Mathematics - 2016-12-23 (page 1 of 5)

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12:00 AM

uhh, I'm not sure, I've never had a topoly course, so I'm kinda in the deep end here :P

The most standard metric?

I just want to get a snippet of an intuition about whether or not this set in connected, not something necessarily rigorous.

Couldn't you write any matrix as a sum of each of its diagonal elements with all $0$s elsewhere? e.g. $$ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} = \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & b \end{pmatrix} $$

Meaning that it'd be disconnected in a sense

It's certainly a disjoint union

I'm not entirely sure though

Hmm, maybe you're right

Wait, why should the sum of some of the points in the space mean anything? Isn't that confusing the sum with the union?

Or am I missing something?

I think I've confused the sum with the union

¬¬

Great, then there is still hope for the connectedness! :D

12:05 AM

It's really late here! Expect me to talk some nonsense :-b

¬¬

Here as well, so likewise! Where are you in the world?

England, sadly

Wish I was somewhere hotter

How about you?

Denmark

Kinda similar timezones-ish :-)

12:07 AM

lol, "twink"

@kari yep!

although, I kinda enjoy the cold.. It's the darkness that gets to me.

Shortest day of the year was yesterday

They'll only get longer from here on in :-)

I know, so nice!

Well, I'll sleep on the connectedness thing, and perhaps post a question in the morning. Thanks for your input, and have a nice christmas!

Happy holidays :-)

I better get going as well

Peace!

2 hours later…

1:45 AM

Since I am trying to get the bounty room going, perhaps a reasonable thing could be to promote it here. So here is link to the relevant meta post and here is the room.

2:24 AM

@GFauxPas Exercise: Spot a major flaw in this proof

good evening

uh oh

what did I do wrong :(

oh

trust yourself

for sufficiently large $t$?

no, that isn't a major flaw

2:26 AM

greetings, @Zach.

And hi, @DHMO, @GFauxPas. (Warning: I may disappear momentarily.)

@TedShifrin and now it appears I have strep.

hi shifrin

Oh no MeowMix :(

Oy, Zach, these things happen when you're stressed and don't sleep :(

Get lots of rest and recover.

DHMO I dont see a problem

2:27 AM

@GFauxPas name me a polynomial of degree 1

I can name several.

$5t$

@GFauxPas another

Throw in a constant for good measure.

$5t + 1$

2:27 AM

@GFauxPas bingo

oh feh

fixable

physics answer: well obviously adding a constant isn't going to change the result

I have no idea what you guys are discussing, but that's just fine with me.

please, start from a degree 0

yes that's what I was about to say

if I start with degree $0$ its fine

@TedShifrin a proof involving polynomials assumed that the base case is n=1 when P(t) = Ct

where it should be P(t) = Ct+D

2:29 AM

Oh, no fair.

it's a proof by induction I wrote for all polynomials and DHMO doesn't like my base case because I forgot about the constant term

He wins.

@TedShifrin if you don't mind, i'm going to save the computations in exercise 25 for tomorrow. i just want to lay down right now.

but Dad the theorem is so obviously true

@Zach: Don't be ridiculous. Go rest. You aren't on any deadline with my exercises. I'm fine if you finish in 2019.

(Or never, I suppose.)

2:30 AM

alrighty then. (hopefully i'll finish them some time)

Is $2^\pi$ rational?

good night, chat.

Night, Zach. Merry Christmas and healthy recovery.

@DHMO: I never remember what Gelfond-Schneider tells us.

non-trivial algebraic a, irrational algebraic b, then a^b transcendental

it doesn't apply here

Clearly not. There are other theorems, but I never remember them.

2:33 AM

You could use this

I don't know if that helps

There's also five exponentials theorem

Plus it's only a conjecture. I don't see where you're headed.

that is a theorem

six exponentials also

the first one is a conjecture

@TedShifrin but point is: do we really need theorems for transcendentals to judge rationality?

@GFauxPas If I can prove that f(x) is order 0 but not any lower, what should I call the result?

look what you made me do DHMO, now I have to prove another theorem that I was going to need to do anyway

I've never found any context where an order lower than zero was important

even though the definition of exponential order allows it

@GFauxPas have you fixed the polynomial? and what theorem?

check the page

2:37 AM

Company here. Need to disappear. Bye.

NO Ted

we want your company here

aren't we your friends

@GFauxPas delete the first line of the proof

the zero polynomial doesnt have a degree

Hello everyone, I'd like to ask for some help regarding something. My analysis class went at the speed of light through things, integration took under a week (differentiation, more like 2). Second quarter we're starting with differential forms and Stokes's theorem, so I'll need to get a better grounding on things than I currently do.

or maybe you want it to be $-1$ or $-\infty$ or whatever but I dont want to do induction on it

2:40 AM

Anyone have recommendations for getting the material down reasonably quickly?

But also still actually understanding it?

I've been recommended Spivak Calculus on Manifolds for that

I hear Spivak's books are good

well there you g

But I've also heard very iffy things about it

Is $2^\pi$ rational?

Oh @GFauxPas

My professor told me that his professor told him that

Spivak's book on Calculus was a masterpiece. Spivak's book on the Calculus of manifolds is not.

