Mathematics - 2022-11-14

copper.hat

00:23

that's a bit vague, if there is a valid (& suitably smooth) one dimensional parameterisation of the constraint space, then minimising the resulting one dimensional function will be equivalent in the appropriate sense.

leslie townes

yeah. maybe some ways of 'solving' g(x,y) = k don't immediately give you that, but the principle isn't wrong.

copper.hat

if i ever show up in somebody's email box i will be in bits...

Ted Shifrin

@copper.hat The mistake students make is that they consider the “substituted” function on the wrong domain. They forget that often the domain has boundary and don’t check boundary points.

This is pretty much what Mike was talking about.

copper.hat

00:38

i see

Ted Shifrin

Obviously, if you parametrize a circle by $\theta$ you’ll be fine, but they often do $y=\pm\sqrt{1-x^2}$ and miss endpoints (not to mention differentiability issues). But a function like $y^2+x$ will work fine unless you forget that $x\in [-1,1]$.

copper.hat

i stand suitable elaborated

Ted Shifrin

01:03

I put that there for anyone else who might be interested :)

leslie townes

munchkin's new thing is to accuse one of us of having a 'tantrum,' which at least in my wife's case sometimes does produce anger.

Ajay

They're not wrong. After all, kids don't lie.

leslie townes

yes

Ajay

However, I do suppose they will reach the age when they lie like politicians do.

Ted Shifrin

01:31

The tantrum produces anger or the accusation does?

leslie townes

the accusation.

munchkin is very good at presenting alternate realities in which authority figures are having tantrums and 'not listening.'

Ted Shifrin

We need to work on munchkin’s (or your) syntax.

She has all sorts of Trompian characteristics at the ripe age of 3.

leslie townes

4!

Ted Shifrin

Oh, then normal.

Can’t wait to see 7.

leslie townes

it's certainly going to be some kind of journey.

Ted Shifrin

01:42

Do you have a child psychiatrist on speed-dial?

leslie townes

nah. we don't want her to be punished for having too many good ideas.

Ted Shifrin

One of those “good ideas” is to enroll her parents in an insane asylum.

leslie townes

01:57

uncle ted is no fun at all.

Ted Shifrin

Nope. Plenty of people here agree with that.

I was about to link my latest response toma homework post, but my iPad refuses.

one potato two potato

03:03

0

Q: Calculating an induced map $\phi_*:H_i(M_2,\Bbb Z)\to H_i(M_2,\Bbb Z)$

Let $M_2$ be a closed orientable genus $2$ surface and $\phi:M_2\to M_2$ be a homeomorphism given by rotation of angle $\pi$ by the $x$-axis. Calculate $\phi_*:H_i(M_2,\Bbb Z)\to H_i(M_2,\Bbb Z)$ for $i =0,1,2$ with respect to a basis of $H_i(M_2,\Bbb Z)$ for each $i$. (Hint: use $\phi$-invarian...

algebraic-topology

Don't know what to do

one potato two potato

05:29

https://math.stackexchange.com/q/4576167/668308
Reposting because of the downvote

Ted Shifrin

05:41

Do not repost. Edit the original post.

And ordering people not to downvote is pretty bad form, too.

BAYMAX

05:55

Suppose if all eigenvalues of a square matrix $A$ are less than one in absolute value, then that implies $\det(I-A) > 0$

Is this true?

copper.hat

06:11

Choose an 8x8 matrix with eigenvalue $i \tan {\pi \over 8}$.

Koro

Showing that [0,1] is not Frechet compact as a subspace of $R_l$ (that is R with lower limit topology):

Consider the set S={1-1/n, n is greater than 1}. 1 is not a limit point of S because {1} is a neighborhood of 1 that contains no element of [0,1].

And it is clear to me that any number outside [0,1] can't be a limit point of [0,1].

one potato two potato

07:25

@TedShifrin I'm not ordering. I usually interpret it as a sign to delete. If someone wants me to revise the post, that person usually leaves a comment.

robjohn

07:53

@copper.hat However, if the matrix is real...

one potato two potato

09:42

I draw everything one by one (the rotation process) and the CW structure does not change at all (obviously..?). Does that imply the induced homomorphism on homology is an identity map?

one potato two potato

09:54

Ah no

Yes it should be like this.

Suddenly everything makes sense

Koro

10:17

In a T1 space, why is limit point compactness equivalent to countable compactness?

Suppose that X is countably compact. Let A be an infinite subset of X. Choose a countable subset {a_i} out of A. {X-a_i} is an open cover of X so there exist finitely many indices 1,2,...,k such that X-a_i cover X.

