Mathematics - 2023-08-26

Peter

00:01

Does anyone know whether there is a program in progress checking whether the $33$ th Fermat number is prime ? A factor search has not yet revealed a result.

MathCrackExchange

00:39

0

Q: Most deceptively simple linear independence question that I've come across!

Define $$(d \mid \cdot) = \begin{cases}1 \text{ if } d \text{ divides } \cdot \\ 0 \text{ otherwise }\end{cases}$$. This is called a divisor indicator function. It's the same exact thing as you'd define "$\chi_{d\Bbb{Z}}$" to be. Now someone mentioned in another thread that to prove: $$ \{ (d \m...

Dannyu NDos

05:01

How should I notate the set of quaternions with integer coordinates? $\mathbb{Z}[i,j]$ maybe?

Ted Shifrin

05:15

No, that is a polynomial ring.

You have to define a notation. Maybe $\Bbb H_{\Bbb Z}$?

Dannyu NDos

I mean... $i, j \in \mathbb{H}$ anyway, and I've never seen variables other than $x, y, z$ for polynomial rings.

Ted Shifrin

You’re wrong. For any complex number $\alpha$, people write $\Bbb Z[\alpha]$ all the time.

Dannyu NDos

D:

one potato two potato

05:41

I sometimes don't read the book linearly but directly read the chapter I'm interested in and go back if there's something I need to know.

wait, maybe I always read in this way

Ted Shifrin

I’m not reading what you say. Maybe I’ll come back to it later.

2

Sourav Ghosh

06:55

Is the operator $T:C_00\to C_00$ defined by $T(x=(x_n)) =(\frac{1}{n}x_n) $ compact?

$T$ is the operator limit of a sequence of (bounded and finite rank) compact operator .

leslie townes

what's the norm on C_00? oh okay then yes

Sourav Ghosh

sup norm

leslie townes

norm limits of finite rank operators are always compact. there should be a proof on MSE for bounded operators on a general banach space

Sourav Ghosh

But C_00 is not a Banach space.

leslie townes

what is C_00

Sourav Ghosh

06:59

Sequences with all but finitely many terms are 0.

leslie townes

okay, you might want to doublecheck your definition of 'compact' for operators on things that might not be banach spaces

but you ought to be able to get this out of the fact that T extends uniquely to an operator on the banach space of sequences going to 0, and that extension is compact

Sourav Ghosh

Correct.

c_00 dense in c_0

leslie townes

or you could do it bare-handed without touching the general results about norm limits of finite rank operators, but that strikes me as maybe not the best approach

okay when you write stuff like c_00 and c_0 just as a word of warning, these kinds of notations are very popular but by no means standard or universal

so in the long run it may save time to specify what you mean by them in advance :)

Sourav Ghosh

@leslietownes Is this result valid for general normed spaces?

leslie townes

i don't know, that seems like it ought to be easy to answer if the answer is yes, though

i basically never think about operators from A to B where at least B isn't complete

maybe all you have to do is fiddle about with completions a little

Sourav Ghosh

07:08

Totally bounded and relatively compact are equivalent in a complete metric space. May be that's why it is easier to deal with Banach spaces :)

leslie townes

at some point it feels like the definition you choose of 'compactness' will start to matter

e.g. all of the things that are generally equivalent will stop being equivalent

the notion of which maps on banach spaces ought to be called 'compact' was itself pretty unstable for a while

Sourav Ghosh

$T:X\to Y$ is said to compact iff $cl(T(B) ) \subset Y$ is compact.

leslie townes

okay but it feels like in general you'd "want" to take the closure of T(B) in something larger than Y, if Y isn't complete

maybe it's worth a MSE question

if the answer is simple, someone will surely provide it quickly

Thomas Finley

07:26

0

Q: Trouble in Proving the Sequential Criterion for Limits at Infinity

Let $A\subseteq R,$ let $f : A\to R,$ and suppose that $(a,\infty)\subseteq A$ for some $a \in R.$ Then the following statements are equivalent: (i) $L=\lim_{x\to \infty}f(x)$ (ii) For every sequence $(x_n)$ in $A \cap (a,\infty)=(a,\infty)$ such that $\lim x_n=\infty$, the sequence $f(x_n)$ conv...

real-analysis sequences-and-series limits functions convergence-divergence

Need help!

