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Semiclassical
Semiclassical
12:00 AM
Hmm, yeah
 
Ted Shifrin
Ted Shifrin
That's what's going on.
 
Balarka Sen
Balarka Sen
Cool stuff.
I feel less bad about forms now, surprisingly.
 
Ted Shifrin
Ted Shifrin
LOL
 
Semiclassical
Semiclassical
That works for taking a 3-form to a 2-form, anyhow
 
Balarka Sen
Balarka Sen
i guess i'm just a moody kid after all
so much for being avant garde
 
Ted Shifrin
Ted Shifrin
12:01 AM
Something like that, grumpy kid.
@Semiclassic: $n$-volume and $(n-1)$-hypervolume :D
 
Semiclassical
Semiclassical
What about 2- to 1- and 1- to 0-form?
 
Ted Shifrin
Ted Shifrin
You can figure those out.
 
Balarka Sen
Balarka Sen
Compute area as length plus an additional vector.
(oriented)
 
Semiclassical
Semiclassical
Hmm
 
Secret
Secret
Is it possible to do a discrete analogue of variational calculus by considering a 'curve' $\lambda$ made of countable points and then vary it to maximise/minimise an infinite sum $\sum_{\lambda} g({\lambda})$?
 
Ted Shifrin
Ted Shifrin
12:03 AM
I would highly doubt it, but who knows.
 
Balarka Sen
Balarka Sen
I think Mike made me figure out Cartan's magic formula by working the LHS and RHS out as derivations commuting with the boundary map in the chain complex. Doesn't make it unmagical to me, though.
 
Semiclassical
Semiclassical
The loose mental image I have is that a line is the intersection of two planes and a point is the intersection of three planes
 
Ted Shifrin
Ted Shifrin
That doesn't account for the bracket term, @Balarka.
I have no idea why you're doing that, @Semiclassic.
 
Semiclassical
Semiclassical
Well, i have in mind here a sequence of interior products taking dx dy dz -> dx dy -> dx
 
Ted Shifrin
Ted Shifrin
That doesn't really make sense. To get that you have to have specific vectors you're contracting against.
And it works left to right, actually, but up to sign ...
$\iota_{(a,b)}dx\wedge dy = -b\,dx + a\,dy$.
 
Topologicalife
Topologicalife
12:09 AM
Standard part function
In non-standard analysis, the standard part function is a function from the limited (finite) hyperreal numbers to the real numbers. Briefly, the standard part function "rounds off" a finite hyperreal to the nearest real. It associates to every such hyperreal x {\displaystyle x} , the unique real x 0 {\displaystyle x_{0}} infinitely close to it, i.e. x − x 0 ...
 
Ted Shifrin
Ted Shifrin
So $\iota_{(a,b)} dx\wedge dy$ is the $1$-form whose value on $v$ is the signed area of the parallelogram spanned by $(a,b)$ and $v$.
 
Semiclassical
Semiclassical
I had in mind contracting with (0,0,1) and then with (0,1,0)
 
Ted Shifrin
Ted Shifrin
You actually have sign issues. You feed in at the left, not the right. But ok.
 
Semiclassical
Semiclassical
Yeah.
Should've done dy dz etc
 
Topologicalife
Topologicalife
Oh, the thirst answer on this link is what I just explained: math.stackexchange.com/questions/1997424/…
But with less motivation.
I think is a great answer to get a intuitive idea of differential.
Also, I think the ultralimits of Tao are a good approximation but I haven't read them deeply.
 
Balarka Sen
Balarka Sen
12:13 AM
That formula for $d\omega(X, Y)$ is really interesting. I have no idea what it geometrically means though.
 
Semiclassical
Semiclassical
And the notion in my head being, for instance, that the intersection of the planes y=0 and z=0 is a flow line in the (1,0,0) direction
 
Ted Shifrin
Ted Shifrin
Think about Stokes's Theorem on a tiny "curvilinear" parallelogram following the flows of $X$ and $Y$. (Of course, the bracket is there because that parallelogram might not close up.) @Balarka
 
Semiclassical
Semiclassical
But it's hard to say this right without a chalkboard
 
Secret
Secret
Just found something might be related to what I pondered: Slides on discrete lagrangian mechanics. The sample points are finite in number though.
 
Semiclassical
Semiclassical
Of course, I also am thinking in 3D, and that in itself is dangerous
 
Ted Shifrin
Ted Shifrin
12:15 AM
I'm not sure you're thinking about this right, @Semiclassic. It only works if you start with a rectangular set-up with your 3 vectors, probably. Maybe I'm misunderstanding.
 
Balarka Sen
Balarka Sen
@Ted Ah!
 
Secret
Secret
(There is also a Hamiltonian version too)
 
Balarka Sen
Balarka Sen
So that's the infinitesimal Stokes' you were talking about. That's very fun.
 
