Mathematics - 2022-11-13

Ted Shifrin

00:03

@Derivative You can find a reference to a U Penn masters thesis if you look on main.

Derivative

00:19

davidbrander.org/penn.pdf

yeah

the original is in German

Ted Shifrin

00:33

That’s why math grad students are supposed to have reading knowledge of math in French, German, Russian.

D.C. the III

First book I've read to actually provide a concrete reason for the exitence of eigenvalues

also just went over three books on multivariables and your's is the only one to provide a proof of the lagrangian multiplier......I do see why this is the case though....

leslie townes

mm, there really is no reason for the existence of eigenvalues. it is a finite dimensional anomaly.

D.C. the III

explain to me adjoints while your at it because all it is is some nebulous object to me. There is some relationship between it and the column space and row space but haven't fully got it yet

Ted Shifrin

@leslietownes LOL

leslie townes

00:49

someone gets my sense of humor.

Ted Shifrin

By adjoint you mean the classical adjoint or the transpose (in the real case)?

D.C. the III

Is the classical adjoint the one defined as $\langle T(x), y\rangle = \langle x, T^*(y) \rangle$ ? The transpose is just a special case of such right?

for linear operator $T$

Ted Shifrin

No, classical adjoint is built out of cofactors. You meant the standard adjoint. You’re talking Ted’s favorite formula. It’s really the map on dual spaces.

Using the dot product we identify the vector space and its dual. That’s the fancy explanation.

D.C. the III

I remember doing the dual space chapter in Friedberg, Insel, co.....wasn't pleasant. The one thing that stuck was the idea of attaching a scalar to the linear maps in the space

the adjoint being the map on dual spaces/

copper.hat

I think of the adjoint as a notational device. Given some operator $A$, a point $x \in X$ and some $f \in X^*$, we can write $f(Ax) = (A^*f)(x)$. there are slight differences between Hilbert & Banach adjoints, but the idea is the essentiall the same.

Ted Shifrin

00:58

You get dual spaces in chapter 8 of my book, starting differential forms.

copper.hat

is that the same as double spacing?

D.C. the III

I'm glad I'm doing your book in conjunction with the linear algebra one.....you give me concrete examples.

Ted Shifrin

I always do examples!

D.C. the III

What field are adjoints and dual spaces discussed more after linear algebra?

@TedShifrin and draw a picture

Ted Shifrin

Functional analysis and, for sure, multivariable analysis and smooth manifolds.

D.C. the III

01:01

So I can relax and not feel out of place because they will appear once grad school is on the horizon...

copper.hat

think of them as transposes.

with benefits.

D.C. the III

at this interval that is how I think of them

with benefits makes them spicy

Ted Shifrin

But you were right. You associate vectors with column vectors and dual vectors with row vectors.

D.C. the III

and the adjoint is the map between the spaces....I'm beginning to see its place in all of this..

one potato two potato

05:33

@TedShifrin Have you ever seen such construction of a chain map?

The problem itself is not hard but never seen such construction.

Ted Shifrin

06:29

There’s a typo in it, of course.

I’ve seen versions of this in cohomology, yes. But didn’t you notice the typo?

one potato two potato

@TedShifrin $t:C_*(Y,\Bbb Z_2)\to C_*(X,\Bbb Z_2)$.

Ted Shifrin

Indeed.

The dual map is called the trace mapping on cohomology. It is a special case of invariant cohomology (under a group action). I do it with deRham, but you can do this problem with singular, I guess.

2

Prithu Biswas

I am having some problems with a galvanometer and shunt resistor exercise.

Lets say a galvanometer has an internal resistance of 20Ω.

With just the galvanometer connected to the voltage source, 1A of current is flowing through it.

Now if we connect a shunt resistor parallel to it, that current reduces from 1A to 0.01A.

What is the resistance of that shunt resistor?

In the answer, it says the resistance of the shunt resistor is 0.2 Ω.

Now I am wondering if the problem is assuming a constant voltage source or a constant current source.

