Mathematics - 2017-06-26 (page 3 of 6)

Zee

08:15

Am too sleepy to be awake and too awake to be sleepy

Akiva Weinberger

Put on shoes and take a walk

Zee

That's a great idea

But no

Leaky Nun

why does everyone in this room either call themselves moron or call themselves prophets?

Zee

@LeakyNun low self esteem in both cases

Akiva Weinberger

I only call myself an idiot in private

Zee

08:20

Why would you call an idiot?

Why in private?

Akiva Weinberger

Sometimes I do dumb things

I don't do it as much anymore (call myself an idiot) but I used to a lot

Zee

Sounds awful...

Well I can tell you aren't an idiot but even if you were, what's the big deal?

In the end, both grothendieck and the idiot end up as dirt

The idiot probably had more fun too

Akiva Weinberger

I dunno, maybe it's like "Remember not to do this mistake next time," or "Remember you're not infallible"

Though not phrased that way

Usually it's "Oh, I'm an idiot. But I knew that already"

As I said, though, I don't really do that anymore

Zee

That's plain old anxiety

Which is the root for all search of knowledge

Akiva Weinberger

I think like 20% of all people have general anxiety disorder

Zee

08:27

Seems reasonable

The interesting thing about anxiety

Is that it feels exactly the same as happiness

Bodily wise

The difference is in the head

Akiva Weinberger

Explain?

Like, faster heart rate, that sort of thing?

Zee

Next time your happy, compare how your gut feels and your heart, it's the SAME as anxiety

Akiva Weinberger

Huh.

Zee

The ONLY difference is the mental interpretation of the feeling

Leaky Nun

or maybe both feelings are concomitant with the release of some hormones that make your gut and your heart feel the same way?

Akiva Weinberger

08:30

I think that's the same thing

Anyone got anything interesting about math?

Zee

Weather the feeling is pleasant or painful depends 100% on your verbal interpretation for the couse of the feeling, at least that's the conjecture

The feeling I get from being in a fight and being in love is the same, but the verbal explanation is different

To end

Try this

When you feel those butterfly's next time, try to convince your self of a positive explanation for this feeling

Akiva Weinberger

I don't get the butterflies all the often

Maybe I just give no fucks

Astyx

What do you mean by anxiety ?

Akiva Weinberger

(Cont'd) It depends on the people around me, really.

Hi, Astyx!

Astyx

Oh you mean like in a fight ?

Zee

08:37

It's not anxiety that matters, but the bodily sensation of it

Which I call the butterflies, but those only come when it becomes strong

Astyx

Hi Akiva !

Zee

Speaking of butterflies, am gonna go zzzzzleep

Astyx

What's up ?

Night @Zee

Zee

lol it's 5 AM

SteamyRoot

Not sure whether you'll find it interesting, but I'm curious how you'd solve the following question (was on a 1st year bachelor exam):
"For which value of $a$ (assume $a > 0$) does the following integral vanish:
$$\int_0^\infty x^3 e^{-ax} \sin(x) dx"$$

Leaky Nun

08:42

shift the problem up one dimension?

SteamyRoot

That's a way to solve it quickly, yes.

Astyx

That also how I'd do it

Akiva Weinberger

You mean, use ${\rm Im}(e^{ix})$?

SteamyRoot

Yup

Akiva Weinberger

I would probably just do out the integration

SteamyRoot

08:45

It's rather easy, but you have to be careful with the boundaries

Akiva Weinberger

Well, now you want the integral to equal a pure imaginary

rather than zero

SteamyRoot

Purely real, actually.

Akiva Weinberger

Oh, yeah, whoops

Astyx

I just f*cked up my french oral exam this morning

SteamyRoot

It's rather straightforward, but you have to be careful that you don't do careless substitutions where the infinity becomes some complex infinity :P

The method expected on that exam was actually to use Laplace transforms, but meh, screw those.

Akiva Weinberger

08:49

Ends up being $a=1$, yeah?

SteamyRoot

Yup

Akiva Weinberger

@SteamyRoot I don't actually remember what those are

Astyx

Heh, Laplace transforms, damn engineers

Akiva Weinberger

(I just used Wolfram|Alpha :P )

Oh, wait

Is that the $\cal L$ stuff?

