Mathematics - 2021-02-10 (page 1 of 2)

anakhro

00:04

@TedShifrin I shifted the goal posts on this question.

I realized it's probably because of the poles of the function in question, and then I get that $\sum\text{zeros} = \sum\text{poles}$.

Mod the lattice, that is.

So now I am looking at poles. Which is probably something like you suggested.

user2103480

@TedShifrin It's probably also due to your specialization. The complex geometry/geometric analysis prof at my old uni held lectures on: ODE & PDE, riemann surfaces, functional analysis, (linear) algebra, all the intro analysis courses, morse theory, analysis on manifolds, and index theory. And I'm sure all of this would have been no problem to you as well

I don't know if every mathematician's specialization allows for such a broad range

anakhro

Well I don't think any professor of pure mathematics is not qualified to teach undergrad level analysis/algebra/ODEs or linear algebra.

user2103480

Probably true. Although I think in bigger departments, this is more compartmentalized

anakhro

In larger departments they prefer to have an algebraist teach algebra, etc. They also like to give some of these undergrad courses to newly hired professors (in my experience).

user2103480

And why would the usual analyst care about teaching group theory and whatnot

...and why would anyone care about teaching ODE

anakhro

00:16

Because they are easy to teach undergrad courses and are much more intriguing than teaching calc 1/2

BigSocks

So say $A$ a ring of dimension $1$. Let $J$ be an ideal of $A$ that can be factored as a product of maximal ideals, $J = P_1^{a_1} \dotsc P_s^{a_s}$ and let $M$ be any maximal ideal of $A$. Apparently $J_M = (MA_M)^{a_i}$ if $M = P_i$ for some $i$, and $J_M = A_M$ if M \neq P_i$ for all $i = 1,...,s$.

The second part is easy enough, but how do you show the first part?

Thorgott

@user2103480 t h i s

funnily enough, here usually an algebraist teaches ODE

BigSocks

$J_M$ would be $J_{P_i}$ so it would be $S^{-1}J$ for $S = A \setminus P_i$ and $(MA_M)^{a_i} = (P_i A_{P_i})^{a_i} = (S^{-1} P_i)^{a_i}$ for $S$ the same as before.

anakhro

@Thorgott I assumed that this would be more along the lines of an ODE analysis course or like following Arnold's book. Not engineer-tier odes.

Thorgott

I wouldn't know the difference

anakhro

00:27

Unfortunate. My school, too, skimped out on budget for a good ODE course.

The next-door school didn't though.

Thorgott

we did ODE for like a solid 4 weeks

anakhro

Poor Thorgott. :(

user2103480

@anakhro there's only analysis II (differentiability in R^n, implicit function theorem, point set topology and such) in germany, which is way more interesting than a usual ODE course (imo)

uh and of course analysis one

but not the calculus sequence

Thorgott

Î'm fine, I learned Picard-Lindelöf in that course

anakhro

@user2103480 You should pick up Arnold's ODE book one of these days. See what you missed out.

Thorgott

00:37

that's all the ODE I need

user2103480

@anakhro I did have an ODE course which fueled my scorn

anakhro

Yes, and apparently it wasn't very good.

So I think it's only reasonable that you be open to there existing better courses than that which you took.

user2103480

I think I saw enough. I find SDEs and (S)PDEs satisfying enough

anakhro

I suppose ignorance is bliss.

If you knew for a fact that ODEs were interesting, you would have one less thing to complain about.

user2103480

One can find interesting things in any area. Claiming ignorance while 3 out of 4 of my current courses are about extensions of those is a bit sensitive too, isn't it?

anakhro

00:48

It's not ignorance of ODEs that I am suggesting, but ignorance of the existence of better ODE courses than that which you took.

user2103480

That one might exist, but I've had exposure to ODEs in a subsequent numerical mathematics course and in probability courses. Still can't find that much aesthetic appreciation, don't think a better ODE course would have changed that. Personal preference does exist

I'll also have to learn some more related stuff for random dynamical systems

Clarinetist

Let $t \in \mathbb{C}$, and let $A_n = e^{i(nx + y)}$ with $n$ an integer, and $y \in [0, 2\pi]$. Could someone please explain to me how to calculate
$$\int_{0}^{2\pi}e^{\text{Re}(\bar{t}A_n)}\text{ d}x$$

Thorgott

sub-infty-groupoid of the jet bundle or something

gottem

user2103480

@Thorgott clearly an ODE

@Clarinetist you know what the real part of $A_n$ is?

