Mathematics - 2024-10-30

think_meaning_builds

01:18

@XanderHenderson ANOTHER Grandslam & the Yankees are still alive!

copper.hat

02:15

still wondering what a grandslam is other than my mother's punishment for talking back

Xander Henderson

03:08

@think_meaning_builds Boo!

think_meaning_builds

04:16

Grand slam (baseball)

In baseball, a grand slam is a home run hit with all three bases occupied by baserunners ("bases loaded"), thereby scoring four runs—the most possible in one play. According to The Dickson Baseball Dictionary, the term originated in the card game of contract bridge, in which a grand slam involves taking all the possible tricks. The word slam, by itself, usually is connected with a loud sound, particularly of a door being closed with excess force; thus, slamming the door on one's opponent(s), in addition to the bat slamming the ball into a home run. == Highlights == === Players === Roger Connor is...

think_meaning_builds

07:35

cf superover

The Empty String Photographer

08:07

@BenSteffan Don’t worry. I was sufficiently careful with my symbols.

Thomas Finley

10:34

Can someone please xlarify the meaning of a non dense subset in a metric space. I thought a subset A in ametric space X is said to be non-dense iff it's not dense, i.e. $cl(A)\neq X.$ But strangely enough, our professor gives a weird complicated definition for the same: A set is said to be non-dense iff there exist for every open sphere $B(x,\epsilon)$ another open sphere $B(x_1,\epsilon_1)\subseteq B(x,\epsilon)$ such that $B(x_1,\epsilon_1)$ is free from any point of the set $E.$

Can someone please clarify which is a relatively standard notion?

In the internet I find no articles about non dense sets. Only the term nowhere dense is there everywhere. That makes me even more confused.

Sine of the Time

There are different definitions of dense subset and they're all equivalent

Ryder Rude

11:17

hi

Jakobian

@ThomasFinley what your professor calls "non-dense" is called nowhere dense

I never heard of "non-dense" but according to this site for example, its another name for nowhere dense, so I guess some people do use it

the idea is that a nowhere dense set is not only not dense, but its not dense in any open non-empty set

@Jakobian @Claudio this formula by the way, it should be true for partial derivatives, not sure what it would mean for gradients I guess there is some version for that too

Thomas Finley

13:14

@Jakobian Ok, I get it. That makes a lot more sense now. Thanks!

Thomas Finley

13:29

@Jakobian In the definition: a set is said to be non-dense iff there exist for every open sphere $B(x,\epsilon)$ another open sphere $B(x_1,\epsilon_1)\subseteq B(x,\epsilon)$ such that $B(x_1,\epsilon_1)$ is free from any point of the set $E.$ we can also replace the open spheres by open sets and even then it seems to coz no problem, isn't it?

Then it becomes: a set is said to be non-dense iff there exist for every open set $U$ another open set $U_1\subseteq U$ such that $U_1\cap E=\phi.$

Imho these two definitions seems to be equivalent, right?

Jakobian

@ThomasFinley no, this is not quite correct

in here both $U$ and $U_1$ need to be non-empty

wait no that doesn't matter

sorry, yes its correct

Thomas Finley

14:16

@Jakobian ok, thanks for the help!

if $B(x,r)\subseteq B(x,s)$ then can we always say that $r\leq s$ in a metric space $(X,d)$ ?

($B(x,r)$ is an open ball of radius r>0 centered at x.)

Sine of the Time

Draw a picture in $\Bbb R$

Thomas Finley

@SineoftheTime In $\Bbb R$ that is trivially true

Also in $\Bbb R^2$ it's trivially true.

In general, for $\Bbb R^n$ it is true.

But I'm talking about any random metric space $(X,d).$

I can prove that this is indeed valid for any metric space $(X,d)$ if the following statement holds: Let $x\in X.$ For every $r>0$ there exists a $y\in X$ such that $d(x,y)<r$

Again, my question is: Is the above above statement really valid?

@SineoftheTime You see the problem, now?

Sine of the Time

@ThomasFinley I misread. I switched the radius with the center :D

Thomas Finley

@SineoftheTime Ah, I see.

