Mathematics - 2024-11-11 (page 1 of 2)

copper.hat

00:00

Are you asking about that specific function graphed?

ahh, i missed that f is zero on the boundary

ok, the global $\min$ occurs either on the boundary or the interior. Since $f$ is zero on the boundary and there is a point with $f(x_0) <0$ we know the global $\min$ occurs in the interior, where the gradient will be zero. Since $x_0$ is the only stationary point, it is the only global $\min$.

@psie probably not much use to you at this stage, but i'm slowly grinding through Folland

Sine of the Time

00:18

@copper.hat I'm planning to study from it in a couple of months, I'll bother you then :)

copper.hat

@SineoftheTime study what? worry, i wasn't paying attention.

Sine of the Time

Study from Folland (Real analysis)

copper.hat

ahh, got you

i am impressed with Folland. it goes into detail that other books omit.

Xander Henderson

@Claudio Gross.

copper.hat

finite element analysis of the stresses and strains of a protective device

my android 15 install is taking forever

psie

00:52

@copper.hat I keep a lot of notes from Folland's text (over 50 pages so far). It'd be cool if I can help someone the way I've received help from others. If you or @SineoftheTime wants to take a look at my notes, let me know. Maybe I can arrange something.

I started reading from the book separate sections here and there, but then I started from the very beginning. It helps reading from the beginning, even though you are familiar with some of the stuff he goes through.

copper.hat

yeah, that's why i'm skipping through from the beginning, to figure out his conventions

pie

03:07

Why The $D$ operator can be treated like this? When solving the ODE $y''+ay'+by=f(x)$ and $f(x)$ is either $e^ax, \sin ax, \cos ax , x^n \dots$ one can make this fraction $\frac{f(x)}{D^2+aD+b}$ , well if $f(x)$ is a polynomial one can turn $\frac{1}{D^2+aD+b}$ into an infinite series and then differentiate.

I have no idea why does this works or how to prove it

should I make a dedicated MSE question?

nitsua60

04:08

6

A: Is there an analytical solution for this summation?

We can begin by using the taylor series for the inverse tangent. $$ \tan^{-1}(x) = \sum^{\infty}_{k=0}\frac{(-1)^k}{2k+1}x^{2k+1} $$ Plugging this in to the infinite sum we get the following. $$ \sum^{\infty}_{n=1} \left( -\frac{1}{n} + \sum^{\infty}_{k=0}\frac{(-1)^k}{k}\frac{1}{n^{2k+1}} \right...

answer needs, in its second equation, a 2k+1 rather than a k in a denominator. It's too small an edit for me to make--anyone wanna help an answer out?

nickbros123

04:24

sheldon axler is good

very fun and all

nickbros123

05:47

If say I have n sequences of vectors $\vec{v_1}^{(j)}$, $\vec{v_2}^{(j)}$ and so on till $\vec{v_n}^{(j)}$, each $v_i$ being in $\mathbb{C}^n$, with each being orthonormal with each other for any fixed $j$. Suppose also that each of these converges to $v_1,v_2 \cdots v_n$. Can I say that this orthonormality is preserved even in passing to the limit? Suppose convergence happens in the euclidean norm

I understand that normality will be preserved, i.e, $\langle v_i,v_i\rangle=1$

but not sure abt orthogonality

Soumik Mukherjee

4

A: Proving the limit of any convergent sequence of orthogonal matrices is again an orthogonal matrix

You proof seems good to me. Using matrix formalism can make it simpler: Let $\{M_p\}_{p\in\mathbb N}$ be the sequence of orthogonal matrices (so $M^T_pM_p=I$) and let $$M=\lim_{p\rightarrow+\infty}M_p$$ Then $$\begin{split} \|M^TM-I\| &= \|(M-M_p)^T M +M_p^T(M-M_p) +\underbrace{M_p^TM_p-I}_{=0}\|\\

nickbros123

06:36

thanks

I proved orthogonality using cauchy schwarz: $|\langle v_i-v_i^{(j_n)},v_k-v_k^{j_n} \rangle| \leq || v_i-v_i^{(j_n)}|| \cdot ||v_k-v_k^{(j_n)}|| $ and taking $n \to \infty$ limit

The Empty String Photographer

07:42

@Derso It’s highlighting the area where the stated condition is true.

