« first day (3906 days earlier)   

12:00 AM
up he goes
what is updog?
the reverse of godpu.
Ok, I'm going to change $l$ to $k_{j}$. And I will say $k_{j} > K$. Now since $x_{k}$ is Cauchy that should work.
godpu = god poo?
I finally got cleared for a date for my undergrad thesis exam
No, @dc3rd. At least I don't see what you're doing. Write it all out carefully and tell me the radius of the ball. I'm going for a walk, so you have 20 minutes or more.
12:12 AM
take deep breaths and enjoy the fresh air
hi chat
hello Lucas
Are you Lucas Podolski the great striker of the German football team?
Oh nvm that's Lukas
12:33 AM
We have a German Lukas here from time to time, but a different one.
or so he would have us believe
Well, true, you earn your living not believing (in) anyone.
12:49 AM
I had forgotten what the * meant in your text and I caved to temptation....I have failed you Jedi Master ...😞....but the idea of every term in the sequence is a realization and not just something abstract has been made more clear.
Unless you got the corrected printing, there is a stupid typo in that solution. Did you catch it?
Was it using $j$ instead of $k$ to generalize?
better than a smart typo.
No. The formula for the radius of the ball has what?
@copper Mostly I leave the smart typos to you.
$1 + \|X_{K + 1}\|$
12:53 AM
i was just wondering, i wonder if there's ever been an argument improved, or even a unintentional theorem proved, by a typo.
That was most definitely NOT what I have (corrected or not).
typos only take. they never give.
I certainly wrestled with my share of errors in proofs (not necessarily typos), @leslie.
oh you meant $min(\|x_{1}\|, \dots , 1 + \|x_{K+1}\|)$
Right. What is the typo in that?
Remember the picture of what's going on here.
Actually, I had a colleague (who made many millions off well-written textbooks) who tried to make sure there was an error on each page of the answers at the back of his calculus and ODE books. I found that amusing.
This particular error should be caught immediately but is still instructive.
1:00 AM
Ugh, I'm going to have to get the Benadryl out. We got SNOW today! SNOW!
I miss snow!
we never get snow where I live
what's good about it
I lived in it for 10 years. Now at my decrepit age, I am happy just to visit it.
But fluffy beautiful white stuff is amazing. Not so pretty in a dirty big city.
First time I went to snow as a kid I just jumped into it and ended up freezing like a moron
it will surprise no one that in my first experience with snow i couldn't stop throwing snowballs at members of my family, even when they told me to stop.
1:11 AM
it seems to me that the error is that it should be $\min(\|x_{1}\| + 1, \|x_{2}\| + 1 \dots , 1 + \|x_{K+1}\|)$
i was given a 'clean up the proofs & do the numerics' paper during my graduate work, and discovered a type like $a \le b$ where the real result was $b \le a$ which invalidated the entire result.
@copper.hat how did you give them the sad news?
that was not the issue. the issue was i had to find a whole new proof. it was pure grind whereas their original was short.
@dc3rd where did the +1 come from in the first place? You are not understanding the argument.
If leslie weren't here, I'd tell you to draw a picture.
@copper.hat good for you!
1:25 AM
I've drawn a picture. may not be instructive though.

My understanding of the +1 is that since the sequence is Cauchy for all $\epsilon > 0$ there will be a $K$ whereby $\|x_k - x_l\|< \epsilon$. In this case we've chosen $\epsilon = 1$ since it is Cauchy this can occur and as such for $k, l > K$ we have $\|x_k - x_{K+1}\|$ where we've taken $l = K+1$
now reverse triangle inequality will give me the specific $\|x_k\| < 1 + \|x_{K+1}\|$ relationship
For all $k>K$, yes.
So you draw a ball of radius 1 around $x_K$ and all the further terms are in there. Now what?
then that would mean $\|x_1 - x_K\|, \|x_2 - x_K\| < 1$ as two examples if I'm interpreting the further terms as you mean them to be.
No. What does further mean?
I was trying to save typing math symbols on my iPad.
ok then further in the sequence is what you mean by further terms. i.e: $x_{k+1}, x_{k+2}$, etc
K, but of course.
1:36 AM
well then each of those is less than $1$. i.e: $\|x_{k+1} - x_K\|, \|x_{k+2} - x_K\| < 1$
So you're convincing yourself of the validity of my sentence. I asked you Now what?
1:52 AM
should I be comparing the distance between the further terms that are in this ball of radius 1?
You have the damn answer, except for a typo. Figure it out. I am not wanting to be mean, but I think my book and Rudin are not realistic goals for you.
2:24 AM
You're entitled to have that opinion based on questions I've asked and what your years of experience have illustrated to you.

