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00:00
@N3buchadnezzar What's the problem?
@DanielFischer Hello Daniel!
How are you? Did you hear the bad news?
@PedroTamaroff You mean Grothendieck?
I just read it here a couple of minutes ago.
@PedroTamaroff The good news is that Grothendieck will be reborn and contribute to math again, hopefully. What seems good may be bad, and what seems bad may be good.
00:01
@skullpatrol They don't render in my browser, but I know that they are the three monkeys...
@DanielFischer The loop stops prematurely
Yes @robjohn classics :-)
@robjohn I have three monkeys now: runny nose, sore throat, dry cough.
@DanielFischer It is amazing how the proof of the Riemann Mapping theorem uses almost all the fundamental yet elementary facts about holomorphic functions.
@DanielFischer I think I figuredit out =)
00:04
does anybody know if he wrote anything at all about his life in isolation?
@PedroTamaroff Is this your first course in complex analysis?
@JasperLoy Yes.
Ooh... the clock has flipped... I can get points again :-)
I would imagine so, @Alexander, but I would also imagine it is not publicly available
@PedroTamaroff Then it must be moving very fast to cover the RMT.
00:05
@N3buchadnezzar What was it? That range doesn't include the upper bound?
Good luck @robjohn
@DanielFischer Well it does except for the last one
@MikeMiller You mean in letters to his friends?
@robjohn I better put the clock back, lol.
@N3buchadnezzar I meant range(1,3) = [1,2].
00:06
i'll tell you one thing, i do not understand that guy's theorems. everytime his name comes up in a class I have to brace myself for some next level confusion.
@DanielFischer scracthes head Does it really do that?
I hope I win something big in tonight's lucky draw.
@JasperLoy Your momma?
@N3buchadnezzar Yes, it is my mum going to the lucky draw, not me, lol.
@N3buchadnezzar Yes. It's the spec.
00:07
He started off as a functional analyst, @Alex ...
@DanielFischer I should add a +1 then
I ordered a whole bunch of books from amazon, decided to return them before they arrived, and interestingly they all got lost by the carrier, lol.
@DanielFischer pastebin.com/7N2LFtZ9 Still the code seems to miss many combinations..
@TedShifrin maybe that's why.
The books about learning languages @JasperLoy?
00:13
@JasperLoy I don't know if they'll get there. I am studying on my own.
@N3buchadnezzar You mean like 3+3+3+3+3+3+3 = 21?
@AlexanderGruber Look. =)
@DanielFischer Yeah
@N3buchadnezzar Since you let the upper bound for the sum of the first six be S - t, which is 14, and the upper limit is excluded in range, you never get anything with g < 8.
@DanielFischer Well each number must be atleast 1
00:20
@N3buchadnezzar But that only gives an upper bound on g, namely 15. The sum of the first six could be as large as (numbers-1)*(S/numbers) (18).
(Note that if numbers doesn't divide S, it's more complicated.)
@DanielFischer I think I figured it out
@DanielFischer pastebin.com/pqd63iu4 This seems to get all the cases.
00:54
@Chris'ssis: I am mired in Si and Ci functions. I think I need a different approach, so I am putting this to rest for a while.
How are you @Skull?
Anyone remembers a bit from logic course? I am stuck on some simple exercise
Fine thanks. How are you pal @Studentmath ?
Fine, a tad exhausted but fine :)
01:09
Grothendieck died today :O
Yep, we heard :'(
He was 86.
Sleep? Never heard of it.
Hey!!! Is every ideal $I=<f(x)>$ a principal ideal?
Oh wrong princaple
@KajHansen And if it is irreducible?
01:17
Oh wait.
I was thinking maximal ideal >_<
Yes, it is principal @evinda.
By definition, any ideal generated by a single element is principal.
"princ"
@KajHansen Even if it is not irreducible?
single
That is correct @evinda.
@evinda What is your definition of principle ideal?
01:19
Moreover, in polynomial rings over a field are principal ideal domains (all ideals are principal).
@Alizter There is no definition in my notes :/
@evinda That is not too good :P
@Alizter Could you give me a definition?
sure
for polynomial rings in one variable
01:21
Drum roll....
The definition I use is as follows:

A principal ideal is an ideal $I \subset R$ that is generated by a single element $a \in R$. Notation-wise, a principal ideal is any ideal that can be written as $<a>$.
Or an integral domain where all ideals are principal
oh wait
@KajHansen @Alizter Could you also maybe at the following exercise?