Atleast one of X-a_i, 1<=i<=k contains infinitely many elements of a_n.

Let the index be k WLOG and call the elements of a_n as $a_{n_k}$'s. Claim: a_k is a limit point of $ a_{n_k}$

Let U be an open set containing a_k. U has to be of the form X-F, where F is finite set that doesn't contain a_k. Clearly X-F contains some $a_{n_k} $.

QED.

But I have a doubt here: is it true that in a T_1 space, every open set has to be of the form X- F, where F is finite set? In T_1, finite sets are closed.

For the other direction: Let U_i be an open cover of X. Assume on the contrary that there is no finite subcover. Choose x_k in X- $ \cup_{i\le k} U_i$.

Let x be the limit point of {x_n}. x is some U_m. U_m doesn't contain u_i with i>m-1.

So U_m intersects x_n at finitely many points.

Lukas Heger

@Koro no that's not true in general. it's not even true in metric spaces

Koro

10:33

Let $\{x_1,...,x_{m-1}\}\cap U_m = $ non empty. But I don't see how to get any contradiction from here. It seems that limit point compactness is NOT equivalent to countable compactness in T_1.

@LukasHeger oh yes, thanks. Not true even in R, take N, the set of naturals for example.

one potato two potato

11:05

0

A: Calculation of induced map on homology by self homeomorphism by rotation on genus $2$ surface

Using the above CW structure, on a chain level the map $\phi_*$ is given by \begin{cases} a\mapsto d\\ b\mapsto c\\ c\mapsto b\\ d\mapsto a\\ \end{cases} Since $a,b,c,d$ are the generators of $H_1(M_2)$ (consider cellular homology for example), we conclude $\phi_*:H_1(M_2)\to H_1(M_2)$ is given b...

Enjoy nonsensical pictures with cumbersome calculations

Koro

13:22

0

Q: The integral $\int_A f$ exists $\iff$ the series $\sum_i \int_A \phi_i|f|$ converges.

The theorem 16.5 in Munkres' analysis on manifolds reads: Let $A$ be open in $\mathbb R^n$; let $f : A → ℝ$ be continuous. Let $\{ϕ_i\}$ be a partition of unity on $A$ having compact supports. The integral $∫_A f$ exists if and only if the series $\sum_i \int_A \phi_i|f|$ converges and in this ca...

real-analysis multivariable-calculus

one potato two potato

14:03

@Thorgott Hello, could you take a look at my solution (my latest chat before this)? I wonder if it makes sense and (whether it's ok or not) if there's an easier way to solve it.

I think the calculation of $\phi_*:H_2\to H_2$ can be improved. I used $\Delta$-complex but it's quite complicated.

Ajay

14:24

Is it okay for me to swear on here?

user193319

Let $F_n$ denote the $n$-th Fibonacci number. I am trying to show that if $n \ge 29$, then $F_n > (1.6)^n$. The hint is to prove this by contradiction and use the smallest counterexample. By counterexample, I take it to mean any $n \in \Bbb{N}$ such that $F_n \le (1.6)^n$. But isn't it true that $F_0 = 0 < 1 = (1.6)^0$, so $n=0$ would be the smallest counterexample? How does that even help? I don't see how to prove this by contradiction (or even induction).

Ajay

Well you can let your base case be n = 29

then work your way from there by induction

user193319

Yeah, I didn't get very far with induction. So, $F_{n+1} = F_n + F_{n-1} > F_n > (1.6)^n$. That's all I've come up with so far, but that inequality isn't good enough.

I don't think we can necessarily say $F_{n-1} > (1.6)^{n-1}$ because it's possible $n-1=28$ which is a counterexample.

@Ajay Do you know how to get the induction to work?

Ajay

Well, you need to use strong induction since it's a fibonacci sequence

Do you know how to do it?

user193319

Yes, I know strong induction.

Ajay

14:38

Then simply apply it here.

user193319

Apply it where?

Ajay

Well, you can "let P(n) be the statement $n \geq 29$, then $F_n > (1.6)^n$"

Then the induction process should flow naturally

user193319

Yes, I am quite familiar with the fundamental idea behind (strong) induction.

How exactly does it flow naturally? I showed you what I tried and it doesn

...doesn't seem to flow naturally.

Ajay

Then let's try something a little easier

$f_{n+1} < \big( \frac{7}{4} \big)^n$ for all $n \geq 1$

We'll work our way up to your question

user193319

$P(n)$ should actually be $F_n > (1.6)^n$. Quantifiers shouldn't be mixed in.