Koro

07:32

voltage is cause while current is the effect.

one potato two potato

07:52

0

Q: Computing induced homomorphisms of covering maps over a punctured torus

Let $X$ be the space $S^1\times S^1 - \{p\}$ for a point $p$. Let $\pi:Y\to X$ be a covering map and let $\pi_*$ be an induced map on $\pi_1$ and $\pi_H$ be an induced map on $H_1$. Compute $\pi_*$ and $\pi_H$ explicitly for a finite covering and an infinite covering of $X$. My attempt: Here's ...

Prithu Biswas

08:31

@Koro I assumed a constant current source and got an answer R = 0.202020...

ILikeMathematics

10:25

> Show that $f(x) = -\frac{1}{2\sqrt{x^3}} > 0$ for all $x \in (0, \infty)$.

How would you go about proving this?
$$\sqrt{x^3} > 0$$
$$\frac{1}{\sqrt{x^3}} > 0$$
$$\frac{1}{2\sqrt{x^3}} > 0$$
$$\implies -\frac{1}{\sqrt{2x^3}} < 0$$

Is that correct?

Overtherainbow

0

Q: How can I show that $X\setminus Y$ is dense in $X$ if $X$ is a normed space?

Let $X$ be a normed space and $Y$ a subspace s.t. $Y\neq X$. I need to show that $X\setminus Y$ is dense in $X$. I'm a bit confused how to do this. I thought about this geometrically, so I thought that it is enough to show that for all open balls $B(x_0,r_0)$ with $x_0\in X$, $r_0>0$ there exis...

real-analysis functional-analysis analysis dense-subspaces

could someone give me a hint?

one potato two potato

10:43

0

Q: Stein complex analysis 6.2

Prove that $$\prod_{n=1}^\infty{n(n+a+b)\over(n+a)(n+b)} = {\Gamma(a+1)\Gamma(b+1)\over\Gamma(a+b+1)}$$ whenever $a$ and $b$ are positive. Using the product formula for $\sin\pi s$, give another proof that $\Gamma(s)\Gamma(1-s) =\pi/\sin\pi s$. My attempt: First of all, I guess the assumption $...

complex-analysis gamma-function infinite-product

several lines of dirty nonsensical equalities.

Jaakko Seppälä

13:16

How can one show that $\sum_{r=1}^{13.7*10^9*365.25*24*3600} 1/r<42$?

Jakobian

13:52

@JaakkoSeppälä I guess the problem is that it converges too slowly. There are ways to "fasten" series so that they converge faster to their limit

originally they were developed as a method of computing $\pi$ iirc

I know this from Sierpiński monograph about infinite series.

anak

@Jakobian hasten is the word you were looking for. :)

Jaakko Seppälä

Oh, I find the solution: ln(13.7*10^9*365.25*24*3600)+1)<42. This is from en.wikipedia.org/wiki/Harmonic_series_(mathematics)

Jakobian

Ah yes. But also, I realized this should just be some logarithm. $$\log(n+1) = \int_1^{n+1} \frac{1}{y}\mathrm{d}y = \sum_{k=1}^n \int_k^{k+1} \frac{1}{y}\mathrm{d}y\geq \sum_{k=1}^n \frac{1}{k+1} = H_{n+1}-1$$

So $H_n \leq \log(n)+1$

@JaakkoSeppälä yep. Same thing

Jaakko Seppälä

14:10

Yes indeed.

Jaakko Seppälä

14:30

Wha is an example of a real sequence that converges but has no maximal term?

leslie townes

-1/n.

Jakobian

It can't happen if we have positive terms converging to $0$, say

Jaakko Seppälä

I see.

Jaakko Seppälä

15:00

Suppose that numbers $x_k$ are positive and $\sum_{k=1}^\infty x_k$ converges. I would like to show that $\sum_{k=1}^\infty x_k/k$ converges. I got a hint that this follows from Young's inequality but how?

leslie townes

it follows more simply from the fact that termwise one has x_k/k <= x_k and a monotone increasing sequence that is bounded above has a limit.

monoidaltransform

tangent space is a embedded submanifold of tangent bundle right?

leslie townes

so young's inequality seems, well, maybe not out of place, but something of a red herring. what is young's inequality? (he has several.)

if it's the first thing here en.wikipedia.org/wiki/Young%27s_inequality_for_products, one instance of it (take p=q=2) tells you that x_k * 1/k <= (1/2) (x_k^2 + 1/k^2), so if you know the series sum x_k^2 and sum 1/k^2 converge, that is another way to do it. but if you know that sum b_k converges whenever 0 <= b_k <= a_k and sum a_k converges (which you would be using even there), you can already do that more directly as indicated above, without young's inequality.