Notice that in $\mathbb{C}[X,Y,Z]$:
$$V(Y-X^2,Z-X^3)=\{ (t,t^2,t^3)/ t \in \mathbb{C}\}$$

In addition, show that:

$$I(V(Y-X^2,Z-X^3))=<Y-X^2,-X^3>$$

Finally, prove that the ideal $<Y-X^2,Z-X^3>$ is a prime ideal of $\mathbb{C}[X,Y,Z]$. Conclude that the algebraic set $V(Y-X^2,Z-X^3)$ is irreducible.


Could we show like that, that in $\mathbb{C}[X,Y,Z]$:
$$V(Y-X^2,Z-X^3)=\{ (t,t^2,t^3)/ t \in \mathbb{C}\}$$
?

$$V(Y-X^2, Z-X^3)=\{(a,b,c) \in \mathbb{C}^3 | b-a^2=0, c-a^3=0 \Rightarrow b=a^2, c=a^3\}=\{(t, t^2, t^3)| t \
that is PID
A principal ideal is an ideal that has multiplicative absorption.
How can you tell so quickly @Alizter?
01:23
a screw it
I am too tired
Try sleep.
Ever heard of it ^_^ @Alizter
@RobJohn I found that my divergent integral indeed should be regularized by the surrounding structure of the problem, and comparing some numerical results it seems like just subtracting the offending divergent piece is the right thing to do. Now I'm trying to prove it though and I have no idea how to make progress.
@Alizter Were you referring to the exercise I asked? :/
@evinda No. I can't think straight.
I need to finish my physics coursework by tomorrow
All the hardness tests look the same to me.
don't tell semiclassical.
@Alizter Hardness tests?
01:30
@KevinDriscoll For materials.
@ALizter Yes, where are you learning about that?
Have somebody an idea if that's what I have tried is right and how I can show that I is a prime ideal?
@KevinDriscoll erm school?
Ya but what subject?
physics?
01:31
UUUUUUGH. That hurts my soul, @Alizter
@KevinDriscoll Why?
They are all "Stick diamond pointy thing in material"
tbf most datas use brinell and convert.
@Alizter There are so many more important things to learn in physics class. Hardness tests sounds like the ultimate waste of time
You are in secondary school, right?
@KevinDriscoll Oh no. You will hate the physics we are supposed to learn. I can't even understand how they made it so boring.
The most complicated thing is solving linear ODEs
@ALizter Well in secondary school you sweep most of the calculus under the rug because even the students who've taken the calculus course aren't very good yet.
@KevinDriscoll On the other side of my physics course we have spent forever on waves.
01:35
But yes id its things like this that you're expected ot learn I would hate it..... There's no better way to make physics seem like a terrible subject than to turn it into just memorizing a bunch of formulas and experimental techniques
We are doing the rayleigh thingy?
With apertures and stuff.
The Rayleigh criterion?
Reasonably interesting
@KevinDriscoll yes.
I am actually finding chemistry a lot more fun though.
MEH it's slightly more fundamental. Wave emchanics is fine. Rayleigh criterion is just sort of arbitrary.
I hope that I will not need a second redbull tonight
01:38
In our 1st semester freshman physics class we spend all of our time on 3 main principles. Newton's 2nd law for translation. Newton's 2nd law for rotation. Work-energy theorem.