Jakobian

09:05

@DannyuNDos I don't understand this conversation

Jakobian

09:43

@AlessandroCodenotti math.stackexchange.com/a/4758785/476484

It actually assumes separable too

Dannyu NDos

@Jakobian I meant usual adjoint rings.

Jakobian

And it proves something a little stronger, namely existence of a complete geodesic metric

So all balls are not only connected but also path-connected

But yeah the setting is only slightly more general from Peano continua

@DannyuNDos oh, I get it now. You're arguing about using $\mathbb{Z}[i, j]$ without further context. But if you add $i, j\in\mathbb{H}$, it should be fine.

Dannyu NDos

Thx!

Thomas Finley

11:00

0

A: Trouble in Proving the Sequential Criterion for Limits at Infinity

We can prove that $(ii)\implies (i)$ in a much simpler way. We show this, by proving that the contrapositive of the statement i.e $\neg (i)\implies \neg (ii)$ is true. The contrapositive translates to the below statement: If $\lim_{x\to\infty}f(x)\neq L$ then $\exists (x_n)\in A\cap (a,\infty)$ ...

Guys! I have just posted an answer. What d'ya'll think bout it?

sunny

11:57

@robjohn I have just stumbled upon a power series in my exercise compendium that you found the radius of convergence of here, i.e. $\sum_{n=0}^{\infty}(2^n+n^2)x^n$. However, I'm a little uncertain about the limit below that you evaluate to $1^0$. What was your reasoning? Here is mine:

$$\lim_{n\to\infty}\left(1+\frac{n^2}{2^n}\right)^{1/n}=e^{\left(\lim\limits_{n\to\infty}\frac{\log \left(1+\frac{n^2}{2^n}\right)}{n}\right)}=e^{0\cdot\log(1)}=1.$$

I'm a little uncertain about the second equality. It holds that $\lim_{n\to\infty} f(x_n)=f(\lim_{n\to\infty} x_n)$ if $f$ is continuous and the limit exists, however, how do you proceed from $$\lim_{n\to\infty}\left(1+\frac{n^2}{2^n}\right)^{1/n}=\lim_{n\to\infty}e^{\log \left(1+\frac{n^2}{2^n}\right)^{1/n}}.$$

Jakobian

I could help you with this instead

sunny

Now, $e^{\log x}$ is continuous, but a priori we don't know the limit $\lim_{n\to\infty}\left(1+\frac{n^2}{2^n}\right)^{1/n}$, so can we still move the limit operator inside the exponential function?

@Jakobian that would be amazing :)

Jakobian

You don't need to use exponentials here

$\lim_{n\to\infty} x_n^{y_n} = x^y$ where $x = \lim_{n\to\infty} x_n$, $y = \lim_{n\to\infty} y_n$ where we assume both $x, y$ are finite and $x > 0$

so this is just continuity

Instead of function of one variable, consider the function of two variables $g(x, y) = x^y$ for $x > 0, y\in\mathbb{R}$. This function is continuous

To see this one can simply note that $x^y = \exp(y\cdot\log x)$, and composition and product of continuous functions is continuous

well I guess I do use exponentials to justify that x^y is continuous

there would be another approach depending on how you define x^y

(as limits of powers of rational numbers)

but that one is more complicated

sunny

12:15

Ok.

Jakobian

@Jakobian yeah, the point is that this is true

sunny

if the limit of $y_n$ wouldn't exist, we would have issues, like $\left(1+\frac1{n}\right)^n$.

Jakobian

12:39

yeah

sunny

The "Short proof" here is pretty neat.

Using continuity of exponentials and log, and the product rule for limits.