I made the choice a while ago to be all in on this endeavor regardless of the hardships I may endure....there is no Plan B. I've been face to face with death so enduring the difficulties of changing my thought patterns is soft.

I did figure out the typo and I figured out the issue with why it took me so long to see it. But I'll chill out on asking you questions for a bit until I got something that would be considered "extremely challenging" in your realm.
i can't even back out what the problem is, although i think i was here. i was mostly in another window.
there's no problem anymore. there's a typo in Ted's book that says $\min(\|x_{1}\|, \dots , 1 + \|x_{K+1}\|)$, but in reality it should say $\max(\|x_{1}\|, \dots , 1 + \|x_{K+1}\|)$
oh, i meant the problem that all of this was about solving. i was trying to back it out from some of the sequence stuff, formed some vague ideas, but couldn't get there.
i'm assuming this comes up in some exercise, if it was just some random inequality who knows
oh the question asked to show that if $x_{k}$ is a cauchy sequence, then all points lie in some ball centered at the origin.
my two mistakes were overlooking the importance of the word all and what that meant for the Cuachy sequence
I can't wrong him for getting frustrated....shrug
For a large summer party 3 large boxes of peaches have been bought. However, while decorating, a mishap happened as some of the fruit fell on the floor, making it unusable. They distributed the remaining peaches evenly on 3 tables. During the party, 4 peaches were eaten at each table. At the end of the party, there were a total of as many peaches left as had previously been on each table.

How many peaches remained intact?

It's 6, isn't it?
2:43 AM
@TedShifrin ok, so trying to take this approach. I start with trying to find $\int{\Psi^* [-ih \frac{\partial}{\partial x}] \Psi}dx$.
Ah screw it will need to wait till i am home, this is too much tex for my phone keyboard skills to handle
Was trying to type out this
But i saw how you all feel about images this morning so sorry
Also i ommited the constant$ih$ because i can, and the limits of integration are positive and negative infinity
And i am clearly not getting what you meant
Why are you even using integration by parts here? @Andrew
I thought it would make life easier
Sorry after all that my lunch break just ended (i think i will post a proper Q in main when I get the chance(
If I give you $\int ff’\,dx$ what is the better approach?
@Zest What is $6$?
3:06 AM
i feel like the guy in the theater who needs to alert the projectionist and shouts FOCUS!
LOL ... to me?
P.S. I've done that before.
no to the blurry photos. if we must have photos i want them well composed.
Oh, you're almost as picky as @robjohn.
did projectors just get more reliable? i guess there's a lot of digital projection now? i don't remember hearing or yelling focus for a long time. it's probably been 15 years.
@TedShifrin @TedShifrin the amount of peaches that remained intact.
3:10 AM
The peaches that fell on the ground are gone. Aren't they asking how many are left on the three tables?
Yes, that's as i understood it as well. But i'm confused because others saight the answer would be 18 peaches.
that first line sounds like the beginning of a sad poem.
where are the peaches on the ground? ay, where are they?
And it is correct,
Which answer is correct @TedShifrin? 6 or 18 peaches?
Nothing like irrelevancy in word problems!
18, of course.
Back to 5th grade math!
3:14 AM
Thank you very much @TedShifrin. I got it wrong since i solved the equation

$3(x-4) = x$ which returned 6
What is $x$?
the amount of peaches that has been put on each table initially.
Always write a sentence defining $x$.
OK. So why is $x$ the answer ?
I looked at it this way: Let $x$ be the amount of peaches put on each table. From each table $4$ peaches have been eaten, so on each table there remain $x-4$ peaches. It's said that "at the end of the party, there were a total of as many peaches left as had previously been on each table."