You know, like actual fundamental physics.
@KevinDriscoll We have done more mechanics in maths.
Which I find really fun.
We are doing moments now.
All very basic.
I think the hardest it will ever get is some fluid mechanicsy type thing.
Moments like moment of inertia? Or moments like moments of proability distributions?
torque?
OK moment of intertia then
Im sure you wont do any kind of serious fluid dynamics because even professional physicists can't do it. But you could do some hydrostatics and laminar flow, perhaps.
@KevinDriscoll I did not want to say inertia because I have not seen the word come up yet.
01:41
Do you do any relativity?
@Skullpatrol It's in our introductory textbook but we sorta ignore it usually. SOMETIMES at the end we do some conservation of energy problems involving radioactive decay
Hmm...
@KevinDriscoll LOL. I looked at it again and I had read it wrong the first time. It was just displacement with water current effects.
@ALizter Ah yes. WE do those sorts of problems as a way of getting more work with vectors.
Second order vector DE are the "hardest" topic.
01:45
I would say difficulty is not the most important thing. It's about accurately communicating what physics is about and giving students some insight into what it means to think like a physicist.
I think economics is really the model we should follow. My experience is that most introductory economics teachers understand that the goal is tohelp students learn to think like an exonomist
@KevinDriscoll I hated economics.
@Alizter Engineering at best......
It was really essay oriented.
01:46
Lots of people don't like economics.
Ya they shy away from the math even more than physics
And memorising of reasons for inflation etc...
The hardest maths we did was percentage calculations.
Well that's unfortunate. If you understand the principles of economics there should be no need for memorization
I am still surprised that Mathematical Chemistry is not a big thing.
It is at the university level, but htey call it Physical Chemistry
Given a compact topological space $(X,\mathcal{A})$, let $C(X)$ denote the space of continuous complex-valued functions defined on $X$, that is, $C(X):=\{f:X\to\mathbb{C}|f\text{ is continuous}\}$. Define $+:C(X)\times C(X)\to C(X)$ by $(f+g)(x):=f(x)+g(x)$ for all $x\in X$. How does one show that $+$ is well-defined, that is, that $f+g\in C(X)$? Surely it must be simple. I know $f^{-1}(V),g^{-1}(V)\in \mathcal{A}$ for all open sets $V\in \mathbb{C}$ and want to show the same for $(f+g)^{-1}(V)$
01:49
@KevinDriscoll Silly naming.
Good evening, guys
There is also Chemical Physics.
Astronuclearthermophysics
as avengers would say
Hello :-)
Indeed, but in chemical physics and physical chemistry basically all serious work is numerical
01:50
@KevinDriscoll I would love just to model chemistry.
Perfect periodic table
bounded PDF's for electron orbits
topologically logical molecules
can someone check my work real fast? The answer is correct, but I don't know if it's by coincidence or not