Ajay

14:41

Yes, I know. Doesn't really matter though, nothing formal here.

user193319

Well, you're wavering between formality and informality. I think formality is essential.

Ajay

Ok fine, i'lll be more formal from now on.

Anyways, let's do the induction problem now.

Ajay

14:55

@user193319 your base case is not 0

it is 29

You first need to show that the 29th fibonacci number is indeed greater than $(1.6)^{29}$

Wait what...

$F_n = F_{n-1} + F_{n-2}$

so $F_{29} = F_{28} + F_{27}$

Which is greater than $(1.6)^{29}$

Now assume that P(k) is true.

SO what we need to show is that $F_{k+1} > (1.6)^{k+1}$

SInce we have that $F_{k+1} = F_{k} + F_{k+1}$

Our initial hypothesis says that $F_k > (1.6)^k$

So $F_{k+1} > (1.6)^k + F_{k-1}$

Ohhhh, but now we need a second base case

uhhhh i'm confused

ANy help?

Koro

15:31

My question has been answered. I still need one clarification though.

2 hours ago, by Koro

0

Q: The integral $\int_A f$ exists $\iff$ the series $\sum_i \int_A \phi_i|f|$ converges.

The theorem 16.5 in Munkres' analysis on manifolds reads: Let $A$ be open in $\mathbb R^n$; let $f : A → ℝ$ be continuous. Let $\{ϕ_i\}$ be a partition of unity on $A$ having compact supports. The integral $∫_A f$ exists if and only if the series $\sum_i \int_A \phi_i|f|$ converges and in this ca...

real-analysis multivariable-calculus

Partition of unity is defined for $\scr A$, a collection of open sets. How should one interpret it in case it is defined for an open set A? It seems natural to me to treat that as follows: For every a in A, there is a ball $B_a$ lying completely in A. So take $\scr A=$ $\{B_a: a\in A\}$.

anak

15:46

@Koro it's unclear what you are asking. Can you elaborate a bit more? Given an open set A, you can find a partition of unity subordinate to each open cover of A. That open cover could be anywhere from just A itself, or something much larger.

Koro

16:06

Hi @anak!!

anak

Hi :D

Koro

'subordinate to' terminology is not used in Munkres (at least upto the point of introducing partitions of unity). But my question is: based on theorem 16.3 (stated in post), one would think of partitions of unity to be defined for a collection of open sets (this collection is denoted by $\scr A$, say). But then, in 16.5, the partitions seems to be defined for an open set A. How does that make sense without any mention of $\scr A$?

To make 'partitions for an open set A' meaningful, I'm trying to assign a meaning to it in my last to last message.

anak

The point is that the integral can be broken up along any open cover of A, using a partition of unity which corresponds to that cover.

If that cover was just A, then the statement is trivial, but still true nonetheless.

It's just saying that integration of the function is independent of way you break it up with a partition of unity.

Koro

16:22

:(.
A being open can be written as union of open sets. So we can consider 'partition of unity' on A in this fashion.

This should make sense as you say -independent of the way I break A.

If there is any justice left in this world, then in all fairness this should be correct.

anak

16:35

@Koro do you have doubts?

Koro

no. I'm just new to this so wanted second opinions and more importantly if I understood it correctly or not.

Is this a PSQ? math.stackexchange.com/questions/4576470/… ?

I want to answer it.

anak

What does PSQ stand for?

Koro

Problem statement question.

People are encouraged on this site NOT to answer PSQs.

anak

Like homework question?

Koro

I think a question that shows no attempt of the asker in trying to solve it.

But sometimes it is difficult to tell if a question is a PSQ or not.

anak

16:43

Well indeed it shows no attempt.

Koro

But it may so happen that the asker has no idea as to how to solve it.

anak

Well it's rare to work on a problem and not have anything to show for it.

My rule of thumb is that you can ask in the comments what they have tried. If you want to answer something, it should always be in the form of a hint, because complete solutions in stackexchange answers are terrible for everyone.

Koro

@anak I think you're right.

BAYMAX

17:03

1

Q: Proving an implication of two dimensional matrix.