Therefore i computed

$(x-4) + (x-4) + (x-4) = x$
Maybe @leslie will draw a picture :)
I understand your equation. It's fine. You ignored my question, and that makes me impatient.
3:19 AM
i'm not a wild person, but this sounds like a dull party. offer your guests something more than peaches.
Martinis would be a good start.
Especially today.
yeah. you have to build to the peaches.
Oh i'm sorry, i'm not sure if i understood you correctly.
At least peach melba!
What does “why is $x$ the answer?” possibly mean?
@copper said I was very patient. I’m thinking those days are disappearing.
Well we have eaten 4 peaches of each table but in the end there are as many intact peaches left as we initially put on each table. That's why i solved the equation for $x$ and obtained $x=6$ as the total amount of intact peaches.
3:23 AM
Total amount? Think, man.
Reread what you said $x$ was.
Ah, of course. Now i see it.
When you define $x$ you should also say what the answer to the question is in terms of $x$.
The answer is not automatically $x$.
You're absolutely right. I somehow confused my answer with the definition of $x$.
Thank you very much @TedShifrin.
@leslie I need a sabbatical. Maybe legal secretary?
Just remember what you’ve learned from this, @Zest.
Yes, indeed. I brushed over it and got confused where i'm wrong. Should have taken a step back.
3:29 AM
motivational speaker. that's what i'm going to do.
It's good to write a sentence saying explicitly what the question wants you to find, once you have defined your variable.
I'm not feeling motivational.
@TedShifrin compared to me you are very patient.
That's indeed a very good advise @TedShifrin. Highly appreciating your help.
You're welcome.
this is going to sound slightly stereotypic, but everybody knows that irish blood boils at very low temperatures
3:31 AM
Not much longer, copper. I'm coming over to your side.
What about Russian/Polish blood?
i usually just fill in the blank with the ancestry but i already used it once on the irish. i got nothing.
also you really don't want to rile some of those folks up. the irish sometimes have a sense of humor about it.
Do some research.
i will moonlight as an ethnographer.
Copper has a sense of humo(u)r?
3:48 AM
Hey Hi Guys Can anyone please tell what really is Maths? What significance it has of it's own and is it just about numbers and no visualisation of what that equation is really signifying ?
@TedShifrin what do you say ?
it means different things for different people. for some it is all numbers. for some the numbers do not play a big role. people do try to figure out what things, equations included, are signifying, although ahem not always through visualization.
4:29 AM
i know there is an irish stereotype regarding tempers, but would say that, drinking situations excluded, irish are more likely to be socially constrained in pressure situations.
of cource, the other stereotype means that the measure of the exclusion is pretty large.
you are definitely more patient in general @TedShifrin. if i think someone is making a genuine effort i will go the extra mile, but otherwise i will switch off quickly. you persist and will continue to engage.
not a lot of tempers on that side of my family. just a bunch of people who mostly never talked to each other
mmm, well, if we are talking family then that is a different story :-).
4:58 AM
@TedShifrin $$ \begin{align} \int vv'\,\mathrm{d}x &=vv-\int vv'\,\mathrm{d}x\\ &=\frac12vv+C \end{align} $$
roundabout, but it works.
oops, I misread the fuzzy image
Not that I'm picky ;-p
a lot of entendres in here
5:16 AM
i don't mind fuzzy images as fuzzy images i just don't like the sensation that i'm losing my eyesight more than i actually am
@robjohn Yes, I realize that works, but I think @Andrew would profit from seeing the immediate substitution as the more natural approach. :P
@TedShifrin I know that. I was just being picky!
5:34 AM
5:55 AM
those optical illusions that had multiple focus solutions used to give me a headache.
if you stare long enough into that FOCUS image, you see a giraffe in 3d
a klein giraffe with its neck... i'll leave that to your imagination.
good Eine kleine Nacht
my mom loved mozart's music
1 hour later…
7:16 AM
@copper.hat I've toyed with the idea of generating some of those to do some 3-D illustrations.
Re: "What is Math"
It's abstraction. Instead of talking about "3 apples" or "3 meters per second" or "3 volts," you just consider the abstract notion of "3" itself. The advantage of abstraction is that it simplifies things, and the abstract result can then be applied to any of hundreds of real world situations.
One of the drawbacks of abstraction is that suddenly you don't have anything to experiment with, which means you have to fall back (hard) on logic and reasoning, so you could also say that math is the rigorous application of logic.
1 hour later…
8:26 AM
> Why was 2019 afraid of 2020?
Because they had a fight and 2021.
1 hour later…
9:43 AM
What is math if not 3 objects persevering... cnet.com/a/img/qVQMRM1NE2KelrmjcUC99uXgXI0=/1200x630/2021/03/05/…
9:57 AM
@LeakyNun u here?
@ShaVuklia ja
cool, kan ik iets vragen ~
hoe kan ik jou helpen
so, my algebra is a bit rusty ;x
but I think the argument should be simple
I don't understand why $K(C)$ is an algebraic extension of $K(t)$
and they mentioned three things
eh, but I don't see why what they say yields the conclusion?
oh, maybe I should give some definitions
$C$ is a projective curve
$K(C)$ is its function field
(ehh, actually I don't know if there is some ambiguity in the definitions?)
are you happy with the fact that K(t) embeds into K(C)
10:03 AM
so t is transcendental in K(C)
but since K(C) has transcendence degree 1, this means {t} is a transcendence basis of K(C), so K(C) is algebraic over K(t)
@LeakyNun I didn't realise this, let me see
t is transcendental in K(t)
this can be seen by "uniformizer"
t is the uniformizer, in a DVR sense, of the local ring of the stalks at P
so t is transcendental
10:07 AM
$(t^n)\neq 0$ for each $n$?
ye I get it, maybe I didn't give the best reason
cool, thx!
hm, maybe I don't see it after all
You're also giving a slightly different argument right? Because they also use that $K(C)$ is finitely generated over $K$ (or maybe they need to mention that to conclude that $t$ is transcendental? since they only mention $t\notin K$)
I really should apologise for my shitty fotos. I should just wait until I have time to talk properly instead of hurridly scribbling things down on paper.