Ah, I very vaguely remember that stuff.

Astyx

Boring tbh

Akiva Weinberger

08:50

Presumably all I need is a table to jog my memory.

Fargle

@AkivaWeinberger The Laplace transform of a function $f$ is given by $\cal L[f](s) := \int_{0}^{\infty} e^{-st}f(t) dt$

Akiva Weinberger

Oh. So that's the Laplace transform of $x^3\sin x$.

Is there a formula for the Laplace transform of $x$ times something, then?

Fargle

Yes.

$\cal{L}[x^nf(x)](s) = (-1)^n\frac{d^n}{d^ns}[\cal L[f](s)]$

Oh God why does that 1 look so weird.

Akiva Weinberger

Got it. And the Laplace transform of $\sin$… it's some rational function, right?

@Fargle \cal has the scope of the entire expression it's in

SteamyRoot

It's $\frac{s}{s^2 + 1}$

Fargle

08:56

Yeah, but I bracketed it.

Oh dear lord.

Akiva Weinberger

No, you didn't.

You need {\cal L}.

Either that, or use \mathcal instead :P

Fargle

True.

Akiva Weinberger

@SteamyRoot So I take the third derivative and find when it's zero? shivers

Fargle

Yep.

SteamyRoot

Actually, wait, no s in the numerator

Akiva Weinberger

09:03

Oh, I think this is how Wolfram Alpha did it, actually

I wonder why "Can irrational numbers be rational in another base" is such a popular question

Leaky Nun

@AkivaWeinberger because we are taught "rational numbers are those that terminate or repeat"

so it is inherently related to base

Akiva Weinberger

Yeah

SteamyRoot

I was always thought "expressible as a fraction", rather than "terminate or repeat"

Akiva Weinberger

Rather than the "ratio of two integers" definition

It just betrays a fundamental misunderstanding of what a rational number is.

SteamyRoot

Well, given that $0.99 \dots = 1.00 \dots$, what's the difference between terminal and repeating anyway :P

Fargle

09:10

I remember once getting into an argument with someone about whether $\pi$ was rational in base $\pi$.

Akiva Weinberger

(This is the question in question)

@SteamyRoot Even without that, it's a repeating zero

@Fargle Of course someone would bring that up…

Fargle

I don't know how much more loudly I could say that being an integer means you are a successor of zero, and this property is independent of base

Akiva Weinberger

Oh, they thought pi was an integer in base pi??

Fargle

Yep.

Akiva Weinberger

And, um, I don't think "successor of zero" is exactly what you meant to write

Fargle

09:12

Because, lol, it's $10_{\pi}$.

Akiva Weinberger

but I get what you mean

Leaky Nun

@AkivaWeinberger some do use "successor of zero"

Fargle

@AkivaWeinberger I mean it in the sense that it is an iterated successor of zero.

Akiva Weinberger

Oh, OK, sure.

Fargle

For those purposes, this was either a troll arguing in bad faith or someone with next to no mathematical maturity, so I used "sum of ones or negative ones".

Right, right.

And then explained that if a number is transcendental, it is irrational, because if a number is rational, it is algebraic, and that transcendentality is clearly base-independent.

Akiva Weinberger

09:14

I still have never heard any reason to care about any transcendental base.

Fargle

There isn't, as far as I know, besides as a mathematical curiosity.

Akiva Weinberger

Things like base phi are interesting because you get $100=11$.

Fargle

Though from what I hear, base $\varphi$ is cool.

Yeah.

Akiva Weinberger

$2=1.11=10.01$

(Thus, $2=\phi+\phi^{-2}$)

Fargle

Which obviously holds from the definition of $\phi$ as the positive solution to $x^2 = x+1$.

Akiva Weinberger

09:16

Of course.

Fargle

You necessarily get that $1/\phi = \phi - 1$, so that $1/\phi^2 = \phi^2 - 2\phi + 1 = \phi + 2 - 2\phi = 2 - \phi$, done.

$\phi$ is really cool. I'm not so enamored with its geometry, but its algebraic properties make it very aesthetically pleasing.