If $A = \begin{bmatrix} x & 1\\ y & 0\end{bmatrix}, B = \begin{bmatrix} z & 1\\ w & 0\end{bmatrix}$, for $x,y,z,w \in \Bbb{R}$. I have observed by considering many random values of $x,y,z,w$ that: If all the eigenvalues of $A^2B$ and $AB^2$ are less than one in absolute value, then $det(AB+A+I)<...

linear-algebra eigenvalues-eigenvectors determinant

anak

@BAYMAX did you try the idea from the comments?

robjohn

@copper.hat However, if we don't use real matrices, we can also use $e^{\pi i/3}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

geocalc33

17:49

$\lim_{s \to \infty} \varphi_s(x) \sim \text{Dirac Delta dist.}$ and $\lim_{s \to 0} \varphi_s(x) \sim \text{Uniform dist.}$

where $\varphi_s(x)$ is a probability distribution

actually it's an "L" shape distribution

for s to infinity

geocalc33

18:08

is there neat distributions on $(0,1)\times (0,1)$ whose density is not 100% but the complement of area under curve + area under curve has total area 100%? As if the distribution function partitions the region into density and complement density that add to 1

normal distribution has a density but complement density is infinite

anak

18:59

w h a t

Ted Shifrin

19:35

seconds anak

Semiclassical

20:04

the area + complement-of-area bit seems to amount to: if i have a function $f:(0,1)\to(0,1)$, then $$\int_0^1 f(x)\,dx + \int_0^1 (1-f(x))\,dx=\int_0^1 1\,dx = 1.$$

(or just "i've divided a unit square into two pieces")

can't really parse the rest of it

anak

@Semiclassical but then how is the area of 1 minus a Gaussian infinite?

Or do they mean over R^2 for that one?

Semiclassical

in the sense that $\int_{-\infty}^\infty f(x)\,dx=1$ but $\int_{-\infty}^\infty (1-f(x))\,dx=\infty$ i guess?

but that'd be an infinite strip not a unit square

anak

yeah, that's what I thought

Semiclassical

the phrase "complement density" doesn't make a lot of sense to me regardless

you can talk about the complementary probability of an event just fine, but evaluating a PDF at a point does not yield a probability

anak

speaking of math, what have you been up to lately, @Semiclassical? Or have you been more physicsing?

Semiclassical

20:14

this semester has mostly been working in the trenches of intro physics

but i'm also preparing lab material for a spring intro course on quantum

some silly probability exercises, some fun polarization physics, and some silly balancing-a-meterstick tasks

anak

Nice, what sort of lab material are you planning?

Semiclassical

1 min ago, by Semiclassical

some silly probability exercises, some fun polarization physics, and some silly balancing-a-meterstick tasks

:P

the probability exercises amount to them playing with simple raffle games to see what distributions of outcomes they get

the tickets have three options on each side. each option gives a +/- result, and the two sides have opposite results for each option. so one side could be ++- and the other --+. then you'll always get opposite results unless one person takes option 1/2 and the other takes 3.

that's simple enough. but there's 4 such ticket types---what happens when you have some mixture?

stuff like that

anak

Oh the labs have probability exercises?

Semiclassical

yeah

anak

That's pretty neat.

Semiclassical

20:21

the balancing-a-meterstick stuff is really an excuse to talk about linear correlation: if you put three masses on a meterstick, then the net torque due to their weights is a linear function of the positions. so if you hold the net torque fixed while varying two of the mass positions, you can do bivariate regression to recover the masses

first varying the masses independently, then discarding some of the pairs in order to get correlation between them

i'd talk more but i gotta run :P

smth

I have one random question, is disc without just center homeomorphic to disc with hole?

Thorgott

depends on what a hole is to you

smth

something like this is disc with hole: {z \in \mathbb{C} | \frac{1}{2} \leq |x| \leq 1}

anak

Annulus is a better word for that.

And the "disc without center" is what? {z in C | |z| = 1}?

smth

Sorry, that is what I meant

no no

anak

20:27

Oh, disk - {0}?

smth

yeah that

anak

Geometrically imagine it in your head, what might such a homeomorphism look like?

Thorgott

the statement is false, but I encourage geometrically imagining it to see what related statement is true

smth

I intuitively see that this two are not homeomorphic but I don't know how to prove that

anak

@smth What's your intuition?

smth

20:36

somehow we need to "strech" point-hole from center into circle-hole

different cardinality there

anak

That's not quite the issue.

Ted Shifrin

Is $(0,1]$ homeomorphic to $[1/2,1]$?

Thorgott

think one dimension lower, perhaps

oh, Ted beat me to it

smth

@TedShifrin one is compact and other is not? so they are not isomorphic

Am I right?