Jakobian

12:55

@Jakobian this is just my proof

my proof is in fact, a one line

robjohn

13:09

@sunny $x^y$ Is continuous at $(x,y)=(1,0)$, so there is no problem. $1^0=1$

lynx_s

13:42

I don't understand how the union of the subspaces of x-axis and y-axis is not a new subspace

in R^2

Jakobian

@lynx_s what do you mean by that

$\mathbb{R}\times \{0\}\cup \{0\}\times\mathbb{R}$ ?

subspace in what sense

it is a subspace of the topological space $\mathbb{R}^2$...

lynx_s

Hold on I'll look it up it was written in my book

I'll translate the whole section what they write about it

Jakobian

well you see, the word "subspace" can mean many things in different settings

so, what is your setting?

I'm guessing this is introduction to vector spaces

I don't need the whole section of the book translated, and this would complicate this

lynx_s

Well the chapter is about vector spaces and we just got the definition of subspace, then we have the definition of span, where then they give some examples and they state that the subset of two subspaces is always a subspace which I get. Then they start about sum and direct sum of subspaces

Jakobian

when you're talking about subspaces on the internet, try to hint at that you're talking about vector spaces

for example you can say "linear subspace"

this makes it easier for people reading to guess the context at hand

Xander Henderson

13:52

Or "vector subspace".

Jakobian

yes, or that

lynx_s

and there they say that subset of subspaces is always a subspace but it's not always true for union of subspaces

and okay will do

Jakobian

@lynx_s do you mean intersection, not subset?

lynx_s

wait not subset

yes

sorry

Jakobian

yes. The union of x-axis and y-axis is not closed under addition for example

$(1, 0) + (0, 1) = (1, 1)$ is not part of it

Lukas Heger

13:54

One can show that a finite-dim vector space over an infinite field is never a finite union of proper subspaces

lynx_s

wait what is closed maybe that's what i'm missing to understand it properly

Jakobian

oh sorry

by closed under operation I mean that

$A\subseteq \mathbb{R}^2$ is closed under addition if $x, y\in A$ implies that $x+y\in A$

so that sum of two vectors still belongs to this set

one potato two potato

@LukasHeger !! I've seen this statement about 2 years ago

Jakobian

this is one of the requirements to be a vector subspace of a vector space

the union of x-axis and y-axis fails this property

Lukas Heger

@onepotatotwopotato it's occasionally useful

lynx_s

13:57

Okay I think I'm getting it but still a little bit vague I'm going to re-read this and my book and if I still don't get it I'll come back, thanks

one potato two potato

@LukasHeger any nontrivial application of it?

Jakobian

Alright, good luck learning

Lukas Heger

@onepotatotwopotato a finite field extension is simple iff there are only finitely many intermediate fields

The thing about unions of subspaces gives you one direction pretty immediately

Note that this can be used to prove the primitive element theorem, but that requires an unconventional order in which one proves things

Jakobian

@LukasHeger but it's not needed to prove this I think, right?

When I was reading Fields and Galois theory book, they didn't use any theorem like this

Lukas Heger

No you don't need it

lynx_s

14:05

@Jakobian when we do $(1, 0) + (0,1)=(1,1)$ do we look if $(1,1) \in U \cup W$ or if $(1,1) \in U+W$ I think here my confusion lies, or even with if $(1,1) \in U$ or $W$

Jakobian

I don't really know what $U, W$ are

Let $A$ be the union of x-axis and y-axis, then $(1, 0), (0, 1)\in A$ but $(1, 0)+(0, 1) = (1, 1) \notin A$

Lukas Heger

3

A: Polynomial takes distinct values at set of points

The set of all possible $(a_1,\dots,a_n)$ is itself an $n$-dimensional vector space over $F$. The subset of $(a_1,\dots,a_n)$ for which the undesired $f(p_i)=f(p_j)$ occurs is a subspace of that vector space of dimension $n-1$. There are finitely many such subspaces, and it's not hard to check th...

Here's another application

lynx_s

@Jakobian Yes sorry should've clarified that I supposed U to be the subspace (a, 0) so the x-axis and W the y-axis

Jakobian

@lynx_s the former then

lynx_s

Okay I get it now, but I do have an extra question now, about the span((1,0), (0,1)) because they do span the whole vectorspace right? but the span takes a subset right? if that's the case I think everythings clear for me

Jakobian

14:14

Yes, and yes

lynx_s

Okay perfect thanks alot

Jakobian

No problem!

