understandable have a nice day

How does one get directional vectors from symmetric form for example $\frac{x}{-2}=\frac{y-4}{-1}=\frac{z}{1}$

i think it's <-2,-1,1> but i don't understand how

Set the equal things equal to $t$ and write the parametric equation.

Ohhh

That's why my prof did it so fast, because the coefficient of t is just the denominator

There you go!

Good for you for making sure you understand!

Truly blessed to have access to all the wonderful scholars of mathematics here as well.

how was normal distribution derived?

apparently de Moivre was goofing around and found it.

1733, no less.

he played a lot of dice games

The point is that when $n$ is very large, the binomial distribution $\binom np$ looks like the normal distribution with $\mu = np$ and $\sigma = np(1-p). It's algebra fiddling.

he was acting the buffoon throwing needles on the ground

@TedShifrin Here is my work on the problem: If $g: U\longrightarrow V$ is a diffeomorphism so by the inverse function theorem $\nabla g(x): T_{x}U\longrightarrow T_{g(x)}V$. Then ${\rm rank}\, H_{f}(x)={\rm rank}\, H_{f\circ \nabla g}(x)$.

It's not just the rank. It's the signature. The number of positive/negative/zero eigenvalues is the same for both. That's the important thing.

So probability is all about playing a lot of dice game and then knowing probability of anything is 0.5. Is this the secret to winning a lot of bets?

Anything? With dice?

Yeah you either get the number or not thus P(getting number 6)=1/2.

The author should publish a one page book for undergraduate students.

Oy.

@TedShifrin Uhm... I did not understand the comment. What I wrote is not correct?

It’s correct, but not the heart of the matter.

What I was trying to do was to then look for a way to relate RHS to ${\rm rank}\, H_{f\circ g}(x)$.

Are you not trying to understand the critical point? What is the ultimate goal?

i heard learning math can burn calories

but i wonder why professors are so fat

i regret saying thast

why is probability of finding that point inside $dA\cdot Pdf(x,y)$?

shouldn't the probability be 0?

how you model these things mathematically or think about integrals is kinda up to you, but however you do that, the answer to your question is likely tied up in the very definition of what Pdf(x,y) is supposed to be/represent

but watch out for grand dA

he's just resting

@PM2Ring this book banning is getting absurd and more insane; way out of hand.

@NotTfue perhaps you can construct a good null hypothesis and check for statistical significance.

as long as he is not taking a P in the pool.

perhaps the forum is memes community now

@User1865345 that is just a bunch of noise for uncritical writers & readers.

one star for the pun

awesome, a star :-)

we need granddads for liberty

@copper.hat these moms rant a lot in those meetings.

Of course for their alleged liberty. ¯_(ツ)_/¯

to be fair, i would be ok with banning social media for any group that i am not a member of

@copper.hat legit point.

the world has lots of flat earthers

Bill Nye and deGrasse Tyson need some feast day taking one flat earther a day.

bring on the posits (as in the chess world, posits are big in numerical analysis at the moment)

the excitement is not confined to the mom's for 'liberty' or flat earthers...

I remember Hawking remembering he was confronted by some old flat earther lady during his talk. These people are lurking everywhere.

i am a magnus fan, however

you cannot battle axioms with logic

Moms needz mores liberty.

burn those clothing elements

Does burning curb liberty? Damn they will ban burning.

we should ban bans

the social implications of unfettered liberty escapes many

good night folks, i am going to ban local wokeness for a few hours.

I wonder why did past mathematicians called quaternion evil instead of probability

Hello ☕

I have a question about sequences and number patterns

I always get confused in the definition of general term. And the difference between it and the general formula.

And the relationship between those two and explicit and recursive formulas ?

Number patterns isn't math

Um... they are under the category sequences and series

Important note. Questions about guessing the next number in a sequence, with no explicit mathematical context, will usually be quickly closed. Consider posting these questions at Puzzling Stack Exchange instead.

From the tag description. Clearly they aren't. It specifically tells you that guessing number patterns is a puzzle, not math

The reason is that no matter what kind of number you come up with, there'll always be some possibility that it's the correct number

And that's why it's not math either

But math is a pattern game

My lecturer said that if $x_0$ is a point of maximum/minimum for $f$ in a set $A$ then $x_0$ belongs to one of the following sets: $int(A)$ or the set where $f'(x_0)=0$ or the set where $f'$ doesn't exist. I was trying to convince myself that there are no other possibilities, so I thought that I can write $A=int(A) \cup \partial A$ and distinguish cases. So if $x_0 \in \partial A$ we have done, if $x_0 \in int(A)$ surely there are only two possibilities ($f'$ exists or not).