$\mathbf{F} = -x^2y \textbf{i} + xy^2 \textbf{j}$, find $ \displaystyle\mathop{\oint}_C \mathbf{F} \cdot \textbf{n} ~\mathrm{ds}$. My thoughts: $\int_{0}^{2pi} -(\cos{t})^2\sin{t} + \cos{t}(sin{t})^2 = 0$
@Zach what is C?
or n?
oh, right
or s?
01:52
@KevinDriscoll $C : \textbf{r}(t) = (1 \cos t) \textbf{i} + (1 \sin t) \textbf{j}, ~ 0 \leq t \leq 2 \pi$
@Alizter n is the unit normal to C, ds is the line element of C
@Guest The composition of continuous functions is continuous. The summation map $\Bbb C \times \Bbb C$ is continuous, and $f$ and $g$ are continuous, so their sum is continuous.
Ok. Stuff I do not know.
runs
trips over rock and cries
then falls asleep
@Alizter o/ sleep well
That wasn't a rock that was me :p
01:53
@ZachSaucier No sleep well. Coursework 24 hrs before the deadline does not create itself.
user105491
Allegedly (see Wikipedia), Grothendieck died today. Anyone know if it's true?
Is there any nice way to prove that if $(\phi)\square (\phi)$ is a sentence, so is $\phi$?
\boxed?
or \square?
@Zach What did you calculate $\hat{n}$ to be?
user105491
@Alizter - my thoughts precisely
01:55
@Alizter thanks!
Square stands for any two-way connection.. for example V
@Studentmath What happens when it is not a sentence. What does this make $\phi$?
or something like that
The sources are obituaries in a pair of French newspapers, @Sanath, Le Monde and Liberation. It's up to you to decide whether or not to trust them.
@KevinDriscoll My understanding was that I was supposed to put it in the form $Pdx + Qdy$, so I didn't have an n directly
@MikeMiller Nobel again?
@Zach Oh, green's theorem>
user105491
01:56
@MikeMiller I don't know; what do you believe?
@KevinDriscoll ya :)
Well, obviously it makes the whole thing not a sentence, but the way of proving why is a bit annoying. I thought about considering all the ways to 'divide' the whole thing so it is a statement and show each is impossible.
But that could be a bit exhausting
I don't see any reason to disbelieve them. MathOverflow seems to believe them.
@MikeMiller If MO does then i am sold.
@MikeMiller What do you mean by "summation map $\mathbb{C}\times\mathbb{C}$"?
user105491
01:58
@Mike I agree with @Alitzer
@Zach you still ahve to calculate what n is though, don't you? HOw else do you know what goes in front of $dx$ and what in front of $dy$?
user105491
It truly is the loss of a great mathematician
@MikeMiller Could you also maybe at the following exercise?

Notice that in $\mathbb{C}[X,Y,Z]$:
$$V(Y-X^2,Z-X^3)=\{ (t,t^2,t^3)/ t \in \mathbb{C}\}$$

In addition, show that:

$$I(V(Y-X^2,Z-X^3))=<Y-X^2,-X^3>$$

Finally, prove that the ideal $<Y-X^2,Z-X^3>$ is a prime ideal of $\mathbb{C}[X,Y,Z]$. Conclude that the algebraic set $V(Y-X^2,Z-X^3)$ is irreducible.


Could we show like that, that in $\mathbb{C}[X,Y,Z]$:
$$V(Y-X^2,Z-X^3)=\{ (t,t^2,t^3)/ t \in \mathbb{C}\}$$
?