Sal Rahman

08:22

I could almost swear, that the only way to guarantee that everyone can agree what a specific line that someone is thinking of, all that someone needs to give is two points on the line, and everyone else can infer the line from those two points alone. But what about other curves? Is there a generic formula to determine what are the exact number of points needed to guarantee that all points lie on that curve? I'm talking about polynomials, by the ways.

nickbros123

08:33

lagrange interpolation requires n+1 points to determine an n degree polynomial, but if you have more information, like derivative information, things can be done with less info

less info as in less points

Sine of the Time

08:47

@psie that'd be useful. I'll ping you if I'll need help :)

pie

why can $\frac{f(x)}{1_D}$ turn into $\sum\limits_{k=0}^\infty D^k$ where $D$ is the differential operator?

$\frac{f(x)}{1-D}$*, sorry.

Also what does $\frac{1}{1-D}$ even means? What is its definition?

Sine of the Time

some kind of geometric series?

pie

09:03

But How? what is even is this? This is part of a course on ODE in my school and there are no proofs whatsoever just memorise the formulas and tricks, I am struggling to understand what is happening here.

Should I make an MSE question?

Sine of the Time

you can

pie

Just making sure that it is not a very known trick or something so the question doesn't get closed immediately

Soumik Mukherjee

09:45

@nickbros123 You can also prove that the set of orthonormal matrices are closed just by observing that it is the preimage of $I_n$ under the map $A\mapsto A^TA$

nickbros123

oh damnnnn

Derso

12:16

Guys, no Lie Group structure is possible on the sphere $\mathbb{S}^2$. What about $\mathbb{S}^2\times \mathbb{R}$? Maybe this is a stupid question...

Thomas Finley

13:30

How do I test the convergence of the series : $\sum_{n=0}^{\infty}\frac{e^{2\pi inz}}{(n+1)^{3/2}}$?

z is any complex no.

Jakobian

@ThomasFinley take the absolute value and see

Xander Henderson

13:53

@ThomasFinley The same way you test the convergence of any series.

Generally, you want to compare your series to something you already know converges or diverges. This is often done using things like the root test or the ratio test.

Jakobian

14:53

The first thing I would do to try to test convergence, is to take the absolute value. Then take ratio or root test

Then if none of those work, it means its just more delicate situation and I need to analyze this further

Of course there are some other more specific convergence tests one might use if the example desires this

@SalRahman So you're thinking about polynomial curves of the form $f(x_1, ..., x_n) = 0$ where $f$ is a polynomial?

Sal Rahman

@Jakobian yep

Jakobian

Actually this is not analogous

Sal Rahman

Let's say someone has had a curve of the form $y - ax - b = 0$. I can immediately know what $a$ and $b$ are if they gave me two points on the curve.

Jakobian

For example a line in 3d is of the form $(x-x_0)/a = (y-y_0)/b = (z-z_0)/c$

I mean you can just take the square of their differences so actually it is in that form

but not a degree 1 polynomial at that point

but yeah I think we can boil this down to, lets suppose we have zeros of some non-zero polynomial, how many do we need to determine it

But even then examples like $xy = 0$ show that we definitely can't have just finite amount of them

polynomials given by lines in 2d have another property of being irreducible that this example doesn't, though

but then how do you represent a 3d line by irreducible polynomials, I guess you need a system of equations instead of just one equation at this point

So perhaps in your setting we have a family of irreducible polynomials $f_1, ..., f_k$ and trying to determine the set $\{(x_1, ..., x_n) : \forall_j f_j(x_1, ..., x_n) = 0\}$ given some points in it

Jakobian

15:17

yeah I don't know exactly what you are going for, you might need help from someone who actually knows algebraic geometry (since I imagine this is related to cubic curves)

Pizza

15:45

@SineoftheTime I managed to solve it thanks to the site you recommended to me

Basically the mistake was that I wrote $dl$ instead of $dx$ and then I had to write $r=\sqrt{x^2+l^2}$

I noticed that I was getting $\int^{l}_{0} \frac{1}{{(l^2+x^2)}^{\frac{3}{2}}}dx$ to integrate

So I used $x = l \tan\theta, dx = l \sec^2\theta d\theta$

$\theta = \text{arctan}(\frac{l}{x})$

So in the end $\sin(\text{arctan}(\frac{l}{x})) = \frac{l}{\sqrt{2l^2}}$

But I think there is a way to avoid evaluating this integral, maybe initially I had to write $x = l\tan(\theta)$

Thomas Finley

17:03

If $\sum |u_n(z)|$ (z is a complex no.) converges, then can we say, that $\sum u_n(z)$ converges?

(I am a beginner in complex analysis and this related to my prev question)

Ben Steffan

what's $u_n$?

in any case, yes

absolute convergence implies convergence for complex numbers

Thomas Finley

@BenSteffan ok cool, so this property remains same as was in real analysis.