Lukas Heger

@lynx_s $U\cup W$ and $U+W$ are not the same, here we're looking at $U\cup W$, which is generally not a subspace unlike $U+W$. There's a connection between the two: $U \cup W$ is the smallest subset containing $U$ and $V$, while $U+W$ is the smallest vector subspace containing $U$ and $V$, so they're analogous

And to connect everything with the span, we always have $\mathrm{span}(U\cup W)=U+W$ if $U$ and $W$ are subspaces of the same vector space.

lynx_s

Oh okay now I see, makes way more sense now, I think I was mixing up all the definitions, thanks for clarifying it

Shadow sparkle

15:44

Hello

John Zimmerman

Hello @Shadowsparkle

Shadow sparkle

I have a question about elementary set theory can I ask ?

John Zimmerman

sure

0

A: Proving $\phi$ is a smooth map and constructing an explicit isometry

I will prove that the mapping $T$ is linear, continuous and smooth. I will not explicitly construct an isometry at this time. We are given a spacetime: $$(\zeta,g)$$ which we recognize as Minkowski space in different coordinates. I will change the notation here slightly but the core idea remains ...

Is my answer correct?

Shadow sparkle

Wow I can't send my question ! Interesting I wonder why

Thanks. When reading about proper and improper subsets. I understood the definition of a proper subset. But an improper subset makes no sense. I used to see A ⊆ B as in, A can be a proper subset of B or equal to B.

But when I searched for the symbol of Improper subset, I understand that they use the same symbol for it. But improper subset means A subset which contains all the elements of the original set is called an improper subset So it is equal to the original set. So yah basically I am confused 🤔

Did it 🙂

Soumik Mukherjee

You can use $\subsetneq$ to denote proper subset and $\subseteq$ to denote any subset( proper or improper)

Xander Henderson

15:56

$A \subseteq B$ means that either (a) $A$ is a subset of $B$, or (b) $A = B$.

It is like $a \le b$ for the real numbers.

I have never seen $\subset$ used to denote a proper subset, and an author who does it is making a mistake (in my opinion).

Typically, authors either use $\subset$ for a proper subset, and $\subseteq$ for any subset, or...

...or they use $\subsetneq$ for a proper subset, and $\subset$ for any subset, or...

...(and this is the best, in my opinion) they use $\subsetneq$ for a proper subset, and $\subseteq$ for any subset.

one potato two potato

I never use $\subseteq$

Xander Henderson

@onepotatotwopotato So you must be one of the $\subsetneq$ / $\subset$ crowd.

Shadow sparkle

Thank you so much so there isn't any symbol for improper subsets?

Soumik Mukherjee

@XanderHenderson I always use this one

Xander Henderson

@Shadowsparkle I just gave you THREE!

Shadow sparkle

15:59

Symbols don't show up in my phone sorry 😔

Xander Henderson

Though I must admid that I have never heard the phrase "improper subset" used before.

Shadow sparkle

I ment on my phone 🤣

Not native sorry

@XanderHenderson Wow really !

Thank you all 💜

one potato two potato

I use $\hookrightarrow$ instead of $\subset$ thesedays. Writing $\subset$ is difficult for me. It looks like $<$ sometimes.

Xander Henderson

"$A$ is a subset of $B$" means that if $x\in A$, then $x \in B$. This includes the case that $A = B$. That is, every set is a subset of itself. Sometimes we want to exclude this condition, so we introduce the notion of a proper subset. "$A$ is a proper subset of $B$" means that $A$ is a subset of $B$ (per the above definition), and $A \ne B$.

So for $A$ to be a proper subset of $B$, there must be some element $x \in B$ which is not in $A$.

@onepotatotwopotato Yuck!

I don't like that.

That looks like an injective map.

one potato two potato

of course I don't use it in a public place.

