Mathematics - 2023-05-16

Ted Shifrin

00:14

nods loudly

冥王 Hades

Heading to the nurse's office. I feel weird

Ted Shifrin

After your altercation yesterday, not a bad idea!

AnArrayOfFunctions

@TedShifrin Yeah sure - the question was closed though.

Not a big - maybe it was a dumb question anyway.

冥王 Hades

00:31

Came up with a fever. Perfect. I'd be asked to rest a bit, then likely sent home.

AnArrayOfFunctions

In any case I did read the suggested link:

https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question/29298#29298

Wolgwang

$$h=\frac{5\cos t}{3}-\frac{\sin t}{3},k=\frac{5\sin t}{3}-\frac{\cos t}{3}$$
Is there a simple way to eliminate trig functions? (I solved for $\sin t,\cos t$ and then squared them)

AnArrayOfFunctions

And added my "motivation" which appeared to be poorly recieved

Ted Shifrin

00:44

@AnArrayOfFunctions Why should any of your choices result in a different result? They all seem equivalent to me in the long run.

@Wolgwang You did it.

AnArrayOfFunctions

Well ok - for that I was convinced - my theory was because if you choice a constant in the beginning and the random number you are predicting all for some chance are different than you predicted number you have no chance of hitting them. (Yeah I agreed with @JMoravitz - who basically said that this is ignoring all the other cases which are balancing it)

(Whilst my original theory is that predicting random you could sort of correct yourself)

冥王 Hades

@TedShifrin To give you an idea of just how stubborn I am, I was prescribed by the nurses to turn my phone off and get some rest after taking a medication, yet here I am typing this on my phone

I am my worst enemy

Ted Shifrin

01:05

STOP. BYE.

leslie townes

04:07

HELLO. START.

PNDas

08:54

I want to find all distributional solutions of the PDE $u''-4u=\delta'$.

It's previous exercise was: Suppose $L=\frac{d^2}{dx^2}+a\frac{d}{dx}+b$. Let $f,g$ be smooth such that $L(f)=L(g)=0, f(0)=g(0), f'(0)-g'(0)=1$. Define $F(x)=\begin{cases}f(x),x\leq0\\g(x),x>0\end{cases}$ . Then show that $-F$ is a fundamental solution of $L$.

So using the previous exercise I found a bunch of solutions. But how do I know they are all?

So $f=c_1 e^{2x}+c_2e^{-2x}$ and $g=c_3e^{2x}+c_4e^{-2x}$. Now any $c_1,c_2,c_3,c_4$ which satisfy $c_1+c_2=c_3+c_4$ and $c_1-c_2-c_3+c_4=1/2$ will lead to a solution.

Then $T_{-F}\ast \delta'=T_{F'}$ is a solution of our original PDE.

Koro

12:39

6

Q: Proving that the outer measure of a closed interval $[a,b]$ is $b-a$

In Sheldon Axler's book, Measure Integration, and Real Analysis, he defines outer measure of a set as $|A| = \inf\big\{\sum_{k=1}^\infty \ell(I_k): I_1, I_2, \dots \text{are open intervals such that} A\subset \bigcup_{k=1}^\infty I_k\big\}$, where $\ell(I)$ for an open interval $(a,b)$ is just $b...

I have one confusion in this. Could you please help me understand why the following doesn't work?: I claim that $|(a,b)|=b-a$. Proof: Given any $\epsilon>0$, choose $A_1= (a-\frac{\epsilon}4, b+\frac{\epsilon}4), A_k=\emptyset \, \forall k\ge 2$. Then, clearly $I\subset \cup_k A_k$. Moreover, $b-a\le \sum_k l(A_k)=b-a+\frac{\epsilon}2<\color{red}{b-a}+\epsilon$. This satisfies the infimum definition hence the result follows. — Koro 5 mins ago

Can anyone please explain this to me? Thanks.

Wolgwang

I have to ask two parts of the list 1. Should I ask separate question or incluude them in one?

geocalc33

12:58

If $(A,B,C)$ is a Mellin triad what do we call $(A,C)$? It's not a Mellin pair. The Mellin pairs are $(A,B)$ and $(B,C)$

geocalc33

13:38

0

Q: Mellin transform *triads*

Mellin transform pairs are ubiquitous in analytic number theory. Maybe the most famous pair is $e^x$ and $\Gamma(s).$ Define a "Mellin triad" by the discrete pre-periodic forward orbit of Mellin transforms of a function $\psi$: $$O_\mathcal{M}(\psi)=\bigg\lbrace\psi, \mathcal{M}^{(1)}(\psi),\math...

complex-analysis analytic-number-theory mellin-transform

Xander Henderson

14:00

Grades are submitted. Yay.