But what d' ya think? It's not valid in general, right?

I'm talking 'bout the validity of this:
if $B(x,r)\subseteq B(x,s)$ then can we always say that $r\leq s$ in a metric space $(X,d)$

Sine of the Time

14:40

Can't you take for example $z\in B(x,s)\setminus B(x,r)$ ?

Assuming $B(x,r)\subsetneq B(x,s)$

Jakobian

@ThomasFinley no

@SineoftheTime the thing is that equality can hold

for example in discrete metric space, or even $\mathbb{R}$ with the metric $d(x, y) := \max(1, |x-y|)$

That for $r < s$ there is a strict inclusion $B(x, r)\subsetneq B(x, s)$ holds only in some particular class of metric spaces

ModularMindset

Are certain groups of mathematicians interested in non-varieties whose singular loci are varieties?

Sine of the Time

14:57

@Jakobian right, with the equality there are problems

Pizza

Hi

Sine of the Time

@Pizza Hi

ModularMindset

Hi

Anyone interested in reading my manuscript?

Pam Munoz Ryan

hi

ModularMindset

hi

Pam Munoz Ryan

15:10

can ayone help me solve a problem

Sine of the Time

@PamMunozRyan Just write the problem

Pam Munoz Ryan

and then post it?

or write it in here

ModularMindset

Just don't ask to ask.

Sine of the Time

write it

instead of asking if someone can help

Pam Munoz Ryan

Let the nonempty sets $A,B$ be with the property that $A \cup B= N^* $ and
they are disjoint. Determine the functions $f:N^* \to N^* $ with the property that
$f(m+n)=f(m)+f(n)$, for any pair $(m,n) \in (A \times A) \cup (B \times B)$.

So I thought of taking the case where both are zero, then we get the fact that $f(0)=0$.

Another interesting thing I discovered is that if we take $n \to -n$ then we have a periodic function $f(m)=f(m-n)+f(0)=f(m+n)$.(idk if this stands cus we have natural numbers but anyway)

Also $f(2m)=2 \cdot f(m)$.

Vladimir Lysikov

15:18

What is $N^*$? Natural numbers?

Pam Munoz Ryan

Natural numbers without zero

Jakobian

wait maybe not, yeah sorry

Governor

How would I prove for two PMFs $P$ and $Q$ on countably infinite support $A$, if $D(P||Q) < \infty \implies \sum_{x} P(x) |\log(\frac{(P(x)}{Q(x)})| < \infty$. I tried to split them over the positive and negative part sums and show each one is finite, but I can't apply the log-sum inequality to do so because I have an infinite support

Defining the set $A^+ = \{x \in A | \frac{P(x)}{Q(x)} \ge 1\}$ and taking $A^-$ to be the complement I can split $D(P||Q)$ into two sums $S_1$ & $S_2$ over each disjoint support. It suffiices to show that $S_1$ (positive sum) or $S_2$ (negative sum) is finite, the easiest way I think is to apply jensens inequality in the form of the log-sum inequality but I can't do that since I don't know if the sum converges absolutely.

Other way I think is to basically show that $S_1$ and $S_2$ can't be infinite, so the partial sums can't be conditionally convergent - however I am unsure of how to start this. I think there is a simpler way...any advice??

$D(P||Q) = \sum_x P(x) \log(\frac{P(x)}{Q(x)})$ - probably want to use that the sum of the pmfs is 1 somehow but I can't see it...

Pizza

@SineoftheTime Can I show you if I'm doing the exercise right?

It's about the residues

Sine of the Time

@Pizza Just ask; don't ask to ask

Pizza

15:33

Ok

$\oint_C \frac{e^z}{\cosh z} dz$ where $C$ is the circle defined by $|z| = 5$

I started by writing $\cosh(z)$ in another form

$\frac{e^z + e^{-z}}{2}$ so the function becomes $\frac{2}{1+e^{-2z}}$

To find the poles I wrote $e^{-2z} = -1$ , so $e^{-2z} = e^{i\pi(2k+1)}$

$z = -\frac{i\pi(2k+1)}{2}$

$\left| -\frac{i\pi(2k+1)}{2}\right| < 5$ so $\quad (2k+1) < 10/π$

$10/π \approx 3.18 , k < 1.09$

So I took $k = 1,0,-1,-2$ , beyond that it did not respect |z| < 5

So I get $z = ± 3πi/2$ and $z = ± iπ/2$

So far Is fine ?