@TedShifrin Thanks for your patience. If I follow your suggestion I think that takes me back to your advice from a few weeks ago, and I need to just find a profitable substitution instead of integrating by parts. I will try take that approach.
10:22 AM
Oh wait, $K(C)$ being finitely generated over $K$ should then yield that the algebraic extension is finite
oh, I see it I think (why $t$ is transcendental)
the order wouldn't be well-defined
okok, I'm good
guys I don't know how to know if partial differential equation is linear or not?
I don't understand that function F(x_1,...x_n,u_1,....,u_n)=0 gibberish
BTW wth is this F(....) thing?
10:51 AM
@LeakyNun I'm afraid I don't see it after all ;x would you mind to elaborate why $t$ is transcendental?
Is it in general true that if we have a field extension $L/K$, and $L$ had transcendence degree $k\geq 1$ over $K$, then any $x\in L\setminus K$ is transcendental over $K$?
@ShaVuklia let the valuation be normalized with v(t) = 1, suppose t^n + a1 t^(n-1) + ... + an = 0. dividing by powers of t, you may assume an != 0, so v(t^n + a1 t^(n-1) + ... + an) = v(t(something) + an) = v(an) = 0
@ShaVuklia take $K = \Bbb Q$ and $L = \Bbb Q(\sqrt2)(t)$
Oh, oops
I was thinking more in terms of powers of the maximal ideal, and I didn't realise that I could also evaluate
@LeakyNun and thx for this as well x')
yeah it's useful to use the valuation
11:12 AM
Ohhhh, (I realise we are clearly in different timezones but I want to say thanks before heading to sleep...I'm not sure how to make the next step (what to do about the complex conjugate), but I think I should muse over it myself until the weekend. For the time being I have this on my sheet of paper:

$\int{f \cdot f' dx}$ let $u = f$, then $\frac{du}{dx} = f'$ so $dx = \frac{du}{f'}$