Ted Shifrin
Ted Shifrin
Oh hell. I don't want power point.
 
Semiclassical
Semiclassical
I suspect what I'm doing only makes sense in 3D where there's a nice correspondence between vector fields and 2-forms
 
Ted Shifrin
Ted Shifrin
12:17 AM
No, I don't think that's relevant to this, @Semiclassic.
 
Semiclassical
Semiclassical
Hmm
 
Ted Shifrin
Ted Shifrin
OK, I have to eat something and then go out for the evening. Bye, all. Good un-sleep, Balarka. Glad you are slightly less grumpy :D
 
law-of-fives
law-of-fives
Is Noah Schweber here?
 
Balarka Sen
Balarka Sen
Enjoy!
 
Ted Shifrin
Ted Shifrin
Never here, @law-of-fives.
 
law-of-fives
law-of-fives
12:18 AM
ok
 
Topologicalife
Topologicalife
Wow, there is so much information on math stackexchange about the differentials.
Why does it confuse a lot of people?
Bye @TedShifrin! Enjoy.
 
Balarka Sen
Balarka Sen
@TedShifrin What's cool is that if $\omega$ is closed, that formula implies $X(\omega(Y)) - Y(\omega(X)) - \omega([X, Y]) = 0$ for a 1-form dual to an irrotational vector field. If I let $X = e_1, Y = e_2$, that gives me back the mixed partial condition for irrotational fields.
 
Adeek
Adeek
hi everyone.
 
Balarka Sen
Balarka Sen
hi
 
Adeek
Adeek
hi @TedShifrin @BalarkaSen
 
Topologicalife
Topologicalife
12:25 AM
Good night guys.
 
Akiva Weinberger
Akiva Weinberger
@law-of-fives Unfortunately, he doesn't seem to ever use the chat
(Shame, he's really fun to talk to)
 
SaitamaSensei
SaitamaSensei
12:55 AM
so this is the chat eh? looks nice :)
 
Daminark
Daminark
Hey there!
 
SaitamaSensei
SaitamaSensei
hey there good sir!
 
Daminark
Daminark
How's it going @Saitama?
 
SaitamaSensei
SaitamaSensei
great ! @Daminark i always wanted to find a place where i can talk about math ! how about you?
 
Daminark
Daminark
This is a nice place to do so for sure!
 
SaitamaSensei
SaitamaSensei
12:58 AM
you bet !
 
Simply Beautiful Art
Simply Beautiful Art
If anyone thinks this should be reopened...
https://math.stackexchange.com/questions/2241955/does-sum-n-1-infty-frac1-logene-n-converge-or-diverge
 
Daminark
Daminark
I mean, it is a very particular sequence, on that I don't think the typical person will come across, and this question has been answered to the satisfaction of the OP
 
Zee
Zee
1:37 AM
Hello
 
CausingUnderflowsEverywhere
CausingUnderflowsEverywhere
1:57 AM
does each transcendal function just have its own rules for everything? is f(x) a trascendental function?
 
Akiva Weinberger
Akiva Weinberger
It really depends on what f is.
If f(x)=x^2, then no, f is not transcendental.
If f(x)=sin(x), then yes, since the sine function is.
 
CausingUnderflowsEverywhere
CausingUnderflowsEverywhere
oh okay
why can sin(30+50) mean something completely different than what would be inferred in regular math of that being equal to sin(80) ? then a different transcendtal fucntion will have its own other rules?
is that the definition of a transcental function?
 
Secret
Secret
2:14 AM
what do you mean, $\sin (30+50)=\sin (80)$?
 
arctic tern
arctic tern
what in the world are you talking about
 
Secret
Secret
Or do you means the fact that $\sin(a+b)\neq \sin(a)+\sin(b)$ in general makes it (insert suitable clause)? This question is missing context or other details
 
TheGreatDuck
TheGreatDuck
0
Q: Are all solutions to an ordinary differential equation continuous solutions to the associated implied differential equation and vice versa?

TheGreatDuckNow I have to heavily emphasize the fact that I have never studied differential algebra or the concept of other types of differentiation (which is what I believe is the concept behind a differential algebra). So, if I am abusing the terminology a little bit, please forgive me. Let us define a di...

differential-equations limits derivatives differential-algebra
maybe someone here knows how to solve this
 
Secret
Secret
Brief observation: if something in one variable has a limit as it approaches to a, then its left and right limit must coincide
Therefore I suspect the set of all ordinary subdifferential equations is a superset of all ordinary differential equations
 
TheGreatDuck
TheGreatDuck
2:30 AM
but im not referring to the subdifferential equations
the subdifferentials include the entire range between the left and right hand limits
whereas mine do not
 
Secret
Secret
Implied differential equations will then be a subset of subdifferential, since $\{a,b\}\subset [a,b]$, so the above observation should still apply
 
TheGreatDuck
TheGreatDuck
depends on whether or not the two subsets intersect
i can say a a subset of integers is the even integers
and so are the odd integers
do they intersect? Nope.
Anyway, they are meant to emulate a differential algebra wherein which piecewise constants are algebraically treated the same as regular constants. I do not believe the subdifferentials have that same quality. They seem to do something else depending on the choice of subdifferential at the discontinuities or sharp corners.
 