Shadow sparkle

16:04

Oh and one other thing why we use this symbol A⊆B we know that A might be a proper subset of B or equal to B. So does that mean we don't know the elements of the two sets to know which is correct ?

When we use *

Xander Henderson

@Shadowsparkle When you write $x \le y$, do you know if $x < y$ or $x = y$?

Shadow sparkle

Nope

Xander Henderson

@onepotatotwopotato PERVERT!

Shadow sparkle

But they are sets not variables, what is we know the elements and I can see that A is a proper subset of B but I still write A ⊆ B is that wrong ?

Soumik Mukherjee

"We give the desert the
leonine topology, in which a subset is closed if it is the whole desert or if it contains no
lions. The set of lions is now a dense subset. Now put an open cage in the desert.
By density it contains a lion. Once we shut the cage we have caught the lion! " I wonder what an open cage is, a cage whose door is open?

Xander Henderson

16:07

If you care that $A$ is a proper subset of $B$, then writing $A \subseteq B$ is a bad idea.

But most of the time, we don't care.

And it is true that if $A \subsetneq B$, then $A \subseteq B$.

Shadow sparkle

Thanks ❤️

Xander Henderson

So it is about keeping track of the things you care about. Sometimes it matters. Sometimes it doesn't.

Please don't emoji all over me. :/

Shadow sparkle

forgive me :(

one potato two potato

@XanderHenderson Keep it to yourself, I once used $n$ instead of $\cap$ and didn't notice any problem before getting a scolding from my math teacher

Xander Henderson

@onepotatotwopotato I think that "Keep it to yourself" is exactly the advice I am giving you, you math pervert. :P

Ted Shifrin

16:16

@onepotatotwopotato That strongly suggests a mapping, rather than a set statement.

Xander Henderson

@TedShifrin That's what I said!

He's a PERVERT!

Ted Shifrin

I said it in a far more erudite fashion.

Xander Henderson

@TedShifrin I said it all fancy, first:

16 mins ago, by Xander Henderson

That looks like an injective map.

:(

Ted Shifrin

Oh, that escaped my eyes. Sorry.

Xander Henderson

:P

Ted Shifrin

16:19

But not just any injective map. It really is inclusion.

Xander Henderson

Oh, right.

Yes. That is what I meant.

Ted Shifrin

My erudition wins. One potato is the big loser.

Xander Henderson

It keeps all the structure, too.

Yay!

Speaking of potatoes, I'm going now. Groceries.

I've been in the office too much for a Saturday. :/

Ted Shifrin

Happy shopping.

Jakobian

@SoumikMukherjee a one dead man

@onepotatotwopotato this is the weirdest opinion on the topic I ever heard

Thomas Finley

16:22

Soumik Mukherjee

@Jakobian I don't get it

Jakobian

because the lion ate him

the cage was open

Soumik Mukherjee

ohh XD

Jakobian

@Shadowsparkle sets are variables that are sets

Thomas Finley

Can anyone help me understand how do they write, "...from which the conclusion follows readily" ?

Soumik Mukherjee

16:27

@ThomasFinley Sandwich theorem

Thomas Finley

@SoumikMukherjee Oh, but is Sandwich Theorem valid when $\lim f(x)=\lim h(x)=\infty$

Shadow sparkle

@Jakobian can you please explain it a bit more ( I again apologize for having such a stupid question )

Oh and I have another question too " this time about probability "

Thomas Finley

I thought only, if $f(x)\leq g(x)\leq h(x)$ and $\lim_{x\to c} f(x)=\lim_{x\to c} h(x)=a$ we can say, $\lim_{x\to c}g(x)=a$

But, now that you mention, I think, $a$ can be $\pm\infty$ besides being real.

Soumik Mukherjee

@ThomasFinley Yes

Thomas Finley

@SoumikMukherjee Thanks! I understand

Soumik Mukherjee

16:34

Welcome!