Thorgott

that looks like a horribly ugly integral

Clarinetist

00:56

@user2103480 Right, so that's just $\cos(nx + y)$. What is worrying me particularly is that we are trying to find the real part of $\bar{t}A_n$.

Thorgott

isn't e^cos not elementarily integrable or sth

Clarinetist

Yeah, when I wrote $\bar{t}$ in polar form, I ended up with $e^\cos$ stuff

I thought maybe I did it wrong

user2103480

Re((Re(t) + i.Im(t))(Re(s) + i.Im(s))) = ... product... = Re(t)Re(s) - Im(t)Im(s)

Clarinetist

Oh, that's an idea

anakhro

@user2103480 again, feel free to look at Arnold's book to see what could have been.

Clarinetist

01:00

So let's suppose $t = a + bi$, thus $\text{Re}(\bar{t}A_n) = a\cos(nx+y) - b\sin(nx+y)$

Now the next problem is integrating that thing...

I bet there's some complex analysis that I don't know that helps with this problem

Thorgott

nah, no way

user2103480

^

Clarinetist

Well, that's unfortunate

user2103480

what's the context of the problem

Clarinetist

It's probability. $A_n$ is a complex random variable with $x, y$ uniformly distributed over $[0, 2\pi]$. I'd like to find its characteristic function.

This would be one part of a double integral that would have to be computed ($y$ would have to be integrated over $[0, 2\pi]$ as well), and then we'd have to divide by $2\pi$.

(oh, and $x, y$ are independent)

So the computation is actually

$$\dfrac{1}{4\pi^2}\int_{0}^{2\pi}\int_{0}^{2\pi}e^{\text{Re}(\bar{t}A_n)}\text{ d}x\text{ d}y$$

If this can't be done, it can't be done. I thought I would give it a shot.

Thorgott

01:10

I seriously doubt that's the right expression

Clarinetist

I'm basing it on what I see here: en.wikipedia.org/wiki/…

$A_n$ is correct as is

user2103480

Uhh isn't the characteristic function defined as $\Bbb E[e^{ik^TX}]$

Clarinetist

For a real-valued random vector, yes. For a complex one, see Wikipedia above

user2103480

There is no difference to a 2-d real variable

Ah wait

Thorgott

your $A_n$ looks completely off to me

it should just be the density of the distribution

user2103480

01:12

You're not doing the uniform distribution alone, sorry

Thorgott

what am I missing?

user2103480

but what you're missing is the density yeah thorgott

The substitution isn't right

Clarinetist

Isn't that provided from the $\dfrac{1}{4\pi^2}$?

user2103480

Yeah too sorry I mean. The A_n is just (x,y)

The density is there, sorry

but during the subtitution you lose the random variable and just integrate over the range of the random variable

rescaled with the density

Note that you can't integrate over the random variable itself if you're not on the probability space anymore

Clarinetist

Sorry, I've lost you there. So we have $A_n = e^{i(nx+y)}$... what do you mean by "losing the random variable"?

Thorgott

01:15

yeah, the integral should just be $\frac{1}{4\pi^2}\int_{[0,2\pi]^2}e^{\operatorname{Re}(\overline{t})}d(x,y)$

Clarinetist

What I'm really confused about is how to handle $\text{Re}(\bar{t}A_n)$

user2103480

duh I'm not being focused enough

Thorgott

do you agree with me?

Clarinetist

I just don't know what happened to the $A_n$ in the integral you provided

user2103480

@Thorgott No not completely. We need an expression of the form Re(t*z) and integrate over the square in the complex plane

Ah no now it makes sense. We don't integrate over a square

I*m just not focused enough smh

Clarinetist

01:18

So if it's not a square, what is it...

user2103480

We integrate over a circle, since that's where the random variable's values lie

Thorgott

huh?

we're uniformly distributed on a square

Clarinetist

Yeah, you've lost me there. We are indeed uniformly distributed over a square.

Thorgott

so the pdf is 1 on the square and 0 else

user2103480

the density is that of the random variable $A_n = e^{i(nx+y)}$

Clarinetist

01:20

Does change of variables not apply in the complex probability world?

