Mathematics - 2017-03-28 (page 1 of 7)

I just mentioned that as a general rule, with infima you should always take a minimizing sequence. He hadn't worked them often with it so it is somewhat fair that this technique be laid out somewhat explicitly.

Balarka Sen

@Daminark I was just pulling legs. You're fine

Daminark

Kek

Ted Shifrin

This isn't a Ted exercise ...

Meow Mix

Balar-karma

Sounds like "calamari"

Daminark

12:35 AM

I sen-se a name pun

Ted Shifrin

Only if your ears are stuffed ...

Balarka Sen

lol @Ted

Meow Mix

Ted exercise rhymes with Fedex arrives

Daminark

Lol @Eric it seems like we have multiple people TeXing notes this quarter

3 people, one is live TeXing

Eric

lol i tried to TeX notes for your class last year but I couldn't read her handwriting and gave up.

Daminark

12:40 AM

We're from the department of redundancy department so yeah

I would except that I can't both take notes and concentrate

Ted Shifrin

Or accept ...

Daminark

So for that reason I just accept it (credits @Ted) and don't take notes

Ted Shifrin

Psychologists have established that you retain more with some (handwritten) note-taking.

Eric

I TeX in algebra and hand-write my notes in classes where pictures matter

and then TeX them later

Ted Shifrin

You'll need lots of macros for diff geo ...

Eric

12:46 AM

i came up with a looooot of macros when i took the smooth manifolds class here

Daminark

Perhaps, but I guess in my experience, I'm not fully concentrated when I write things down, counterintuitive as it may seem

And I'm not terribly quick to process the stuff, if I were to take the time to process and write, I'd miss the next few things said

Mike Miller

\M gives you M

most important macro

Daminark

Also @Eric how was Lawler's class?

@Mike Lmao

Ted Shifrin

M is irrelephant.

Mike Miller

I guess in these modern times most 3-manifolds are Y.

Balarka Sen

12:59 AM

Funny, indeed I have seen Y used more

Daminark

I wonder Y that's the case

Eric

@Daminark he's a fine lecturer, I probably won't be sitting in on the class long enough for him to do anything

John P

Guys, if I take a curve, and at each point I move in the direction of it's unit tangent by a fixed length - does the curve I obtain have a special name?

And what's it curvature relative to the original curve? I've tried to solve that, but got a very long expression, so I'm probably missing something.

Ted Shifrin

1:21 AM

@JohnP: Not a well-known curve (evolute, involute, etc.). Is this a plane curve?

John P

No, a curve in space.

Meow Mix

That's what I was thinking about @John, except with a normal rather thaan a tangent

because with the normal, it kind of "Expands out"

Ted Shifrin

Yeah, you're gonna get a mess for the curvature, involving curvature, torsion, and the fixed length.

Meow Mix

and, instead of going fixed in one direction, i'd take it over $n$ fractional parts of the length we want, each time taking the normal over again, and then increase $n$ for more accuracy

Ted Shifrin

Normal gives you involute ... if I remember correctly.

Semiclassical

1:23 AM

If you do that for a circle, you'll get a family of concentric circles depending on the fixed distance.

No idea for an ellipse.

Ted Shifrin

Oh, involute won't be a constant length. Never mind.

Semiclassical

I may play around with it in Mathematica.

Meow Mix

So, the reason I take it over $n$ fractional parts is because otherwise it won't stay continuous

Ted Shifrin

Huh?

John P

Thanks @TedShifrin , I guess I'll just hand in the answer I have.

Ted Shifrin

1:26 AM

Just don't forget chain rule.

Mike Miller

:o

Meow Mix

Let me show you a terrible drawing of what I mean

Daminark

That if every chain is bounded above, there is a maximal element?

Meow Mix

The gray is what you get when you go $2$ times the radius, all at once

it's not differentiable, I mean

Ted Shifrin

puts Demonark on permanent ignore

Meow Mix

1:27 AM

but, if you do $1$ times the radius, twice, it will stay differentiable

Semiclassical

Ted Shifrin

Nah, Zach. Any constant works.

John's question for a plane curve is the path of the front wheel of a bike if you're given the path of the rear wheel. This is an exercise in my diff geo text.

Meow Mix

What?

Semiclassical

The construction being: If $\vec{r}(t)$ is some parametric curve, then let $\vec{r}(t;s)=\vec{r}(t)+\vec{r}'(t)\cdot s$

I guess I should probably be rescaling $r'(t)$ to get the unit tangent.