Jakobian

@Shadowsparkle by variables you probably think of elements of sets, but when you write something like "let $A$ be a set" then $A$ is a variable in this setting too. Moreover, in ZFC we assume everything is a set.

the difference between $A\subseteq B$ and $x\leq y$ is mostly that in state of mind

Shadow sparkle

We know that the formula to calculate the probability of an event in an experiment is ( number of desired outcomes in our event / the total outcomes ), now I have always seen events as sets containing elements that are related, but what do we do if our desired outcomes are not related? We can't put them in the same event right? So do we have to calculate the probability of each one and then add them up ?

Sorry again 🥲

Jakobian

the thing you're talking about is very basic version of probability called classic probability

in practice we actually work with something called random variables

it doesn't matter if they are related or not

$P(A) = |A|/|X|$ where $X$ are all outcomes and $A$ is the set of outcomes we measure probability of (according to your formula)

it's irrelevant what $A$ is, as long as it's a subset of $X$

Shadow sparkle

Yes you are right it is irrelevant thank you so much

@Jakobian do you know any sources that I can read more about this ?

Jakobian

set theory?

Shadow sparkle

16:42

I am already studying that but till now I haven't heard of the thing you talked about earlier

Jakobian

which part of it

Shadow sparkle

Maybe it will come up later thanks

Thomas Finley

@SoumikMukherjee Just a sec, what if, $f(x)$ tends to infinity then how to show, $g(x)$ tends to infinity as well ?

Ted Shifrin

@ThomasFinley Forget sandwich. That is not relevant.

Thomas Finley

@TedShifrin Then what?

Ted Shifrin

16:49

If $\frac12 f(x)<g(x)$, what do you conclude?

Thomas Finley

@TedShifrin if f(x) tends to infinity then so does g(x)

Ted Shifrin

Well, then ...

Thomas Finley

Also, as g(x) tends to infinity then so does f(x) as g(x)<3/2f(x)

Ted Shifrin

Right.

Thomas Finley

Is this the reason?

Ted Shifrin

16:51

That's it.

Thomas Finley

Thanks , I understand

Shadow sparkle

@Jakobian can I ask you a question ⁉️ when did you study this at 11th grade cause it is a bit hard for me to understand.( I just started 11th grade )

Jakobian

I don't understand American and British education

Shadow sparkle

Ohhhhh where are you from ?

Jakobian

Poland

Shadow sparkle

17:02

Oh wow I have always loved to visit Poland, and thank you for your help

Ted Shifrin

@Shadow I don't understand your question. Can you explain that exact circumstance you're talking about?

Shadow sparkle

Which question ? I asked two questions :)

Ted Shifrin

It looked like one question. What do you do if the outcomes are not related?

I'm asking what exercise you're trying to do.

Shadow sparkle

Oh I was just studying and this question came inside my mind. I examined some events and I saw that the outcomes are all related so I was like what if I want the probability of two outcomes that are not related. I thought maybe we should consider them as two different events and then calculate the probability and then add them up. ( I know that because the sample space is same for both it doesn't matter, I can put both of those in a set and then calculate it's probability)

but I wanted to see how to be more accurate since the outcomes of an event are all related to each other

Ted Shifrin

I want a concrete example. I still do not understand what you mean by outcomes that are related or not.

Eventually, you will get to something called conditional probability, where you compute the probability that event A occurs given that event B has already occurred. This is why I'm wondering what you're actually thinking of.

Shadow sparkle

17:19

Ok for example Randomly choosing one ball from each of four bags, each containing one black and one white ball, each equally likely to be chosen, is a 4-trial probability experiment with 16 outcomes.

So, this experiment's sample space, which comprises the experiment outcomes, is
{BBBB,BBBW,BBWB,BBWW,BWBB,BWBW,BWWB,BWWW,WBBB,WBBW,WBWB,WBWW,WWBB,WWBW,WWWB,WWWW}.

If I say let's calculate the probability of outcomes with at least one white in them ( we know what to do ) and see all these outcomes of my event are related cause they have one white in them. But what if my desired outcomes are unrelated for example I have these two desired outcomes {wwww} and { bbwb} what should I do ? I can't put them in the same event cause they are not related so should I calculate the probability of each and then add them up

This was the example that I came up with :(

@TedShifrin oh you are right !!!

