Mathematics - 2016-12-09 (page 3 of 7)

Akiva Weinberger

3:00 AM

All points of that form

In any case, a geodesic on this.

The helix is a limiting case in the same way a line is the limiting case of a circle

G. Bergeron

you mean infinite radie

Akiva Weinberger

Yeah. Leave the last two coordinates alone, shift and grow the circle defined by the first two coordinates

in words

G. Bergeron

but a circle is not the same kind of limiting case of a helix

What I meant by not a regular torus is that it has the same topology but not the same geometry

Akiva Weinberger

Ah.

Right, the usual torus in R^3 does not have constant zero Gaussian curvature

G. Bergeron

only cylinder

ok need to go...

Akiva Weinberger

3:05 AM

Me too. Bye!

Adeek

hi @TedShifrin

hi everybody

GFauxPas

math.stackexchange.com/questions/2047334/… k cleaned up my question

TheGreatDuck

3:22 AM

Doesn't Euclidean Geometry work upon the surface of a torus? It's just a thought...

Jessy Cat

1

Q: Fourier Transform: If $H(\alpha)=e^{-t\alpha^{2}}F(\alpha)$, what is $h(x)$?

I am trying to solve the following problem from my textbook: If $f(x) \in L^{2}(-\infty, \infty)$ has Fourier transform $F(\alpha)$, then what is the function of $x$ whose transform is the function $H(\alpha)=e^{-t\alpha ^{2}}F(\alpha)$? ($t$ is a positive real constant). The answer giv...

I'm having a horrible time with Fourier transforms. I hate that I have to teach myself pretty much everything in this class...

GFauxPas

@JessyCat I'm using a book by Vretblad and it's very accessible

Jessy Cat

Unfortunately, I doubt I'll get a chance to get my hands on a copy before my final exam.

GFauxPas

it's for free online as a pdf

Jessy Cat

oo.

GFauxPas

3:32 AM

iujfk.files.wordpress.com/2013/04/…

Also, are you actually a cat

Jessy Cat

How many cats usually come in here and strike up a conversation?

GFauxPas

at least 2 others

Jessy Cat

You might be able to help with that question I posted a link to up there. If it makes you more amenable to do it, I'll be a cat. meows

Well, there goes the neighborhood.

TheGreatDuck

@GFauxPas he is obviously not a cat. hes a human.

GFauxPas

I'm still working on Laplace transforms

TheGreatDuck

3:37 AM

for what class?

Jessy Cat

@TheGreatDuck and you are obviously not a duck. Although if you look like a duck and quack like a duck...

GFauxPas

self study

TheGreatDuck

@GFauxPas what subject

Jessy Cat

<-- taking a PDE course with the flakiest professor ever.

TheGreatDuck

@JessyCat Are you implying I am a quack, cause to some degree I am.

Jessy Cat

3:38 AM

@TheGreatDuck to be honest, the thought never crossed my mind.

GFauxPas

@TheGreatDuck Laplace transforms! I found them interesting when studying ODEs but my ODE class only covered them in passing]

Jessy Cat

But you know, cats eat birds...just sayin'.

GFauxPas

so im going further

Jessy Cat

Also, not a he.

As for how human I am, that's a matter for debate.

TheGreatDuck

@GFauxPas know any obscure transforms?

GFauxPas

3:40 AM

Obscure Laplace transforms, or obscure transforms other than Laplace?

The $\mathcal Z$-transform is the equivalent of the $\mathcal L$-transform for sequences

TheGreatDuck

obscure laplace transforms

any regarding floor?

GFauxPas

The Laplace of $\ln$ is fun to figure out

TheGreatDuck

or any interesting/exotic functions that someone in a regular ODE diff eq class would be aware of?

hmm

GFauxPas

you'll need proofwiki.org/wiki/… if you want to figure it out yourself

TheGreatDuck

now that I think about it, we never did cover it.

GFauxPas

3:43 AM

Because it's a weird one

TheGreatDuck

oooh

I remember seeing that one

GFauxPas

doesn't give you a rational function or a power function

TheGreatDuck

it gives you gamma

which is the real value extension of factorial

(might also be a product integral)

GFauxPas

$\mathcal L \{ t^q \}$ gives you gamma of something over something, where $\operatorname{Re} q > -1$.