Ted Shifrin

15:07

@XanderHenderson Mazltov!

user977780

15:21

Who needs a superhero when you have an amazing brother to support you always.

geocalc33

15:42

hello

geocalc33 here

D.C. the III

@TedShifrin Have a minor question. I was finding the integral $\int_S \frac{x}{y}$ over the region bounded by $y = x, y = 2x, xy = 3, xy = 1$. Instead of choosing $v = \frac{x}{y}$ as my second choice of variable I used $v = \frac{y}{x}$. Doing it this way I ended up with the negative of the correct answer ($1/2$). I went over the work I did and tried to adjust things, the only difference between what I did and letting $v = x/y$ was the limits of integration.

In my form the limits for $v = y/x$ are $(1,2)$, meanwhile if I let $v = x/y$ the limits are $(1/2, 1)$. I know it's something minor I"m messing up but can't seem to find it.

Ted Shifrin

Did you remember the absolute value of the Jacobian?

D.C. the III

Yes. That part of it is fine. I ended up with $1/2v^2$

So the integral "my way" was $\int \frac{1}{v} |\frac{1}{2v}|dA_{uv}$

Ted Shifrin

16:05

I ordinarily use $y/x$, anyhow. So what is your entire integral?

D.C. the III

$$\int_1^2\int_1^3 \frac{1}{2v^2} dudv$$

The answer is $\frac{1}{2}$ which I would've gotten if I had done $x/y$ as the setup. In this set up I got $-1/2$

Ted Shifrin

First, that integral can’t be negative. Second, where did the integrand go?

You’re missing $\frac xy$, aren’t you?

D.C. the III

The integrand itself was $\frac{1}{v}$

Ted Shifrin

Oh, no, you have it.

So how did you get a negative integral when you integrate a positive function?

D.C. the III

$\int_1^2 \frac{1}{v^2} = \frac{1}{v} \bigg]_1^2$

AHHHHH

the minus sign!!!

when I took the integral.

Ted Shifrin

16:11

Yup.

D.C. the III

HGFHUEEF***$&$&$UJFHDUJF*&$*Y$*$*$

THat's me cursing

Ted Shifrin

Common sense should make you find that error!

A positive function has positive integral.

D.C. the III

I was feeling too proud in myself and thought the integration could not be the issue

hence why one must take their time

back to work I go. SOrry for the disturbance.

I'll bother you later with another common sense thing

KZ-Spectra

Hey guys, I wanting to find the critical point of my functional. I know its upperbound and when that upper-bound is negative. mathb.in/75324 I'm not really sure how to obtain the crit point though. Ideas?

Wolgwang

16:29

4 hours ago, by Wolgwang

I have to ask two parts of the list 1. Should I ask separate question or incluude them in one?

robjohn

16:48

If they are related, I'd post one question. If they are two separate aspects, I would post two.

shintuku

i'm about to drink a post-midday coffee and it's not even friday

idgaf i live dangerously

Ted Shifrin

It's not even Monday, either.

geocalc33

Lakers game tonight

Who's watching?

Hello

Q: A function $\psi : (0,\infty) \to \mathbb{C}$ has Mellin transform $\mathcal{M}(\psi)(s) = \int_{0}^{\infty} \psi(x) x^s \, \frac{dx}{x}$ for $s \in \Bbb C$ for which this integral converges absolutely.

I got a comment from user Peter Humphries that in general it is not possible to take the Mellin transform of the Mellin transform of a function.

My question is about this exactly

I have 2 examples of a Mellin triplet/triad

But I get the impression that these are very rare

These are rare $$O_\mathcal{M}(\psi)=\bigg\lbrace\psi, \mathcal{M}^{(1)}(\psi),\mathcal{M}^{(2)}(\psi) \bigg\rbrace$$

rarer than these $$\bigg\lbrace \psi,\mathcal{M}^{(1)}(\psi)\bigg\rbrace$$

the latter has an example $\psi(x)=e^{-x}$ and $\mathcal{M}^{(1)}(\psi)=\Gamma(s)$

So you have to start with products of Gamma functions and take iterated inverse transforms if you want a Mellin triplet.

And that is why I'm interested in triplets. If Mellin pairs are so important, are Mellin triplets also important, and how?

Wolgwang

17:34

@robjohn Uhm I have posted one. I thought I was wrong but the gives answer was wrong. So I edit another part in that question, Is that against the rules?

This is my situation. :/

shintuku

18:28

Let $\mathfrak m$ be the maximal ideal of a local ring $B$. Let $\mathfrak m[x] \subseteq \mathfrak m'$, where $\mathfrak m'$ is a maximal ideal of $B[x]$. Prove $\mathfrak m' \cap B = \mathfrak m$.