Ted Shifrin

No, @Semiclassic. You need unit tangent.

Right.

Semiclassical

1:31 AM

Oh, weird.

Ted Shifrin

See my comment re bicycles.

Semiclassical

Neat.

That was with $s\geq 0$, I should note. But if I make it negative it just reverses the figure.

Now I wonder what would happen if the original curve had less symmetry.

This is what I get if I plug in a Lissajous figure:

Ted Shifrin

Interesting challenge for you @Semiclassic: go backwards. Given the path of the front wheel,

Semiclassical

Ted Shifrin

find the path of the back.

Quite hard!

Semiclassical

1:37 AM

Now, is it just me, or does that look like lingerie?

I can't unsee it now.

Daminark

Can we be physicists and give within an $\epsilon$ range? And by $\epsilon$ I mean 2 meters?

Mike Miller

not until you said it

Daminark

Oh god @Semi why did you do this?

Semiclassical

lol

arctic tern

hrm

Akiva Weinberger

1:52 AM

Hrmm.

Battle.

Meow Mix

Hi @Akiva!

Akiva Weinberger

Hi! I slept for a few hours and now I'm hungry

Meow Mix

I'm always hungry

Akiva Weinberger

"That's my secret, Cap."

Meow Mix

And tired

Akiva Weinberger

1:55 AM

(Whoa, apparently I got the quote wrong for a few years)

(I thought he said 'captain')

Meow Mix

I watched Pulp Fiction.

logical123

2:28 AM

I'm attempting to find a closed form for the following partial sum: $\sum_{n=1}^{k} (n+100) 1.3^{n/16}$

Which is a sum of the form $\sum_{n=1}^k (n+\alpha ) \beta^{\gamma n}$

Semiclassical

Probably you want to further condense that by defining $x=\beta^\gamma$.

logical123

$\sum_{n=1}^k (n+\alpha ) \beta^{\gamma n} = \sum_{n=1}^k n\beta^{\gamma n} + \alpha\sum_{n=1}^k \beta^{\gamma n}$

right, i was thinking i needed to do a change of variables, but i'm kinda at a loss how to do that

Semiclassical

Well, once you introduce that $x$ you've got $\sum_{n=1}^k (n+\alpha)x^n$.

logical123

derp

Semiclassical

For the second term, you've got a geometric series which can be summed in closed form.

logical123

2:33 AM

totally blanked

$1.3^{n/16} = 1.3^{(1/16)^{n}}$

Semiclassical

noooo

$1.3^{n/16}=(1.3^{1/16})^n$.

logical123

thats what i tried to type

Semiclassical

Ah.

logical123

not 1/16 to n

Semiclassical

gotcha.

For the first term, I'd notice that $\sum_{n=1}^k n x^n = x\sum_{n=1}^k nx^{n-1}=x\dfrac{d}{dx}\sum_{n=1}^k x^n$.

logical123

2:35 AM

i got it now, dunno why i wasnt seeing that simple equivalency

Semiclassical

In which case, yay for geometric series again.

logical123

yeah thats straight out of thermodynamics for me

Akiva Weinberger

That's the arithmetico-geometric series, isn't it

Semiclassical

Ah, the joys of the partition function.

logical123

yessiree!

Akiva Weinberger

2:36 AM

Yeah, the classic way to do that is with Semi's trick with the derivative

logical123

Thermo and Stat Mech was my favorite Physics course by far, even more than quantum.

Akiva Weinberger

Wiki has a page on it

Semiclassical

Ya.

The trick with differentiating something of the form $e^{-xt}$ to get something more useful is something that comes up a ton in physics.

Heck, you even rely on it when you get into QFT and you want to compute stuff.

logical123

It's everywhere, generating all the different special polynomials in a similar form as well

Semiclassical

Of course, at the level of field theory you get quite a bit of confluence between QFT and stat mech (temperature = imaginary time)

Yeah, generating functions are great.

logical123

2:40 AM

I have a couple text pdfs on QFT that I want to crack open, but I've been focusing on math and cs recently, trying to proficiency out of courses in grad school.

Semiclassical

nice.

My own favorite thing with generating functions is that, since they're power series, you can obtain the individual coefficients from the GF itself via complex contour integrals

BAYMAX

hey@Semiclassical

Semiclassical

In which case, you're able to do stuff like saddle-point approximation in order to get bounds/estimates.

hey @BAYMAX. How goes Poisson stuff?

BAYMAX

GF?

Semiclassical

generating function

Have you seen that before?

logical123

2:45 AM

that's really slick

I haven't

Semiclassical

It's pretty cool.

logical123

Do you have a resource?