Secret
Secret
More concretely, (unless there exists highly nontrivial examples) if $f$ satisfy some ODE $D(y)=c$, then it must jointly satisfy $D_+(y)=c$ and $D_-(y)$ where $D_+$ and $D_-$ are your one sided differential operators
But you are right, subdifferentials are only defined for convex functions, while one sided ODE can be defined for any jump discontinuity function
So I suspect Satisfy ODE $\Rightarrow$ Satisfy corresponding one sided ODE, but not the other way around (because the latter, as you have noticed, can contain discontinuous function solutions not necessary convex)
 
TheGreatDuck
TheGreatDuck
no i mean mine is a subset of the subdifferentials
i accept that
but to claim that because the subdifferentials are a superset of the differentials makes mine a superset of the differentials is a bit of a leap from a purely set theoretic perspective.
plus, it's not just that they are a superset. Proving that might be trivial depending on the knowledge known by other people. It's that the entire continuous subset is the set of differential solutions is what would be harder to solve.
note, I am probably speaking a weird in that last paragraph. :p
i'm sure you can get what i mean.
 
Secret
Secret
I don't really get what you meant by continuous, unless you mean any solution in this solution set can be continously transformed to another solution in the solution set by some map
 
TheGreatDuck
TheGreatDuck
2:44 AM
im saying you have some differential equation
find the general solution with the implied differential
those solutions might be a subset of the solutions given by the subdifferential
and might even be a superset of the solutions given by the ordinary differential.
however, the question is whether the subset of the implied differential consisting of continuous functions is equivalent to the set of solutions given by the ordinary differential.
i.e. the function found as a solution is continuous.
 
Secret
Secret
Ok, so you mean: Let $S_{D} , S_{D_{\pm}}, S_{D_{[a,b]}}$ be the solution sets of the ODE, the implied/one sided ODE and the subODE. We knew that $S_{D_{\pm}} \supset S_{D}$ and $S_{D_{[a,b]}}\supset S_{D}$, but there may be nontrivial intersection $S_{D_{[a,b]}}\cap S_{D_{\pm}}$

and you are interested in only the continuous functions (in the usual sense) within these sets on whether $S_{D}\vert_{\mathcal{C}}=S_{D_{\pm}}\vert_{\mathcal{C}}$
where $\mathcal{C}$ is the set of all continuous functions
 
TheGreatDuck
TheGreatDuck
Yeah, i think so. I don't have the mathjax on so a bit hard to read. Looks right
to be more direct than that though, I'm interested in whether or not this differential algebra works as a valid of solving the ordinary differential equations: by solving them and then solving the resulting functional equation to determine the general continuous solution.
so in essence, the true meaning of the proof is to verify a manner of solving differential equations involving step functions
 
Secret
Secret
In that case, a continuous function within the domain $\textrm{Dom}(f)$ is defined to be $\forall x,a\in \textrm{Dom}(f), \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=f(a)$. Now take appropriate limits on both sides:
 
RE60K
RE60K
Hello, I need some help with number theory
 
TheGreatDuck
TheGreatDuck
@RE60K fire away. I haven't taken a number theory course but the proof we've been doing is number theory themed and I've been messing with some stuff.
ring theory or field theory?
 
Secret
Secret
2:58 AM
$\lim_{h\to 0^+}\frac{\lim_{x\to a^+}f(x+h)-\lim_{x\to a^+}f(x)}{h}$

$\lim_{h\to 0^-}\frac{\lim_{x\to a^+}f(x+h)-\lim_{x\to a^+}f(x)}{h}$

$\lim_{h\to 0^+}\frac{\lim_{x\to a^-}f(x+h)-\lim_{x\to a^-}f(x)}{h}$

$\lim_{h\to 0^-}\frac{\lim_{x\to a^-}f(x+h)-\lim_{x\to a^-}f(x)}{h}$

Now using linearity of limits and assume that all one sided limits exists for each case:
 
RE60K
RE60K
I am trying to prove that for negative discrimanat there is exactly one reduced form
 
TheGreatDuck
TheGreatDuck
@RE60K hrmm... are you trying to prove that if the discriminant of a polynomial is a negative number then there is exactly one REAL number solution?
 
