\cos(nx)$, you want $x\ne\pi k$ I think)

Liad

$x \notin \Bbb Z$

and it is $cos (2 \pi n x) $

Akiva Weinberger

@Liad So $e^{ix}=\cos(x)+i\sin(x)$. What's $e^{-ix}$, then?

@Liad Ah, OK. So that's good

Liad

4:46 PM

it is $cos(x) - sin(x)i$

:P

Akiva Weinberger

Now add them together

and divide by two

Liad

so if you sum it you get $2cos$

yea ok

Akiva Weinberger

So $\cos(x)=\frac{e^{ix}+e^{-ix}}2$. And $\sin(x)$, by the way, is $\frac{e^{ix}-e^{-ix}}{2i}$.

(The denominator for sine is $2i$, not $2$. It's annoying)

In any case, substitute it in and you get a geometric series

Liad

Ok, so far im with you :P

Akiva Weinberger

Well, really, two geometric series added together

Semiclassical

4:50 PM

It's funny how $i\sin x$ is sorta easier to work with than $\sin x$ itself.

Akiva Weinberger

This all follows from the fact that, if $|z|=1$ (and $z\ne1$), then $\sum z^n$ is bounded.

Semiclassical

Also, regarding the problem I gave earlier: Some Mathematica manipulation makes it appear that the bound is only going to be sharp as $x\to 0^+$.

Akiva Weinberger

If $z$ is almost $1$, say $z=e^{0.001i}$, then the partial sums of $\sum z^n$ follow this large circle

Semiclassical

5:02 PM

Hmm: $e^x\geq 1+x\implies x\geq \log(1+x)\implies \log(1-x^n)\leq x^n$?

Akiva Weinberger

Maybe Taylor series?

Of The logged version

@Liad Oh

Semiclassical

I feel like I'm doing something silly.

$\log(1+x)\leq x$, so if I replace $x\to -x^n$ that's $\log(1-x^n)\leq -x^n$. oh, I was right initially

Akiva Weinberger

You could use $\cos(x)={\rm Im}(e^{ix})$

That's probably a lot simpler

Balarka Sen

I don't understand the third step.

Akiva Weinberger

So $\sum\cos(2\pi nx)={\rm Im}(\sum e^{2\pi inx})$

Balarka Sen

5:04 PM

Yeah that's more like it.

Akiva Weinberger

Wait, today's tau day.

Semiclassical

My main motivation here is that the inequality only appears to be effective at small $x$.

Akiva Weinberger

${}={\rm Im}(\sum e^{\tau inx})$

Semiclassical

So I'm seeing what happens if I use the most obvious approximation.

Balarka Sen

But you're just getting a $\leq$

We want $\geq$

Akiva Weinberger

5:05 PM

Tau is $\#e^{i\tau}$!

Semiclassical

Hmm. That's true.

Akiva Weinberger

$e^x\ge1+x+\frac12x^2,~x>0$

etc

Maybe that's useful

Semiclassical

Yeah. I guess one may need the second-order form.

user76284

Is it correct to say $\sup \{ r \mid f \text{ is analytic in } \lvert z \rvert < r \}$ is the radius of analyticity of $f$?

Akiva Weinberger

I wouldn't know, but I'd be very surprised if it weren't

Semiclassical

5:09 PM

Problem is that the Taylor series approximations of $\log(1-x^n)$ give bounds from below.

Balarka Sen

$\log(1 - x)$ is $-(x + x^2/2 + x^3/3 + \cdots)$ right

Semiclassical

Right.

Sha Vuklia

@Akiva would you maybe have time for one more "ggd" question?

Balarka Sen

So if you need to bound that above, you need to bound $\sum x^n /n$ from below.

Akiva Weinberger

$\tau\!\!\pi\stackrel?=\frac13\tau$

Semiclassical

5:10 PM

hmm.

Balarka Sen

Which is a little strange because $x < 1$

Semiclassical

I don't think this is the right approach.

Balarka Sen

The natural bounds are from above.

Akiva Weinberger

$\frac1\tau\stackrel?={\perp}$

Semiclassical

Avantgarde

5:14 PM

$\cos(x)=Im(e^{ix})$?

Semiclassical

The red line is the RHS, and the blue lines are the LHS for n=1,2,3 (highest is n=1)

SteamyRoot

@Liad You could also use $| \operatorname{Re} z | \leq |z|$ for any complex $z$.

Akiva Weinberger

@Avantgarde And by $\rm Im$ I mean $\rm Re$

hides

Avantgarde

:P

Semiclassical

Oh. I've also squared both sides b/c whatever that's what I did.

Balarka Sen

5:17 PM

I have to write down my essay-sized chemistry practical reports.

Semiclassical

god help you

SteamyRoot

So $$\left| \sum_{i=2}^N \cos(2\pi n x)\right| \leq \left| \sum_{i=2}^N \exp(2\pi i n x)\right|$$

And then you just add and substract the first two terms, pull out the $n$ as a power (which is allowed because it's an integer), use geometric series, and triangle inequalities everywhere

Akiva Weinberger

You could also just look up the formula for the LHS of that ^ if you feel sick of deriving it

This is how you derive it

SteamyRoot

and you should probably end up with something like this thing being smaller than $$\frac{2}{|1-\exp(2\pi i x)|} + 2$$

which is independent of $N$, and since $x \notin \mathbb{Z}$ the denominator isn't zero.

Akiva Weinberger

It's also kind of intuitively obvious when you think of what the partial sums of $\sum z^n$ look like when $|z|=1$ and $z\approx1$

It follows a big circle

Like driving with your steering wheel at a constant angle.