Ben Steffan

yeah

you might want to think about a proof

Thomas Finley

@BenSteffan in the meantime I was able to prove it :D

Ben Steffan

nice :D

Thomas Finley

17:07

Your answer made me realize that my proof was not nonsense

:))

Now, coming back to my question: "How do I test the convergence of the series $\sum_{n=0}^{\infty}\frac{e^{2\pi inz}}{(n+1)^{3/2}}$?"

Firstly, I tried applying the ratio test, @XanderHenderson and @Jakobian

Sine of the Time

@Pizza you can use $$ \int \frac1{(x^2+a^2)^{3/2}}\mathrm dx=\frac{x}{a^2\sqrt{a^2+x^2}}+k$$

Thomas Finley

If $u_n=\frac{e^{2\pi inz}}{(n+1)^{3/2}}$ then, $$\lim_{n\to\infty}\frac{u_{n+1}}{u_n}=\lim_{n\to\infty}e^{2\pi i z}(\frac{(1+1/n)}{(1+2/n)})^{3/2}=|e^{2\pi i z}|.$$

Now, my question is, is the inequality, $exp(2\pi iz)\leq 1$ always true?

Ben Steffan

the inequality does not make sense

Thomas Finley

@BenSteffan I thought so, so that means ratio test is inconclusive.

Root test don't seem to work either

Ben Steffan

it does not make sense

it's not well-formed

Thomas Finley

17:16

@BenSteffan which one?

Ben Steffan

I only see one inequality

Thomas Finley

Yeah, this inequality :$exp(2\pi iz)\leq 1$ is fake:P

Sine of the Time

1 min ago, by Ben Steffan

it does not make sense

Ben Steffan

you're trying to compare complex numbers

anyways, $|\exp(2 \pi i z)| \leq 1$ is also false in general, as you can easily verify :)

Thomas Finley

@BenSteffan ughh... sorry for the huge typo, I was thinking about $|\exp(2 \pi i z)| \leq 1$ this whole time.

Ben Steffan

17:19

I guessed as much

Thomas Finley

@BenSteffan thanks for pointing it out tho

Now, I can assume, z=x+iy

Using this, $|u_n|=|\frac{e^{2\pi inz}}{(n+1)^{3/2}}|=|\frac{e^{-2\pi ny}{(n+1)^{3/2}}|.$

Again, $\lim_{n\to\infty}e^{2\pi ny}=\infty$ if $y>0$

This means, $$|u_n|=|\frac{e^{-2\pi ny}{(n+1)^{3/2}}|\leq 1/(n+1)^{3/2}$ for all $n\geq K$ for some $K\in\Bbb N$

Xander Henderson

Hint: $\lvert \mathrm{e}^{i z} \rvert = \mathrm{e}^{\Im(z)}$

Thomas Finley

Now, using comparison test (the version for real analysis), $\sum 1/(n+1)^{3/2}$ converges and so does $\sum |u_n|$ from where we can say, $\sum u_n$ converges if y>0, i.e. Im(z)>0

@XanderHenderson is my method correct?

Xander Henderson

@ThomasFinley You've messed up your TeX in multiple places, making it very hard to read what you are trying to convey.

Thomas Finley

@XanderHenderson ok, let me fix them... give me a sec...

Derso

17:32

@Jakobian Actually there's a nasty trick: you can represent a line in $\mathbb{R}^3$ with a single irreducible polynomial equation by putting something like $(ax+by+cz-d)^2+(ex+fy+gz-h)^2=0$, where the two polynomials $ax+by+cz-d$ and $ex+fy+gz-h$ have no common component. Of course this trick won't work on $\mathbb{C}^n$.

Thomas Finley

Here it is:

I can assume, z=x+iy
Using this, $|u_n|=|\frac{e^{2\pi inz}}{(n+1)^{3/2}}|=|\frac{e^{-2\pi ny}}{(n+1)^{3/2}}|.$

Again, $\lim_{n\to\infty}e^{2\pi ny}=\infty$ if $y>0$
This means, $$|u_n|=|\frac{e^{-2\pi ny}}{(n+1)^{3/2}}|\leq 1/(n+1)^{3/2}$$ for all $n\geq K$ for some $K\in\Bbb N$

Now, using comparison test (the version for real analysis), $\sum 1/(n+1)^{3/2}$ converges and so does $\sum |u_n|$ from where we can say, $\sum u_n$ converges if $y>0,$ i.e. $\text{Im}(z)>0.$

@XanderHenderson Up until this point, does it seem valid to you?