Sine of the Time

it seems ok

Pizza

15:52

Ok , each pole is a simple pole

$\operatorname{Res}(f, z_k) = \lim_{z \to z_k} (z - z_k) f(z)$

I'll try to calculate the limits for a moment

Oh no maybe it was better to do it this way

$f(z) = \frac{g(z)}{h(z)}$

$\operatorname{Res}(f, z_k) = \frac{g(z_k)}{h'(z_k)}$

Sine of the Time

well, it's the same formula at the end of the day

Pizza

16:10

oh yeah, anyway I managed to find 8πi luckily

But is there a way to check these kinds of exercises on wolfram?

Sine of the Time

wolframalpha.com/…

Pizza

@SineoftheTime Do you know other ways to solve the exercise?

Oh anyway thanks for the link

Sine of the Time

@Pizza these exercises are solved using Cauchy formula or residue theorem

same thing

Pizza

More than anything I didn't quite understand if that part where I approximated, can be done in some other way, I don't know

Sine of the Time

you have to compute the modulus, so no I don't think there's a different way

Pizza

16:21

ah ok

psie

17:01

Consider the inverse function which we all know and love. It says that if the function is $C^1$ and the determinant of the Jacobian matrix does not vanish at a given point in the domain, then we are guaranteed a $C^1$ inverse in a neighborhood around that point. Now suppose the determinant of the Jacobian matrix does not vanish at each point of the domain. Does this mean we can form a global inverse of all the local inverses we obtain at each point?

Pam Munoz Ryan

Determine the functions $f:N^* -> N^*$ with the property that:$\frac{3}{f(1)+f(2)+...+f(n)}= \frac{4} {f(n)}- \frac{4}{f(n+1)}$ for any $n \in N^*$

The first thing that came in my mind is that formula that says that $\frac{1}{k}-\frac{1}{k+1}=\frac{1}{k(k+1)}$

psie

EDIT: I meant the inverse function theorem.

Pam Munoz Ryan

And maybe we can take all these functions in a member and process them. I tried doing it but to be honest in got to nothing useful

Claudio

@psi I'm studying it right now and I think no

psie

I think so too, but why?

Claudio

17:14

take the ring $A = \left{ 1<\sqrt{x^2+y^2}<3} \right}$

and consider $\psi(x,y) = (x^2-y^2,2xy), \psi \in C^1(A)$

$det(J\psi(x,y))>0 \text{ for } (x,y) \in A$

but obviosly the function is not injective

psie

@Claudio do you think you can post the ring again? You need to put \} whenever you want to use curly brackets.

Or \{

Claudio

psie

Ok. 👍

Claudio

this is an example from my book

but obv the theorem gives you local invertibility so u can take a small ball around either of those two points and you're fine

@psie u can find ulterior conditions to obtain global invertibility

I haven't studied them though

psie

hmm, ok

Claudio

17:22

it's a long proof tho

I mean very complicated

I'd link you the book but it's in Italian only unfort

psie

@Claudio isn't there a typo in the example? Shouldn't the determinant be $4x^2+4y^2$?

Claudio

yeah hahahah my bad :p

thanks

Sine of the Time

@Claudio are these notes from a course you're following or are you studying on your own?