Though I suppose, to many ancient Greeks, its algebraic properties were in some sense its geometric ones.

Leaky Nun

and I've heard that it is consistent with the Peano axioms that there is a non-zero natural number which is not a successor of zero

Akiva Weinberger

And the Lucas numbers ($L_0=2$, $L_1=1$) satisfy $L_n=\phi^n+(-\phi)^{-n}$, meaning that the phinary expansion of the even-indexed Lucas numbers $L_{2n}$ is really clean

@LeakyNun Yes, but you can't write that in the language of Peano arithmetic!

Leaky Nun

@AkivaWeinberger indeed.

or else it wouldn't be consistent with the Peano axioms

Akiva Weinberger

Nonstandard models FTW.

Leaky Nun

09:20

so, are the Peano axioms inadequate?

Akiva Weinberger

And, since you can't prove that PA is consistent in PA, it's consistent with PA that PA is inconsistent

which means there's a model of PA in which there exists a proof of PA's inconsistency!

Of course, the length of that proof is a nonstandard number (not a successor of zero), but it doesn't know that!

(This is all assuming PA is consistent.)

Fargle

rubs temples

Akiva Weinberger

(Meaning, if PA is consistent, then it's consistent with PA that PA is inconsistent.)

Fargle

Number theory is really cool, by the way, as an aside.

@Daminark and I just proved that the product across any Pythagorean triple is divisible by 60, which was a neat fact I'd never known before.

Akiva Weinberger

They're all multiples of $3\cdot4\cdot5$?

Cool. I suppose you'd start with the $(m^2-n^2,2mn,m^2+n^2)$ formulation

Fargle

09:26

Nah, not at all.

Perfect squares are $0,1$ mod 3, $0,1,4$ mod 5, and $0,1,4$ mod 8.

Leaky Nun

@Fargle interesting

Akiva Weinberger

Ah. They can't all be nonzero in each projection

Fargle

Right, right.

Akiva Weinberger

That's very cool.

Fargle

You have to use mod 8 rather than mod 4 because 4 isn't prime, but if a perfect square has 3 factors of 2, it must have 4, so its root is divisible by 4.

Akiva Weinberger

09:28

Ah, I see.

$8|p^2\Rightarrow 4|p$

$4|p^2\nRightarrow 4|p$

Fargle

Indeed. 4 is the easiest counterexample.

Akiva Weinberger

@Fargle Where did this problem come from?

Fargle

@AkivaWeinberger Weil's Number Theory for Beginners.

Akiva Weinberger

Ah, yeah, you wanted to do that over the summer

Fargle

The well-named one, not the badly-named one.

Akiva Weinberger

09:32

looks at watch, sees that it is indeed summer

@Fargle (Basic Number Theory?)

Fargle

Yep, that one.

Felix.C

Just to be sure, every complex differentiable function is analytic right?

Akiva Weinberger

Yes

Felix.C

thx

Fargle

@Felix.C If by differentiable you mean differentiable on the whole plane.

Akiva Weinberger

09:36

Every complex differentiable function is integratabtle, infinitely differentiable, analytic, magic, and shits rainbows

7

BAYMAX

:)

Avantgarde

hah

BAYMAX

how is everyone?any health concerns ?

Secret

09:51

Even with the knowledge of $E^h=e^{hD}$ as per previous discussion, it is still not helping. It is easy to prove $[E^{\frac{1}{2}},P(x)]\implies [E,P(x)]$ but to prove the converse, currently I am kinda out of ideas except trying to brute force it by computing $e^{hD}P(x)Q(x)$, $P(x)e^{hD}Q(x)$

and trying to match up coefficients of what exactly is the explicit form of the polynomial $P(x)$ has to be in order for it to commute with $e^D$. While for RHS it is still ok as what happens is that Q(x) is effectively differentiated n times and thus $e^DQ(x)=\sum_{k=0}^{n}\frac{1}{k!}Q^{(k)}(x)$ w…

Avantgarde

health concerns?