TheGreatDuck

which would make sense

int(ln(x)) = ln(prod(x))

where int and prod are the integral and product integral

GFauxPas

3:45 AM

Oh I don't know anything about product integrals

TheGreatDuck

do you know of product integral?

ooh

GFauxPas

but $\int \ln x \, \mathrm dx = x \ln x - x + C$

TheGreatDuck

you've seen it believe it or not. they just do not tell you.

GFauxPas

Which is a fun thing to prove

write it as $1 \cdot \ln x$ and integrate by parts

TheGreatDuck

remember diff eq of the form: y' + f(x)y = g(x)

GFauxPas

3:46 AM

I don't remember how to do any ODE's except for linear ones

TheGreatDuck

and how you do that weird exponential logarithm integral of f to find the factor to divide by?

O.O

well it is linear

GFauxPas

like $\dddot{y} - 5 \dot{y} + y - \cos t + \sqrt{t} = 0$

those kinds

Semiclassical

hi chat

GFauxPas

I mean, I could reteach myself the others, I'd just need to invest time

hi @Semiclassical

TheGreatDuck

$user image$

tutorial.math.lamar.edu/Classes/DE/Linear.aspx

Semiclassical

3:49 AM

The joy of integrating factors

TheGreatDuck

@GFauxPas the right hand side is the product integral of p(t)

GFauxPas

Why would you need to use product integrals for that

oh yeah but I mean

TheGreatDuck

shrugs

GFauxPas

I don't feel like dealing with "Riemann products"

or however youd define it

Semiclassical

I tend to prefer doing a general solution + a particular solution

when doing inhomogenous ODEs

TheGreatDuck

3:50 AM

someone probably has a good reasoning why the product integral is the integration factor, but I cannot say.

Semiclassical

@TheGreatDuck You can derive it without -too- much trouble

GFauxPas

That's the semi-classical way to do it @Semiclassical

Semiclassical

hah

TheGreatDuck

@Semiclassical I mean just a really simple one-sentence reasoning why the product integral yields it. Atm, it's only good as a memory-tool for me.

GFauxPas

Oh you mean, you're asking the motivation behind it?

TheGreatDuck

3:52 AM

I'm not asking anything.

Semiclassical

well, you've got the ODE $y'+f(x)y=g(x)$

GFauxPas

Oh

then I'm not answering anything.

TheGreatDuck

you asked why. I just said, Idk.

GFauxPas

well if ydk idk either

Sophie

product integral? Is that the same as integration by parts?

TheGreatDuck

3:52 AM

and said it's probably obvious, but idk.

@Sophie addition is to integral as multiplication is to product integral.

GFauxPas

No @sophie it's a generalization of the idea of a definite integral to a limit of a "Riemann product" or something like that

TheGreatDuck

^^^

GFauxPas

I think

TheGreatDuck

except it's to the power of dx rather than multiplied by dx

GFauxPas

except that you have to figure out what that means

TheGreatDuck

3:53 AM

@GFauxPas what do you mean by that?

I didn't make it up. I read it on wikipedia for two seconds once and remember it cause I got a good memory.

GFauxPas

"mutliplying by $\mathrm d x$ can't be defined the same way as the multiplication of numbers

you need to figure out how to define it

Kaj Hansen

@Null, not, uh, nicotine if that's what you're referring to

Semiclassical

well, you want to arrange things so that the above ODE is equivalent to $d(W(x)y(x))=g(x)W(x)\,dx$

GFauxPas

I'm not saying its wrong

I'm just saying it's not obvious what it means

TheGreatDuck

@GFauxPas dx comes from delta_x. I'd imagine it's just a restructuring of the riemann sum....

Semiclassical

3:55 AM

Typically, multiplying by $dx$ is just a notational shorthand

TheGreatDuck

raise it to the power in the limit calculation rather than multiplying.