Jaakko Seppälä

Yes. I got a strange hint to use Young's inequality.

Jakobian

If the assumption were that $\sum_{k=1}^\infty x_k^2$ converges, then it would be more useful

Jaakko Seppälä

15:15

Oh. That was the assumption.

ILikeMathematics

15:35

A company is selling a product. The production cost per 1000 products is described by $K(x) = 0.02x^3 - 0.6x^2 + 6.5x$, $K(x)$ in $1000$ euros.
$K'(x) = 0.06x^2 - 1.2x + 6.5$ is approximately the cost for the next product at $x$ products.

One unit of $1000$ products is sold for $2500$ euros.
When does the company make profit?

So if $K(x) < 2.5x$, then they should make profit, right?
But when $K'(x) < 2.5$, then the production cost for one more product is less than it's sold for, so that should be correct too, no?

$K'(x)$ is approximately the cost for the next 1000 products (one unit)*

Jaakko Seppälä

16:15

If $\lim_{n\to \infty}nx_n\to 0$ then is it necessary that $\sum_nx_n$ converges?

Jakobian

no

leslie townes

ilike it may help to think general principles here. profit(x) = revenue(x) - cost(x), here you have revenue(x) = 2.5x and cost(x) = k(x). you have written down the condition for profit'(x) to be positive, i.e., for there to be at least some incremental benefit in selling slightly more items when you are selling x items.

"when does the company make profit" seems to just be asking when profit(x) is positive. it doesn't seem to be asking when profit(x) is maximized (which may occur at a value where profit'(x) = 0).

Jakobian

Take $x_n = \frac{1}{n\log(n)}$

leslie townes

as a side project maybe try to envision a business setting where a seller could be capable of deciding how many products they will actually sell (as potentially distinguished from how many products that they want to sell), particularly if items are sold at a fixed cost.

Jaakko Seppälä

16:33

If $\lim_{n\to \infty}n^2x_n\to 3$ then is it necessary that $\sum_nx_n$ converges?

leslie townes

hell yeah bro

Overtherainbow

Could maybe someone help me here?

For $p\in [1,2)$ I need to show that $l^p$ has empty interior in $l^2$.

I know that I need to show that there is no open ball $B(y,r)=\{x\in l^2: ||x-y||_2<r$ in $l^p$ with $y\in l^2$ and $r>0$.

My Idea was to assume that there exists $y\in l^2$ and $r>0$ such that $B(y,r)\subset l^p$. Now I wanted to find $x\in B(y,r)$ s.t. $\sum_{n\geq 1} |x_n|^p=\infty$ because then I would get a contradiction.

Is there a possibility to find such an $x$?

Jakobian

16:50

@JaakkoSeppälä For big enough $n$, $x_n$ are positive and $x_n\leq \frac{4}{n^2}$

leslie townes

@JaakkoSeppälä this is a special case of something called the 'limit comparison test' with a_n = x_n and b_n = 1/n^2 check it out it rocks en.wikipedia.org/wiki/Limit_comparison_test

Jaakko Seppälä

Thanks!

Jakobian

@leslietownes well, the limit comparison test is when terms of $x_n$ are non-negative

leslie townes

a hypothesis implied by the hypothesis of the above. but thanks!!!

'eventually nonnegative' can be built into the hypotheses of the LCT. but thanks!!!!!!!!

anak

hi leslie!!!!!!!

leslie townes

16:59

i didn't actually read the hypotheses of the wiki LCT before linking to it

so i do appreciate these spot checks

anak!!!!!!!!!!