Ted Shifrin

I still don't see what you mean by "are not related." These are two possible events with the criterion that there is at least one W.

Shadow sparkle

Maybe my understanding is wrong!

Ted Shifrin

I misspoke. These are two possible outcomes which appear in the same event.

I don't understand why your outcomes are unrelated.

They have the same characteristic that you're trying to study.

Shadow sparkle

@TedShifrin how can they be in the same event ! They are not related 🥲 like consider BBBB AND BBBW. What did I understand wrong while studying, I hate that I'm confused

Like see what the relationship between BBBB and BBBW in your opinion

There is no relation ! Except for the fact that they are both outcomes of the same experiment

I am so sorryyyyy for bothering you with my question

Like if our sample space is for another experiment 1,2,3,4,5,6 if I say I want the probability of event A ( even numbers ) I know what to do but if I say I want the probability of outcomes 2 and 5, do I consider them as different events ?

copper.hat

17:36

An event is juts a subset of the possible outcomes. You can define the event whatever way you want.

$\{2\}, \{5\}, \{2,5\}$ are three different events.

Ted Shifrin

@Shadow You don't need to apologize; I am the one who asked you to elaborate. I'm trying to understand how outcomes could possibly be related, other than by being different possibilities for a given event. I could make up different experiments/events that have the same possible outcomes. For example, I might ask for at least two B's (rather than asking for at least one W).

Shadow sparkle

@copper.hat Really well I know that an event is a subset of the sample space but if we think about the word event it means sth happening and that sth can have different outcomes when I say (even numbers) that is my event and based on the sample space I have different outcomes like 2,4,6 but when we have two things that are unrelated what should we do ?

Ted Shifrin (wait do you see experiments and events as the same thing?)

copper.hat

Sorry I do not understand what you are saying/asking. shorter sentences would help

i do not understand what you mean by 'have two things that are unrelated'

Ted Shifrin

That's where I started, copper :)

No, sets of events are outcomes of experiments. Events all live in the sample space. I have been guilty of sloppy vocabulary.

copper.hat

A depressing quote (from a mathematical perspective) AE Housman,"Perfect understanding will sometimes almost extinguish pleasure."

Shadow sparkle

17:48

@copper.hat 🥲

Nice quote

I loved it

Ted Shifrin

I have learned a lesson this morning: I jumped into a triple integral problem without having read it carefully. I assumed the region was a "standard" one that I've taught/assigned literally dozens of times. But no — they had given a different inequality. How utterly embarrassing.

And now I'm messing up probability vocabulary trying to understand Shadow. Maybe I should go back to bed :P

Shadow sparkle

@TedShifrin oh dear, ouch that hurts. No it is not embarrassing we are all sometimes like this well for me is always 🤣

It is*

@TedShifrin Can I talk to you tomorrow then? Till then I will think about my own question more

Ted Shifrin

Anyhow, both copper and I have yet to understand what "related" or "unrelated" is supposed to mean.

Shadow sparkle

@TedShifrin Are you in university or college? what are you studying?

copper.hat

It think a much omitted part of teaching is to make mistakes and fix them real time.

Ted Shifrin

17:53

To make things more complicated (or perhaps not), what if your experiment consists of completely unrelated portions? Toss two coins and throw a die. What is the probability that the number of heads plus the result showing on the die is even? :D

@Shadow I am a retired university professor :D

Shadow sparkle

Oh my god 🥲 I am sorry professor for bothering you, I hope you don't mind 🥲

Ted Shifrin

Probability was something I never studied, but I taught the course my final year because I was bothered by the fact that I hadn't taught it in 36 years of teaching. It was actually a wonderful course, but I worked hard :)

Shadow sparkle

Wow :) awesome

Ted Shifrin

But, yes, I had to keep the vocabulary clear and correct. But that was 8 years ago.

@copper It looks like Anne posted her solution to that even function problem. It looks similar to, but easier than (I think), the Dieudonné exercise.