@Semiclassical WRONG. it implies each element is scaled into an infinitely tiny contribution to the sum. Like an additive accumulation.

GFauxPas

so if $\lim_{\Vert \Delta \Vert}\sum f(x_i) \Delta_i = \int f(x) \, \mathrm dx$,

Semiclassical

e.g. $f'(x)=\frac{df}{dx}\implies df=f'(x)\,dx$ really means $\int f'(x)\,dx=\int df.$

GFauxPas

and even that actually needs additional details for it to make sense

Semiclassical

If you have a specific meaning you want to attach to $dx$, go right ahead.

TheGreatDuck

3:57 AM

@Semiclassical that is the meaning of dx.

GFauxPas

right, what S.C. said. You have to figure out a meaning, and if it makes sense, prove it makes sense.

TheGreatDuck

why do I need to prove anything? It's the definition.

XD

mikeonly

@KajHansen Are you a bit into fundamental groups and algebraic topology?

Semiclassical

Not if you're talking differential 1-forms, it's not. @TheGreatDuck

GFauxPas

Because you want the definition to be consistent with the results you expect

TheGreatDuck

3:58 AM

@Semiclassical We're talking about regular integrals on real numbers. Not abstract math on dogs and cats. :p

GFauxPas

If I say "I define the derivative of a product of functions to be the product of their derivatives", that definition wont be consistent with results that make sense

Kaj Hansen

Don't know anything about that @mikeonly; wish I did

GFauxPas

integrals arent on numbers, they're on functions, or on intervals

Semiclassical

I really, really wish Ted was here right now. :/

Adeek

hey @KajHansen just want to verify here we consider the $e_i$ as coloumn vector and think of the permutation as coloumns ?

Kaj Hansen

3:59 AM

I know a tiny bit of homological algebra. Things like the five lemma. I think that comes up some in that

Adeek

mikeonly

@KajHansen Interesting stuff. Thank you anyway. :)

TheGreatDuck

@GFauxPas I never said that though. I just said the product integral uses the riemann product where the individual term is raised to the power of delta_x...

Akiva Weinberger

What's happening? Someone mentioned product integrals?

TheGreatDuck

how is that confusing to you?

GFauxPas

4:00 AM

@TheGreatDuck the riemann sum definition of the integral is hardly trivial

TheGreatDuck

@GFauxPas yeah, functions of real numbers.

Kaj Hansen

They have them as rows there @Adeek, but columns works fine too

Akiva Weinberger

This thing? https://en.wikipedia.org/wiki/Product_integral

GFauxPas

for example, if the value of the integral depended on the partition of the interval used in the riemann sum, that would be very bad

Adeek

yeah I see.

TheGreatDuck

4:00 AM

@GFauxPas um... it's a simple equation. do you mean to say evaluating it is hard?

GFauxPas

Because riemann sums are equivalence classes of sums

TheGreatDuck

facepalm

GFauxPas

I'm saying the hard work has been done for you

but don't take it for granted

TheGreatDuck

that's an issue of integration theory

GFauxPas

every time you use the fundamental theorem of calculus you're using the combined efforts of thousands of man hours of mathematicians making sure that you're allowed to do what you're doing

TheGreatDuck

4:01 AM

"I'm saying the hard work has been done for you
but don't take it for granted" what work? All I did was tell you the product integral was used to extend factorial and that you use in diff. eq. without realizing it. um... what is the point of this?

@GFauxPas and you're telling me this... why?

GFauxPas

Because you called the idea simple

Semiclassical

Well, for one, because I've never heard of the product integral and I've been able to do ODEs via integrating factors just fine.

GFauxPas

and I didnt like that

Kaj Hansen

Last night, I noticed that a post of mine was 1 upvote away from winning me 2 medals. I found an excuse to edit the post (I did legitimately improve it though) knowing that this would bump my post in the "active threads" tab and likely get me the upvote I wanted. Is this abusing the system / frowned upon around here do you (guys) think?

TheGreatDuck

The riemann sum equation itself is simple, I mean.