Thorgott

@psie you were already given a counter-example, but perhaps the most instructive counter-example is the complex eponential $\mathbb{C}\rightarrow\mathbb{C}^{\times},\,z\mapsto\exp(z)$

Claudio

I'm re-following the course since I passed the written exam last time and I thought I could maintain the vote for the next session and then do the oral part of it but I actually couldn't. But the course material expanded and changed a bit (they added a small measure theory part)

and so here I am, but actually these notes are mine not my professor's, this is just the same example of the book which I kind of expanded with some drawings and additional things hahahha

Sine of the Time

17:33

nice

the course seems well done

Claudio

yeah the professor is super nice and explains wonderfully

I need a little of measure theory anyways, especially for some more rigourous QM

psie

ah yes, the complex exponential, I will study it in closer detail

Claudio

maybe some operator theory would be nice but time is limited hahaha

Sine of the Time

There are operator theory courses in the math department (master degree usually)

Claudio

yeah I know that's why I firmly believe Ill never get to it

Sine of the Time

17:40

:(

Claudio

maybe one day I will finally read some parts of Teschl's book

Thomas Finley

18:13

@Jakobian ok. I get it.

Thomas Finley

18:24

My professor's says, that: a subset $A$ of a metric space $(X,d)$ is said to be compact in X iff every sequence in $A$ has a convergent subsequence that converges to a point in X which is not necessarily in A. However, a subset A of a metric space $(X,d)$ is said to be compact iff every sequence in $A$ has a convergent subsequence that converges to a point in A.

Is this distinction between the phrases, "a compact subset of a metric space" and "a subset is compact in a metric space X" very standard?

Ben Steffan

These are two definitions of the same thing

In other words, no

Also the first definition is simply wrong

are you sure you're quoting it correctly

it would be right if you assume that $A$ is closed, say

Thomas Finley

18:44

@BenSteffan umm yes I think...

Ben Steffan

that doesn't exactly sound like you're certain :)

Thomas Finley

Our professor gives an example : (a,b) is compact in [a,b] where a<b and [a,b] is a subspace of the metric space R equipped with the absolute value metric.

@BenSteffan :?)

Ben Steffan

!?

Thomas Finley

@BenSteffan ?

Ben Steffan

if you tell pretty much anybody that $(a, b)$ is compact you will be laughed out of the room

I have no idea what your professor wants with this, but it is very much non-standard

Thomas Finley

18:50

@BenSteffan That's the reason why I'm posting it in here. But note the phrase (a,b) is not compact but it's compact in [a,b]

Did that make any sense?

@copper.hat Hey how are you? Long time no see...

I hope ur doing good

Ben Steffan

I see the distinction, but it's still just horrendous :/

Thomas Finley

@BenSteffan I agree LOL

Ben Steffan

no idea why anybody would do this

but yeah, it's very much non-standard

the standard definition here is "a subset of a metric space is compact if every sequence has a subsequence converging in the subset"

Lurie is very funny. "The construction of $\mathrm{Mod}_A^\mathcal{O}(\mathcal{C})^\otimes$ is fairly straightforward" proceeds to give a highly technical definition 2 pages long

Jakobian

@ThomasFinley makes no sense

Relatively compact subspace

In mathematics, a relatively compact subspace (or relatively compact subset, or precompact subset) Y of a topological space X is a subset whose closure is compact. == Properties == Every subset of a compact topological space is relatively compact (since a closed subset of a compact space is compact). And in an arbitrary topological space every subset of a relatively compact set is relatively compact. Every compact subset of a Hausdorff space is relatively compact. In a non-Hausdorff space, such as the particular point topology on an infinite set, the closure of a compact subset is not necessarily...

but compare with this concept

Thorgott

19:09

@BenSteffan classic Lurie

Ben Steffan

how am I going to talk about modules in my talk

Thorgott

I've been torturing myself for the majority of today and am this close to understanding what it means to state that an $\infty$-topos is a presentable $\infty$-category in which all colimits are van Kampen :)

Ben Steffan

:)

Thorgott

it's a mixture of absolute torture and genuine beauty

I learned yesterday that an edge is cartesian for the codomain fibration $\mathrm{Fun}(\Delta^1,\mathcal{C})\rightarrow\mathcal{C}$ if and only if it classifies a cartesian square in $\mathcal{C}$

Thomas Finley

@Jakobian hehe... but guess what we have to follow our professor coz he will be grading in our exams...