Thorgott

oh wait wait wait

so $X,Y$ are uniformly distributed rvs, $A_n=e^{i(nX+Y)}$ is the random variable we're looking and we want the characteristic function of that?

Clarinetist

Yes

Thorgott

yeah ok, no chance

Clarinetist

Law of the Unconscious Statistician was what I used to get to that integral

Thorgott

yeah, I agree it's the right integral now that I understand what you're trying to do

my advice is giving up

Clarinetist

01:22

Okay, great. And we're in agreement that there's no amount of complex analysis learning I could do to solve that reasonably in a day or 2

Lol, k

Just from observing that definition... having that $\text{Re}(\bar{t}X)$ term in the characteristic function looks like an utter pain

user2103480

Ok so I'll now do it detailed. We have a random variable with values in the unit circle in the complex plane. The density $f(z)$ I'd have to think about, but maybe one can bypass this somehow using the structure of the uniform RVs. Anyways the integral is $$\int_{S^1}e^{i\mathrm{Re}(t^\ast z)}f(z) \, \mathrm{d}z$$

Clarinetist

Sure, but have fun finding that density

I don't even want to consider what that two-to-one-variable transformation would even look like shudders

Semiclassical

Here's a basic calculus problem that I ran into just now that I don't remember how to show

Suppose $f$ is some smooth function such that $f(t+T)-f(t-T)=2T f'(t)$ for all $t,T$. Must $f$ be quadratic in $t$?

pretty sure the answer is yes but my face is tired

Clarinetist

Gosh, I wish I had taken differential equations in my undergrad

(or maybe not)

dc3rd

@Clarinetist, so you're insinuating I'm wise for planning on taking ODE's and PDE's in my undergrad then...... :p

Semiclassical

01:30

geometrically, this is the source. Suppose you estimate the first derivative of a function $f(x)$ evaluated at $x_i=x_0+i \Delta x$ by drawing secant lines at points $x_{i\pm 1}$ and computing the slope

user2103480

@Clarinetist A wild guess would be that the argument of $e^{inX}$ is uniformly distributed on the circle on $[0,2\pi]$ (technically we need to exclude $2\pi$ but measure zero something something)

Clarinetist

Hi @TedShifrin. If $A_n = e^{i(nx + y)}$ with $x, y \in [0, 2\pi]$ and $n \geq 1$ an integer, would you agree the following is impossible to solve given some $t \in \mathbb{C}$?

$$\dfrac{1}{4\pi^2}\int_{0}^{2\pi}\int_{0}^{2\pi}e^{\text{Re}(\bar{t}A_n)}\text{ d}x\text{ d}y$$

Semiclassical

If f(x) is linear, that certainly just gives back $f'(x_i)$ exactly. But it seems that it also works if $f(x)$ is quadratic

Ted Shifrin

What do you mean impossible to solve?

Clarinetist

@TedShifrin Well, you end up with some $e^{\cos(\text{stuff})}$ if you convert $t$ to polar form

Semiclassical

01:31

@Clarinetist This looks like some bessel function shenanigans

Ted Shifrin

So you mean the integral cannot be done in elementary terms?

Semiclassical

c.f. en.wikipedia.org/wiki/Jacobi%E2%80%93Anger_expansion

Clarinetist

@TedShifrin Yeah, sure

user2103480

Then the probability that the argument of the product of $e^{inX}$ with $e^{iY}$ lying in a an arc on the circle is the same as the convolution lying in one of the periodic intervals

dc3rd

I'll let you finish with Clarinetist first @TedShifrin, then I'll finish discussing what I was thinking about with regards to the problem befor eI had to go run my errands.

Semiclassical

01:32

Jacobi-Anger is like...the only bit of Bessel function stuff which I actually enjoy

Clarinetist

What the heck is that @Semiclassical O_o

Semiclassical

it's not actually as strange as it looks

$e^{i x\sin \theta}$ is a periodic function in $\theta$

Clarinetist

Lol, considering I learned what Fourier series were barely 24 hours ago, it looks strange

Semiclassical

therefore it must have an Fourier expansion in complex exponentials

the amazing part is that the complex Fourier coefficients are just Bessel functions

i generally don't like Bessel functions but I always think that's crazy neat

a more appealing version of what I wrote above: For what smooth $f$ is it the case that $\dfrac{f(y)-f(x)}{y-x}=f'(\frac{x+y}{2})$ for all $x,y$?