SteamyRoot

Another day closer to death :D

Secret

and what I am fearing are number theoric looking "nonlocal" cancellations is what govern the criteria that $P(x)$ commutes with $E$ (what that means is that LHS=RHS is not because of various coefficients of the polynomials vanishes, but that all coefficients are nonzero and then conspire together to make the whole expression just add to zero, one reason I don't understand number theory because such a result is almost like mathematical coincidence)

But anyway, here's what I had so far after the brute forcing:

$$e^DP(x)Q(x)=\sum_{j=0}^{n+m}\sum_{k=1}^n\sum_{l=1}^m\frac{a_kb_l}{j!}D^j(x^{k+l})$$

$$P(x)e^DQ(x)=\sum_{j=0}^{m}\sum_{k=1}^n\sum_{l=1}^m\frac{a_kb_l}{j!}x^kD^j(x^l)$$

As mentioned earlier, the nilpotent properties of polynomials kills off the infinite series that forms $e^D$ thus giving a finite series which might make things easier. I am now trying to figure out what it means for the coefficients if these two expressions are equal

Akiva Weinberger

You know, $\{1,x,x^2,\dots\}$ might not necessarily be the best basis to use here

Secret

10:08

In that case, I felt like it might be something I don't know about cause I don't remember what special basis set I can use for polynomials that simplifies things.

If I recall, my knowledge about algebraic properties of polynomials are cayley hamilton theorem, how minimal polynomials can factorise a given polynomial and characteristic polynomials in matrices

Akiva Weinberger

I never said that alternate bases are actually used in my solution…

But there does exist a specific basis set that's specifically associated to $E$.

Pascal's rule says that $\binom{x+1}n=\binom xn+\binom x{n-1}$

Or, rather, that $E\binom xn=\binom xn+\binom x{n-1}$

Again, no guarantees that this is actually relevant to my solution.

Secret

Ugh, number theory, I should have thought about pascel triangle given how I notice the same pattern when trying to write $E$ under the standard basis

Fargle

10:32

Okay, just read that $S_2$ and $S_6$ are the only "incomplete" symmetric groups; that is, all other symmetric groups have trivial center and trivial outer automorphism group (i.e. all automorphisms are inner).

$S_2$ makes sense: its automorphism group is trivial, but its center is the whole group. But why $S_6$?

Or more precisely, why does $S_6$ have non-trivial outer automorphism group? What is it about 6-ness that makes this happen?

Akiva Weinberger

It's very weird

I found a PDF about it earlier

I don't think it was this, but it seems relevant: math.stanford.edu/~vakil/files/sixjan2308.pdf

"A DESCRIPTION OF THE OUTER AUTOMORPHISM OF $S_6$, AND THE INVARIANTS OF SIX POINTS IN PROJECTIVE SPACE"

Secret

11:04

\begin{align}
e^DP(x)Q(x) & =\sum_{j=0}^{n+m}\sum_{k=1}^n\sum_{l=1}^m\frac{a_kb_l}{j!}D^j(x^{k+l})\\
& =\sum_{j=0}^{n+m}\sum_{k=1}^n\sum_{l=1}^m\frac{a_kb_l}{j!}(x^kD^j(x^l)+D^j(x^k)x^l)\\
& =\sum_{j=0}^{n+m}\sum_{k=1}^n\sum_{l=1}^m\frac{a_kb_l}{j!}x^kD^j(x^l)+\sum_{j=0}^{n+m}\sum_{k=1}^n\sum_{l=1}^m\frac{a_kb_l}{j!}D^j(x^k)x^l\\
& =\sum_{j=0}^{m}\sum_{k=1}^n\sum_{l=1}^m\frac{a_kb_l}{j!}x^kD^j(x^l)+\sum_{j=0}^{n}\sum_{k=1}^n\sum_{l=1}^m\frac{a_kb_l}{j!}D^j(x^k)x^l\\
& =\sum_{j=0}^{m}\sum_{k=1}^n\sum_{l=1}^m\frac{1}{j!}a_kx^kD^j(b_lx^l)+\sum_{j=0}^{n}\sum_{k=1}^n\sum_{l=1}^m\frac{1}{j!}D^j(a…

I am not sure how to deal with the case where $e^{hD}$ and $h$ irrational, I guess I need to think about how to express $E$ in the "pascel basis" that you hinted, which might take a while cause I am bad at combinitorics

Bottomline: I still don't quite understood number theory and I am still bad at counting

Fargle

This second construction is very helpful. It also makes sense that it can't be generalized in quite the same way: the only complete graph on $n$ vertices you can $2$-color and wind up with $2$ $n$-cycles is the one on $5$ vertices.