Ted Shifrin

Not homeomorphic. Right.

smth

20:38

homeomorphic*

Ted Shifrin

Your two spaces are, however, homotopy equivalent, whatever that means :)

anak

@smth Look at your annulus definition with that in mind, smth

smth

oh so annulus is compact and disc without point is not I am dumb

Ted Shifrin

What if you took a half-open annulus, $\{1/2<|z|\le 1\}$?

smth

then they are homeomorphic?

what if I have something like this:

geocalc33

20:41

@anak @Semiclassical i just meant the normalised area enclosed by the gaussian curve and the x axis is 1 and if you take all the area on the plane besides that enclosed area it's infinite

smth

@TedShifrin disc without center and {x^2 + y^2 =1 | 0 < z \leq 1}

homeomorphic?

Ted Shifrin

Oh, you're doing polar coordinates. Yes. You can do a cylinder or you can think of polar coordinates in the plane. Yes, so now what?

Master

Anyone familiar with Regularised Least Squares methods?

anak

@geocalc33 If the distribution is finite over any non-null measurable subset, then the "complement" is automatically infinite, is it not?

smth

Ted Shifrin I am not so familiar with polar coordinates, cylinder is what I meant, how to prove that these two are homeomorphic?

Ted Shifrin

20:48

Wow. You haven't learned polar coordinates $(r,\theta)$ or $(r,\phi)$ in calculus?

smth

I had okay but what with that here?

Ted Shifrin

Your choices of letters is very awkward. Parametrize your unit circle by $\theta$ and use elementary trigonometry to give $(x,y)$ in the punctured unit disk in terms of $\theta$ and $z$ (which I would ordinarily call $r$).

smth

is parametrization: x = z cos theta , y = z sin theta 0<z\leq r and 0< theta \leq 2pi

@TedShifrin btw thank you so much for helping me

Ted Shifrin

You got it. Now you should be able to see that the half-open annulus works, too. You’re most welcome; you figured most of it out yourself. :)

geocalc33

21:16

@anak the distribution's parameter space fills only R=(0,1)^2. fun fact there are no dist. like this covered in undergrad probability courses

the support is bounded and the range is also bounded

*besides the uniform distribution

the uniform distribution is the only one covered in those courses

Master

$x_0 = c$
$x_i = ax_{i-1} + bv_i$ where $a,b,c \in \mathbb{R}$ I want to express $min_{v \in \mathbb{R}^N} ||x||^2 + ||v||^2$ in terms of $v$.

Whenever I try to model this recurrence relation I get a non-invertible matrix, how can I do this?

user123234

Does someone know about the notation $\Bbb{E}_\nu$ in probability theory when we speak about markov chains?

Master

does it not denote expected value with respect to that parameter

user123234

but $\nu$ is a probability measure

Jakobian

Then with respect to that measure

Master

21:29

^

Its a discrete time Markov chain?

user123234

but does this then mean that if $X$ is a discrete random variable with values in some discrete set $S$, then $\Bbb{E}_{\nu}(X)=\sum_{x\in S} x ~~\nu(\{x\})$?

Jakobian

@user123234 precisely

user123234

ah okey, so we denoted by $\Bbb{E}_x(X)=\Bbb{E}_{\delta_x}(X)$ where $\delta_x$ is the dirac measure. But then would this mean $\Bbb{E}_x(X)=\Bbb{E}_{\delta_x}(X)=\sum_{y\in S} y\delta_x(y)=x$?

Master

Any techniques to "invert" this non-invertible matrix $\begin{bmatrix} 0 & 0 & 0 &\dots & 0 & 0 \\ -a & 1 & 0 & \dots & 0 & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ 0 & 0 & 0 & \dots & -a & 1 \end{bmatrix} * \begin{bmatrix} c \\ x_1 \\ \vdots \\ x_N \end{bmatrix} = d \begin{bmatrix} 0 \\ u_1 \\ \vdots \\ u_N \end{bmatrix}$. The goal is to find an expression for $\vec{x}$ in terms of $\vec{u}$. But the use of that constant in my recurrence relation starts messing things up.

I know the value of $c$ and $d$

Ted Shifrin

21:53

Can't you just solve the system of equations directly?

$ac+x_1=du_1$ gives $x_1=du_1-ac$. Then $-ax_1+x_2=du_2$ gives $x_2$, etc.

Master

I'll have a go, I do want to find an expression for $||x||^2$

ohhh it works

<3 thank you

Ted Shifrin

Sure thing :)

Jakobian

22:38

@user123234 ye

Master

Is this condition sufficient for invertibility of $A^T A + \lambda D^T D$

$Null(A) \cap Null(D) = \{0\}$

Jakobian

What if $\lambda = 0$ and $A = 0$

Master

Oh assume $\lambda > 0$ and $A$ is non trivial

Ted Shifrin

22:58

So take $x$ in the nullspace of that matrix $C$ and play the usual $Cx\cdot x$ game.

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