Xander Henderson

Seems like overkill.

And you should hope, at the end of the day, to get some estimate on $z$.

Which you don't seem to be doing.

Thomas Finley

@XanderHenderson In other words, my method seems to be working fine to you (,right?)

@XanderHenderson I am not sure what you mean by this.

Xander Henderson

@ThomasFinley No. That isn't what I said.

You seem to be estimating $u_n$, in terms of $n$. I don't understand.

Thomas Finley

?

@XanderHenderson you mean this line: $$|u_n|=|\frac{e^{-2\pi ny}}{(n+1)^{3/2}}|\leq 1/(n+1)^{3/2}$$ for all $n\geq K$ for some $K\in\Bbb N$

Xander Henderson

17:40

And I don't understand why you are using the comparison test, as this isn't going to give you everything you need (i.e. it can tell you cases in which the series converges, but it might not find all cases, and it can't tell you when it diverges, at least not without comparison to another series).

You are looking for some domain of convergence---I would really recommend doing what I suggested in the first place, and employ a standard result, such as the ratio test.

Thomas Finley

@XanderHenderson I did...

Xander Henderson

Not in the bit I replied to...

Thomas Finley

If $u_n=\frac{e^{2\pi inz}}{(n+1)^{3/2}}$ then, $$\lim_{n\to\infty}\frac{u_{n+1}}{u_n}=\lim_{n\to\infty}e^{2\pi i z}(\frac{(1+1/n)}{(1+2/n)})^{3/2}=|e^{2\pi i z}|.$$

This is what I get applying ratio test.

But how do I take it from here?

Xander Henderson

You haven't applied the ratio test.

You have computed the limit of the ratio.

The ratio test tells you something about the series, based on the value of that limit.

Thomas Finley

If $|e^{2\pi iz}|=L<1$ the series is converging

If $L>1$ then the series is diverging.

Xander Henderson

17:43

And so...?

Thomas Finley

You mean these conclusions?

Xander Henderson

Yes.

Thomas Finley

@XanderHenderson but what if L=1?

Xander Henderson

The ratio test is inconclusive.

But you generally don't care.

Convergence on the boundary typically doesn't matter.

Thomas Finley

@XanderHenderson yea... but the original question asked me to find the region of convergence.

Xander Henderson

17:44

What is your definition of "region of convergence"?

Usually, it is taken to be an open set.

Thomas Finley

@XanderHenderson My defn assumes it can be a closed set, i.e. boundary points may get included.

Xander Henderson

What is the precise definition?

Thomas Finley

@XanderHenderson let me write it up...

Xander Henderson

That said, you generally aren't going to get convergence on the boundary, as the terms are going to oscillate around some circle.

Thomas Finley

@XanderHenderson actually, I am following Schaum series and it seems they haven't precisely defined it. But in some example problems of this type, they are examining the convergence at boundaries.

@XanderHenderson so how to prove that its divergent there?

I am not sure about this either. What I mean is, it might so happen that for some boundary points it may converge and for others it does not.

But I doubt whether this will be the case at all...

Any ideas?

If L=1 then, it means, $|exp(2\pi i z)|=1$

So, $$|u_n|=|\frac{e^{-2\pi ny}}{(n+1)^{3/2}}|=1/(n+1)^{3/2}$$ if L=1

Now, $\sum 1/(n+1)^{3/2}$ is convergent.

So, $\sum |u_n|$ is convergent by comparison test.

Hence, $\sum u_n$ converges

@XanderHenderson Did I get it correct?

user10478

18:02

In an (unconstrained?) optimization problem, is there an interpretation of the function you get if you set only some of the variables' partials to zero and plug the result back into the objective function, as opposed to the entire gradient?

Thorgott

18:38

@Derso It also does not possess the structure of a Lie group

I feel like there should be an elementary argument for this, but I actually don't have

one (hopefully overcomplicated) way of seeing it is that if $S^2\times\mathbb{R}$ were a Lie group, it would be homeomorphic to a product $K\times\mathbb{R}^n$, where $K$ is a maximal compact subgroup. in particular, $K\simeq S^2$. now, $K$ is simply connected, so $n\le1$, and $S^2\times\mathbb{R}$ is non-compact, so $n>0$, hence $n=1$ and $K$ is a $2$-manifold, but then classification of $2$-manifolds implies $K\cong S^2$, but we already know $S^2$ does not possess the structure of a Lie group

copper.hat

19:30

@user10478 that's probably a bit broad to given any conclusive answer

Derso

@Thorgott So, "every Lie group is homeomorphic to a product $K\times \mathbb{R}^n$ with $K$ is a maximal compact subgroup". I lack these well-known results. Thank you!