RE60K
RE60K
So I assumed that if there are two such forms $\{a,b,c\}$ and $\{a',b',c'\}$ with $-a<b\le a<c$ or $0\le b\le a=c$ then WLOG $a\le a'$. let the transformation taking $\{a,b,c\}$ to $\{a',b',c'\}$ is $\begin{vmatrix}r&u\\s&t\end{vmatrix}$
 
TheGreatDuck
TheGreatDuck
this is the first I've heard of discriminants outside of the quadratic equation so I'm pretty stumped.
*WOLOG
 
Secret
Secret
$\lim_{h\to 0^+}\frac{\lim_{x\to a^+}f(x+h)-f(x)}{h}$

$\lim_{h\to 0^-}\frac{\lim_{x\to a^+}f(x+h)-f(x)}{h}$

$\lim_{h\to 0^+}\frac{\lim_{x\to a^-}f(x+h)-f(x)}{h}$

$\lim_{h\to 0^-}\frac{\lim_{x\to a^-}f(x+h)-f(x)}{h}$

Next:
 
RE60K
RE60K
3:01 AM
$F(x,y)=ax^2+bxy+cy^2$, $\Delta=b^2-4ac$
 
TheGreatDuck
TheGreatDuck
@RE60K hold on a sec
@Secret wait? Are you actually proving it? O.O
 
RE60K
RE60K
pg.172 Edmund Landau Elementory Number Theory
 
Secret
Secret
@TheGreatDuck I don't know if I am proving it, I am just trying to understand how it behaves for continuous function and it just happens the process look like a proof to you
 
TheGreatDuck
TheGreatDuck
ah fair enough. I'm just partially distracted as I'm helping this guy. :)
@RE60K sounds like I have a lot to learn with number theory. Our experience has mostly been number theory in the ordinary integers as well as numbers of the form $a + b\sqrt{3}$ where a and b are integers.
 
RE60K
RE60K
NP
 
TheGreatDuck
TheGreatDuck
3:04 AM
i've been doing the same with 3 replaced with solutions to arbitrary second order polynomials
if you got any questions about that or modular arithmetic or prime number stuff in those sets I could help but sadly i only know the basics.
@Secret pretty sure that for differentiable functions the implied derivative gives the regular results. The unusual things are the step functions. Might want to keep that in mind. :) (If the implied derivative acts weird for differentiable functions then that means my formulae is flawed and doesn't replicate the concept I wished to emulate with it).
 
MMM
MMM
Anyone interested in Maple software can join us here
http://area51.stackexchange.com/proposals/107315/maple
 
TheGreatDuck
TheGreatDuck
@MMM not personally interested but thanks for the memo!
:-)
 
MMM
MMM
@TheGreatDuck Thx for your comment
 
TheGreatDuck
TheGreatDuck
@RE60K I'd recommend posting a question if you haven't yet.
 
RE60K
RE60K
Thanks!
 
TheGreatDuck
TheGreatDuck
3:08 AM
just make sure to write it like you would expect to see on your homework and explain its a homework problem you would like help on
idk if you're new here or not. :p
nvmd
 
Secret
Secret
$\lim_{h\to 0^+}\lim_{x\to a^+}\frac{f(x+h)-f(x)}{h}$

$\lim_{h\to 0^-}\lim_{x\to a^+}\frac{f(x+h)-f(x)}{h}$

$\lim_{h\to 0^+}\lim_{x\to a^-}\frac{f(x+h)-f(x)}{h}$

$\lim_{h\to 0^-}\lim_{x\to a^-}\frac{f(x+h)-f(x)}{h}$

Using what we start with about the continuity of $f$, we get:

$\lim_{h\to 0^+}\lim_{x\to a^+}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0^+}\lim_{x\to a^-}\frac{f(x+h)-f(x)}{h}$

$\lim_{h\to 0^-}\lim_{x\to a^+}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0^-}\lim_{x\to a^-}\frac{f(x+h)-f(x)}{h}$

Now:
 
TheGreatDuck
TheGreatDuck
you've been around the block a few times
@Secret Well, I'm not really sure what you are doing aside from manipulating the limits. If you see anything useful feel free to let me know. I might head over to simply arts chat in a bit to see how they're doing. :-)
 
Secret
Secret
Uh... might be missing something. I need to figure out how to handle the case where $f$ is a weistrass function (example of nowhere differentiable function)math.stackexchange.com/questions/930780/…
 
TheGreatDuck
TheGreatDuck
To be honest, I think this is one of those things in math that stick around a while. probably not years but definitely something that feels like it needs some stuff built up first. I think the key to this will be to find a transformation operator for the implied differential that turns their equations into algebraic ones like the laplace transforms and use that to solve the most general Nth order equation. Then the solution comes from showing that it's solution and the ordinary version's...
solutions are equivalent.
@Secret I can answer that easily.
 