I did the proof manually checking additive closure and multiplicative absorption of $\mathfrak m' \cap B$, and convinced myself in this way, but is there any faster way to see this?

i.e., it's not immedately obvious $\mathfrak m' \cap B$ is an ideal at all

D.C. the III

Having trouble finding the region to integrate over in the $u-v$ plane given by the following in $x-y$ terms: $x \geq 0$, $ y + x^2 = 0$, $x - y = 2$, and $x^2 - 2x + 4y = 0$. As well we were given to consider the following change of variable$ x = u + v$ and $y = v- u^2$.

I've done the algebraic setup , but getting the limits of integration in terms of $u-v$ is becoming a problem. As of now for expressions I have: $u \geq -v$, $ v^2 + 2v + 2uv = 0$, $u^2 + u - 2 = 0$, and $v^2 + 2v + 2uv - 2u - 3u^2 = 0$. None of the regions I get from graphing these is properly bounded by the four indivi…

@TedShifrin

Ted Shifrin

19:04

The issue is that $g$ is far from one-to-one, so you can’t do this with mindless plug and chug.

D.C. the III

Yes I noticed this.

I guess that prevents me from drawing a region in the $u-v$ plane?

Ted Shifrin

There are different regions that map to the given region. You want the simplest.

D.C. the III

I was hoping to get a box...lol.......when I draw it, there isn't really a simple region I've found. As of now it would be eaiser I think to just integrate the region given in terms of $x-y$, but that would be against the point of the exercise

Ted Shifrin

I didn’t try. Can you really do it?

D.C. the III

I'm not sure, but the region is not that outlandish in the $x-y$ plane. You might have to split it up into two integrals, but it doesn't look that bad.

shintuku

19:14

oh the slash used to denote quotient rings is the opposite one of setminus

Ted Shifrin

You didn’t figure that out when I asked you a week ago why setminus?

D.C. the III

shintuku

no i just now realized that that's why you asked me heheh

D.C. the III

That's what the region looks like in the $x-y$ plane

Ted Shifrin

What part of it?

shintuku

19:19

what's latex for:
rhs = lhs_1
\tab = lhs_2
\tab = lhs_3

it's begin{iforgot} \end{iforgot} edit: oh no i switched lhs and rhs

D.C. the III

The portion between $(0,2)$, above $y = x-2$ to the right of $y= -x^2$

Ted Shifrin

Where did you figure out $x-y=2$ in the new coords?

D.C. the III

no this is in the $x-y$ coords

Ted Shifrin

I’m asking a new question.

shintuku

it's not \begin{array}

D.C. the III

19:22

In the new coords I figured out $x-y=2$ and got $u^2 + u -2 = 0$

shintuku

ah, \begin{align}

Ted Shifrin

So simplify and use your brain.

Same for $y=-x^2$.

D.C. the III

Yea for that set I got $u = -2$ and $u = 1$

and then simplifying $y = -x^2$ I got the expression $v^2 - v + 2uv = 0$

Ted Shifrin

Once we know the mapping isn’t one-to-one, brains are needed.

shintuku

zombiemaths

D.C. the III

19:26

Well those are hard to come by in these expensive times..........

I had done some other playing around but wasn't comfortable with what I got.

Using the $v^2 - v +2uv = 0 $ expression, I can isolate $u$ and get $u = \frac{-v}{2} - \frac{1}{2}$. I could use the set of $u$'s I found, and possibly find $v$'s, but that doesn't seem right

Ted Shifrin

You’re making a silly error. What did you just assume?

So you need to use one of the choices, not both. In both cases.

D.C. the III

Hmmm.....

Well I do have that $x \geq 0$ which means $u \geq -v$

Ted Shifrin

That’s the least important, but helps in making the choices, I guess.

D.C. the III

The other thing that comes to mind that I wasn't all the way cool with was dividing out one of the $v$'s when I isolated $u$

Ted Shifrin

Precisely. A ninth-grade error.

D.C. the III

19:40

So I would use one of the previously solved $u$'s and putting it in $v^2 - 2v +2uv = 0$ will give me a quadratic which should be easy to solve, but how do I know which $u$ to choose?

Ted Shifrin

Huh? What happened to $v=0$?

D.C. the III

$v = 0$? from where?

Ted Shifrin

Sigh.

D.C. the III

What I was asking about was if I say choose to use $u = -2$, then I will eventually get after simplifying is $v^2-3v = v(v-3) = 0$. So I do see $v = 0$ coming from there, but I was asking how to know whether to use $u =-2$ or $u = 1$

and after working it out they give me the same thing

I was thinkg about it in my head without writing it down

or not exact but very similar

shintuku

20:03

square, circle are closed; line, parabola are open

what's a strict way to say this

D.C. the III

level set?

shintuku

i'd like a 3d parabola to be open, and a 3d sphere to be closed

need to see if there's a way to state this as a property of a variety

D.C. the III

So using $u = -2$, I end up with $v = 0, 3$. So would those be 3 of the bounds in the $u-v$ plane?