user10478

19:52

@copper.hat Hmmm, not sure in which way exactly. For example, if I have a function z of x and y, and I just take a partial wrt x and set it to the constant 0, I'll get some relation between x and y which I can substitute into the objective function to reduce it to either z of x or z of y.

copper.hat

again, there are many presumptions there. is the 'coordinate-wise' stationary point a $\min,\max$ or other? is it unique? etc, etc. at some level, Lagrangian style optimisation (augmented Lagrangians, etc) does this.

coordinate-wise descent basically tries to cycle through stationary points to reach an optimum.

its a pretty open question is what i meant.

open as in broad

user10478

Okay, I'll look up those terms, thanks.

Claudio

@copper.hat I looked at your comments on yesterday problem: I asked my professor for the solution pdf and she gave it to me. It seems the solution is more complicated than what I expected, I had a similar reasoning to yours but it seems it's not quite as simple

copper.hat

@Claudio yeah, i think i presumed the existence of a global $\min$.

Claudio

$f(x,y) = (x^2-(y+1)^2)y e^{-y}$ and $D = \{(x,y) \in R^2: y\ge 0, |x| \le y+1\}$. The only stationary point in the interior is $x_0 = (0,1+\sqrt{2})$ and $f|_{\partial D} = 0$ and $\lim_{\lVert (x,y)\rVert \to \infty} f = 0$

copper.hat

20:03

no wait. give me a minute to find your original question.

Claudio

I just added the actual function and domain

the problem is the same

I forgot to include yesterday that the limit above is valid for $(x,y) \in D$

and $f(x_0)<0$

copper.hat

it is possibly i am misunderstanding your notation.
from my understanding, the set $\{ (x,y) | f(x,y) \le {1 \over 2} f(x_0,y_0) \}$ is compact, hence there is a global $\min$ (on $D$).

sometimes you use $(x,y)$ and also $(x_0)$, so i may be missing something.

Claudio

my definition of compact is: a closed and bounded set

copper.hat

yep.

Joe

I am reading the following definition from an algebraic geometry book. In what follows, $k$ is an algebraically closed field, and $V\subseteq k^n$ is an algebraic set. Let $U$ be an open subset of $V$. A function $f:U\to k$ is regular at $P\in U$ if there exists $g,h\in k[V]$ with $h(P)\neq0$ such that $f(Q)=g(Q)/h(Q)$ for all $Q$ in some neighbourhood of $P$.

Question: when we say $U$ is open, do we mean open relative to $V$ (with the subspace topology), or open relative to $k^n$? Similarly, what topology is intended when we say "neighbourhood of $P$"?

copper.hat

20:10

@Claudio if $f \to 0$ at the boundary and as the parameters get large, and there exists a point with value $<0$ then a global $\min$ must exist.

Ben Steffan

@Joe relative to $V$

here everything is supposed to happen relative to $V$

Claudio

@copper.hat if the set I'm working on is compact then I can conclude that $f$ has global max and min in said set

the problem is that I must show that $x_0$ is indeed that global min

copper.hat

by global, i mean on $D$. but $D$t is not necessarily compact, but it is closed.

however, the set of points that have value $\le {1 \over f(x_0)}$ (which is negative) is compact (given your assumptions).

if not, there would be an unbounded sequence whose value is $\le {1 \over f(x_0)}$ which would contradict the assumptions.

Claudio

@copper.hat $f(x,y) \le a, a = const$ is closed. But you're saying that in $D$ it also becomes compact?

copper.hat

$a$ is not arbitrary, it must be $<0$ (strict).

but yes, if $f$ is continuous, then $f \le a$ is always closed.

$f(x_0) <0$ is the important thing here.

Claudio

20:19

it must be $< 0$? Is this because $f|_{\partial D}=0$?

copper.hat

? i thought that $f(x_0) < 0$ was in the original question?

Claudio

yep

it's just I'm not good with working with sequences...

copper.hat

because $f$ 'goes to' zero at the boundary and $\infty$, any set of the form $f\le a$, with $a<0$ is bounded.

you have the assumption $\lim_{\lVert (x,y)\rVert \to \infty} f = 0$.

Claudio

I see it I see it, so I do not need to say that if it wasn't then I could find a sequence whose....

your last two messages are enough I suppose

copper.hat

there always exists a global (on $D$) infimising (as opposed to minimising) sequence, but if this is unbounded then the value must have values that get close to zero, which contradicts the fact that the infimising value must be $\le f(x_0) < 0$.