$$V(Y-X^2, Z-X^3)=\{(a,b,c) \in \mathbb{C}^3 | b-a^2=0, c-a^3=0 \Rightarrow b=a^2, c=a^3\}=\{(t, t^2, t^3)| t \in \math
@KevinDriscoll I suppose I did it wrong then :P I just plugged in cos(t) for x and sin(t) for y
@Guest $(z_1,z_2)\mapsto z_1+z_2$
01:59
@Zach Ya I suspected as much, that is part of what you need to do, but you still have to calculate n and take the dot product
@Zach I think you will find that your vector field is divergence-free $\nabla \cdot F = 0$ which is why the answer is 0
@KevinDriscoll Can't I just find $nds$? Not just n?
@Zach how do you propose to do that?
isn't $nds$ just the normal? So $nds = (dy, −dx)$ ?
I suppose it wouldn't work because it's dotted
02:06
@Zach Normal to what, though?
to the function they give us for F I thought
@Zach Ah no, the normal is the unit normal vector of the curve $C$
I see
that helps my understanding
@MikeMiller Ok thanks. So if I get it right, if we define $+:\mathbb{C}\times\mathbb{C}\to\mathbb{C}:(z_1,z_2)\to z_1+z_2$ and $h:X\to\mathbb{C}\times\mathbb{C}:x\to (f(x),g(x))$ then $+\circ h=f+g$. I can show that if $h_1:X\to Y$ and $h_2:Y\to Z$ are continuous then $h_2\circ h_1$ is continuous. Also the topology on $\mathbb{C}\times\mathbb{C}$ is just the cartesian product of open sets of $\mathbb{C}$ (since we have a direct product of a finite number of topological spaces there is no choice).
@Zach Yes, think about the plain-English meaning of the integral you wrote. If n is a vector normal to the field $F$ at point $(x,y)$ then the dot product always gives 0 ebcause you're dotting a vector into a vector normal to it!
02:10
mmhm
Yeah, @Guest.
02:33
It has been proved in my Functional Analysis course that $(\mathcal{B}(E),\|\cdot\|_{\infty})$ is a Banach space, where $E$ "is a set", $\mathcal{B}(E):=\{f:E\to\mathbb{C}|\sup_E f<\infty\}$ and $\|f\|_{\infty}:=\sup_E f$. However the proof begins with: Let $(f_n)_{n\geq1}$ be a Cauchy sequence in $\mathcal{B}(E)$. I wonder if the case $E=\emptyset$ vacuously gives a Banach space?
Hi. someone could help me is measure theory I only want to check if what I get is correct. This exercise says: let $f_n = 1_{[0,n]}/n^2$ a family of function. Is there a Lebesgue integrable function which dominates the $f_n$. I think that $g(x)= 1_{[0,1)}(x)+ 1/{\lfloor x \rfloor} 1_{[1,\infty)}$ is the function.
We have the homomorhism $\phi: \mathbb{C}[x,y] \to \mathbb{C}$ with $\phi(z)=z, \forall z \in \mathbb{C}, \phi(x)=1, \phi(y)=0$. I have shown that for $p(x,y)=a_0+\sum_{k,\lambda=1}^m a_{k \lambda} (x-1)^k y^{\lambda}$, we have $\phi(p(x,y))=a_0 \in \mathbb{C}$. How can I find the kernel of $\phi$ ?
03:01
@Guest A vector space has an identity element.
@MikeMiller Well if $E=\emptyset$ then $\mathcal{B}(E)$ contains the so-called "empty function", which acts as an identity element? But if $E=\emptyset$ then I think the norm $\|\cdot\|_{\infty}$ is not well-defined since the $\sup$ is taken over an empty set...
I misread your comment. I would probably agree that $\mathcal B(\varnothing)$ is a Banach space, though I guess you would have to define the norm specially in this case...
03:50
I think the $g$ I choose is correct, $g$ is measurable as is in $ (-\infty, 0)$, $ [0,1)$ and in the each interval $[n,n+1)$ with the relative sigma algebra.
Also dominates the function $f_n$ for each $n$, and its integral is $1+\sum_n 1/n^2<\infty$
I'd appreciate if someone could say me if all these is correct and I don't make a stupid mistake. Thanks.
$$\boxed{ \boxed{\boxed{\boxed{\boxed{\text{R.I.P Grothendieck}}}}}}$$
It is better to have a good category with bad objects than a bad category with good objects.
To Grothendieck, a problem was not truly solved until it was viewed from the "right" general perspective, from which it could be solved effortlessly, from which it became in a sense obvious, from which it fits naturally into a larger conceptual framework. As Grothendieck himself explained, rather poetically:
"The ... analogy that came to my mind is of immersing the nut in some softening liquid, and why not simply water? From time to time you rub so the liquid penetrates better, and otherwise you let time pass. The shell becomes more flexible through weeks and months—when the time is ripe, hand pressure is enough, the shell opens like a perfectly ripened avocado!

A different image came to me a few weeks ago. The unknown thing to be known appeared to me as some stretch of earth or hard marl, resisting penetration ... the sea advances insensibly in silence, nothing seems to happen, nothing moves,
04:06
Someone? please I believe is correct but I prefer if someone could check what I have so far.
04:17
How would I do this? Integral (sec^3 x dx)?
0
Q: The proper and easiest way of doing an integral with derivative?

DemCodeLinesI have this integral: $\int{sec^3x dx}$ I don't understand how I would solve this. Google and YouTube videos don't help me understand much, other than just giving the answer. Is it possible to explain step-by-step how this would be solved, assuming that this is the first time I'm seeing it?