Now, here's a general question:

Thorgott

19:14

so if $\mathcal{C}$ has pullbacks, the codomain fibration is cartesian; and it is always cocartesian

if you now pull this back along a map $\Delta^1\rightarrow\mathcal{C}$ classifying a morphism $f\colon x\rightarrow y$, you have a bicartesian fibration over the interval, which, by straightening-unstraightening both ways, classifies the pushforward-pullback adjunction $f_{\ast}\colon\mathcal{C}^{/x}\leftrightarrows\mathcal{C}^{/y}\colon f^{\ast}$

strikingly elegant imo

Thomas Finley

I found the following definitions for a bounded set in a metric space: Let A be a subset of a metric space (X,d).

Defn 1: A is bounded iff the diameter of A is finite (The defn of diam of a set is the supremum of all d(x,y)s such that $x,y$ are elements in the set)

Defn 2: A is bounded iff $\exists p\in X$ and a $B\in \Bbb R$ such that $d(x,p)\leq B,\forall x\in A.$

Are these two defns equivalent?

Soumik Mukherjee

@ThomasFinley if the professor is giving incorrect definition then you should not follow them

@ThomasFinley what do you think?

Thomas Finley

@SoumikMukherjee I think they aren't equivalent. Am I right?

Coz nowhere in the 1st definition we are concerned with the elements in X. But the second defn takes it into account.

Soumik Mukherjee

@ThomasFinley I won't say

Thomas Finley

@SoumikMukherjee ugh.... why not?

Soumik Mukherjee

19:24

@ThomasFinley are we really not concerned with the elements? you even wrote the definition of diameter

Ben Steffan

@Thorgott that's really cool :o

Thomas Finley

@SoumikMukherjee We're concerned with the elements of only A in the 1st case

In the second case, an element of X is being taken into account.

Soumik Mukherjee

@ThomasFinley If you think that the definitions are not equivalent then give an example where something is bounded according to one of the definitions and not bounded according to the other

@ThomasFinley so?

Thomas Finley

@SoumikMukherjee that's my Achilles hill. I'm bad at giving examples.

@SoumikMukherjee So they doesn't seem to be equivalent...

Ben Steffan

didn't realize you've moved on to another definition, sry :)

Thomas Finley

19:30

@BenSteffan :D

Soumik Mukherjee

Few more removed messages and it'll seem like some heated arguments were going on :D

@ThomasFinley or maybe there aren't any examples to give?

Thomas Finley

@SoumikMukherjee no no everyone knows here I am an easy goin' guy, with a laid back approach in life.

:D

@SoumikMukherjee maybe. Let me try proving them to be equivalent.

Soumik Mukherjee

👍🏼

Thomas Finley

Ok, I am able to show that Defn 2 implies Defn 1 by an application of trianjle inequality.

Showing definition 1 implies dedinition 2 seems difficult @SoumikMukherjee ?

Oh wait...

Ok, if diam (A) is finite then $\exists B\in \Bbb R$ such that for all x,y in A we have $d(x,y)\leq B$. Now, let $a\in A.$ This means that $a\in X\implies d(x,a)\leq B,\forall x\in A$. So, $\exists a\in X$ such that $d(x,a)\leq B,\forall x\in A.$

Conversely, if $\exists k\in X,B\in \Bbb R$ such that $d(x,k)\leq B$ for $x\in A$ then, $\forall x,y\in A$ we have, $d(x,y)\leq d(x,k)+d(k,y)=2B.$

@SoumikMukherjee Isn't this a valid proof of their equivalence?

Soumik Mukherjee

19:45

Yes, except where you wrote 'a\in X\implies'.

Thomas Finley

@SoumikMukherjee why?

Did you mean, to say that "a\in X is not the reason why d(x,a)\leq B"?

Ok, then I get it.

It was a typo...

It's because a\in A but I just wanted to note that a\in X and $d(a,x)\leq B$

The implies should have been replaces by 'and'. Did I make any sense @SoumikMukherjee?