Secret

To be checked: The taylor series of $x^{\pi}$

Fargle

You'd have to, for example, 3-color the heptagon, and I'm inclined to think there are more than 8 ways to 3-color the heptagon into 3 7-cycles.

Leaky Nun

12:01

@Secret what?

Secret

if you try to naively plug that into the taylor series formula, you get an infinite series with factorial looking coefficients of the form $\prod_{i=1}^n (\pi-n)$

Leaky Nun

at where?

SteamyRoot

around x = 1 most likely

Secret

$0^{\pi}=\lim_{n\to 0}e^{\pi \ln n}\to 0$

$1^{\pi} = e^{\pi \ln 1}=1$

$x^{\pi}=e^{\pi \ln x}$

O, I can just use exponentials, my bad...

SteamyRoot

What are you even trying to do? o.O

Secret

12:10

It's related to that exercise Akiva gave me, I am trying to see how the proof can be generalised to $e^{hD}$ where $h$ is irrational

and for that I need the taylor series of $x^h$

SteamyRoot

Then why take $\pi = h$ ? There's nothing that makes this value of $h$ special...

Secret

That's true, any irrational (preferrably, transcendental) $h$ will do, I just have a preference to select $\pi$ to test the approach first

Whenever I want to test something involving irrationals, I tend to pick $\pi$ as its trancendental nature will mean it will not accidentally solve polynomials that might pop up in the draft proofs, resulting in coincidental cases that make the proofs work

only when it works (or in rarer cases, does not work) will I go directly to $h$

and I do have a history back in my undergrad of constantly hitting the rare counterexamples when I want to prove something, and never hit counterexamples when I tried to find a counterexample to a proposition

Simply put, I am often unlucky when it comes to doing my maths exercise, often bumping into cases which solves things accidentally and giving a false positive

SteamyRoot

I'm not sure whether you should call that "unlucky"...

Semiclassical

@LeakyNun with regards to idiots versus prophets, I tend to agree. If I'm going to disparage myself (or another, I suppose) I prefer to say I'm being silly rather than stupid.

On the other hand I really don't understand the frequency with which we get people proclaiming themselves as prophets.

It tends to say a lot more about the speaker's psychology than it does about their actual accomplishments.

Secret

Re: $I^n (1- (n+1) f(x))= 0$
Consider orthogonal functions $\langle f,g\rangle=0$ over some interval $(a,b)$. This should not happen for the scenario shown here unless $f(x)=e^{ax}$. However $e^{(ax)^2}$ diverges under integration hence it is not integrable if the interval is partially unbounded.

If we want the indefinite integral to hold for all intervals, then 0 should be the only solution, otherwise there might be nontrivial solutions involving functions that are not differentiable at some points or even discontinuous ones as Leaky suspected. Will investigate this later, not enough back…

Felix.C

12:26

hey does someone know if there is a question here about the proof that if we have a function $f: U \subset \mathbb{R}^n \to \mathbb{R}$a and this function is n times partial differentiable and the (n-1)-th partial derivative is total differentiable, then we can change the order of partial derivatives, like a weaker form of the theorem of schwarz

Ivan Ivanov

Hello is Hamilton cycle reducable to Vertex cover?

Secret

5

Q: Geometric or intuitive proof of the symmetry of second partial derivatives

What was given in my calc book is a "consider the function" proof. That is, the author gives a function out of the blue and would deduce all the nice properties from it. I'd prefer a proof which is motivated (perhaps, intuitive) - you see how the proof is crafted in the mind of the person. So my ...

multivariable-calculus

?