Shadow sparkle

@TedShifrin I'll think about my question again to understand what I am missing but if I have any other question can I ask you please ? Later maybe

Ted Shifrin

17:58

Sure, @Shadow. That said, I'll do better with topics other than probability :)

copper.hat

@TedShifrin I finally cleaned up my answer.

Ted Shifrin

Oh, I'll have to go look.

copper.hat

I struggled to express the answer compactly.

Ted Shifrin

I've now removed two garbage answers of mine in the last two days. This is not a trend I'm proud of.

copper.hat

Don't be hard on yourself.

Your contributions are innumerous

Ted Shifrin

18:05

Well, some of the annoyance is deserved.

copper.hat

I understand. I beat myself up all the time.

Often deserved.

user 1591719

18:47

@robjohn Hi. How is it going lately? :-)

robjohn

19:26

@user1591719 Doing okay. How are things? I see a sequel has come out?

user 1591719

19:52

@robjohn True. I wanted to share a kind of problem you would probably enjoy a lot (good to try when you have time) $$\displaystyle \log^2(2)=\lim_{n\to\infty}\sum_{k=1}^{n} \left(\frac{H_{k+n}}{k+3n}-\frac{n H_k}{(k+n) (k+2n)}\right)$$.

@robjohn Nice to hear from you. Long time no see. :-)

robjohn

@user1591719 thanks, I’ll take a look at it later. We’re getting lunch soon.

user 1591719

@robjohn At any time you want to. Pretty much away from the noisy places of the world (mostly) for a long time in order to be capable to listen to my deepest ideas.

robjohn

20:10

@user1591719 The best of luck. My life is pretty noisy these days, but hopefully it will quiet down again soon.

冥王 Hades

20:25

@TedShifrin when I have to do that I just rip the page off and eat it

not kidding about the eating part

That reminds me, paper tastes better with some PDE seasoning.

John Zimmerman

dam nobody looked at my question nor answer to my own question

no votes at all

just one unwarrented 'close' vote

Jakobian

unwarranted

John Zimmerman

unwarranted

冥王 Hades

“””unwarranted”””

Jakobian

My differential geometry knowledge disallows me the judgement of your question and answer

but I'd like to drink something tbh

冥王 Hades

20:35

Water

John Zimmerman

Water

冥王 Hades

I wanna eat a materwelon

John Zimmerman

Did you exchange the 'm' and 'w' there?

冥王 Hades

I did?

John Zimmerman

shouldn't it be watermelon

Jakobian

20:37

ice tea

冥王 Hades

I type too fast

Jakobian

I'm going to put it in a freezer after

for 5 minutes

冥王 Hades

3 take it or leave it

Jakobian

or 8 maybe

John Zimmerman

I just got an upvote

user 1591719

20:40

@robjohn Thanks!

Jakobian

21:02

opening window was a bad idea

mosquitos came

冥王 Hades

Guys I think my hairline is receding

I don’t wanna look like a divorced 45 year old

Xander Henderson

@冥王Hades C'est la vie.

John Zimmerman

21:19

What do you call two solutions to the like heat equation whose product are a solution to a nonlinear partial differential equation?

$f(t,x)$ and $f(t,y)$ are solutions to the like heat equation for example

s.t. $f(t,x)\cdot f(t,y)$ is a solution to a highly nonlinear partial differential equation.

highly nonlinear (by the way)

wow it's already 5:24

$$2\sqrt{s} \frac{\partial}{\partial s} \sqrt{\Omega_s(x,y)}=\sqrt{x}~\sqrt{-\frac{\partial}{\partial x}\Omega_s(x,y)}+ \sqrt{x}\sqrt{-\frac{\partial}{\partial y}\Omega_s(x,y)}$$

square root cityy

Yai0Phah

22:53

@Jakobian Off topic, but there is a thing called fly screen which is in general effective (I am using tesa.com/en/consumer/tesa-insect-stop-standard-for-windows.html).

Xander Henderson

23:31

@Yai0Phah the product you have linked seems to be for Windows only.

Is there a Mac version?

:P

Transcript for

$Mathematics$ Mathematics