Claudio

20:27

if I now work in $C = \{f(x,y) \le 1/f(x_0)\, (x,y) \in D\}$ then $x_0$ is the global min(this is easy to prove)? But is it also in $D$?

copper.hat

Sorry, I mean $(x,y) \in D$ too.

So $C \subset D$.

I should have written $\{ (x,y) \in D | f(x,y) \le {1 \over 2} f(x_0,y_0) \}$

The line of argument is (1) a global (on $D$) $\min$ exists, (2) we know the value $<0$ and so the $\min$ occurs at an interior point, (3) this point must be stationary, (4) there is at most one stationary point; hence we conclude that $x_0$ is the global (on $D$) minimiser.

The crucial element of (1) is that $C$ is compact.

Claudio

@copper.hat I'm confused about many things: I think what you proved is $x_0$ is global min on $C\subset D$

copper.hat

Let $L= \{ (x,y) \in D | f(x,y) \le {1 \over 2} f(x_0,y_0) \}$, do you see that this must include any global (on $D$) minimiser of $f$?

If there was a point of lower value in $D$ then it must be in this set, by definition.

It think the specific values may be confusing you, the important thing is that I pick some level set $L=\{ (x,y) \in D | f(x,y) \le \alpha \}$ for some $\alpha<0$ such that the set $L$ is not empty.

Claudio

I think I understand: my objection is that we should take $f\le f(x_0,y_0)$ because you're excluding a part of possible candidates

Oh lol

yeah my bad $f(x_0) < 1/2 f(x_0)$ since we're working with negative level sets

lol

yeah yeah it makes perfect sense

I was confused as to why your $L$ was non-empty :P

this is much easier, dang

@copper.hat thanks btw

copper.hat

20:48

glad to be able to help :-)

Claudio

I don't know if you're interested, but the pdf solution is more contrived, that's the only word I can think of. I could've never come up with something like that on my own. I don't know why they write these unnecessarily longer proofs which the average student can't obv come up with (I stress average)

maybe they are showing off :P

copper.hat

21:20

@Claudio :-) beauty is in the eye of the beholder.

it takes a lot of work to provide simple, clear solutions, and then folks say, "wow, that was easy".

Ben Steffan

chances are apparently higher if the beholder is fluent in italian :)

copper.hat

i wish i could master a single language other than English. 5 years of French, many more years of Irish, 5 classes of Spanish and still I have none.

Ben Steffan

who ever masters a language

copper.hat

i would settle for a lot less than mastery too

Ben Steffan

I think my german is pretty good, not to brag but

copper.hat

21:23

i suspect necessity helps?

Ben Steffan

in my case being a native speaker has actually helped a lot :^)

copper.hat

lol

Ben Steffan

my english is still kind of rough

perpetually unhappy with it

copper.hat

lol really doesn't convey the extent of amusement :-)

one good thing about English is that it is used by so many groups that one does not need to be formally correct

besides, what is correct is time varying...

Sine of the Time

miss the times when mathematicians used to communicate using latin

Claudio

21:26

@SineoftheTime what you think of that solution

Ben Steffan

...said the italian

copper.hat

i do not miss the Latin mass

Ben Steffan

I like latin, but not enough to do mathematics in it

Sine of the Time

@Claudio which solution?

copper.hat

i like Latin & modern Greek, purely for the structure

Ben Steffan

21:27

I think we should only use classical chinese

Claudio

the one I posted above, in Italian. I consider it contrived but I am no mathematician :p

Ben Steffan

at least that would solve all our notational issues

copper.hat

@SineoftheTime about page up

Claudio

@BenSteffan Ah yes, classical chinese, well-known to be amongst the easiest languages out there to master

Ben Steffan

would be a good fit for mathematics, well-known to be amongst the easiest subjects out there to master

copper.hat

21:29

I do have a smattering of Mandarin

Claudio

I might have to star this @BenSteffan

Ben Steffan

do what you must

@copper.hat so do I

mandarin is fun as long as you don't have to actually speak it

as soon as you open your mouth it becomes an uphill battle of a single man against four tones

Claudio

@BenSteffan It is starrable but without my message it doesn't make sense, so I will refrain from doing it

copper.hat

I spent a month or so in China, it was mostly street level stuff for food, directions, accommodation. this was in '96

Ben Steffan

damn, hard to make a dollar in the star economy these days

we're in a star recession

@copper.hat nice, that sounds fantastic

copper.hat

21:32

regarding nice solutions. i used to help one of my son's friends with high school mathematics. smart kid, poor teaching. years later he thanked me. he said, "I thought that I was stupid, but when I watched the way you solve problems, I realised that I am not stupid" :-)
I was hoping he was complimenting my incredible teaching, but instead my swarm of mistakes approach to solving problems was what got him over the hump.