05:05
Here is what I have so far. Let $f_n=1_{[0,n]}$ and $g(x)= 1_{[0,1)}(x)+ 1/{\lfloor x \rfloor} 1_{[1,\infty)}$. To show that $g $dominates each $f_n$ we have to show that $f_n\le g$ and $g$ is absolutely integrable. That $f_n\le g$ is easy considering each of the possibilities for $x\in \bf{R}$. On the other hand, $g$ is measurable since is measurable in $(-\infty, 0), [0,1)$ and $[n,n+1$ for each $n\in \bf{N}$ therefore is measurable in $\bf{R}$ given that is in the partition defined above.
Now let $F_N(x)=1/{\lfloor x \rfloor} 1_{[0,N+1)}(x)$ then is not difficult to show that $0\le F_N \uparrow 1/{\lfloor x \rfloor} 1_{[1,\infty)} $ and using the monotone convergence thm we have that $$\int 1/{\lfloor x \rfloor} 1_{[1,\infty)} = \lim_N \int \sum_N 1/N^2 1_{[N,N+1)}= \sum_n 1/n^2< \infty$$ so by linearity of the integral $\int g = 1+ \sum_n 1/n^2<\infty$ which shows that $g$ is absolutely integrable and $g$ dominates $f_n$ as claimed.
I believe the above is correct but I'd appreciate the opinions of others and also any suggestions. Thanks in advance.
@JoseAntonio You've been asking about this for most of the day now; I certainly can't comment, and it looks like nobody else can. I think you should post it on main!
@DemCodeLines Just do $\int \sec x \sec^2 x \mathrm dx$ using byparts.
$$\int \sec^3 x \, \mathrm dx = \frac{1}{2}\sec x\tan x + \ln \sqrt{\sec x + \tan x} + \mathcal C$$
@JoseAntonio Post it on the main site. Your concern is worthy of it. The worst come worst situation is that you are absolutely right and no one has anything more to add.
05:37
@NIck, @MikeMiller Thanks for the advice. I already posted
@JoseAntonio My prayers are with you =) Godspeed, man.
06:11
Curses!! Class XII students will not be eligible to appear for RMO/pre-RMO from 2014 onwards.
Well, there's go my dream of ever having a chance to get to the IMO,
Mhh, maybe it's better not to have a chance than to have a chance and fail =)
You can still try.
@skullpatrol I can still try the out the papers unofficially for my own personal satisfaction, you mean :D Ofcourse I'd do that.
Yep, make it your own personal Olympics ;-)
@robjohn You should be sleeping, lol.
@skullpatrol hugs that's a wonderful idea (sometimes we all need someone who can reaffirm our ideas inorder for us to act based on it. Thanks for being that someone)
06:27
Anytime pal :D
Don't l.o.l. in the middle of the night.
For it might give someone a fright.
Especially when they turn on the light
and see you @Jasper; a blinding blue sight!
@skullpatrol Even at say, 5:30 AM on May 11, 2020 ?
@skullpatrol You are way too free if you can schedule things that far away on your calender.
Have you ever heard the song Born Free?
07:06
hello
07:45
Hello
08:47
On September 26, 2014, 43 students from the Raúl Isidro Burgos Rural Teachers' College of Ayotzinapa went missing in Iguala, Guerrero, Mexico. According to official reports, they had travelled to Iguala that day to hold a protest against what they considered to be discriminatory hiring and funding practices by the Mexican government. During the journey local police intercepted them and a confrontation ensued. Details of what happened during and after the clash remain unclear, but the official investigation concluded that once the students were in custody, they were handed over to the local Guerreros...
09:07
who can help me on homotopy equivalence ?
if someone have an idea :math.stackexchange.com/questions/1019919/… thank you in advance
you already got an answer and helpful comments!
09:25
I am looking at the following exercise:

Prove that for all sets $A,B$, there are the sets:

$$\{ A \cap x: x \in B \} \text{ and } \{ A \cup x: x \in B \}$$
$$y: \exists x \in B (y=A \cap x) \rightarrow y \in \mathcal{P}(A \cap (\bigcup B))$$

How do we conclude that $y \in \mathcal{P}(A \cap (\bigcup B))$ ?
09:50
@user2179021 the answer is fals
10:05
?
false
10:38
If $x \neq y $, define the sets $\bigcup \langle x,y \rangle , \bigcup \bigcup \langle x,y \rangle$.