By substitution I get $\int{f \cdot f' dx} = \int{u \cdot f' \frac{du}{f'}}$,
The $f'$ cancels, and now I get $\int{u}du$.
1 hour later…
12:26 PM
@Leaky ;oo
mind if I ask another thing?
@ShaVuklia hoe kan ik jou helpen
I also posted it on stack:
But let me recap it here
I don't see why their $\phi$ maps into $C_2$ in the first place
(and I also don't see why not all the $g_i$ can be constant, but I worry less about that, and more about the first question)
12:47 PM
Oh, I also think there is a typo: it should say $C_2\subset\mathbb P^N$
@ShaVuklia if $C_2$ is $y^2 = x^3 + 1$ (i.e. $Y^2 Z = X^3 + Z^3$) then $\iota(y)^2 = \iota(x)^3 + 1$
something like this should work, sorry i'm busy
Right, that helps, thanks!
1 hour later…
2:16 PM
does calc 4 exist
2:38 PM
If $f:D^n\to D^n$ is continuous map such that $f|_{\partial D^n}$ is a homemomorphism, then $f$ is a surjection?
nvm my calc 4 question it is silly
but i have heard that some universities refer vector calculus to as calc 4
@love_sodam you mean homeomorphism onto its image?
Ah, $f|_{\partial D^n}:\partial D^n\to \partial D^n$
aha, that's a useful assumption
convince yourself you can assume $f\vert_{\partial D^n}=\operatorname{id}_{\partial D^n}$ WLOG
2:55 PM
ok, you don't actually need this tbf
why can I assume like that?
the above step reduces this to the fact that a compact manifold can't retract onto its boundary, but you can also just directly do a similar proof here by looking at the top homology
the hint for that WLOG assumption is radial extension
Ok I get it thanks
3:36 PM
@Andrew My humble apologies. I had forgotten you were working with complex functions and the complex inner product. So of course you are right that you need to differentiate $|f|^2 = ff^*$ by the product rule. Undoing the product rule is, of course, integration by parts.
@Thor: How was your lecture?
4:22 PM
it went fine
sadly my time management was as terrible as always, so I didn't manage to get to the examples
they're in my writeup, though, so it's not like they were worked out in vain
Hi @TedShifrin
@Thor: Examples should always be a high priority. I often do them before proofs. :)
Hi, Karim.
I don't think this was a scenario in which discussing examples first is the better order, but I do regret not being able to present them
in other news, the Hodge index theorem is really cool
4:47 PM
Remind me which theorem?
signature of an even-dimensional compact Kähler manifold is computed by $\sum_{p,q}(-1)^ph^{p,q}$
Oh, I forgot that.
Is there an easy way to see why $\operatorname{deg}(\phi)=5?$ By definition, the degree of $\phi$ is the degree of the field extension $\overline K(\mathbb P^1)$ over $\phi^*(\overline K(\mathbb P^1))$
Wait, maybe I can figure this out
Ok, so I basically have to show that $[K(X):K(X^3(X-1)^2)]=5$ (where K is some (algebraically closed) field)
5:22 PM
you can guess the minimal polynomial
5:48 PM
@Thorgott Ye, that should be $T^5+2T^4-T^3-X^3(X-1)^2$, but I'm a bit out of it to see why it's this one, and not one with a smaller degree
But I guess I can try to write it all out
If A,B are R-algebra (R is commutative ring with 1) R-algebra homomorphism f is just a R-linear map with f(aa')=f(a)f(a') for a,a' in A right? wiki described K-algebra homomorphism using term 'K-linear' but not on R-algebra homomorphism
ye ok, I think I see it
@love_sodam and f(1)=1
an R-algebra is something that's both a ring and an R-module in a compatible way and a map between those is a map that respects both structures, i.e. somethings that's both a ring homomorphism and R-linear
6:03 PM
If A is finitely generated R-module, then A is isomorphic to a quotient of polynomial ring R[x_1,...,x_n]. But we can also write $A = R[a_1,...,a_n]$ where a_i in A yes?
This has been upvoted to 11, but hasn't received any comments or answers. Would it be frowned upon if I cross-posted it on Math.SE? What can be proven regarding the differences in power between unary ECMAScript regex functions and primitive recursive functions?