Secret
Secret
... while $f$ is continuous, its derivative is not necessary continuous (think |x|), and I need to account for that in the above two equations
 
TheGreatDuck
TheGreatDuck
3:13 AM
for that function if we assume it is a fractal of the absolute value
then there exists a step function such that when removed gives us some sequence of +-1
 
CausingUnderflowsEverywhere
CausingUnderflowsEverywhere
I guess so
 
TheGreatDuck
TheGreatDuck
so the implied derivative would be +- 1 everywhere
granted
 
CausingUnderflowsEverywhere
CausingUnderflowsEverywhere
5^5x is the chain rule
 
TheGreatDuck
TheGreatDuck
@CausingUnderflowsEverywhere dude... wrong room.
 
CausingUnderflowsEverywhere
CausingUnderflowsEverywhere
laugh out loud I am too in the correct room
 
TheGreatDuck
TheGreatDuck
3:15 AM
step functions have to have a measurement in the step length or something so I guess that the Weirstrauss Function might push that to its limit and break the one-sided differentials.
@CausingUnderflowsEverywhere then why are you posting irrelevant stuff?
 
CausingUnderflowsEverywhere
CausingUnderflowsEverywhere
'for both general discussion & math questions alike.' I'm alike
where should I go to post then?
 
TheGreatDuck
TheGreatDuck
how does "5^5x is the chain rule" at all make any sense. That's clearly a response to something yet it doesn't clearly pertain to anything we were saying.
 
CausingUnderflowsEverywhere
CausingUnderflowsEverywhere
it's 5 to the exponent of 5 to the exponent of x. that is mathematics in clarity
 
Secret
Secret
Ah, that good ol' $x\sin (\frac{1}{x})$ (as well weierstrass) will screw it, cause one sided derivatives don't even exist for countable or uncountably many points
3
Q: Does continuity imply existence of one sided derivatives?

MarekFrom what I understand a derivative may not exist at a given point if the function is not continuous or the right and left side derivatives are not equal. Does that imply that if a function is continuous, the one sided derivatives exist at it's every point?

calculus real-analysis derivatives
Ok, so we now have our first constraint
 
TheGreatDuck
TheGreatDuck
facepalm
y' = ___
for either of those has no solution
so that's pretty tangential unless I'm missing a point you're making?
perhaps i should start smaller with something like the second or third order differential equations with a heaviside forcing term.
 
Secret
Secret
3:22 AM
Well, in a sense, I guess one reason my above thought process look like a proof is because I tend to think top down, and proofs look like top down things.

My idea is that if my thought process fail somewhere, that will introduce a constraint that can show where the property we want will fail. Rinse and repeat and we get enough constraints so that the thought process works, this can then be worked into a proof with conditions supplied as the constraints
 
CausingUnderflowsEverywhere
CausingUnderflowsEverywhere
for n, such that 0 < n < 1 ; n - 1 is equivelant to -(n)
 
TheGreatDuck
TheGreatDuck
well... you're just showing that the differential doesn't evaluate those functions
that doesn't say anything about it's antiderivative which is what differential equations are primarily concerned about
 
Secret
Secret
yeah, I guess that is not very relevant, since they won't even solve any ODE anyway, let alone solving possible subsets of them. Let me think again...
 
TheGreatDuck
TheGreatDuck
:-)
i mean, if the weirstauss came out a solution somewhere that would give a counter example... but let's work with something more basic.
i mean, while this proof would allow for us to justifiably solve much more exotic equations than before... that sort of thing seems to messy to worry about until it becomes relevant.
plus, I can tell you that the discontinuous function giving the weirstrauss function it's sharp corner absolute value like shape is everywhere discontinuous and kind of like an indicator function (probably something similar could be made with the indicator function of the rationals or irrationals).
those are not step functions, so my expected behavior (presumes step functions are constants and acts the same everywhere else) prescribes they are nowhere implied differential.
;-)
course that's not a proof but my intended design prescribes that result
so to not get that result indicates an error in the operator itself and means my definition is flawed.
make sense?
not that there is anything wrong with looking at those. I just wanted to let you know that those were expected results and not to waste your time with them if you didn't wish to. :)
 
Daminark
Daminark
3:55 AM
This room feels enchanted...
Hey @Tim!
Also hey @Thegreatduck!
 
CausingUnderflowsEverywhere
CausingUnderflowsEverywhere
@Daminark is it due to the recent discovery of how 0.3 - 1 = -1 * 0.3 ?
 
Daminark
Daminark
Tim is Tim the enchanter :P
And also wat?
 
TheGreatDuck
TheGreatDuck
4:15 AM
@Daminark hi. Ignore the other guy. He's a little odd.
 