According to my notes, it is like that:

$$\langle x,y \rangle= \{ \{x\}, \{x,y\} \} $$

$$ \bigcup \langle x,y \rangle=\{x,y\} $$

$$ \bigcup \bigcup \langle x,y \rangle=x \cup y$$

Why is it $ \bigcup \bigcup \langle x,y \rangle=x \cup y$ and not $\bigcup \bigcup \langle x,y \rangle=\{x,y\}$ ?
Could somebody explain why it is like that?
11:10
first question is why doesn't this chatroom render latex!
11:41
hey guys
I have a meta-type question
on the new History of Science and Mathematics SE, I just asked a question related to a remark in the book by Lee, the question can be found here
Now, I suspect that Lee himself may have something to say about this, perhaps he'd even be able to write a complete answer
Is there any way to draw Jack (John?)'s attention to this question without coming across as annoying?
(also, it may very well be that other people who know something about differential geometry are able to answer this question adequately; please feel free to do some if you can!)
11:59
@Danu leave the question for a while. Then email him directly with a polite email asking what you want to know. You can mention the question in the email but don't suggest you are asking him to reply to it there
Now that Grothendieck is dead, will his papers be released to the public, if any?
a good question
It is also possible that his final years were spent in dementia.
@user2179021 I have emailed Lee a few times to ask about his books. He always replies, lol.
Of course, I don't email them math questions.
@JasperLoy He seems to be a super nice guy, judging from his math.SE posts
@Danu Yes, just like me, lol.
12:22
Grothendieck died. I wonder how many mathematicians were waiting for this to happen.
I wonder where and what he will be reborn as.
He won't. People don't get reborn.
You and I have different beliefs.
@JasperLoy do you have a specific belief about being reborn or something general?
I mean, in what sense is the reborn creature you?
can it for example remember the previous life?
@user2179021 In the sense that the original elements of consciousness combine with other elements to form new life. Most people do not recall previous lives.
12:30
@JasperLoy ok so in every practical way you are not reborn?
@JasperLoy you just have a theory about untestable and unseeable common elements between animals?
@user2179021 Yes, for certain definitions of practical, lol.
@user2179021 Not just animals, but other beings too.
@JasperLoy ok so given that it is unobservable.. why would you believe such a thing?
@JasperLoy just it follow logically from something?
@JasperLoy or did someone important tell you it must be true?
@user2179021 Because I believe in most of the things the Buddha supposedly said. This seems to be a real possibility as well.
@user2179021 There are stories of people who recall their previous lives, though they may not be true.
@JasperLoy ok.. so someone told you about the Buddha and then you thought, well I will believe everything the Buddha said from now on?
@JasperLoy I have weird beliefs. Anything not proven scientifically is either wrong or yet undetermined.
12:34
@UserX the problem with this is how do you know if it is proven scientifically?
@UserX I mean you don't do the experiment yourself nor do you have the time to verify them
@user2179021 No, I do not believe in everything he supposedly said. But like I said, rebirth seems very real to me somehow.
@UserX so you just hear that someone has proven something and then chose to believe it?
@JasperLoy how?
@user2179021 I trust the hate between academic parties.
@UserX That is a reasonable position.
@UserX so you think it must be true or some other academic would be pointing it out?
12:35
@user2179021 Let me think how, lol. You seem very interested in this conversation, lol.
@user2179021 Yes
@UserX but there are always people who disagree with anything. So who are you believing? The majority?
@UserX people at the most prestigiuos university?
@user2179021 Who, in the scientific community, dissagrees with everyone without pointing out anything?