2

Q: Symmetry of second (and higher) order partial derivatives

We've learned today that if $f$ has second order partial derivatives that are continuous at some point $a$, then they're all equal to each other at that point. Then there's a short remark that says this holds for higher order as well, i.e. $$D_1D_2D_3f= D_3D_2D_1f$$ What's the proof of this? I ...

partial-derivative

Felix.C

These rely on the second order partial derivatives being continuos no? I want the first partial derivative being total differentiable without further restrictions about the second

en.wikipedia.org/wiki/… of twice-differentiability

the part wth the headline: Sufficiency of twice-differentiability

Secret

In that case, I am not sure

Simply Beautiful Art

12:53

@Secret Have I missed anything? Haven't checked in for a while.

Secret

nothing much, mostly trying to solve Akiva's exercise, some integral algebra attempted formulation and more chat crawling

Simply Beautiful Art

Cool

I'm sort of comprehending recursively inaccessible ordinals and such in my time

The $\alpha$th recursively inaccessible ordinal is the $\alpha$th ordinal which is both an admissible and a limit of admissible ordinals. I suppose I could then make the $\alpha$th $\beta$-recursively inaccessible ordinal, where $\alpha$th $0$-recursively inaccessibles are the $\alpha$th admissibles and $\alpha$th $\beta$-recursively inaccessibles are $\gamma$-recursively inaccessible and a limit of $\gamma$-recursively inaccessibles for all $\gamma<\beta$.

Secret

sounds like some kind of fixed point map applied to the context of admissible and limit admissibles

Btw, some interesting stuff from messing around with powerpoint:

Simply Beautiful Art

in This is the Realm of Simply Beautiful Art, 1 min ago, by Simply Beautiful Art

Oh, so I was trying to prove the following statement:
$$f''(a)>0\iff\exists\delta>0\left( \forall(x_1<x_2\land |x_1-a|<\delta\land |x_2-a|<\delta )\implies \frac{f(x_1)-f(a)}{x_1-a} < \frac{f(x_2)-f(a)}{x_2-a}\right)$$

Ivan Ivanov

Hi is feedback node set reducible to feedback vertex set .I know that feedback vertex set is NP-complete, I think they are because they seem to me pretty much the same because if we have minimum edges we got mininum vertexes and vice versa?

Semiclassical

13:11

hi chat

Ivan Ivanov

hi

Semiclassical

@akiva re: the dodsy vs. collatz thing you noticed

One can make the correspondence explicit as $g(x+1)=f(x)+1$, where $f(x)$ is the Collatz map as you wrote it and $g(x)$ is Dodsy's map.

or, more suggestively, $g\circ T = T \circ f$ where $T(x)=x+1$.

one then immediately has $g^n\circ T = T \circ f^{n}$ (intended as n-fold composition)

Akiva Weinberger

Ah, I see, they're conjugates

Semiclassical

Right.

Akiva Weinberger

Right, of course.

Semiclassical

13:13

I guess $g=T \circ f \circ T^{-1}$ is also pretty suggestive.

That should also work for more generic $T$, though in that case one has to worry about whether $T^{-1}$ actually exists.

(in particular, whether it exists as a map from integers to integers)

Pretty cute regardless, though.

How chat feels sometimes: smbc-comics.com/comic/angles

user84215

For a point of a regular surface with positive Gaussian curvature, if it is simultaneously a point of local maximum and a point of local minimum for the principal curvatures functions, then why is it an umbilical point ?

Secret

SteamyRoot

@aminliverpool If a point is both a local maximum and minimum of a function, then said function is locally constant, no?

user84215

not for a one specific function. they are two different functions

Semiclassical

hmmmm

on the one hand, the paper I'm looking at describes something I want to understand.

SteamyRoot

13:28

@aminliverpool What do you mean with that?

I understood your question as the point $p$ being both a local minimum and maximum of each of the two principal curvature functions.

Semiclassical

on the other, the authors make it sound like in doing so they did something incredibly novel...rather than, well, pretty darn obvious.

Which instantly makes me more skeptical.

SteamyRoot

Making it sound like something incredibly novel is possibly just a way to get published :P

Semiclassical

Maybe.

user84215

for one of its normal curvature it is a local maximum and for the other one it is a local minimum

Semiclassical

But when the article is from 2003 and you can easily find similar remarks from papers about 10 years older...hrm.

Astyx

13:32

Yo chat

Semiclassical

hi @astyx

SteamyRoot

ohi

Astyx

What's up ?