I never took complex analysis, only self taught the basics of it

most simple of contours and such

Semiclassical

For generating functions, the obvious one is Wilf's generatingfunctionology. But for the asymptotic stuff, see Flajolet's book on Analytic Combinatorics. (both are available on their authors' websites in pdf form).

BAYMAX

ohk, yes I have seen it in probability theory

Semiclassical

And this really doesn't require a lot of complex analysis off the bat, just the Cauchy integral formula.

Akiva Weinberger

2:47 AM

I like the fact that it's not capitalized or anything

logical123

oh not bad, I will search now

Akiva Weinberger

Just 'generatingfunctionoloy'

Hell of a title

Semiclassical

Do you recall what $\displaystyle \int_C \dfrac{dz}{z^k}$ comes out to when $C$ winds around the origin?

BAYMAX

I hope this is generating function because it generates the moments like mean,median,variance..is it that thing ?

Semiclassical

It can be related to moments, yeah.

Akiva Weinberger

2:48 AM

Well, the coefficients generate the series you're thinking about

Semiclassical

In fact, that's basically what the partition function does in statistical physics (as @logical123 was alluding to above).

Akiva Weinberger

I don't know how to do median, but you can do partial sums by dividing by $1-x$, yeah?

And then means by... I dunno, integrating or something like that

Semiclassical

I don't think median fits quite well into this.

BAYMAX

Yes then I am on the same track

Akiva Weinberger

'cause that'd divide it by $n$

Topologicalife

2:49 AM

Hi.

Semiclassical

Typically, you'd have some discrete probability mass function $p_{n\geq0}$.

Topologicalife

Is $ A=\{m/(n+m) : n,m\in\mathbb{N}\}$ bounded?

Semiclassical

In which case the moment generating function is $\mathbb{E}[e^{-N t}]$, I think? Been a while.

Topologicalife

I think it is bounded below by 0 and $\sup A =\infty$

Akiva Weinberger

That's $1-\dfrac n{n+m}$, innit

'cause it's $\dfrac{(n+m)-n}{n+m}$

Topologicalife

2:51 AM

Mm.

Right.

Semiclassical

I'm guessing the natural numbers don't include zero here.

Akiva Weinberger

So I'd wager it's bounded above by $1$

Semiclassical

(Some conventions do, so it seems worth checking.)

Akiva Weinberger

You can see that just from the original $m/(n+m)$ thing, actually, since the numerator's smaller than the denominator

Topologicalife

I have a little question: when we write something like $\{\mbox{blabla} : n,m\in\mathbb{R}\}$... what are the rules to choose $n$ and $m$?

Semiclassical

2:53 AM

There's probably some geometric interpretation of that.

There aren't any rules as written. As it stands, n and m could be -any- real numbers.

Topologicalife

Ah, nice.

Semiclassical

You'd need to add more in order to introduce those.

Topologicalife

Yeah, the upper bound must be 1 (I was trying to see it in a geometric.mode)

Akiva Weinberger

You can only make it exactly $1$ if you allow $0$ to be in $\Bbb N$

(because that's what you get if you set $n=0$)

Parth Kohli

Help my dad: is there a way to solve normal distribution problems without integrals?

Maybe they teach something in stats class that gets rid of integrals. I don't know.

Semiclassical

2:56 AM

Depends on what you're after exactly.

If you want a specific probability, probably not.

Parth Kohli

just a normal distribution problem.

mean = 500, standard deviation = 200.

(i) find the percentage of the distribution above 580
(ii) find the percentage of the distribution between 390 and 610
(iii) find the percentage of the distribution between 350 and 410

Topologicalife

Yeah I see.

Parth Kohli

This is what he's trying to solve, and I highly doubt that it could be solved without integrating.

Semiclassical

Well, I should amend that. In principle you need integrals. But often a textbook on stats will have tables in the back.

Akiva Weinberger

Oh, actually, doesn't matter, does it @Topologicalife

Semiclassical

2:57 AM

And those tables will give specific probabilities.

Akiva Weinberger

because the supremum is gonna be $1$ either way

Semiclassical

lemme find an example online

Akiva Weinberger

because setting $n=1$ and taking $m\to\infty$ gives you $1$ also

Topologicalife

Yeah, I think you meant max.

skill patrol

Use a table.

Semiclassical

2:58 AM

Yeah, here: stat.ufl.edu/~athienit/Tables/Ztable.pdf

Akiva Weinberger

Yeah @Topologicalife

Parth Kohli

So it is integrals, then.