We were trying to prove Baye's theorem, Teacher did like this: $P(A|B)P(B)=E(\chi_A|\sigma(\chi_B))(B)P(B)\overset{?}=E[E(\chi_A|\sigma(\chi_B))\chi_B]=E(\chi_A\chi_B)=E(\chi_{A\cap B})=P(A\cap B)$. I'm trying to prove second equality,

$E[E(\chi_A|\sigma(\chi_B))\chi_B]=\int_B E(\chi_A|\sigma(\chi_B))\,dP$

I don't know what to do now

Is $E(\chi_A|\sigma(\chi_B))(x)=E(\chi_A|\sigma(\chi_B))(B)$ for $x\in B$?

Also $E(\chi_A|\sigma(\chi_B))$ is a $\sigma(\chi_B)$ measurable function from $\Omega\to\mathbb R$, right? We have $P(A|B):=E(\chi_A|\sigma(\chi_B))(B)$. Then isn't RHS a set? and LHS a number?

I'm confused

@geocalc33 what about the pullback are you having difficulties with?

@anak just having difficulty using this information to obtain the pair $(D^3, w)$

@geocalc33 if you adjust the setup so you start with a closed disk $D^2$, then it's actually really easy to find an $e$ and $w$. I don't know about the open disk case, though....

@anak okay I'll try further. $w$ is a 3-vector field that is defined everywhere in the 3-disk

Where are your accumulation points on the open disk?

anywhere, as long as they are at the furthest distance apart

So the accumulation points aren't on the disk, but are off of it on the boundary?

@Mad what's your definition of a manifold and what have you tried

@anak no from how i was understanding it, the accumulation points are infintesimally close to the boundary, and that they sort of limit to the boundary point in some sense

I don't understand what infinitesimally close to the boundary means.

limit points

That would lead me to believe the accumulation points are on the boundary.

Oh i think you're right, i was misunderstanding that. Can accumulation points become undefined at the boundary?

I have it in my head that they ought to be sources/sinks: is that what you are imagining?

That is, are you wanting them to be stable/unstable points of the vector field? If you extend the vector field to the boundary in this case, they become zeros, I think.

yes that's what I'm imagining

@ZaWarudo Be very careful. You have something backwards here. If $x_0$ is an interior point and $f'(x_0)=0$, you do not know that $x_0$ is a max/min. The converse is true. What's a counterexample to your statement? Your lecturer's statement is wrong. If $x_0$ is a max/min point, then either $x_0$ is an interior point — in which case either $f'(x_0)=0$ or $f'(x_0)$ does not exist — or $x_0$ is a boundary point.

@geocalc33 I think you need more restrictions, e.g. starting on the closed disk. It doesn't seem possible to even extend a vector field on the open disk to the closed disk in every case.

An example is just taking the vector field $v(p) = \frac{1}{\|p\|}\frac{\partial}{\partial x}(p)$.

Sorry, $r - \|p\|$ in the denominator. :P

Where r is the radius of the disk

@anak gotcha! I'll start on the closed disk. I could even add the restriction that both the vector field on the closed disk and closed ball preserve some metric

Preserves in what sense?

preserve the continuous isometries

What does that look like mathematically?

Flow is a 1-parameter family of isometries or something?

@TedShifrin you're right, the function $f(x)=x^3$ with $x_0=0$ is a counterexample because $f'(x)=3x^2$ is zero for $x_0=0$ but $0$ is not a point of maximum or minimum for $f$. Thanks for the correction. But I still don't get why that reasoning exhausts all the possibilities because of the equality $\bar{A}=int(A)\cup \partial A$. It would convince me fully if it was $A=int(A)\cup \partial A$, but it seems it isn't true. Maybe it holds for open sets?

@anak yes exactly

So just that it is Killing, I see.

Do you have a particular metric in mind, @geocalc33?

@ZaWarudo why does it seem it isn’t true?

But what if $k=n$, Mad?

not yet for the closed disk, I just have a metric for the open unit square @anak

Well, the V intersection R^n is just R^n so h( u intersection M ) = R^n x {0}

There is no $0$.

In my definition, we do have it. but i am guessing only in case k smaller than n

Exactly, i dont

hahaha

Nevertheless, so h will map to all R^n and its diffeomorph.

I think your prof probably means an open set in $\Bbb R^k$, not necessarily all.

Yes, so U and V are open sets

By requirement

I meant on the other side of the equation.

Alright well i dont know about that and i cant answer that

An open set in $\Bbb R^n is diffeo to itself.

Which is probably going to be my answers in the exam

Alright i did not know that.

By the identity?

as a diffeomorphism

You should ask your prof. I think that definition makes things more difficult than intended.

Yes.

Yes thats not an option we dont have semester now

Anyways thanks

Should have studied this a lot sooner!

@TedShifrin in my textbook, the boundary of a set is defined as $\partial A=\bar{A} \setminus int(A)$, hence it should follow that $int(A) \cup \partial A=\bar{A}$. But, if I am not wrong, for that reasoning above I need $A=int(A) \cup \partial A$.

With all of $\Bbb R^k$ there, I’d like to see the proof that a circle or sphere are manifolds. Turns into a royal pain.

Why do you need that, Za? You only need containment.