SteamyRoot

Enjoying the fact that the new Windows 10 Creators Update allows you to install Ubuntu inside Windows 10 and use bash :P

Semiclassical

Oh, also the fact that this paper dates back to 2003 and has 3 citations. :S

Astyx

13:35

Inside Windows ?

What's this sorcery ?

SteamyRoot

Yup... Unlike a virtual machine, you can access all of your files under Windows (all drives are mounted in /mnt)

Semiclassical

Two of which share an author with this paper, lol.

SteamyRoot

So you can run anything that works on Ubuntu natively now :D

Semiclassical

Nooooot a very high quality source, I think.

SteamyRoot

@Semiclassical I think you're going to run out of red flags at this rate

Semiclassical

13:37

Probably.

Or I would if I was going to bother to keep looking at it.

Ivan Ivanov

Hello is someone really good at lower bounds or minimum questions to ask in a game,I have a complicated question to whisper him ?

Astyx

14:06

Ask your question anyway @IvanIvanov

Ivan Ivanov

I have posted my question here math.stackexchange.com/questions/2336832/…

Astyx

You're not using the word set correctly : what you are refering to is a tuple

You're basically asking for a sorting algorithm (I'm assuming you know what values the $x_i$ can take, otherwise A cannot have sufficient information to solve the problem)

These can be done in $\mathcal O(n\ln n)$, although I don't know the exact number of questions

($\mathcal O(n\ln n)$ is better than $n^2$)

Ivan Ivanov

I have the solution and it is not with graph it says that the minimum asymptotic bound is omega(n^2) and it is not contradiction with the fact that the lower bound for sorting is O(nlogn)

Astyx

I'm not sure I understand your question then

Oh are the questions all asked at the same time ?

Ivan Ivanov

for example you write them on a list and give them to A and he tells you the answers

Astyx

14:22

Yeah, Ross's answer is what I had in mind

So it's option 2, in which case it's $\mathcal O(n)$

Be sure to accept the answer if it satisfies you

Akiva Weinberger

@Secret Hint: You can have infinite series in $D$ without worrying about convergence, because polynomials are annihilated by higher powers. Unfortunately, that's false for $E$. However, you can define $\Delta=E-I$, and now you can have infinite series in that because $\Delta f(x)=f(x+1)-f(x)$ has lower degree than $f(x)$, so high enough powers kill polynomials as well.

And in general you could define $\Delta_h=E^h-I$

Ivan Ivanov

@Astyx If I create n nodes and after A tells me the answers for each pair I will build directed edge for example if there is edge (u,v) then u>v ,now in order to find the sorted order of the nodes I must have the graph connected,by having it connected I will have path from every node meaning I know which is bigger between 2 nodes,but in order to find if graph is connected we need omega(n^2).Isnt this proof correct?

Secret

14:40

@AkivaWeinberger But isn't $E=e^D=\sum_{j=0}^{\infty}\frac{D^j}{j!}$ thus I will expect if the polynomial $P(x)$ has degree n, then all terms with $D^{j}P(x),j>n$ must vanish regardless,and is only a problem if it is a generic function which has an inifnite series thus the $D^j$ cannot annihilate it into a finite series?

Semiclassical

I find it a little weird to treat $E$ as the basic object rather than, say, $E^t=e^{tD}$. The latter has a radius of convergence.

Astyx

I'm not sure, I don't have time right now @IvanIvanov

Akiva Weinberger

@Secret As I said, it's fine to make an infinite series with $D$. That's what you're doing.

Astyx

Although my comment about accepting the answer still stands

You can ask Ross through a comment on his answer if you want

Akiva Weinberger

It's not fine to have something like $E+E^2+E^3+\dotsb$ (as it diverges pretty horribly even on constant functions).

That's all I'm saying.

Secret

14:46

Ah now I get what you mean

Ivan Ivanov

@Astyx who is Ross

Astyx

The one who answer your question on main @IvanIvanov

Ivan Ivanov

ok

Akiva Weinberger

:38375665 That fails on $x$. Try evaluating it on $x$ at $0$.

Semiclassical

is that re: $E^t$?

Or is that re: the thing I deleted because it was silly/wrong?