Topologicalife

then $\inf A= 0$

Semiclassical

Yeah. It's just that the numerical values have been tabulated.

I should also point out that you can't actually compute the values of the normal distribution -by hand-.

Akiva Weinberger

Yeah, I'd think so @Topologicalife

Semiclassical

2:59 AM

The integrals have to be done numerically.

Parth Kohli

Oh, yes.

Semiclassical

So either you use a numerical integration tool, or you use a table.

Akiva Weinberger

You know which one of us is the largest element in the chatroom right now?

@BAY$\max$

Semiclassical

If I remember right, the z-value is pretty simple to calculate.

Whereas I am middle of the road, being merely -semi-classical.

2

BAYMAX

largest element in chatroom @AkivaWeinberger

ha ha

:)

Akiva Weinberger

3:01 AM

Not calling you fat or anything! :)

Topologicalife

Thanks guys.

BAYMAX

I am a healthcare companion ..

ok

Akiva Weinberger

I understand that reference

BAYMAX

ya..

Evaluation homomorphism

Topologicalife

Say this to your tutor if you want to kill him stealthily: $1=\sqrt{1}=\sqrt{-1\cdot-1}=\sqrt{-1}\sqrt{-1}=i\cdot i=-1$. Since $1=-1$, $2=0$ by addition. Dividing by $2$ yeilds $1=0$. Since $2+2=4$, we can add this to both sides to get $2+2=5$ for sufficiently large enough values of $2$. QED

Akiva Weinberger

3:11 AM

round(2.4+2.4=4.8)

2+2=5

Semiclassical

heretics, all of you.

Akiva Weinberger

You actually can prove 2+2=5 in naïve set theory

since it's inconsistent 'cause of Russell's paradox

logical123

or just think about Goedel's theorems and call it quits

Akiva Weinberger

That's incompleteness, that's different

Semiclassical

I prefer things to make sense.

logical123

3:15 AM

i know incompleteness isn't the same as inconsistent

Akiva Weinberger

Or do you mean, like, think about the theorem and just give up on it all

Semiclassical

$\frac{1}{1-x-x^2}=1+x+2x^2+3x^3+5x^4+\cdots$ is more my style

Akiva Weinberger

Despair

logical123

exactly lol

the dispair

Akiva Weinberger

(Despair, dat pear)

Semiclassical

3:16 AM

(dat pear, dat pear)

Akiva Weinberger

I can see we've reached the hour of the night at which things stop making sense.

Semiclassical

Stick with generating functions and it'll all be fine.

Akiva Weinberger

@Semiclassical If you multiply out $(1-x-x^2)(1+x-x^2)$ and then substitute in $x\mapsto\sqrt x$ you get a recurrence relation for $F_{2n}$, fun fact

Semiclassical

Sounds right.

I mean, that's basically just doing $\frac{1}{2}\left[F(\sqrt{x})+F(-\sqrt{x})\right]$ to get the even part of $F(x)$ as a series in $x$.

Akiva Weinberger

Ye

It's symmetric, actually. You end up with the same recurrence relation backwards and forwards. Not sure if there's a particular reason.

Semiclassical

3:20 AM

Probably.

Akiva Weinberger

But, IIRC, it's like $F_{2(n+1)}-3F_{2(n)}+F_{2(n-1)}=0$

Rubertos

I remember there was a post listing facts that are true(false resp.) that you thought it was false(true resp.). I cannot find it now.. if someone knows, please let me know the link! Ty

Akiva Weinberger

@Rubertos There's a Facebook page like that

Rubertos

I saw it on SE though

Akiva Weinberger

@Rubertos This maybe? http://math.stackexchange.com/questions/820686/obvious-theorems-that-are-actual‌ly-false

Fawad

3:34 AM

Random question: for a 2×2 non-zero matrix A , $A^0$ is?

Rubertos

This is right! Thank you :)

Akiva Weinberger

@Fawad One would assume $I$...

Fawad

Ok

Rubertos

@Fawad if it is the matrix over a r"i"ng, then it is 'defined' as the identity matrix.. If it is just rng, that notation not make sense

Akiva Weinberger

I see no reason to assume it's over anything other than $\Bbb R$ or $\Bbb C$ unless explicitly stated otherwise...

Fawad

3:56 AM

What is $i^i$ value?

@Anonymous Pi Actually, $i^i$ has more than one value. In fact, there are an infinite number of values that range from $-\infty$ to $+\infty$. — Dr. MV Mar 9 '15 at 19:21

Semiclassical

Well, recall that $e^{i\theta}=\cos\theta+i\sin \theta$.