Clarinetist

There's one thing that going through measure theory + probability has taught me. It's that I could not handle doing a math PhD. I have a ton of respect for people who know a field so well that they can talk about a subject, and draw connections from topology, complex analysis, measure theory, Fourier analysis, and functional analysis in a short amount of time, but I am not one of those people

Semiclassical

01:37

Bessel functions are something you inevitably deal with in first-year physics grad school

because they show up when dealing with boundary value problems with cylindrical symmetry

(which suuuuuuuuuck)

BigSocks

0

Q: When localizing an ideal in a ring of dimension 1, how do you show that you get this?

Let $A$ be a ring of dimension $1$ and let $J$ be an ideal of $A$ that can be factored as a product of maximal ideals $J = P_1^{a_1} \dotsm P_s^{a_s}$. Let $M$ be any maximal ideal of $A$. Then $J_M = (MA_M)^{a_i}$ if $M= P_i$ for some $i$, and $J_M = A_M$ if $M \neq P_i$ for any $i = 1,...,s$. T...

here's kind of an elementary ring theory question I put on main

Clarinetist

I'd love to help, but I haven't even thought about rings since... 2013

Semiclassical

ideally i'll never do anything with ideals again

2

BigSocks

understandable, I also was avoiding them since around that time, but I had a change of heart

Ted Shifrin

Sorry, Semiclassic. I was thinking about your question, and then my sister and bro-in-law texted to say a present I had just sent them arrived slightly broken — so I'm a bit distracted.

Semiclassical

01:39

oof

BigSocks

@Semiclassical I hope your ring homs are always injective

Semiclassical

lol

Ted Shifrin

I was going to suggest you write out the Taylor poly of degree 2 with remainder.

Semiclassical

yeah

Clarinetist

Utterly random question: my nonparametric stats prof never really explained this in detail. So $e^x$ for example: you can write it as $$e^x = 1 + x + o(x)$$ or $$e^x = 1 + x + O(x^2)$$
both as $x \to 0$. Why would you choose one over the other?

Semiclassical

01:44

listening to Dan Deacon feels like I'm going pleasantly mad

user2103480

@Clarinetist Let $\rho$ be the density of a sum of uniform variables with values in $[0,2\pi]$. This has support on $[0,4\pi]$. Let an arc $A$ on the circle, for simplicity between angles $0<a<b<2\pi$, be given Then I think that roughly $\Bbb P(e^{i(nX+Y)} \in A) = \Bbb P(arg(e^{inX}e^Y) \in [a,b]) = \Bbb P(arg(\text{uniform in }[0,2\pi]) + arg(\text{uniform in }[0,2\pi]) \in [a,b] \cup [a+2\pi, b + 2\pi]) $

Clarinetist

Now the problem is finding the density

user2103480

The density for the arguments then is $\rho(x) + \rho(x+2\pi)$

Semiclassical

@Clarinetist I don't suppose you have the argument of this exponential written out somewhere above

getting rid of the Re in favor of explicit functions

user2103480

Since the arguments have values in $[0,2\pi]$ instead of $[0,4\pi]$, this still sums up to one

Ted Shifrin

01:50

@Semiclassic: Better idea. Look at third degree T.P. (centered at $t$) with remainder. I think that does it.

Clarinetist

@Semiclassical Well, let's see... write $t = re^{i\tau}$. We get $\bar{t} = re^{-i\tau}$ so that $$\text{Re}(\bar{t}A_n) = \text{Re}(re^{-i\tau}e^{i(nx+y)}) = \text{Re}(re^{i(nx+y-\tau)}) = r\cos(nx+y-\tau)$$

So you would have to integrate $e^{r\cos(nx + y - \tau)}$ with respect to both $x$ and $y$ over $[0, 2\pi] \times [0, 2\pi]$

Semiclassical

@TedShifrin I think I get it. Suppose $f(t+T)-f(t-T)=2 T f'(t)$ for all $T,t$. Then regarded as a function in $T$, we have $g(T):=f(t+T)-f(t-T)-2T f'(t)=0$ for all $T$. In particular, we must have $g'''(0)=2f'''(t)=0$ for all $t$.

third derivative, so same basic idea as what you're saying (though probably less rigorous)

@Clarinetist hmm, that doesn't seem terrible

though I'd need to work out what $e^{i x \sin(\theta-\phi)}$ looks like to be sure

Ted Shifrin

I like it, @Semiclassic.