Jakobian

For the one-sided limit comparison test, as they call it, i.e. when the limit is $0$, this assumption is actually necessary. That's why I mentioned it

leslie townes

i should amend to say, for the LCT in my head, as opposed to the one that i linked to, it's a special case

:D

very tempted to edit the shit out of this LCT page rn

Jakobian

@leslietownes I do the same things with my lecturers

hopefully they don't hate me for constantly correcting them

leslie townes

you absolutely need someone like you if you're lecturing or people will write the false things you said and not the true things you meant

silly people

Jakobian

People think it's rude, it probably is

my lecturer from game theory went mad at me once because of it
and I just didn't want to go on his lectures anymore

leslie townes

17:10

the only thing that annoyed me when i taught was people who wanted to suggest improvements (not find errors) in something i'd done in class

because it could waste untold amounts of class time for no purpose

yes, the hypothesis "if __ then sum a_n converges" can be improved

email me or something

i missed the part where i said in the syllabus that every if was going to be an if and only if

Jakobian

I did that pretty often for my topology course

leslie townes

booo

Jakobian

I didn't treat like improvements, more like a comment to what's already written down

leslie townes

i guess even then i wouldn't be annoyed if it was about a hypothesis that was never used

it is often good pedagogy to state one thing and then say, oh weirdly we only used something less than that, so cool, new theorem

but if you are using the stated hypothesis and there's just some wider world of stuff out there that the class could be about but isn't

pointing to that can be annoying

Jakobian

I'm not going to stop commenting anyway

it's boring enough without me saying anything

leslie townes

17:16

do what you do, they don't have to like it

Jakobian

I mean, it's kind of funny too

when you think that they can get annoyed

it's like a prank of some sorts, but isn't

ILikeMathematics

@leslietownes Thank you, but what's wrong with $K'(x) < 2.5$? The rate of change of the cost at x, so approximately the cost of the next product unit, is smaller than what you would get for selling it, no? That sounds like profit

leslie townes

profit'(x) can be positive without profit(x) being positive. profit'(x) only tells you, it might increase profit (whether positive or negative) to sell more.

e.g. if you are selling at a loss it might get you closer to profitability to sell more (profit'(x) positive) even if you do not become profitable by doing so.

to maximize profit you need to look beyond local stuff detected by the derivatives.

ILikeMathematics

@leslietownes Oh, and here we only care about when there is profit, not for the long term

leslie townes

here's what profit(x) appears to be in yoru example. wolframalpha.com/…

notice this huge early chunk (roughly x <= 10) where profit is negative. profit'(x) is positive for some of that chunk. then this interval between 10 and 20 where we're profitable.

ILikeMathematics

17:26

@leslietownes Yes, between 10 and 20 unit products, they make profit

Thank you

leslie townes

good example because setting profit'(x) = 0 gives you two solutions, one of which definitely does not correspond to maximizing profit.

let's go into business together. we just need to decide what the thing x is that we're selling.

ILikeMathematics

Haha

Koro

17:42

@PrithuBiswas ok, I'm not sure then. Please ignore my last comment.

Koro

18:23

can anyone please explain why one point compactification of R is homeomorphic to a unit circle?

copper.hat

pick one point on the circle to represent $\infty$, the added point.

let $f(x) = e^{2i \arctan x}$ for $x$ real, and $f(\infty) = -1$.

Koro

I understand the idea but was having difficulty in creating the function.

copper.hat

18:39

another is to sit a circle on the real line, so the bottom of the circle is at 0. pick a point $x$ on the real line and draw a straight line from $x$ to the top of the circle. then $x$ maps to the first intersection, then $\infty$ maps to the top of the circle. this is more visually appealing.

Koro

hmm, $f(x)=(\cos (2\arctan x), \sin (2\arctan x)), f(\infty)=-1$ works.

thanks copper. :-)

copper: yes, that's visually appealing. I remember using that argument to convince myself that [a,b] and R have the same cardinality.

user123234

19:06

what does it mean that a set $B$ is closed in $l^2$?

Alessandro Codenotti

19:20

What is $I$? In any case presumably $I$ is a topological space and $I^2$ has the product topology

user123234

$l^2=\{(x_n)_n: \sum_{n} |x_n|^2<\infty\}$

Alessandro Codenotti

Oh, $\ell^2$

user123234

ah sorry I didn't know there is a letter for this.