Any point is either in interior or in boundary. NOT the converse.

oh yeah, this definition is phrased in terms of adapted charts

curious

That’s ok, but it’s having all of $\Bbb R^k$ that’s annoying.

why?

@TedShifrin you mean that since $A \subseteq \bar{A}=int(A)\cup\partial A$, if I check both $int(A)$ and $\partial A$ I check all $A$ as well because it is contained? Maybe I am (slowly) getting this, thanks for the help!

@ZaWarudo It depends also on what your lecturer means by $f'$ doesn't exist. The function $f(x) = x$ on $[0,1]$ has $\max$ at $x=1$, and certainly $f'(x) = 1$ qualifies as a valid derivative (in my mind, at least).

@copper.hat he said that it doesn't exist in a point $x$ if $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ is not finite or doesn't exist

I think it is sloppy wording on the lecturer's part, what point were they trying to make?

If $f$ has a maximiser on $A$ then it must either be in the interior or the boundary. There are no other choices.

"Mmm, , table dots are much the same from my perspective" @copper.hat Well it could take longer to first make a table, compute the function values and only then draw the points and plot the graph

But it might still be faster than to right away plot, since you will need to read off the y-values from the graph to compute the sum

Depends ig

Would you personally first make a table with x and f(x) values and then plot, or do it right away?

i don't distinguish between plotting dots and making a table. i fix a few points and then use the form of the function to guess the rest and then do a few checks.

How do I construct a countably many random variables (mutually independent) on $([0,1],\mathbb B,m)$ which take only two values $0$ and $1$ with probability $\frac12$?

Where $\mathbb B$ is Borel sigma algebra and $m$ is Lebesgue measure.

For two r.v.s it is easy. But I can't extend it.

e.g. $X=\chi_A,Y=\chi_B$ where $A=[0,\frac12]$ and $B=[\frac14,\frac34]$

Rademacher functions

at least that will give you the idea.

Sorry pairwise independent

@copper.hat Btw, when the function is of the form (x + 3)^3, do you imagine (-3, 0) as your new origin and go left and right from there?

Or do you imagine x^3 and shift all the points

yes, it is just $x^3$ shifted.

i think you are overthinking this. find what works for you and use it. there is no global optimal technique

@copper.hat Thanks, I didn't know about that. Any idea for pairwise independent case? I think there should be a simple solution.

i'm not sure what you ae asking, they are pairwise independent.

Yes, but this looks difficult

What I'm asking is part of a problem

The nest part asks me to find the distribution of $\sum \frac{X_{2i}}{2^i}$

Thanks, after plotting the graph, it looks simple

Does $\ker T = F_\ann$ imply $\ker T^{**}\subseteq F^\ann$?

Any ideas for finding the distribution?

$X_n$ looks like $\chi_A$ where $A=\bigcup_{0\leq e\leq 2^n-2}\left[\frac e{2^n},\frac{e+1}{2^n}\right]$, $e$ is even.

@Thorgott Giving a diffeo onto the whole thing is a pain. For example, you can’t with a multiply-connected open set.

Correction: Does $\ker T = F_\perp$ imply $\ker T^{**}\subseteq F^\perp$, here $F\subseteq X^*$?

I'm thinking of Hahn-Banach with $\overline{\ker T}^{w^*}\subseteq \ker T^{**}$ since the latter is $w^*$-closed

@TedShifrin there still is a $V$ in the definition, isn't that all we need

@Thorgott Oh, with the bad typing, I totally missed it. Good. I'm happy.

2 answers both are crappy, both upvoted...

Nevermind I'm wrong. The second answer is fine

Ok so in the context I'm considering image of $T$ is finite-dimensional so $(\ker T)^\perp = T^*(Y^*)$ and so $(\ker T)^\perp^\perp = \ker T^{**}$. And $F_\perp^\perp = \text{span} F$

My notes say "Every vector field on a manifold can be restricted to a vector field on a submanifold (embedded or immersed) S". If it were a differential form I would just pull-back using the inclusion map $S\hookrightarrow M$ but for vector fields one cannot generally define a pull-back as far as I know. I think the original statement is false, any hint?

that is just false

you can make it a vector along the inclusion $S\rightarrow M$, if that's terminology you're familiar with

but you can't restrict, that would require the vector field to be tangent to $S$ at points of $S$

picture the radial vector field on $\mathbb{R}^2\setminus\{0\}$, for example, it makes no sense to restrict that to a vector field on $S^1$

(in the presence of a given Riemannian metric, you could do an orthogonal projection, but that's hardly a form of "restriction" anymore)

Thor be right.

a welcome occasion

Oh, like sections along curves, alright. That must have been the meaning of the mysterious note, then. Thank you.

right, a vector field along a smooth map $f\colon N\rightarrow M$ associates to each $p\in N$ a tangent vector in $T_{f(p)}M$ in a smooth manner

if you know about vector bundles, it's the same thing as a section of the pullback bundle $f^{\ast}TM$

Although I know vector bundles, the only pullback I know is the one acting on k-differential forms. I'll look it up