GFauxPas

4:02 AM

and btw the factorial has an infinite number of extensions, the gamma function is not unique

arctic tern

@KajHansen meh

TheGreatDuck

and it's simple to see that the riemann sum gives area (or is intended to give area).

at least, in hindsight.

not saying it was not hard to create

Kaj Hansen

@arctictern, I figured that might be the response. But I felt really dirty doing it

TheGreatDuck

just that it's not hard to understand or learn

arctic tern

I've done it a bunch in previous incarnations

TheGreatDuck

4:04 AM

heck, multiplication giving area is a miracle

GFauxPas

it's "not hard to understand or learn" if you're willing to take certain results for granted

TheGreatDuck

whoever figured that out deserves a medal.

Semiclassical

It's a heck of a lot easier to learn calculus than to invent it.

2

TheGreatDuck

@GFauxPas um... are you implying that you have to learn the entire derivation of calculus to learn it, lol?

GFauxPas

no, what I'm saying is

Kaj Hansen

4:04 AM

haha, yeah

Semiclassical

I find this conversation confusing, though.

TheGreatDuck

^^^

GFauxPas

that if your professor says something like the value of the integral doesn't depend on which riemann sum you use among the infinite number of possible rieman sums you can choose

you should think about it rather than accept it at face value

Semiclassical

The product integral seems like something which is interesting but not really necessary.

TheGreatDuck

^^

GFauxPas

4:06 AM

when they draw a picture of you cutting up the area into rectangles

arctic tern

@Semiclassical Here's an interesting question. Suppose you have a differentiable path on a sphere. Between two sufficiently close points on the sphere, there is a "most efficient" rotation of the sphere that sends the first to the second. If we partition the path into n points and multiply the small rotations together, then take the limit, what is the result? The result is actually the product integral of elements of the rotation group's lie algebra (so, infinitessimal rotations).

GFauxPas

if they draw the rectangles with equal with

know that they don't have to be

TheGreatDuck

my point was merely to make a comment about it existing and this somehow turned into a massive debate about riemann sums

Semiclassical

@arctictern Hmm, fair.

GFauxPas

I'm not debating

I'm philosophising

TheGreatDuck

4:06 AM

yeah, you are.

facepalm

Semiclassical

Yeah, that's kinda claptrap.

Philosophizing is argument by its very nature.

GFauxPas

That's arguable

see what I did there

Semiclassical

Yes, you demonstrated my point.

GFauxPas

yes

TheGreatDuck

@GFauxPas "that if your professor says something like the value of the integral doesn't depend on which riemann sum you use among the infinite number of possible rieman sums you can choose" if you're professor says that raising e to the power of a sum of natural logarithms is the same as a product are you going to argue with them day in and day out or accept it as something trivial to understand if you just pull the logarithms out?

GFauxPas

4:08 AM

Duck I thought your avi was the yeti from that skiing game on old windows

Sophie

is there a way to obtain a tight upper bound on $\sum_{h=l+1}^\infty h!^{-n}$? I tried Stirling's approximation but got a very weird sum

arctic tern

@GFauxPas lol

TheGreatDuck

@GFauxPas umm.... ooook.

i dont even know what yer talking about.

arctic tern

not something I ever thought I'd remember again

@Sophie like, how tight?

Semiclassical

Of course, all of this amounts to just the old standard that $\displaystyle e^x=\lim_{n\to\infty}(1+x/n)^n$

arctic tern

4:10 AM

right

GFauxPas

SkiFree

SkiFree is a computer game created by Chris Pirih and released with the Microsoft Entertainment Pack in 1991. It is a simple game in which the player controls a skier avoiding obstacles on a mountain slope. It is often remembered for its Abominable Snow Monster, which pursues the player after they finish a full run. == Gameplay == The player controls a skier across a white background representing snow on a mountainside. The object of the game is to ski down an endless slope and avoid the obstacles. The game includes three modes: slalom, free style, and tree slalom. In slalom, players ski between...

Semiclassical

Including the thing arctictern was saying (since the exponential map is what relates Lie algebras to Lie groups)

Kaj Hansen

.https://youtu.be/yYDmaexVHic
Pretty dope track right here

Sophie

@arctictern well considering I don't have almost anything, anything asymptotic should do

Semiclassical

@Sophie That is one painful looking sum.