ModularMindset

Define the cohomology of $X$ as a direct sum over the strata: $H^k(X) = \bigoplus_{i=0}^n H^k(X_i),$ where each $H^k(X_i)$ represents the cohomology of the $k_i$-dimensional manifold $X_i$. Does $H^k(X)$ in some fashion, encode contributions of the cohomology over the individual strata, or is this an innacurate conclusion?

Ben Steffan

@ModularMindset you defined it to encode this data perfectly

well, not the relations, but all of the strata cohomology

I don't know why you would make this definition

it seems redundant

if you want to measure how the information of the strata relates to the whole, you should consider something like $\varprojlim_i H^k(X_i)$. ...but then $\varinjlim_i H_k(X_i)$ strikes me as the better option

Xander Henderson

20:04

If I put the $k$ here, it's homology, but if I move it around, it's COHOMOLOGY! I am so SMRT! Yay!

(so gross...)

Ben Steffan

yes and the arrow then goes the other way around :)

Xander Henderson

Look, ma! I'm turning arrows around!

Ben Steffan

beat ya to it

Thorgott

"dualize, dualize, dualize", they chanted

Xander Henderson

Whoop-de-****-do.

Ben Steffan

20:05

dual eyes dualize dual lies

Xander Henderson

Oh, the arrows go the other way! Look at me! I'm so cool!

Ben Steffan

#deep

@XanderHenderson yes, and that's important

it's no good when the arrow goes to the left :(

Thorgott

@BenSteffan duel lies..

Xander Henderson

@BenSteffan Right-pointing arrows are an abomination.

Ben Steffan

my favourite pastime activity

Xander Henderson

20:07

Burn them with fire.

Ben Steffan

I believe this is the core idea of condensed mathematics :^)

Thorgott

it's kind of bizarre how computing $\infty$-colimits in $\mathrm{Cat}_{\infty}$ is in a way easier than computing $1$-colimits in $\mathrm{Cat}_1$

Ben Steffan

is it?

Thorgott

20:23

perhaps not if you scrutinize enough, but the fact that there's a workable description of how to compute colimits in $\mathrm{Cat}_{\infty}$ at all is still crazy

ModularMindset

20:34

@BenSteffan Yes that is a fair point - I did calculate that the Euler char. is zero which should tell me that the cohomology (number of holes) is in some sense "balanced." Although I'm not sure what "balanced" really means yet

I do know about the betti numbers in this context

fwiw that Euler char. calcuation hints at a toroidal like structure

Ben Steffan

Is $X$ closed? What's the dimension?

ModularMindset

@BenSteffan $X$ is closed and is dimension $2$

Pizza

@SineoftheTime Have you studied Laurent series?

ModularMindset

it can be generalized to $n$-dimensions but I haven't even tried to formally work in higher dimensions

Ben Steffan

Then $X$ is a torus.

ModularMindset

20:48

ah okay that makes sense I guess

Xander Henderson

20:58

@BenSteffan You're a torus!

Sine of the Time

@Pizza yes

Ben Steffan

@XanderHenderson a circle, in fact

Sine of the Time

Are you having trouble with it?

ModularMindset

@XanderHenderson You're a torus

psie

Here's a problem from Spivak's Calculus on Manifolds, which comes right after the inverse function theorem (IFT).

> Problem Let $A\subset \mathbb R^n$ be an open set and $f:A\to\mathbb R^n$ a continuously differentiable injective function such that $\det f'(x)\neq 0$ for all $x$. Show that $f(A)$ is an open set and $f^{-1}:f(A)\to A$ is differentiable. Show also that $f(B)$ is open for any open set $B\subset A$.

Suggested solution For every $y\in f(A)$, there is an $x\in A$ with $f(x)=y$. By the IFT, there is an open subset $U\subset A$ and an open subset $V\subset\mathbb R^n$ such that $x\in U$ and $f(U)=V$. ...

Question Why does the IFT imply that $f(U)=V$? It says that $x\in U$ and $f(x)\in V$, but why would $f(U)=V$? (see the IFT below)

Ben Steffan

21:10

it's sort of implicit in the statement: If $f(V) \neq W$, then you of course cannot have an inverse $f^{-1}\colon W \to V$ :)

psie

true

@BenSteffan that does explain it, thanks! :)

Ben Steffan

welcome :)

Pizza

@SineoftheTime Yes, we did it today, but I didn't understand much

Sine of the Time

@Pizza what's not clear?