In particular, for $\theta=\pi/2$ you get $e^{i\pi/2}=i$.

What happens if you take both sides to the $i$ power? @Fawad

Simple

interesting, $i^i=e^{-\pi/2}$

Fawad

^

Semiclassical

Right.

On the other hand, you could just as well have taken $\theta=2\pi+\pi/2$.

In which case, you'd get a different value.

Fawad

$i^i=i^{-1}$

Akiva Weinberger

4:03 AM

The function $f(x)=x^\alpha$ is multivalued for all noninteger $\alpha$. These have no analytic extensions to the entire complex plane.

(Ignoring the stuff at $0$)

Semiclassical

More generally one has $i^i=e^{(i2\pi n+i\pi/2)i}=e^{-2\pi n}e^{-\pi/2}$.

Akiva Weinberger

That said, all possible values of $i^i$ are real, which is interesting

Semiclassical

Not sure how one gets a negative value out of that, though.

Fawad

$i^i$ is a transcendent number…this make me remember of DHMO

Akiva Weinberger

...Technically, they are all in the range from $-\infty$ to $+\infty$

Simple

4:07 AM

how do you get a negative value?

Semiclassical

Apparently there's a quote from de Morgan which references i^i being real:

"Imagine a person with a gift of ridicule. [He might say] First that a negative quantity has no logarithm; secondly that a negative quantity has no square root; thirdly that the first nonexistent is to the second as the circumference of a circle to the diameter."

Samuel Schlesinger

Hey there, I have been examining the set function, particularly its size, which is detailed here [github.com/SamuelSchlesinger/math-questions/blob/master/q1/…. Essentially I'm just trying to compute f(n, k, z) efficiently, or if it somehow makes it special, just f(n, k, 0) efficiently, as that was my original question anywho.

Fawad

@Simple who?

Simple

?

Akiva Weinberger

See the MO quote above

14 mins ago, by Fawad

@Anonymous Pi Actually, $i^i$ has more than one value. In fact, there are an infinite number of values that range from $-\infty$ to $+\infty$. — Dr. MV Mar 9 '15 at 19:21

Fawad

4:12 AM

Oh,yes,how can we show negative value?

Akiva Weinberger

Yeah I think he messed up

Fawad

If $i^i=i^{-1}$ is true than you may know $i^{-1}=-i$

Akiva Weinberger

Where did you get $i^i=i^{-1}$ from?

Fawad

By computing $i^i=e^{-\pi/2}=$\dfrac{1}{e^{\pi/2}}=\dfrac{1}{i}$

Akiva Weinberger

You're confusing $e^{\pi/2}$ with $e^{i\pi/2}$, I think

Fawad

4:17 AM

Oops,sorry

Akiva Weinberger

4:41 AM

@Semiclassical A video I think you'll like: Is This What Quantum Mechanics Looks Like?: Silicone oil droplets provide a physical realization of pilot wave theories by Veritasium

It just made me think of you since I know you're into quantum stuff

(Apologies for the clickbaity title)

今天春天

6:08 AM

Hi there, I know about the CLT and what it does. Now I want to use it to proove: $lim_{n\rightarrow \inf} S_n / n^p = 0$. $S_n$ is a sequence of independent RV with E(X) = 0 and Var(X) = C, C < \inf. Any hints on this?

Simple

consider $S_n=\sum_{k=1}^nX_k$ where $X_k$ are iid with mean zero and variance $c$

今天春天

okay...

Then I know that the sum of the sequences of the RV will approximate a normal distribution N(0,C) as n grows

user400188

7:03 AM

I have a question about the following quote:
"In mathematics an existentially quantified variable may represent multiple values, but only one at a time. Existential quantification is the disjunction of many instances of an equation. In each equation there is one value for the variable."
https://en.wikipedia.org/wiki/Curry's_paradox

My current understanding of the existential quantifier $\exists$ was that is was the disjunction of many instances of the same formula; like they have stated in the quote, however I was unaware that only one instance could have a value at a time.

BAYMAX

8:10 AM

Hi , any idea where I can ask for advice on research ideas on a particular mathematical subject ?

2

is it academia SE?

But it is of mathematical nature so asking here may be good I think.

skill patrol

Math research is the specialty of MathOverflow.

BAYMAX

Can I ask there for advice too,I thought it is a place for advanced research mathematics?

like only mathematical questions

skill patrol