Akiva Weinberger

4:11 AM

I've managed to confuse myself on the non-simply connectedness of the rotation of the sphere

Semiclassical

What's the source?

Akiva Weinberger

Maybe I'm too tired, since this is something I've already understood in the past

Kaj Hansen

@AkivaWeinberger, is the topology on the set of matrices that correspond to rotations?

Akiva Weinberger

Yeah

Semiclassical

Eugh, that reminds me of Euler angles. Don't remind me of Euler angles.

Kaj Hansen

4:12 AM

How do we define matrix topology again?

Akiva Weinberger

@KajHansen The interesting thing is that one full rotation can't be contracted but two full rotations can be

arctic tern

@KajHansen Mn(R) is R^(n^2)

yes, the plate trick

Kaj Hansen

Defined by what?

Sophie

@Semiclassical for the sum? I got it from a MSE question. I'm trying to prove $\sum_{h=0}^\infty h!^{-n}$ is transcendental

Semiclassical

That seems difficult.

Sophie

4:13 AM

I'm going to bury it

arctic tern

@KajHansen well, R^(n^2) is a coordinate vector space, so use the dot product. (same as the frobenius norm on Mn(R))

Semiclassical

(well, except for $n=1$.)

GFauxPas

or $n=0$

TheGreatDuck

@Semiclassical if I give you a coordinate basis can you find a set of euler angles that produce it?

GFauxPas

if you want the sum to stop making sense

Semiclassical

4:13 AM

@GFauxPas Eh, not transcendental in that case

TheGreatDuck

in general?

arctic tern

@Semiclassical who are you talking to? exp(-1) is transcendental

Sophie

infinity is transcendental?

GFauxPas

infinity isn't a number

TheGreatDuck

sorry but it's for something I've been stuck on for a while.

GFauxPas

4:14 AM

well

Akiva Weinberger

Ohh I see my mistake

GFauxPas

depends

Semiclassical

Couldn't help you @TheGreatDuck

Akiva Weinberger

(Regarding the plate trick thing)

TheGreatDuck

and I must use a coordinate basis and I must use angular rotations

Akiva Weinberger

4:14 AM

Well, goodnight

TheGreatDuck

darn

GFauxPas

laila tov

Semiclassical

And wouldn't, frankly. I haven't used euler angles in years and don't want to do them again.

TheGreatDuck

...

Akiva Weinberger

@GFauxPas Laila tov ^_^

arctic tern

4:15 AM

computing and visualizing homotopy groups of rotation groups don't really involve euler angles

TheGreatDuck

that's harsh.

Semiclassical

Harsh on Euler angles. Nothing to do with you.

TheGreatDuck

@arctictern here's the problem really. I have a coordinate basis that represents where I'm moving in space. And I can only rotate some degrees about an axis of rotation. I need to rotate the player to match the coordinate basis

hence I need the euler angles

Semiclassical

Speaking of exponentials.

The lack of response to this bounty I gave is disappointing: math.stackexchange.com/q/2036252/137524

...and now I notice the comment to the main answer. Welp

arctic tern

heh

Sophie

4:18 AM

do people usually browse the "featured" tab? I think it's too discouraging to not even understand most questions

arctic tern

sometimes

usually understand a few

Semiclassical

I do occasionally.

TheGreatDuck

i think they are over-hyped

:p

Kaj Hansen

lol, I never click on it @Sophie. But I would if I were more knowledgeable. I just know that basically everything I know, lots of other people on here do too.

TheGreatDuck

@KajHansen what do you know?

whatwhatwhat

4:21 AM

hello everyone

Semiclassical

Hmm, I like this one: math.stackexchange.com/q/2040726/137524

Sophie

answer mine please

TheGreatDuck

@Sophie not right now I'm busy.

Semiclassical

The sum one?

I don't have anything clever to say on that one. Stirling still seems sensible.