Daminark
Daminark
Hey @Mike!
And lol
 
CausingUnderflowsEverywhere
CausingUnderflowsEverywhere
@TheGreatDuck I will report you soon for cyber bullying if you do not cease your behaviour. that's right, I'm from the U.K.
 
TheGreatDuck
TheGreatDuck
What does you being in the U.K. have to do with me saying your posts are nonsense?
 
Mike Miller
Mike Miller
hi
 
Daminark
Daminark
How's it going?
 
TheGreatDuck
TheGreatDuck
4:27 AM
@Daminark I don't know how the IT department is going.
 
Tim The Enchanter
Tim The Enchanter
@Daminark Hey Dami, sorry was afk.
 
TheGreatDuck
TheGreatDuck
@CausingUnderflowsEverywhere also, when i said you were in the wrong room i meant that it seemed like you were t replying to nonexistant posts. Hence, I presumed you were in another math room in a completely separate conversation. Excuse me if it looked that way. No need to be rude about it.
 
Daminark
Daminark
@TheGreatDuck Darn, I was hoping on getting an update on their status, I've not been getting many reports
How's it going @Tim?
 
TheGreatDuck
TheGreatDuck
:-)
got a tricky problem I'm working on.
wanna see?
 
Tim The Enchanter
Tim The Enchanter
@Daminark Was pretty sick a few days ago but fine now. What about you?
 
TheGreatDuck
TheGreatDuck
4:32 AM
oh there's an easy way to get rid of a cold
just hold your breath as long as you can.
if you hold it long enough, the disease suffocates within your lungs
 
Tim The Enchanter
Tim The Enchanter
@TheGreatDuck You cant have a cold if you're dead. 11/10 can't try again. :P
@TheGreatDuck I'm interested
 
Daminark
Daminark
Rekt by stairs, but otherwise good
For sure @thegreatduck!
 
TheGreatDuck
TheGreatDuck
@TimTheEnchanter no i meant that one can actually hold ones breath and do it without dying
only have to hold it for a minute-minute and a half
and it's actually feasible
-_-
 
Tim The Enchanter
Tim The Enchanter
@TheGreatDuck Well the common cold is caused by rhinoviruses, which don't breathe, but I guess its worth a try :)
@AkivaWeinberger Hey Akiva
 
TheGreatDuck
TheGreatDuck
@TimTheEnchanter perhaps it's pneumonia then which was going around a little and is an aerobic bacteria...
but anyway, it worked and it went away almost immediately after trying a few times every hour or so
 
Daminark
Daminark
4:39 AM
Hey @Akiva!
 
Tim The Enchanter
Tim The Enchanter
@TheGreatDuck Well luckily, I had neither. I was just throwing up a bit. Good to hear it worked for you.
 
Daminark
Daminark
And this is interesting @thegreatduck!
 
TheGreatDuck
TheGreatDuck
@Daminark thanks!
@TimTheEnchanter well Idk what I had. I had what felt like a cold but it could've been bacterial. Or maybe it just made my lungs and stuff calm down and it cured the symptoms. I'm not a biologist so I cannot really say. :)
 
SAWblade
SAWblade
recently i had the experience of solving quite a nice problem
and who knows, it might come in handy one day when i least expect it
 
TheGreatDuck
TheGreatDuck
@SAWblade I think mine will be quite useful as a stepping stone to solving a lot of other problems. I hope yours does the same for you. :)
 
SAWblade
SAWblade
4:45 AM
Thank you very much! :) May I ask what your problem was? :0
 
TheGreatDuck
TheGreatDuck
in a nutshell: proving that a differential algebra equivalent to ours but with step functions treated as constants has differential equations whose continuous solutions are the solutions to our differential question. Essentially, it provides a convenient and trivial method to deal with step functions in differential equations.
 
Tim The Enchanter
Tim The Enchanter
@TheGreatDuck Looked at your problem, never thought of something like that, good stuff.
 
TheGreatDuck
TheGreatDuck
@TimTheEnchanter thanks. Took me a lot long longer than i am comfortable with to realize that double sided limit thing did what I waned to do but maybe with an actual definition (instead of just rambling words sounding like a crazy person) it will start to grab some traction and get resolved. :-)
by all means, I hope it is true. It would deal with a lot of equations.
some of which might not even be solvable by other means.
 
SAWblade
SAWblade
Hm. I don't think I'm well-read enough to understand your problem statement. xD
 
TheGreatDuck
TheGreatDuck
there's a differential operator other than the derivative
so you can have differential equations
suppose you have the continuous solutions
all of those solutions are the entire general solution of the corresponding differential equation using the derivative we're all used to.
get it now?
 