Pizza

like here

Expand : $f(z) = \frac{1}{(z - 2)^2}$

in a Laurent series valid for $(a) |z| < 2, (b) |z| > 2$

Sine of the Time

21:24

I'm rusty in those things, give me a minute

Pizza

That is, when the sign > < changes, the expansion also changes

@SineoftheTime Don't worry, I took this exercise from the book you recommended anyway

Sine of the Time

Exercise 6.27 is the standard one and very instructive. pag.188

Pizza

I'll try to see that for a moment

Sine of the Time

Once you understand ex 6.27 you'll have no trouble doing the other exercises

Pizza

21:47

Maybe I solved it

For case 1, I need to expand the function for values of z close to 0

Sine of the Time

That's too vague

Pizza

so i write f(z) = 1/(2-z)²

@SineoftheTime What do you mean?

anyway then I would have applied the geometric series

Sine of the Time

The Laurent series in the first case should be valid in the disk $|z|<2$

@Pizza So which series do you find for $(a)$ ?

Pizza

$\sum^{\infty}_{n=1} \frac{nz^{n-1}}{2^{n+1}}$

@SineoftheTime yes

Sine of the Time

@Pizza how did you find it? Do you already know the Laurent series of $\frac{1}{(z-1)^2}$ ?

Pizza

21:57

no, it was necessary to derive

Sine of the Time

How did you find it?

Pizza

$\frac{d}{dz} \left( \sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}} \right) = \sum_{n=1}^{\infty} \frac{n z^{n-1}}{2^{n+1}}$

Sine of the Time

nice!!

Pizza

If it was 1/(//)³ then I also had to calculate the second derivative

Sine of the Time

Correct

So, $|z|<2$ is equivalent to $\frac{|z|}2 <1$. In $\frac{1}{(2-z)^2}=\frac 1{(2(1-z/2))^2}$

So, a part from the $1/4$, you're interested in the LS of $\frac 1{(1-z/2)^2}$. The formula you found differentiating the geometric series works when $|z|/2<1$

So you write the series and you're done

For $|z|>2$, note that $\frac 2{|z|}<1$ and then you manipulate $f(z)$ to be in the conditions to apply the geometric series when $\frac 2{|z|}<1$

Pizza

22:11

$\frac{1}{(1 - w)^2} = \sum_{n=1}^{\infty} n w^{n-1}, \quad$ for $|w| < 1$

$f(z) = \frac{1}{(2 - z)^2} = \frac{1}{(2(1 - \frac{z}{2}))^2} = \frac{1}{4} \cdot \frac{1}{(1 - \frac{z}{2})^2}$

Sine of the Time

@Pizza correct

Pizza

did you say that?

So now I can apply that formula

For the other case I understood that I have to write like this: $f(z) = \frac{1}{(z - 2)^2} = \frac{1}{z^2 \left(1 - \frac{2}{z}\right)^2}$

Sine of the Time

Correct

Pizza

$\frac{1}{\left(1 - \frac{2}{z}\right)^2} = \sum_{n=1}^{\infty} n \left(\frac{2}{z}\right)^{n-1} = \sum_{n=1}^{\infty} \frac{n \cdot 2^{n-1}}{z^{n-1}}$

and then I multiply by 1/z²

Sine of the Time

yep

Pizza

22:23

$\sum_{n=1}^{\infty} \frac{n \cdot 2^{n-1}}{z^{n+1}}$

Sine of the Time

Is it clearer now?

Pizza

Yes, thank you very much :)

Sine of the Time

glad to be helpful

psie

23:22

In my book, they say that if $T$ is a bounded linear map between normed spaces $X$ and $Y$, then $\|Tx\|\leq\|T\| \|x\|$ holds. Here the norm on $T$ is the familiar operator norm, i.e. the $\sup$ over the unit ball of $\|Tx\|$. Looking at the proof of the inequality, it really seems to depend on the operator norm. E.g. does the inequality hold for the norm $\|T\|_\infty=\max\limits_{1\leq i\leq n}\sum_{j=1}^\infty |T_{ij}|$?

copper.hat

@ThomasFinley Hi Thomas, I drop by sporadically to see the quiet room.