Kaj Hansen

A broad smattering of undergrad topics @TheGreatDuck, but none in any great depth. Real/complex analysis, combinatorics, linear algebra, abstract algebra (groups/rings/fields/galois), differential geometry, general topology, Ramsey theory, elementary number theory, cryptography, dynamical systems

Semiclassical

4:25 AM

I'd be more interested if I knew a clever way to represent $\frac{1}{h!^n}$. When $n=1$ that amounts to the reciprocal gamma function, and I know fun stuff for that.

But otherwise I dunno.

Sophie

@Semiclassical I I mean another one. I want solutions to $1+2^2+3^2\dotsb n^2=a^b$ where $a,b,n$ are integers. I have reduced that to solving equations like $x^n-2y^n=1$

whatwhatwhat

How do I make the Latex generate in the chat?

Semiclassical

Oh, Pell's equation stuff? No clue on that.

I haven't done any of that.

@whatwhatwhat See the LaTeX in chat link in the room description.

Sophie

well that's a Pell equation when $n=2$

Semiclassical

If it's worse than Pell's equation, then I'm definitely not the person to ask

whatwhatwhat

4:27 AM

@Semiclassical thanks

Semiclassical

@AkivaWeinberger Oh. I still have that hydra game running from earlier.

It's up to 28500 heads cut off :)

TheGreatDuck

@KajHansen you know way more than I know.

Kaj Hansen

Don't worry GreatDuck. I've forgotten most of it ;)

Sophie

by Stirling's I get $\displaystyle\sum_{h=l+1}^\infty \exp\left(nh\ln(h)-nh-\frac{n}{2}\ln(h)\right)$ so if I could get an upper bound for that I'm done. To prove transcendentality I need that to be like $O(l^{-2})$ or somthing

whatwhatwhat

I have a question: somewhere, my understanding of chain rule/partial derivatives off. I have a function $f(y(x)+\eta (x) \alpha,y(x)'+\eta (x)',x)$ and I need to take the derivative w.r.t $\alpha$. How do I do this?

TheGreatDuck

4:30 AM

@KajHansen um... it was a compliment. You still know way more than me.

Kaj Hansen

haha, I know it was....I was trying to be modest. Although I truly have forgotten a lot

arctic tern

@whatwhatwhat Is there something you don't understand about my answer and the comment? For example, if $f(u,v,w)=u^2-vw$ would you be able to compute the derivative of $f(p+q\alpha,r+s\alpha^2,t)$ with respect to $\alpha$?

Sophie

Kaj, you're a mathematician?

Kaj Hansen

If you count someone who's completed undergrad a "mathematician" @Sophie

I don't have any original work to speak of though

TheGreatDuck

@KajHansen that's surprising.

whatwhatwhat

4:32 AM

@arctictern yes, but in the past I've been told (by the site and by people) that further comments should go in the chat. So voila, her eI am.

arctic tern

@whatwhatwhat We're not to the point of "further comments" but whatevs

Kaj Hansen

I pretty much always ignore the "go to chat" warning on MSE posts

TheGreatDuck

^^^

Kaj Hansen

Usually the convo is almost done anyways

TheGreatDuck

usually I'm trying to get the person to stop replying to me and being a jerk.

i gotta go

cya

Kaj Hansen

4:35 AM

Have a good one

whatwhatwhat

@arctictern Ok so the derivative would be $\frac{\partial f}{\partial \alpha}=2(p+q \alpha )(q) - (2s\alpha)$, right?

arctic tern

forgot t at the end of -2sa but yeah

whatwhatwhat

Is this equivalent to (using your notation): $\frac{\partial f}{\partial \alpha}=f^{(1,0,0)}q-f^{(0,1,0)}$ ?