Alessandro Codenotti

Then it has a norm which induces a metric $d(x,y)=\|x-y\|_2$, and closed means with respect to this metric

user123234

Ah so I need tho show that if I take any convergent sequence $(y^{(k)})_k$ in $B$ then its limit lies also in $B$.

or is there like an easier way to prove it?

Lukas Heger

19:34

that depends on $B$

user123234

So my $B$ is $\{(x_n): \sum_n |x_n|^p<R\}$ for some fixed but arbitrary $R>0$ and $p\in [1,2)$

copper.hat

19:45

try something simple like $x_n(1) = (R-{1 \over n})^{1 \over p}$ and zero elsewhere.

user123234

@copper.hat sorry I don't get it, how does this helps me

copper.hat

$x_n \to (R,0,...)$ which is not in $B$.

robjohn

$x_n \to\left(R^{1/p},0,\dots\right)$

copper.hat

oops, thanks @robjohn

user123234

@copper.hat so you mean that I should write the set as the preimage of a continuous map?

copper.hat

19:59

that is not what i meant, i am not sure i follow that leap?

user123234

ah sorry, I'm a bit confused, I don't see the why your $x_n(1)$ should help me. Sorry I'm a bit confused today. So what property of closedness do we want to use in order to show that it is closed

copper.hat

in a normed space, a set is closed iff it contains all its accumulation points.

(true in a metric space, more generally.)

Ted Shifrin

I always encouraged participation (suggestions, corrections, questions). The exception was the annoying kid (who flunked!) who had to ask about stuff that wasn't even related to what we were doing at the time ... presumably his way of showing off, but I think there were more serious issues.

user123234

@copper.hat ah but then I need to pick an accumulation point of $\ell^2$ and show that it is in $B$?

copper.hat

that doesn't really make sense.

how do you characterise closed?

user123234

20:13

one is that the complement is open. Then we also said that the limit of every convergent sequence lies in the set. Or also that it contains all it's cluster points.

And we had one using nets

copper.hat

try the convergent sequence one. that is where i started.

user123234

So this means that I need to take an arbitrary convergent sequence in $B$, i.e. something of the form $(x^{(k)})_k=((x^{(k)}_n)_n)_k$ and I need to show that it's limit $(x_n)_n$ as k tends to infinity lies in $B$?

copper.hat

i gave an explicit example of a sequence above. wht not try that?

user123234

because I don't see why one sequence is enough. I thought that I need to show my statement for an arbitrary convergent sequence

copper.hat

what if the set is not closed???????????

you need to do some work here.

user123234

20:23

hmm in the exercise they wrote that I need to show that $B$ is closed

that's why I'm asking

where $p\in [1,2)$

copper.hat

that is not what you asked originally.

user123234

"So my $B$ is $\{(x_n): \sum_n |x_n|^p<R\}$ for some fixed but arbitrary $R>0$ and $p\in [1,2)$" why is this not the same?

copper.hat

you need to do some basic review.

user123234

ah I forgot the $\leq$

Sorry what do you mean I need to review

copper.hat

you asked why $<$ and $\le$ are not the same.

user123234

20:30

no I made a tipo

i wanted to write $\leq$ in my original question.

but would you now still proceed by taking an arbitrary sequence?

10GeV

21:22

I'm working on a proof where I'm using the fact that, for arbitrary integers a,b, and n, if gcd(a,n)=gcd(b,n)=1, then gcd(ab,n)=1. I know that one can show this by expressing 1 as linear combinations of a & n and also of b & n, and then using those expressions to show that 1 may be expressed as a linear combination of ab & n. I'm wondering, however, if this also follows directly from a&n and b&n having no common prime factors. Or is more needed?

anak

Do you have the fundamental theorem of arithmetic at this point? It's a rather elementary problem. One that seems like it would come much sooner than the proof of the FTA.

Ted Shifrin

FTA comes quite early. Anyhow, it’s easier. If $p$ is a common factor of $ab$ and $n$, then it is a common factor of either $a$ and $n$ or $b$ and $n$. Why?

$p$ prime, of course.

leslie townes

the bezout based proofs are often the 'lowest tech' even if conceptually they are less intuitive than other proofs.