Also, if anybody just wants to comment here on the things discussed in there, I'd welcome it.
i don't know the applicable etiquette but i doubt it would be more likely to get answers on math.SE. there are obvious and close connections to math but there's a lot of regex knowledge required here. math.SE also tends to like one question per post.
if you could distill one of your questions into something perhaps a little simpler than that, maybe it would be more likely to attract attention on math.SE. but you do seem to have targeted the question appropriately already.
in my very humble opinion :)
Okay, thanks :)
side note, they are very interesting questions.
Thank you.
6:18 PM
Yeah i also think so
i would conjecture that the answer to your question 1 is 'no,' but i have no explanation for this, it's just a vibe.
That was the first one I'd consider asking on Math.SE.
conjecture is too strong a word. guess, but not wildly guess.
i asked it once but i am asking again: can someone recommend a good book for cryptography
neither too basic nor too complicated
also a very good question. all of the ones i have seen are one or the other.
6:31 PM
There was a chapter on cryptography in one of my nt books
I liked it and became interested
7:10 PM
can someone give two examples of non metrizable topological spaces?
the weak star topology on the dual of a banach space
*infinite dimensional banach space
pick two non-isomorphic examples of such and i just gave you two examples
okay thanks
i was slightly kidding there. there are probably more natural examples but infinite dimensional vector spaces and their duals provide a host of them.
@Deadcode I assume you've seen FRACTRAN
a lot of examples proceed by first proving that metrizability implies some more abstract property. like first countability, or other stuff about countably many balls being enough to do x or y.
then you go to your library of examples of bad spaces that don't have that property.
wow, i really like nate eldridge's answer here: mathoverflow.net/questions/52032/…
it's like he crawled inside my brain, went back in time, and wrote my answer. we might as well call it my answer.
7:22 PM
take sth non-hausdorff
there's the more natural answer.
1 hour later…
8:36 PM
By that, I take it this would be wrong:
$\frac{d}{dy}\int{f^* [\frac{\partial}{\partial x}] f}dx$ (where $f$ is a function of $x,y$, and $f^*$ the complex conjugate of $f$)
let $u$ = $f$
then $\frac{du}{dx} =? \frac{\partial}{\partial x}f$ (should the LHS be a partial derivative?)
so by the same logic as before: $\frac{d}{dy}\int{u^*}du = \frac{d}{dy}[\frac{{u^*}^2}{2} + C]$
which then becomes $\frac{d}{dy}[\frac{{f^*}^2}{2} + C]$
Yup, wrong. I missed the conjugate.
okay, well at least I am getting some cognative excercise with this one
@TedShifrin Trying to learn spectral theorem for bounded self-adjoint operators
I was so pleased with myself for that and evrything. I seam to be going in circles with the integration by parts: Apply the formula $\int{uv'} = uv - \int{vu'}$
So my case I let $u = f^*$, then $u' = \frac{\partial}{\partial x}f^*$, so $v' = \frac{\partial}{\partial x}f$ and $v = f$.
then I get $$f^*\cdot f - \int{f \frac{\partial}{\partial x}f^*}$$
also, side note $u$ and $v$ are terrible symbols to stick side by side, I get it the follow alphabetically but the glyphs look way too similar, I'm gonna use other ones in the future, $j$ and $k$ seam way better!
And because I was moaning about glyphs I forgot the $dx$ at the end and missed the edit window
8:54 PM
Did you notice my product rule comment earlier? That's all that’s going on.
You have to be careful with infinite limits to make sure things converge appropriately.
So I have a problem. Here it is:
Let $g(x)$ be the generator polynomial of a binary cyclic code $C$. Let $h(x)=\left(x^{n}-1\right)/g(x)$ be the parity check polynomial.
Show that if $a(x)$ is a polynomial satisfying gcd$\left(a(x), h(x)\right)=1$, then $a(x)g(x)$ is a generator of $C$.