Semiclassical

for $\phi=0$ or $\phi=\pi/2$ one gets known Bessel function expansions

@TedShifrin same

dc3rd

Ok @TedShifrin, so after thinking about things I built up from the ideas of the original questions:

In the previous question I only had to deal with the second component $y$ in relation to the $x-axis$, building up from that. I now face a scenario where both components' distances are affected.

So before tackling the problem I asked myself: "what would the distance from the point $\mathbf{a} = (a,b)$ to the point on $y=x$ look like?..... Well using $y = a$ or $y =b$, you would still get a distance of $b-a$

Semiclassical

01:54

this is based on trying to work out what a certain analysis program was doing to estimate first derivatives

Clarinetist

I'm just going to assume it's not doable at this point if we're just looking at elementary functions

Semiclassical

@Clarinetist yeah, definitely not

but bessel functions aren't -that- special

I'd rather run into ordinary Bessel functions than a generic hypergeometric function

Ted Shifrin

@dc3rd This is back to basic geometry again (and you've learned the projection stuff early in Chapter 1, I suppose). Of course, you don't need the optimal proof, but the crucial thing is the distance from $\mathbf a$ to the line $y=x$.

Semiclassical

oh. tangentially related to the first derivative stuff, here's another interesting geometry problem my students ran into today

Ted Shifrin

Pun intended @Semiclassic

Semiclassical

01:56

caught me

I had them using their phone cameras to take videos of them taking small balls and tossing them into the air

Ted Shifrin

To prove that we don't live in a vacuum?

Clarinetist

There are many things I'm looking forward to once this semester is done:

* (Probably) resigning from my adjunct position
* Getting vaccinated, possibly seeing people again
* Re-learning measure theory at my own pace
* Taking the stats PhD classes

Semiclassical

hah

the idea is that the height of the ball should be a quadratic function of time, with the second derivative being free-fall acceleration $g$

Ted Shifrin

@Clarinet: Unfortunately, we can't see people blithely even after vaccines. Especially with the new mutants.

Semiclassical

@TedShifrin based on the balls today, i can't actually prove that :P

Clarinetist

01:58

Yeah, those variants... sigh

Ted Shifrin

Ah, @Semiclassic: Trying to replicate Newton :P

Semiclassical

(even the foam balls displayed the expected behavior, rather than seeing any effect of air resistance)

anyways

dc3rd

Yes @TedShifrin, drawing the picture I chose the distance I chose from using the right triangle: i.e: $(a-x)^{2} + (b-y)^{2} = r^{2}$, letting $x = a$ which means $y = a$, so $r = (b-a)$

Semiclassical

what you'd expect in particular to see, upon plotting the (numerically estimated) first derivative is linear behavior

Ted Shifrin

I don't think that's quite right, @dc3rd.

Semiclassical

01:59

and most people did

but a few people were seeing curvature in these plots

Ted Shifrin

Very cool @Semiclassic. Did they fudge like we all did in HS science?

Semiclassical

ehh, not too much I think. videos don't permit that as much

Clarinetist

I'm just glad that I've finally decided in my head that I can't continue this adjunct job past this semester. Creating a course from scratch with no (great) textbook, teaching it for 2 years, and then teaching it online is not an experience I'd like to go through again.

Semiclassical

@Clarinetist oof

Ted Shifrin

Yeah, teaching is hard enough with a course in existence and a good book.

Semiclassical

02:00

i don't consider my teaching job to be sustainable, but at least i'm not expected to create a new course on my own

Clarinetist

I think some of the faculty think I might teach the freshman-level course of the class I currently teach... it's going to be a no for me there

Semiclassical

anyways. after working it out, I think the curvature is because it's hard to toss the balls straight up. inevitably, there's some deflection side-to-side.

Clarinetist

I am really happy with how it turned out, though. Getting over imposter syndrome and having sophomore-level community college students competing against graduate-level students and placing in data competitions is something I'll be proud of, but I cannot continue trying to keep up with the changing tech.