Ted Shifrin

Well, what I’m suggesting is just the fundamental characterization of primes.

user123234

if I have $\sum_{n\geq 1} |(x_n-y_{n,k})+y_n|^p$ is there a way to find an upper bound s.t. $(x_n-y_n)$ and $y_n$ are separate? because applying the triangle inequality would not help me much since I then need to apply the binomial theorem. Is there an easier way?

leslie townes

21:30

i was not commenting on the current discussion, but the underlying vibe of the current discussion.

Ted Shifrin

But the proof I give of that fundamental fact uses Bezout.

leslie townes

certainly also more helpful in general to use something being prime, if it is known/assumed to be prime, than to think in terms of 'prime factors' of integers, which is already leveraging a larger apparatus.

vibes math.

Lukas Heger

Hello @Ted !

I suppose you might have something more down-to-earth to say about this

there's an amusing group-theoretic proof of FTA, of the nuking-a-mosquito kind

it follwos it follows from the Jordan-Hölder theorem applied to a finite cyclic group

@Ted (sorry if the question disinterests you)

Ted Shifrin

21:53

I pass. Odd to write (in the title) $C^1$ for $S^1$.

Lukas Heger

I agree

@Ted I'm doing some complex geometry right now :) we're still at the beginning, reaping consequences of the Weierstrass preparation theorem etc. to study analytic algebras

but we will eventually defined complex analytic spaces

but I suspect you prefer complex manifolds and the like over this algebraically-flavoured approach because it's more differential-geometric

Ted Shifrin

Yup. You suspect correctly.

Curvature in the complex setting is fascinating. Curvature of a Hermitian submanifold is always at most that of the ambient manifold. Very different from real.

Lukas Heger

yeah I'm a bit studying the diffgeo stuff on my own right now. I think curvature is really intriguing even though I'm an algebraist

I'Ve been reading Well - Differential Analysis on Compelx Manifolds and it's quite nice

I wonder if you can do more differential-geometric stuff with non-smooth (non-reduced) complex-analytic spaces. For a good homology theory, you probably need intersection homology, but what about metrics? I mean you can still define a tangent bundle so maybe it works out

the tangent bundle might not be a bundle, that seems like a problem

Thorgott

22:13

the complex world is too scary

Ted Shifrin

There’s all sorts of stuff on characteristic classes and curvature on singular spaces. I even have a little paper on MacPherson Chern classes of the Whitney umbrella. Even a non-reduced structure to the Nash blow-up if I remember.

Lukas Heger

@TedShifrin sounds intriguing!

I'm still learning the basics though

Thorgott

iirc you can get away with a lot of the theory when you generalize to complex varieties because the non-smooth points are always codimension 2 or something

but algebraic geometers are annoyingly imprecise about stuff like this

Ted Shifrin

No, that’s very wrong. Only for normal varieties are singularities in codim at least 2. You know plenty of singular curves or other hypersurface singularities.

Thorgott

22:33

real codimension 2, i mean

geocalc33

Say you have a function $f(x)$ and it's mellin transform $g(s)=M(f).$ Perform $h(x)=\sum_{n=1}^\infty f(n^x)$ and also perform $j(x)=\sum_{n=1}^\infty g(n^x).$ Are $h$ and $j$ related?

Ted Shifrin

@Thorgott who cares about real? Yes, that’s important for why the smooth locus of an irreducible variety is connected, etc.

mme

23:44

@LukasHeger This book will be good for your soul, especially the chapter on Hodge theory which --- to my eye --- is the best introduction for someone willing to get their hands dirty and who can handle the abstraction of differential operators on closed manifolds.

One has as a lovely corollary of the work in that chapter that any closed complex manifold has finite-dimensional biholomorphism group. Exercise: show that the additive group of holomorphic functions C -> C embeds into Biholo(C^2), so closed-ness is truly essential here.

Lukas Heger

glad to hear that

mme

@TedShifrin This subtlety is rarely discussed in the standard texts. They also love to tell you that if you're minimizing some function with respect to a constraint g(x,y) = k, you can use the constraints to do "substitutions" and minimize a 1-variable function. This is wrong: you may well miss all critical points corresponding to a spot where grad g projects to 0 on the corresponding coordinate axis.

I've complained about this before.

OK, see you all in two years unless you show up in my email inbox.

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