Sine of the Time

Veni vidi vici

psie

@psie Ehm...I didn't mean to write infinity in the sum there, it should be $n$.

leslie townes

psie yes the proof is specific to the operator norm. note e.g. that if \| \| is one norm on a space c > 0 then c \| \| is also a norm on the same space. so if X and Y and T are given but the norms are "up for playing around with" you can potentially get whatever you want as values for the ingredients in the left and right hand sides of that inequality.

copper.hat

@psie not exactly sure what you are asking, given a norm (or norms, really, for $X,Y$) then there is an induced norm on operators. However, the norm you gave is an operator norm for $\|x\|_\infty$.

leslie townes

23:27

psie: to ask whether the inequality holds or not in some altered set of facts, you should specify all of the norms, not just the norm you plan to use on T

psie: note that what "the operator norm" is depends on what norms you choose on X and Y, and that inequality would hold for any choices of norms on X and Y as long as T is bounded and you use the corresponding operator norm for ||T||.

psie

ok, so which norm induces the norm $\|T\|=\sup\{\|Tx\|_b:x\in X,\|x\|_a\leq 1\}$?

leslie townes

we have talked elsewhere about how context is often used to supply meaning for symbols like \| \| that would otherwise be overloaded with subscripts, and that there are good reasons for this. but again, if you find yourself in a situation where you find yourself focusing on this stuff, i would feel very free to invent your own notation and decorate accordingly.

psie

Here I equipped $X$ with $\|\cdot\|_a$ and $Y$ with $\|\cdot\|_b$.

leslie townes

e.g. [as one choice not a universal choice] writing \| \|_V for the norm on the normed space V, and letting B(X,Y) denote the space of bounded linear maps, and understanding this notation to include not just the set of maps but the operator norm induced from the norms on X and Y, you'd get: \|Tx\|_Y \leq \|T\|_{B(X,Y)} \|x\|_X true for any normed spaces X, Y

psie: i'm not sure i understand what X and Y even are in that example. I only understand that T, an operator from X to Y, apparently has (entries?) T_ij associated it, with i from 1 to n and j from 1 to infty.

ILikeMathematics

23:45

Ben Steffan

@ILikeMathematics you will have to explain what all of these things are

ILikeMathematics

chi_A is the characteristic polynomial of A

mu_A is the minimal polynomial of A

Other than that I think everything is clear

Ben Steffan

I don't follow the argument. How does $\mu_A(A) = 0$ show that $\chi_A(A) = 0$?

This does not follow from $\mu_A \mid \chi_A$ alone

...wait no, it does

sorry, I'm a bit tired

:)

ILikeMathematics

It's 0:48AM after all, haha

If the argument works out, this seems to be pretty elegant after having proven the first two statements

Ben Steffan

It seems suspiciously simple to me, but maybe the essence of the proof is in proving that the zeroes of $\mu_A$ are the eigenvalues

ILikeMathematics

23:56

@BenSteffan Yeah that's possible, the proof of that doesn't take long (1/4 page maybe) and it still seems to be more elegant than the proof Wikipedia gives using adjugates

Claudio

I have developed all the calculations finally: $$ g = x^TA^TAx-2b^TAx \Rightarrow \nabla g = \partial_{\mathbf{x}}g = 2x^TA^TA-2b^TA, \text{ for } A \ne [0]_{m \times n}, b \ne \mathbf{0} $$

ILikeMathematics

Proving $\chi_A(\lambda) = 0 \iff \lambda$ is an eigenvalue takes maybe 1/8 of a page

Claudio

the problem is how do I find $x_0 \text{ s.t. } \nabla g = 0$

the result is in accord with this answer

however, my professor's solution is different, hence she's able to find an easy formula for $x_0$

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