Semiclassical

that's fine if said horizontal motion remains the same distance from the camera

but if said motion brings the ball away from/closer to the screen, then there's an issue of perspective

same distance in reality looks smaller if placed farther away

dc3rd

well let's see, the distance from $\mathbf{a}$ to any point on the line $y=x$ would be $(a-x)^{2} + (b-x)^{2} = r^{2}$ correct?

Semiclassical

02:04

so the acceleration looks faster/slower when the ball is closer/farther from the screen

and since the motion is smooth, I think that makes the acceleration look as though it varies when it doesn't

going to test that with some linear perspective calculations i think

Ted Shifrin

For the right $x$.

Semiclassical

hmm, maybe i can make it even simpler

dc3rd

So then the question now becomes, what is that right $x$?............let me try some stuff. 3 mins....

If I let $x =a$ or $x = b$ I still end up with $b-a$, but you're saying that won't help me.............

user2103480

@Clarinetist And thaat just turns out to be $1/2\pi$. But I'm now left with $\frac{1}{2\pi} \int_0^{2\pi} e^{i(t_1 \sin(\phi) + t_2 \cos(\phi))} \, \mathrm{d} \phi$

Semiclassical

@user2103480 that's where bessel functions come in

Clarinetist

02:15

It's fine, thanks for your help @user2103480. I've given up on that approach for now.

user2103480

Wolfram alpha returns that the real part of this is the bessel function of the square root of $t_1^2 + t_2^2$

Semiclassical

yeah

I can explain that much easily

user2103480

and the imaginary parts are zero

Semiclassical

Let $T=\sqrt{t_1^2+t_2^2}$. Then we can write $(t_1,t_2)=(T\cos\theta,T\sin\theta)$ for some $\theta$

user2103480

so that is nicer than I thought modulo bessel functions

Semiclassical

02:17

at which point $$t_1\sin \phi+t_2\cos\phi=T(\cos\theta\sin\phi+\sin\theta\cos\phi)=T\sin(\phi+\theta)$$

But it's a periodic integral, so one can eliminate $\theta$ by shifting $\phi\mapsto \phi-\theta$

at which point the integral is $\frac{1}{2\pi}\int_0^{2\pi} e^{ i T\sin \phi}\,d\phi$

and Jacobi-Anger tells you that's just $J_0(T)=J_0(\sqrt{t_1^2+t_2^2})$

So not horrid

and not utterly shocking, either: Bessel functions can show up when doing Laplace transforms

Ted Shifrin

@dc3rd Seriously, are you drawing pictures? Use projection stuff from Chapter 1.

user2103480

Yeah it seems okay. And the only nontrivial assumption I made in the probability part of this was assuming that $e^{inX}$ has the same distribution as $e^{iX}$ for $X$ uniformly distributed on $[0, 2\pi]$

Semiclassical

couldn't tell you about that, though I see the sensibility behind it

user2103480

May be wrong lol gotta think about this

Semiclassical

i mean

user2103480

02:22

something something wrapping n times around the circle

dc3rd

@TedShifrin, I am drawing pictures, but I guess I'm not gleaning from it what I'm supposed to.....😔

user2103480

... where did this problem come from @Clarinetist

Semiclassical

it amounts to "if I take a uniform r.v. between 0 and 1, multiply by a particular integer $n$ then take the fractional part, do I still get a uniform r.v. between 0 and 1"

user2103480

this seems insidious for homework

Semiclassical

and I can't see why not

Clarinetist

02:23

@user2103480 It's a problem I made up, just in the interest of learning about characteristic functions for complex random variables.

user2103480

lmao what

Semiclassical

adjunct life

Ted Shifrin

Closest point on the line?

user2103480

you should say that beforehand, I spent like an hour on this in the belief that this has a nice closed form solution

Clarinetist

Yeah, I've not calculated the characteristic function of a complex random variable before, and I naively thought it would be just as easy as it is in the real case.

user2103480

02:25

which it incidentally had but that was extremely unlikely

Clarinetist

I have definitely learned my lesson

user2103480

and again, nice modulo wolfram alpha or modulo physicist

Although I gotta hand it, that was a smart problem statement of yours by choosing the right interval of the uniform distribution

Did you do this on purpose, thinking about going around the circle a random number of times?