@UserX take global warming
@UserX or string theory
@UserX or evolution
I have believed in so many things that I have forgotten how I believed in them, lol.
12:38
@UserX also it seems an odd way to live, believe anything you are told until someone says they don't agree with it.
@UserX can everyone live like that?
OK @user2179021 one reason for my belief in rebirth is that the Buddha supposedly talked about the big bang, and how beings were reborn into certain realms along with it. And as we know the big bang is a scientific theory today.
The Nash equilibrium of my strategy on beliefs is definitely the current scientific consensus. But Nash equilibrium isn't always the best choice(see prisoner's dillema). However it's the safest.
@UserX how do you decide what the consensus is?
@user2179021 Majority of scientists agree on something.
@UserX So it's belief by vote :)
12:40
Better than blind belief, worse than finding out myself. I don't have the time though.
do you really mean the majority or do you weight them by how fancy their job title is?
are 3 MIT profs worth more than 6 local college profs?
A belief doesn't have to be accepted as fact though. It is just a belief to the person who believes.
@user2179021 Not really. I do not think highly of MIT professors.
I don't trust academia. I trust the hate between academic parties. That should answer your question.
12:41
@UserX my point is that your system is completely non-scientific
@UserX it's like believing the pope because he is important
@UserX the hate doesn't help you. Because if 6 local college profs disagree with 3 MIT profs you are going for the MIT profs
because they are paid more
If 3 MIT professors are wrong, more than 6 local professors will object
ok so there is a weighting
Quoting something I read in a book "My philosophy, in essence, is the concept of man as a heroic being, with his own happiness as the moral purpose of his life, with productive achievement as his noblest activity, and reason as his only absolute."
I no longer believe that depression or OCD medication works.
I think the serotonin theory is pretty much nonsense.
@UserX that's lovely but completely irrelevant :)
anyone want to discuss math?
12:44
@JasperLoy please don't become a crackpot. I liked you.
@UserX I am not a crackpot. Please do some research into the serotonin theory before saying that.
Speaking of crackpots. If someone wants to downvote / close posts by "Thierno M. SOW" clones, here is one and two.
@user2179021 that last sentence is revelant. My reasoning concludes itself in those beliefs I described above.
@Rafflesiaarnoldii oh dear
who is upvoting this?
12:46
@userx Whether you are a crackpot or not, I will still like you, lol.
@user2179021 read "clones"
oh double dear
My dry cough is still very bad. I hope it dies soon.
@JasperLoy I got it too...
@JasperLoy but it will then be reborn!
:) .. sorry
12:48
By the way, I have done lots of research into the serotonin theory, lol.
is there anyone familiar with general relativity?
I am not familiar with my relatives, generally.
I know the rough idea and how the field equations where formed(roughly)
But I haven't immersed into the math of it so I probably can't help you.
ok thanks
no problem :)
@user2179021 It is possible that the sentient beings in the phlegm get reborn, yes.
12:53
@JasperLoy this is soooooo far off topic
@PDRX that's quite a stretch
@user2179021 Well, you mentioned the reborn part first about the cough.
@PDRX do you have a specific math question?
@JasperLoy second.. you mentioned it first :)
@user2179021 Are you an undergrad?
you mean in this life? :)
ok.. no more non-math talk :)
@user2179021 no thanks. it was just a thought
12:56
@PDRX ok
I am full of thoughts.
@JayeshBadwaik How is your schoolwork?
Oh math talk. I got a question. For the roots $r_1,r_2,r_3,\dots$ of the polynomial $P(x)=a_n x^n+a_{n-1}x^{n-1}+\dots+a_1 x+a_0$ it holds that $\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\dots+\frac{1}{r_n}=-\frac{a_1}{a_0}$

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