Oh right

Semiclassical

@sophie Found this question on the quartic case: math.stackexchange.com/q/1981082/137524

arctic tern

It's equivalent to $$\frac{\partial u}{\partial \alpha}\frac{\partial f}{\partial u}+\frac{\partial v}{\partial \alpha}\frac{\partial f}{\partial v}+\frac{\partial w}{\partial \alpha}\frac{\partial f}{\partial w} $$ or with tuples, $$ u'f^{(1,0,0)}+v'f^{(0,1,0)}+w'f^{(0,0,1)}$$

Semiclassical

4:39 AM

The answer references a theorem of Thue and a Wikipedia page

arctic tern

So do you understand the formula $$\frac{df}{d\alpha}=\frac{\partial (y+\eta(x)\alpha)}{\partial \alpha}f^{(1,0,0)}+\frac{\partial(y'+\eta'(x)\alpha)}{\partial \alpha}f^{(0,1,0)}+\frac{\partial x}{\partial\alpha}f^{(0,0,1)}$$ I wrote in my comment on main?

Sophie

@Semiclassical I proved there are finitely many for any fixed $n>2$

Semiclassical

Ah. That matches the theorem they mention.

But finitely many doesn't say how many, of course :/

Sophie

and it doesn't help that we don't know anything about the continued fraction of $2^{1/n}$ for any $n>2$

Semiclassical

Huh. That's kinda weird.

Sophie

4:43 AM

we know literally nothing. I read some posts at MO saying even at research level we know nothing

Semiclassical

Not entirely surprising, though. Continued fractions are nice for representing solutions to quadratics. No reason they should be nice for solutions to $x^n=2$.

Sophie

there's no a priori reason why continued fractions should be nice for quadratic irrationals

Semiclassical

uh

real quadratic irrationals have periodic continued fractions.

that's pretty nice.

Sophie

that's very nice. It allows you to find solutions to Pell equations quite easily for example $838721786045180184649^2-397\times42094239791738433660^2=1$ would be very hard to find if you didn't know that

Adeek

hey kaj you know manifold theory ?

@KajHansen

Semiclassical

4:46 AM

Do you have a reference to those MO posts? Now I'm curious.

Sophie

18

Q: Is there any pattern to the continued fraction of $\sqrt[3]{2}$?

Is there any pattern to the continued fraction of $\sqrt[3]{2}$ ? Wolfram Alpha returns for cube root of 2: $\sqrt[3]{2}=$ [1; 3, 1, 5, 1, 1, 4, 1, 1, 8, 1, 14, 1, 10, 2, 1, 4, 12, 2, 3, 2, 1, 3, 4, 1, 1, 2, 14, 3, 12, 1, 15, 3, 1, 4, 534, 1, 1, 5, 1, 1, 121, 1, 2, 2, 4, 10, 3, 2, 2, 41, 1...

nt.number-theory continued-fractions

Semiclassical

The lack of response to your problem, then, might just be that it's very very hard.

Sophie

that heuristic I wrote in the post has convinced me that a solution is very, very unlikely

Semiclassical

hmm, Wikipedia has the following in their page on Hermite's problem (which is about continued fractions of cubic irrationals)

>In 2015, for the first time, a periodic representation for any cubic irrational has been provided by means of ternary continued fractions, i.e., the problem of writing cubic irrationals as a periodic sequence of rational or integer numbers has been solved. However, the periodic representation does not derive from an algorithm defined over all real numbers and it is derived only starting from the knowledge of the minimal polynomial of the cubic irrational.

Sophie

wait what. $x^{-n\left(x+\frac{1}{2}\right)}e^{nx}$ has a fixed point under variation of $n$

Semiclassical

4:55 AM

That'll diverge to infinity or go to zero depending on whether $x^{x+1/2}$ is bigger than $e^x$

Sophie

I'm thinking of that as a function of x and n is a parameter

Semiclassical

The crossover seems to be $x\approx 2.26924$.

whatwhatwhat

@arctictern ok this makes sense. Yes, I think I understand your answer now. Sorry for being so dense.

Sophie

is the point of the solution of $\ln(h)(h+1/2)=h$

Semiclassical

right.

Sophie

4:58 AM

can we prove that $x^{-n\left(x+\frac{1}{2}\right)}e^{nx}x^{\alpha}$ goes to 0 as $x\to\infty$ for any $\alpha>0$?

Semiclassical

With $n$ fixed?

Sophie

yes and positive

Semiclassical

I think so, yes?

x^x grows faster than e^x, seems the essential point.