Ted Shifrin

20:24

I doubt it. You have four curves to deal,with.

@shintuku open is wrong, of course. You mean the algebraic set (which is closed) has no points at infinity? I have no idea what you have been learning.

D.C. the III

Since I have let's use $u = -2$ and $v = 0$, what do I do next find the curves? I should know this but it is alluding me

D.C. the III

21:34

@TedShifrin whenever you're back and not busy. I'm still struggling with the region. I put it aside for now and am working on other stuff, but I can't get anything "significant". THis is what it looks like on Desmos (where x = u and y = v:

https://www.desmos.com/calculator/nfqjnyjnhq

Semiclassical

21:51

Sometimes I wish there was a good way to respond to a comment with just “huh?”

B/c sometimes I just do not understand what they’re asking

geocalc33

22:17

check out this cool plot I made

and this one

noballpointpen

what's the data?

shintuku

@TedShifrin the algebraic set has no points at infinity, that's a good one

geocalc33

22:32

@noballpointpen I used the "random()" function in SageCell to generate 10,000 points

noballpointpen

@geocalc33 why cool then

shintuku

at Ted: how about compact vs. noncompact?

need to add a topology on those however

sigh

so much left to learn

D.C. the III

THe learning never stops and never enough time....😥

Ted Shifrin

@shintuku This refers to affine space sitting inside projective space. Ordinarily, alg geo works in the Zariski topology, which is not at all what you’re used to.

shintuku

on towards learning what zariski topology is, then

Ted Shifrin

22:51

This won’t help with this question .

shintuku

there's no characterization of compactness in zariski topology?

Alexander Gruber

@geocalc33 that's some handsome noise my friend

Ted Shifrin

Every variety in affine space is compact,

@D.C. All right. Here's the summary. I suggest you draw this stuff in the $uv$-plane. You recall that you noted that $x\ge 0$ corresponds to the half-plane $u+v\ge 0$. So you want to be positioned in that half-plane.

D.C. the III

23:08

right

shintuku

even $y=0$?

D.C. the III

I got that bound

Ted Shifrin

$y+x^2=0$ becomes *either* $v=0$ *or* $2u+v=-1$.
$x-y=2$ becomes *either* $u=1$ *or* $u=-2$.
$x^2-2x+4y=0$ becomes *either* $v=u$ *or* $3u+v = -2$.

Your goal is to select a region in $u+v\ge 0$ which maps one-to-one to our region bounded by the three curves.

D.C. the III

Ok. I'm going to work them out just to see how you reasoned them

Ted Shifrin

I did high-school algebra and factored.

D.C. the III

23:11

Had to put the "high school" in there huh?....😓

Ted Shifrin

Since "algebra" in this room means abstract algebra or beyond, yes.

We have people dabbling in commutative algebra.

D.C. the III

Fair enough.

I'll go work this out. Thanks for the help

Ted Shifrin

I've never in all my years teaching this course had a student try to do this problem. I think I always put it on my "work but don't hand in" list, but we know what students do with that list.

D.C. the III

Really?

Ted Shifrin

I think I made it up when I was in college and wrote my first set of "notes" on multivariable calculus. But it's a tough problem to sort through and I would certainly never put it on required homework or an exam.

D.C. the III

23:15

I always start with the "work but don't hand in" because I assume that to be the "warm up" to the marked questions

Ted Shifrin

Nah, I put those there because there are answers at the back.

D.C. the III

Exactly, so I figured "answer in back" -> easier than the others

Ted Shifrin

NO, not so.

D.C. the III

hmmm.... the more you know....

Ted Shifrin

I did, however, put on WebWork and exams problems like #11 and #12, where it's a contrived problem made up just to be doable.

Anyhow, if you remember your half-plane, it pretty much forces you into the easiest solution here. I mean, looking at those options, would you pick the hardest ones every time?

D.C. the III

23:18

No thank you. I'm going to go write them up rigt now though.

leslie townes

23:36

i heard that ted is still under a court order preventing him from creating even trickier, weirder problems than the ones in his text.

D.C. the III

Well I figured out the bounds now......I need to stop trying to "outsmart" the question or question asker........I didn't perform the "high school" algebra because I thought Ted treated us as sophisticated and above having to use quadratic formulas and such...

If I just went with the "reasonable" thing to do.smh

Ted Shifrin

No quadratic formula here, but definite factoring. Too low-brow for serious math? Nope.

Has Munchkin been thrown in month-long detention yet?

D.C. the III

I had to use the quadratic formula to get the third one (after putting in u and v)

D.C. the III

23:52

If I drew things correctly, then I should be integrating over a right triangle?

leslie townes

not yet.

Ted Shifrin

Basic factoring is a lost art.

@leslietownes I'm sorta surprised.

Aurel-BG

Hi

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$Mathematics$ Mathematics