Ben Steffan

I have to correct an exercise now for which only 3 out of 10 groups handed in a solution

and I haven't come up with a solution myself yet

wish me luck

copper.hat

:-). good luck

Claudio

@BenSteffan pain, I can sense it through the screen

@copper.hat you are indeed a good teacher, since you managed to help me, and I'm terrible at this math thing

copper.hat

@Claudio You are not, I have just seen this sort of thing many times.

i took measure theory from a Berkeley prof named Paul Chernoff, the guy correcting my homework complained that just reading my solutions was painful because of the length.

i still have his comment somewhere in my mess of a basement.

Claudio

@copper.hat May I ask if you're a professor?

lol engineer

just read your bio, my bad

copper.hat

21:37

engineer :-)

i love mathematics, but am not a mathematician

a dangerous siren

I have a number of Italian connections, the most famous being this guy en.wikipedia.org/wiki/Alberto_Sangiovanni-Vincentelli

And I know the family of the most famous Fish & Chip shops in Dublin :-) rte.ie/archives/2016/0525/790821-the-oldest-fish-and-chip-shop

well, i know one of the family.

my house was owned by a 3rd generation Jacuzzi. The Jacuzzi family used to be a world wide conglomerate that produced all sorts of things including aircraft.

Ben Steffan

my biggest claim to fame is that I have a friend who met Werner Herzog once

leslie townes

imagine working for the weapons division of Jacuzzi and getting prank calls from bored teens about spa baths

copper.hat

lot of lol here atm

Ben Steffan

@leslietownes imagine shooting a Jacuzzi

copper.hat

i'm thinking of heading to Moe's in a while to search for your Folland

very impressed by the book. wished i discovered it decades ago

before Durrett

and Rudin

Claudio

21:44

@leslietownes lol

copper.hat

Countless hours lost with them

They were a pretty amazing family

Sine of the Time

@Claudio It reminds me of a similar result in one variable calc (a sort of Weierstrass theorem for infinitesimal functions). Basically your professor choose a compact set to prove the existence of a min point and then argued that it's a global min

copper.hat

but the tensions between the various international Jacuzzi families became so intense that their lawyers advised them to sel it all off, which they did. slightly inaccurate encapsulation of history, but roughly right

Claudio

@copper.hat I hereby award you with a temporary Italian citizenship pass (valid only on MSE and correlated sites for 2 hours max)

copper.hat

:-) i'm on my way to visit my son in Milan...

i wish

leslie townes

21:46

copper: and one home loan from saddam hussein later, and you were the proud owner of laundered jacuzzi property

copper.hat

my da used to negotiate with gadaffi, before he rose in the ranks

Claudio

@leslietownes you might be onto something no jokes

copper.hat

telex machines and all that

leslie townes

copper: so does your house even have one? it's OK to brag

Ben Steffan

who is jack cuzi

copper.hat

21:49

noooo. i only found out about the Jacuzzi connection decades later

our house is falling apart, i do not have enough $ to fix it. poor financial management here...

Claudio

@SineoftheTime I understand your point, but I still think they could've come up with something shorter maybe

copper.hat

need your driveway repaired? i am your man.

there is a Jacuzzi street in Albany (California)

Claudio

@copper.hat if you're coming to Milan, remember that italian$\ne$ the Sopranos' italian

copper.hat

and it had my favourite cafe granmilan.com

@Claudio I could say similar if you were to visit Ireland :-)

Claudio

hhahahah

copper.hat

21:51

but thankfully the IRA never reached the same degree of movie fame

i have been to Milan a few times. sgs thompson, if i remember correctly

Sine of the Time

@Claudio maybe, or at least explaining the idea behind the computation

copper.hat

My son is working a farm near Milan atm.