If that constant angle is 0, you go forever unboundedly

but anything else, you end up in this large (but bounded) circle

Semiclassical

5:25 PM

with that angle dictating the curvature, yeah.

(for small angles, anyway)

Liad

@SteamyRoot thanks, i am doing it right now, but is it true that $S_n = \dfrac{a_1 (q \ ^ n -1 )}{q-1}$ also when we deal with complex numbers? (we did not learned it yet but it sounds right)

Akiva Weinberger

It is.

The same proof works

Semiclassical

So long as $|q|<1$, sure.

Akiva Weinberger

@Semiclassical This is the finite sum, not the infinite sum

Semiclassical

Woops.

Akiva Weinberger

5:31 PM

It's valid everywhere -

Semiclassical

So long as $q\neq 1$, then :P

(yes yes removable)

Akiva Weinberger

well, not $q=1$ because of the denominator

but I am Semi with a second's delay.

Semiclassical

lol

SteamyRoot

Yup. And again this case is excluded because of the condition on $x$ ^^

BAYMAX

Latex chat room is sleepy

Akiva Weinberger

5:33 PM

You know, in the field $\Bbb R(x)$, I think it's perfectly valid to evaluate $\frac{x^n-1}{x-1}$ at $1$

BAYMAX

so can I ask a latex query here ><

Krijn

@Steamy do you know what is meant by the vertical component of a divisor on a surface?

Akiva Weinberger

@BAYMAX Go for it

BAYMAX

Like I wa searching how to write a system of equations in latex
so i did that using \textbf{systeme} package but out of that one of the equation is missing
why so?

SteamyRoot

@Krijn Not a single clue.

BAYMAX

5:33 PM

Like this
\systeme{

\partial_{t}v_{1} + \epsilon_{1} v_{1} = -k \frac{dT_{1}}{dy}

\hat{A_{1}}\frac{\partial{T_{1}}}{\partial{t}} +a_{T}v_{1}\frac{dT_{1}}{dy}} + M_{s1}\frac{dv_{1}}{dy} = -P + \frac{gE}{P_{T}}.(R+H)

\hat{B_{1}}\frac{\partial{q_{1}}{\partial{t}} +a_{q}v_{1}\frac{\partial{q_{1}}}{\partial{y}} - M_{q1}\frac{dv_{1}}{dy} = -P +\frac{gE}{P_{T}}

}
But the last equation is not displayed any help?

SteamyRoot

It sounds like algebraic geometry so I'm staying away from it, too :P

Akiva Weinberger

I have no idea

I've always used \begin{align*} and \end{align*}

BAYMAX

here

Liad

@SteamyRoot i got $\le \dfrac{2}{|e \ ^ {2 \pi ix}- 1 |} +1 $, thanks :)

SteamyRoot

+1? :O

BAYMAX

5:35 PM

ok

SteamyRoot

If you want to align at multiple positions, you can always shove an array inside

or use the (deprecated) eqnarray

Liad

we have $|\sum_1^{N} e \ ^ {2 \pi i nx} - e \ ^ {2 \pi i x}|$

$\le \dfrac{2}{|..|} + |e \ ^ {2 \pi i x}| \le 2/.. + 1 $

Akiva Weinberger

The actual bound ends up being $|\csc(\pi x)|$, I think.

$|\csc(\frac12\tau x)|$

Liad

i did a mistake?

Akiva Weinberger

@Liad You're not going for the strictest possible bound, I don't think

SteamyRoot

5:38 PM

Ah, no, that seems fine.

Akiva Weinberger

You're just trying to show that it's bounded

SteamyRoot

I worked with the sum from $0$ to $N$, so I had two trailing terms

Liad

@SteamyRoot you sounded surprised when i wrote +1

SteamyRoot

@AkivaWeinberger pukes

Liad

huh

fine

Akiva Weinberger

5:39 PM

@SteamyRoot Happy tau day!

SteamyRoot

@AkivaWeinberger I'm totally aiming for you next time I puke.

Too bad it's a bit far.

Akiva Weinberger

The bound for $\sum\sin(2\pi nx)$ is the same, actually. Well, for $x\in\Bbb Z$ it's zero, but otherwise…

@SteamyRoot In the general direction of the United States :P

Semiclassical

"I ralph in your general direction."

Akiva Weinberger

You know, if you try walking in a straight line without seeing any landmarks, you'll invariably start to curve to the left (I think)

You end up in this weird spiral

Semiclassical

I'm a bit skeptical that the bias in that sense would be universal to all humans.

Akiva Weinberger

5:43 PM

I guess humans are symmetrical enough for most things, but there's no particular evolutionary drive to make us perfectly symmetrical

Semiclassical

I can well believe that with any person you'd start to see that, bu that it's always to the left?

Akiva Weinberger

Strange that we'd all have the same asymmetry, though

@Semiclassical I think so

SteamyRoot

making us symmetric would be problematic for any organs we only have one of

Semiclassical

I don't buy it :/

Akiva Weinberger

I might have it wrong, though

SteamyRoot

5:45 PM

Either we need to become Time Lords with a heart on each side, or we need some kind of ribcage-extension at the center to protect the heart

SevenSidedDie

@AkivaWeinberger I've heard the factoid about walking in a spiral, but I don't remember the universal bias to the left. Maybe it was only statistical? I could see a statistically significant bias to the left in a population.

Semiclassical

Oh hey, there's a nature article about it: sciencemag.org/news/2009/08/why-we-walk-circles