Now what I am not understanding is why I must show that $a\left(x\right)$ must satisfy gcd$\left(a(x), h(x)\right)=1$.
As long as $a(x)$ is of least degree and a divisor of $x^n -1$, the product $a(x)g(x)$ should be a generator for $C$, correct?
There is something obvious my tiny brain is probably missing.
I did, $f^* f = |f|^2$ which is a product, so we need to use integration by parts to undo. I'm about to run off to work. I guess the bit that is troubling me is how that works when I need to differentiate one of those functions first.
I'll keep thinking about this / puzzling over it in my head. and will try to resist the overwhelming urge to write down where I'm at on a scrap of paper at lunch :P
9:23 PM
@Andrew: Ultimately, you should perhaps explain to picky @robjohn and me what your goal is here.
9:40 PM
@TedShifrin Perhaps my pickiness is too much for him.
Well, I was wrong, and you were right! But I'm not sure what he's trying to do with the Schrödinger equation.
Hi robjohn, Ted
Ah, Astyx gave robjohn top billing. Surely that'll placate him for a while.
Hi, @Astyx.
@TedShifrin A very strategic decision
Hi, @Rithaniel
9:47 PM
@robjohn Yep, that's cool, but not really similar. Its exponents can grow without bound, go either up or down, and loop without bound (i.e. a program has the option of never halting).
@Deadcode Yes, old west programming. (1)
@Astyx hi there.
10:03 PM
@TedShifrin Hey Ted
Hope you're doing ok, @Rithaniel. Haven't seen you in a while.
10:31 PM
hey does anyone know what The general form of n-th order ODE is given as F(x, y, y’,…., yn ) = 0
what is function F()?
Anything you want.
Typically a continuous or many times differentiable function.
so I can say x+y+y'=0
That last one wasn't very interesting, but you certainly could have $yy'' + \sin(x)y^2 + e^y = 0$.
Now that's a general partial differential equation.
10:34 PM
ah wrong picture
I was quite confused about linear partial differential equation
Just as with ordinary differential equations. $y$ and all its derivatives appear linearly, no products, no powers.
I was confused about those partial differentiation in terms of some variable in suppose function y
I do not understand.
$\partial^{(2)} y(x,y)/\partial^{(2)}(x)+\partial^{(2)} y(x,y)/\partial^{(2)}(y)+\partial^{(1)} y(x,y)/\partial^{(1)}(y)+\partial^{(1)} y(x,y)/\partial^{(1)}(x)=f(x,y)$ is second degree partial differential equation right?
because it got many variable unlike ordinary differentiation and linear ordinary differential equation only differentiate in terms of one variable instead of multiple variable
so I am curious how does general linear partial differential equation looks like
sorry I found the définition online
No, what you're writing doesn't make sense.
10:46 PM
@TedShifrin ah i get it now thank you
@TedShifrin yeah I admit that. 😅
Hi all, I have a question,
Isn't all quadrilateral shapes with the same side lengths, have the same area?
@mshwf no
rhombuses can have varying areas, depending on their angles.
It's true for rectangles, though, @mshwf.
But if we take crazy quadrilaterals, there's not even a formula with just the side lengths.
I'm imagining having 4 sticks (with different lengths) configured in quadrilateral shape, I think shrinking an angle stretch another maintaining the same area
Well, many shapes won't even be able to be changed, depending on the lengths.
But area generally changes. Robjohn's suggestion about rhombuses is the best thing to think about.
10:55 PM
@TedShifrin so I'll need to get two angles to be able to calc the area?
Two opposite angles will do, yes.
@TedShifrin if the opposite angles are supplementary, one can use Brahmagupta's formula.
@TedShifrin Eh, I'm okay. My mom recently got a feeding tube put in, so she's been able to get food in her, which she wasn't able to for a while. Myself, I've been wandering about, distracted as I usually am
11:14 PM
that's tough for everyone. best wishes.
My best wishes to your mom and family. I’m sorry! 😢
Not to worry, she's actually able to walk around because she has energy. It's a great improvement
Also, we (or, rather, I) have been saying she's a cyborg, now

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