Thorgott

@Ben fun fact: if $f\colon X\rightarrow Y$ is a map of simplicial sets, $Y'\subseteq Y$ a subset and $f'\colon X'\rightarrow Y'$ the pullback, then if you apply the small object argument to obtain factorizations $X\rightarrow Z\rightarrow Y$ resp. $X'\rightarrow Z'\rightarrow Y'$, then $Z'\rightarrow Y'$ is again a pullback of $Z\rightarrow Y$

copper.hat

oh,i forgot, this is a mathematics chat room :-)

Thorgott

(probably true in any topos)

Claudio

21:53

@copper.hat lol

Ben Steffan

@Thorgott that's neat

Sine of the Time

Did you eat pandoro or panettone? @copper

Ben Steffan

I will probably never use it but

Thorgott

if you're ever at the bottom of p.287 in HTT and wonder why $K^0\rightarrow\mathcal{K}^0$ is inner-anodyne, this is a reason why

copper.hat

@SineoftheTime i don't have a sweet tooth really, but they have a range of savoury items. and i like the folks who work there

the latter being more important to me than the food :-)

Ben Steffan

21:55

since you're here, do you see a concise argument for the following @Thorgott
Let $X$ be a compactly generated Hausdorff space, $A, B \subseteq X$ subspaces such that the inclusions of $A$, $B$, and $A \cap B$ into $X$ are cofibrations. Then the inclusion of $A \cup B$, too, is a cofibration.

Sine of the Time

In this season, you'll find a lot of sweet food

November/December

Thorgott

I know an argument, concise might be too much to ask

Ben Steffan

I'd like to hear it nevertheless, if you have the time :)

Thorgott

it's a theorem by Joachim Lillig, I know the proof is in ch. 5 of tom Dieck, but I'd have to read it again to explain it precisely

Ben Steffan

ah

well that's reassuring

because that's the proof people have been citing in their homework submissions :)

Thorgott

22:06

it works by explicitly verifying the HEP by taking extensions relative to $A$ and $B$, homotoping them on $A\cap B$ and then interpolating them to yield an extension on $A\cup B$

Ben Steffan

I guess I'll have to read it

thanks for the pointers

Thorgott

the proof in tom Dieck is pretty short (just not super intuitive), Lillig's original theorem is actually a bit more general

cause he allows the cofibrations to be non-closed, in which case you need the ridiculous additional hypothesis $\partial A\cap\partial B\subseteq A\cap B$

there's also a paper "Mayer-Vietoris-Prinzipien für Cofaserungen" by Jürgen F. Kraus where he combines Lillig's theory with Dold's partitions of unity to prove some more generalizations of these questions, but I think this type of problem stopped being investigated after

Ben Steffan

we're very much not interested in non-closed cofibrations

@Thorgott good luck getting this to be true in the wild lol

Thorgott

@BenSteffan I've wasted a non-trivial amount of time on them and assure you that you're not missing out on anything

Ben Steffan

my desire to do much homotopy theory at this level is low to begin with, but teaching is teaching

getting a refresher on these things is not a bad idea anyways

Thorgott

22:15

this old-fashioned homotopy theory is pretty nice at times, but it's also just basically point-set topology at other times

Puppe's original papers are really nice

Pizza

22:32

@SineoftheTime I didn't understand what formula you used

That is, I used the substitution

I mean, how did you get to this result?

Sine of the Time

@Pizza your substitution with the tangent is fine

Pizza

22:47

@SineoftheTime yes, but I don't understand how you did it?

maybe that's what you meant, did I misinterpret the message?

Sine of the Time

@Pizza one way is subsituting $x=a \tan u$, but there are books with these results

I don't do these integrals each time

there are tables with these standard results

Pizza

@SineoftheTime Yes but it's not enough that for example instead of 3/2 I wrote 5/2 in the exponent, the result changes

Maybe that was a known integral?

Sine of the Time

@Pizza $\int (x^2+a^2)^{-5/2} dx$ ?

Pizza

@SineoftheTime Yes, when I have for example a²+x² i can use x = tan u, for x²-a² i can use x = a sec u and for a²-x² , x = a sin u

@SineoftheTime yes

Sine of the Time

yes, these are the standard substitutions. What I'm saying is that there are tables with these integrals, so I usually use them instead of doing the computation with the subsitution

These are well known results, even if no one usually knows the antiderivative by heart

Pizza

22:56

ah ok

Sine of the Time

Lists of integrals

Integration is the basic operation in integral calculus. While differentiation has straightforward rules by which the derivative of a complicated function can be found by differentiating its simpler component functions, integration does not, so tables of known integrals are often useful. This page lists some of the most common antiderivatives. == Historical development of integrals == A compilation of a list of integrals (Integraltafeln) and techniques of integral calculus was published by the German mathematician Meier Hirsch (also spelled Meyer Hirsch) in 1810. These tables were republished...

since I'm lazy but I know how to solve them, I put on wolfram. You must do them since you can't use wolfram during the exam :)

I think that, when someone has enough experience, can skip those details and write the final result

Pizza

I could have discovered that list earlier :O

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$Mathematics$ Mathematics