SAWblade
SAWblade
4:55 AM
Ah, and this method lets you deal with a bunch of otherwise intractable or needlessly nasty differential equations?
Or so you hope?
 
TheGreatDuck
TheGreatDuck
i already know it would, period
it's just a question of the validity of it
 
SAWblade
SAWblade
That seems rather powerful. :0 Do you have a proof?
 
TheGreatDuck
TheGreatDuck
either it is true or false
 
SAWblade
SAWblade
Ah I see. xD
 
TheGreatDuck
TheGreatDuck
that's the question
the question is literally what I just said to you
 
SAWblade
SAWblade
4:58 AM
My apologies, it took a little bit to sink in. xD
 
TheGreatDuck
TheGreatDuck
it wouldn't surprise me if this proof is either solved before in some weird twisted way, in which case great or it has never been solved and it ends up sticking around for a decade or two.
at least, assuming it doesn't leave MSE and become a target of legitimate mathematical interest in proving.
 
SAWblade
SAWblade
The problem with the expansiveness of math research is that all the interesting problems have either been solved already or won't be solved in your lifetime.
Which is a shame. xD
 
TheGreatDuck
TheGreatDuck
precisely
well it depends
sometimes the truly meaningful proofs have been solved
but nobody has applied them
 
SAWblade
SAWblade
Ah true, the theorists do dislike applications.
 
TheGreatDuck
TheGreatDuck
like I said: it might be proved, but it is proved by some statement nobody thought to use in that manner
for instance, if the subderivatives (which is an operator whose solutions are broader than my operator) can be proven to have continuous solutions only within my operator that might be sufficient
assuming that the subderivative has been proven to be a superset of the ordinary derivative's solutions to equations
because if subDerivative > mythingy > derivative
well... transitivity ftw!
so it might not be a meaningful proof so much as a clever application of current knowledge.
 
Tim The Enchanter
Tim The Enchanter
5:02 AM
It should be
Subderivatives have a nonstrict inequality
 
TheGreatDuck
TheGreatDuck
@TimTheEnchanter what should be?
 
Daminark
Daminark
Rehi @Akiva!
 
Tim The Enchanter
Tim The Enchanter
@TheGreatDuck The subderivative being a superset of the derivative
 
TheGreatDuck
TheGreatDuck
that the subdifferential equation solutions are a superset of the differential equation solutions? Granted, has that been proven?
oh ok.
well then I can understand that comment on my question then
 
Tim The Enchanter
Tim The Enchanter
"If f is convex and its subdifferential at x 0 {\displaystyle x_{0}} x_{0} contains exactly one subderivative, then f is differentiable at x 0 {\displaystyle x_{0}} x_{0}.]" is what wikipedia says
 
TheGreatDuck
TheGreatDuck
5:04 AM
no no no
I'm saying if you write a subdifferential equation
are the solution sets of the subdifferential equation a superset of the corresponding differential equation
unless of course that's what you just said in a roundabout way. :p
the route to proving my statement though, would be to claim that if the solution to a subdifferential equation is continuous then it is a solution to an implied differential equation
that at least establishes they are supersets...
 
SAWblade
SAWblade
I wish you much luck in finding a proof! :)
 
TheGreatDuck
TheGreatDuck
but then the question would be "are all continuous solutions part of the solution set of the differential equations"
 
SAWblade
SAWblade
I'll bask in my combinatorics for a bit more. xD
 
Tim The Enchanter
Tim The Enchanter
AFAIK, a derivative of a function is one of the subderivatives, or am I understanding nothing of what you are saying.
 
TheGreatDuck
TheGreatDuck
granted, once the superset portion has been proven, the other might follow naturally from a proof by contradiction of some kind.
@TimTheEnchanter let SD denote the subderivative. As an example: are all of the continuous solutions to SD(SD(y)) - y = 0 solutions to y'' - y = 0?
though in my case i am talking about a weaker operator which only uses the endpoints
however, it would be good to know if that has already been established
@TimTheEnchanter i'm not comparing the operators themselves so much as the solutions to their corresponding differential equations... to see if the continuous solutions of the two are one and the same
 
Tim The Enchanter
Tim The Enchanter
5:10 AM
@TheGreatDuck Ah, I thought you were talking about the converse of what you just said.
 
TheGreatDuck
TheGreatDuck
nope
 
Tim The Enchanter
Tim The Enchanter
oops
 
TheGreatDuck
TheGreatDuck
though it's not really an if statement. :p
at least, not explicitly if i remember right
wait... if the statement is "if f is a continuous solution to a, then it is a solution to b" wouldn't the contrapositive be trivial to solve by proving "if f is not a solution to b, it is not a continuous solution to a?"
that seems fishy but yet... it feels trivial to assert...
nvmd
i was thinking that that it not being solution to b implied discontinuity
facepalm
be back in a bit
 
SAWblade
SAWblade
5:41 AM
Given some sequence of $1$s and $0$s, we call this sequence $k$-balanced if for all consecutive subsequences, $|\# 1 - \# 0| \leq k$.
Question: What is $a_{n,k}$ - the number of all $k$-balanced sequences of length $n$?
 