Akiva Weinberger

14:48

Re: $E-E^2/2+E^3/3-\dotsb$, which you've since deleted.

Semiclassical

yeah, not going to defend it.

Secret

Is the taylor series of irrational powers really just $\sum_{j=0}^{\infty}\frac{1}{j!}\prod_{k=0}^j(h-j)x^j$?

(obtained by keep differentiating $x^{h}$ by $x$ where $h$ irrational)

SteamyRoot

"Yes, but".

Akiva Weinberger

$\sum E^n/n!$ should converge, though it'll take $\Bbb Z[x]$ to $e\cdot\Bbb Z[x]$, I'm pretty sure

SteamyRoot

Your series has a radius of convergence of, well, $0$.

Semiclassical

14:50

@AkivaWeinberger hmm.

Simply Beautiful Art

in This is the Realm of Simply Beautiful Art, 58 secs ago, by Feeds

I was banned from the airliners.net photography forum by concerned moderators after the end of my lens started brushing against planes as they flew by.

Semiclassical

Converse: If you have a good enough lens, you don't need to add any converters or extenders.

Secret

Uh, in that case, it seems my jordan normal form solution will not work for $e^{hD}$ since lacking a power series to express $x^h$, we cannot have positive integer powers of $e^D$ in the series expansion whcih we can take advantage of the commutativity with $P(x)$ to bubble $P(x)$ out to complete the proof

(Now, to figure out how to use the forward finite difference form that akiva suggested...)

actually wait a sec...

$$\sum_{j=0}^{\infty}\frac{1}{j!}\prod_{k=0}^j(h-j)x^j$$

Semiclassical

There's a story you can tell with that $\Delta$ operator which leads you to the Riemann hypothesis, interestingly. Or rather, it would if it was actually rigorous.

Secret

if we plug $x=D$ then maybe the fact we are dealign with polynomials will truncate off the series enough to be finite due to the annihlating properties of $D$ on the polynomials, however

Semiclassical

14:55

Can be found here, though I'll warn that it's written by physicists: arxiv.org/pdf/1608.03679.pdf

Secret

there's a possibility that give its radius of convergence of zero, the resulting finite series will still not converge to our desired result

Semiclassical

It starts from writing $\Delta f(x)=f(x)-f(x-1)=(1-e^{-D})f(x)$ (If I'm translating right.)

Akiva Weinberger

@Secret What is this a Taylor series of?

Secret

@AkivaWeinberger $((\text{some number})+x)^h$ where $h$ irrational

Akiva Weinberger

@Semiclassical Or $(e^D-I)$ for the forward difference

Semiclassical

14:57

Which at least formally gives $\Delta^{-1} = \frac{1}{1-e^{-D}}.$

Akiva Weinberger

@Secret Oh, yeah. $(1+x)^h=\sum\binom hnx^n$.

@Semiclassical Except that $\Delta^{-1}$ is only defined up to adding constants…

Semiclassical

(Pretty much everything in what I say should be predicated on "don't trust this rigorously")

SteamyRoot

@Semiclassical Oh, that paper

Semiclassical

lolyep

Akiva Weinberger

(Like $D^{-1}\stackrel?=\int$)

Semiclassical

14:58

Could not tell you tbh.

Secret

and by irrational, I mean transcendental, so I am not sure if $\binom \pi n$ makes sense

SteamyRoot

It's always tiring to see newspapers and "media officers" of related universities claiming "physicists have made major breakthrough towards Riemann Hypothesis" and such >.<

Secret

Indefinite sum

In mathematics the indefinite sum operator (also known as the antidifference operator), denoted by ∑ x {\displaystyle \sum _{x}\,} or Δ − 1 {\displaystyle \Delta ^{-1}\,} , is the linear operator, inverse of the forward difference operator Δ {\displaystyle \Delta \,} . It relates to the forward difference operator as the indefinite...

This operator $\Delta^{-1} = \frac{1}{1-e^{-D}}$ actually makes sense. It's called the finite backward antidifference

Akiva Weinberger

@Secret Of course it does. $\dfrac{\pi(\pi-1)\dotsb(\pi-n+1)}{n!}$

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$Mathematics$ Mathematics