Equivalently, you want to show that $nx+\alpha\log x-n(x+1/2)\log x\to-\infty$ as $x\to\infty$ when $\alpha>0$

Sophie

I got it with a transformation like $x\to ex$

Semiclassical

Ah.

The expression I just gave seems to be dominated by $-n x \log x$, so presumably one can show that $x^{-n(x+1/2)}e^{nx}x^\alpha$ is asymptotic to $x^{-n x}$ as $x\to\infty$.

...maybe

arctic tern

5:16 AM

For real inner product spaces $V$, we can identify $\Lambda^2(V)\cong\mathfrak{so}(V)$, and if $V$ is oriented with $\dim V=4$ there are canonical subspaces of $\Lambda^2(V)$ of left and right isoclinic rotations (up to scaling). Essentially, $a\wedge b+c\wedge d$ is left isoclinic if the orientation from the ordered basis $\{a,b,c,d\}$ is correct. Apparently if $P$ is tracefree self-adjoint and $x\in \Lambda^2V$ is isoclinic then $Px$ is also isoclinic but opposite-handed.

$P(u\wedge v)$ being $(Pu)\wedge v+u\wedge (Pv)$

Dunno a good way to see this. Maybe use fact self-adjoint implies linear combination of orthogonal projections onto orthogonal subspaces.

Mike Miller

5:39 AM

best to think of those as the eigenspaces of the Hodge star operator, at which point you're asking to prove the identity $*P+P*=0$, which seems like it should follow straightforwardly from def'ns

Wildcard

Random question: Does anyone know if the Pacific Tech Graphing Calculator is actively developed?

pacifict.com

I have it—it's *excellent*—but the copy I have hasn't been updated for a retina display. Makes me wonder.

whatwhatwhat

Is there a time when one should kinda just reconsider a math-heavy field as a career? I'm 24 and still working on a bachelor's degree (in physics; reasons for the delay involve family, legal, and financial troubles). I now see what they say about how a person's mathematical ability must be well-developed around the age of 23. If not, there's not much hope for developing it going forward. How true is this statement?

I'm might be unlucky in asking this question here because it's highly likely that most people here developed their math skills early on and have built careers out of these abilities.

Adeek

5:58 AM

hey @arctictern

I was wondering why here is $dz_i(\partial/ (\partial z_j) = \delta_{ij}$?

Kaj Hansen

Hey I'm back

Adeek

hey kaj

I need your help understanding this

I don't understand how do we get this last two lines

@KajHansen

Hatshepsut

I'm trying to take the expectation of a squared standard normal distribution. Can I narrate my progress and someone'll let me know if I'm going wrong?

Kaj Hansen

I'm afraid I can't really help

Adeek

oh ok @KajHansen

DHMO

6:26 AM

@Hatshepsut go ahead

Hatshepsut

6:42 AM

@DHMO Ok here goes

$X ~ N(0,1). EX^2 = \int_-\infty^\infty f_X(x)dx$

f_X(x) = 1/\sqrt(2\pi) e^(-x^2 / 2)

DHMO

well, E(x^2) should be int x^2 f(x) dx

Hatshepsut

right

E X^2 = 1/\sqrt(2\pi) \int x^2 e^(-x^2 / 2)

proceed with integration by parts

u = x^2 , du = 2xdx , dv = e^(-x^2 / 2)

i dont know how to take that antiderivative, but its the main part of the standard normal pdf, so it's gonna end up having an integral from -inf to +inf of $\sqrt{2\pi}$

I mean the antiderivative of dv

I'm not sure if it's legit to use at this point because i'm just supposed to be taking an antiderivative not evaluating it at limits

@DHMO How should I be thinking about that

DHMO

7:39 AM

@Hatshepsut try v = e^(-x^2/2)

Hatshepsut

7:53 AM

@DHMO So if I say dv = x e^(-x^2/2)dx, v = -e^(-x^2/2)

Then uv goes to 0 in the limit

so we just have 1/\sqrt 2pi \int e^(-x^2/2) evaluated from -inf to inf

which is the integral of the standard normal pdf

X

so its 1

which is the answer

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$Mathematics$ Mathematics