Clarinetist

It is related to a HW problem I have, so that's where the interval came from

Semiclassical

is this a HW problem out of a textbook

or from the prof's brain

Clarinetist

It was inspired by a prof's homework problem

Semiclassical

02:29

gotcha

dc3rd

https://i.sstatic.net/Osnog.jpg

@TedShifrin

Clarinetist

Combination of a prof's problem, plus me naively thinking that characteristic functions of complex random variables would be easy

Semiclassical

lol

user2103480

we got nerd sniped

for reference: xkcd.com/356

Ted Shifrin

So what point on the line $y=x$ is closest?

dc3rd

02:37

Well visually, it is the point that would be under the "right angle" box I drew......so right now I'm just thining of how to describe it mathematically

Ted Shifrin

Man, you have zero high school geometry.

You have a 45-45 right triangle.

dc3rd

I have been working through a high school geometry textbook since our discussion last week. So I'm working on it. ...but with regards to this, if I have an angle it means I should be able to get the length of that line then

robjohn

@user2103480 That's been me working on some of the last questions I've worked on.

Ted Shifrin

@dc3rd Yup. You know the hypotenuse of that right triangle. Here comes your $\sqrt 2$ you're in love with.

dc3rd

02:53

If I'm looking at it correctly....the hypotenuse is $(a-x)^{2} + (b-x)^{2} = h^{2}$....for hypotenuse of the right triangle, and then taking square root

user2103480

@Clarinetist btw, this bessel function $J_0(\| t \|)$ should be the characteristic function of all random variables of the form $\exp(i(a_1X_1 + ... + a_n X_n))$ with $a_i$ integers and $X_i$ independent uniform on $[0, 2\pi]$

since these should all be uniform distributions on a circle

dc3rd

03:22

Something went a wry @TedShifrin..................so in my quest to find the point I set up the following:

$sin(\frac{pi}{2}) = \frac{\text{opposite}}{\text{hyp}} = \frac{(a-z)^{2} + (b-z)^{2}}{(a-x)^{2} + (b-x)^{2}}$, where I let $z$ be the value of the desired point.

which yielded:

$1 = \frac{(a-z)^{2} + (b-z)^{2}}{(a-x)^{2} + (b-x)^{2}}$, now if my diagram is correctly drawn then $x = a$ and so my denominator simplifies and with a rearrangement I have:

$(b-a)^{2} = (a-z)^{2} + (b-z)^{2}$

Ted Shifrin

04:04

@dc3rd: This is an isosceles right triangle (both legs have the same length). The hypotenuse has length $b-a$, so both legs have length ....

Balarka Sen

Hi again, @Ted.

dc3rd

@TedShifrin. I don't know what it is explicitly, but I know whatever that length is, it will be the radius I need for my ball. As for finding that length....Well I will give you an explicit answer in a day or so as I continue my geometry review and I'm getting to the chapter on triangles..........this geometry issue of mine is so damn sad, but I'm glad it was discovered now as I'm in self study instead of when I begin live courses again in the summer.

So I'll leave that as it is for the moment. You've helped plenty without giving me an answer....I owe it to myself to work on it from here.

Well an idea did pop up in my head to find that length.....law of cosines, or parallellogram law.....one of the two...I can't remember which one...

123

04:46

Hello World..

dc3rd

05:22

length of side is $\frac{(b-a)}{\sqrt{2}}$.............

Balarka Sen

06:18

Hi @Semiclassical

10 Replies

Is there a theorem that proves this true

seems kinda sus to me tbh

Balarka Sen

No, because it's false.

10 Replies

ok

if I have X=AB, how do I solve for B?

assuming all are invertible matrices

and all NxN

oh wait

so A^-1X=B ?

@BalarkaSen

Balarka Sen

Is there a question here which you cannot answer by thinking for more than 2 seconds?

2

10 Replies

i am small brain

i need to rubber duck

but if X=AB, B=(A^-1)X, right?

just want to make sure i'm not losing my mind

I'm gonna assume that that is true

what exactly are I_n and I_m

I think 'I' is typically used for the identity matrix

are these just identity matrixes with different dimensions?

dc3rd

06:40

Have you even read your textbook?

10 Replies

bold of you to assume i can read

I didn't buy a text book

Just full sending it

Semiclassical

@BalarkaSen o/

Balarka Sen

07:21

@Semiclassical How do you think of the spin group

copper.hat

08:17

never heard of them, do they have a youtube video?

Balarka Sen

lol

copper.hat

i got bored reading my quantum computing for everyone.

porridgemathematics