Tim The Enchanter
Tim The Enchanter
Reminds me a bit of the catalan numbers
Lemme look I think I recall a problem on mse like this one
 
SAWblade
SAWblade
oh ive solved this
Just thought it would be a cool teaser. P:
 
Tim The Enchanter
Tim The Enchanter
Oh nvm then
 
SAWblade
SAWblade
It has a neat proof - one of those cases where going more general is key.
 
Hatshepsut
Hatshepsut
How can i show that y>1 => y^n is increasing with in?
 
Tim The Enchanter
Tim The Enchanter
5:52 AM
@Hatshepsut Look at the ratio of $y^{n+1}$ and $y^n$
 
Hatshepsut
Hatshepsut
@TimTheEnchanter oh nice, thanks
 
Tim The Enchanter
Tim The Enchanter
@SAWblade Just to be clear, by "consecutive sequences" you mean the sequences {$d_1,d_1 d_2 , d_1 d_2 d_3 ...$} where $d_i$ is the i'th digit of the main sequence?
@Hatshepsut No problem
 
SAWblade
SAWblade
Given a sequence say $d_{1}d_{2}d_{3}d_{4}d_{5}$, two examples of consecutive subsequences are $d_{1}d_{2}d_{3}$ and $d_{3}d_{4}d_{5}$.
Any subsequence $a_{n_k}$ where $\forall i, n_{i+1}-n_{i} =1$.
 
Tim The Enchanter
Tim The Enchanter
@SAWblade Thanks, nearly went chasing down the wrong alley.
 
SAWblade
SAWblade
if you'd like, there's an entry on the OEIS for $2$-balanced equations of length $n$.
 
Twink
Twink
6:06 AM
could someone give me some hints to solve this problems please?
 
 
1 hour later…
Alessandro Codenotti
Alessandro Codenotti
7:13 AM
Hi chat
 
Studentmath
Studentmath
7:48 AM
Hi all
@Twink noticed you have three 90 degree triangles at 2?
 
Danu
Danu
@arctictern actually, in that isomorphism I don't see why $1/\det$ rather than $\det$
 
Studentmath
Studentmath
Anyone here is into some functional analysis?
If so and could take a look at the following, would appreciate:
https://math.stackexchange.com/questions/2246784/convergence-of-linear-operator-in-l2
Any tips will be extremely appreciated (rather that than full answer, trying to figure out the subject)
 
RE60K
RE60K
8:17 AM
DOES binary quadratic forms with same discriminant equivalentr?
 
Alessandro Codenotti
Alessandro Codenotti
Just asked a topology question if someone knows an answer
 
 
2 hours later…
Krijn
Krijn
10:31 AM
Eyoo
 
Balarka Sen
Balarka Sen
heya
 
Krijn
Krijn
I might have solved something crucial today
Which had been bugging me for two weeks or so
Alas, it was a trivial application
 
Balarka Sen
Balarka Sen
well at least you figured it out!
most of the things I am stuck on these days turn out to be just a picture
 
Krijn
Krijn
Yeah, I hope my advisor agrees
What kind of geometry are you doing
Riemannian, wasn' t it?
 
Balarka Sen
Balarka Sen
Foliation. I should do Riemannian but I am not :P
 
Krijn
Krijn
10:34 AM
Foliation?
As in, rocks?
 
Balarka Sen
Balarka Sen
That's the origin of the terminology. It's about slicing up a big manifold into a bunch of "parallel" submanifolds.
Easiest example is $\Bbb R^3$ foliated by hyperplanes $z = c$ for various $c$'s.
Locally foliations are modeled by those things. But it can be really complicated.
 
Krijn
Krijn
And what' s the purpose?
 
Balarka Sen
Balarka Sen
That's hard to answer. How I think about it is that a foliation is an extra structure on the manifold, just like a Riemannian metric.
It enables you to do some things, like a version of parallel transport, holonomy, etc
 
Krijn
Krijn
Does it give invariants or insights?
 
Balarka Sen
Balarka Sen
10:52 AM
Oh yeah, you can understand a manifold's topology by studying the foliations on it.
If an oriented 3-manifold admits a foliation with an $S^2$ leaf (leaf = one of the submanifold in the decomposition) then it's $S^2 \times S^1$
That's a baby example.
 
Krijn
Krijn
Ah okay
So what's the good stuff
Like, a theorem that's pretty interesting or a famous result that has much more insight using foliations
 
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