¯_(ツ)_/¯

I started going through it and got distracted

2:42 AM

@GFauxPas If I can prove that f(x) is order 0 but not any lower, what should I call the result?

but my other professor said I should read it over and over

I cant even wrap my head around exponential order negative right now DHMO

I'm not asking you to prove

I'm just asking for a nomenclature

um

what's the $f$

constant mapping

Hello chat

2:44 AM

I'm not sure

at least zero?

are you sure lol

i may apply the nomenclature broadly

I'm not sure the concept is useful at all

of a negative order

that I would bother to prove thigns with it

but I dont know, it's late and Im tired

thank you for sprucing up my proofs DHMO I appreciate it

proofwiki.org/wiki/…

Spot the oxymoron on the first line of the theorem

oh, well,

what I mean is that it's an exponential function from $\mathbb R$ to $\mathbb C$

but good call

You'd say it's good for that?

Sorry for the delay, internet died

2:52 AM

DHMO a quick google search shows that negative exponential order is a thing

ok

in one source

the book about transforms I'm using, Vretblad, doesn't even name the concept

just uses it without calling it anything

I'm not sure what to call the thing you're asking about DHMO, maybe sleep will have me think of somethign

ok

thanks for your help with my proofs I appreciate it

wait no now the polynomial proof is broken

congratulations

3:00 AM

the induction step is calling on a result that is somewhere else aaaaaaa

i'll fix it tomorrow hopefully

{{WIP}}

@GFauxPas where is the problem?

$P_{n+1} = P_n P_1$

I didn't prove $P_1 \in \mathcal{E}_\epsilon$

oh

i can fix that for you

go ahead, I'm going to bed

are you a student like me DHMO or are you a full-grown scholar

i'm just grade 12

3:03 AM

nice

@GFauxPas done

good job

still have to write up how exponential order a implies exponential order b for b > a

good night

lol DHMO

you created a loop

where?

oh, god

XD

i'll fix it, just go to sleep

3:11 AM

just use the rational power one

proofwiki.org/wiki/…

no, that's a loop also

wait

i'll create one for natural number

lol

okay good night my online young mathematician friend

bye

1 hour later…

4:36 AM

Please someone take a look at this

5:15 AM

Hello

6:03 AM

-1

Q: Find value of $k$ for a function to be continous

The given function and the domain of $g(x)$ is $(0 ,\pi/2)$. where $[\ \ ]$ denotes the greatest integer function. Find the value of $k$, if possible, so that $g(x)$ is continuous at $x =\pi/4$. I thought about it alot , but could not get any start . Please help me in this

Could anyone help me in this

e^ax eventually grows faster than x^r for any a>0 and r>1

exercise: prove ^

also: x! eventually grows faster than e^ax for any a>0

x! is referring to the shifted gamma function

6:19 AM

$\Pi(x)$, lol

@Brody is there a name?

@DHMO (extended) factorial, pi function [kinda ambiguous], shifted gamma. all the same

I was just being cheeky

I'm genuinely looking for a name

@Brody how would you prove that it grows faster than e^ax?

@DHMO I don't think there's a universal/standard name, but those mentioned earlier should be easily understood in context

Don't know how to prove that

any comparison method?

6:23 AM

Perhaps use calculus on the integral (definition)?

Pi(x) = int_0^\infty t^x e^{-t} \ \mathrm dt right

think so, but I'm afraid to make a hasty conclusion :/

Pi'(x) = int_0^\infty t^x \ln t e^{-t} \ \mathrm dt right

why?

Ke^ax = int_0^\infty ??? \ \mathrm dt

@Brody no idea

6:27 AM

The integral definition is right. I forgot how to differentiate >_>

how the hell am I supposed to prove that

@Brody do you know any other famous functions that grow faster than e^x?

@DHMO famous? not really

I only know Pi(x)

pi?

$\Gamma(x)$ equivalently

and there's the ambiguity

6:30 AM

?

$\Pi(z)=\Gamma(z+1)$, the LHS notation is not very clear

i know a famous function that grows faster then e^x

@TheGreatDuck ?

$e^{x^2}$

not famous enough

6:32 AM

its integral has no closed form

...

the only thing famous about this thing is the Gaussian Integral

you're just looking for a function that grows faster right?

how does $x^x$ compare? not that it's really famous...

@Brody indeed it is

@TheGreatDuck sure

@Brody since x^x = e^(x \ln x)

surely e^{2x} grows faster then e^x?

6:33 AM

and \ln x is unbounded above

:p

@Brody how would you name "x^x"?

what sort of growth are you looking for?

@TheGreatDuck by "faster than e^x" I mean "faster than Ke^ax" for any a>0 and K>0

oooh

are you talking about laplace transforms?

6:34 AM

yes

ah

alright

@DHMO I don't know any conventional terminology for that. You can borrow some (not very popular) words from tetration stuff

pretty much any nonlinear function with a derivative greater than a constant in e^{____} cannot work for laplace transforms

@Brody lol never mind

@TheGreatDuck thanks

supersquare :P

6:35 AM

...

how about superpower?

(for x^x)

Sad. They have super-roots and super-logarithms, but not super-powers.

lol

actually, they do. still far from well used

apparently this question is unclear. anyone have any clue what the unclear part is? I don't see it. :p math.stackexchange.com/questions/2045828/…

Ah, how's the goof-off convention? :) Happy holidays to you all.

6:40 AM

Hello @TedShifrin. How is your knight going?

thanks

happy holidays to you too

My knave ran away with the swords, so no knights.

Ah, DogAteMy has awakened from his cross-oceanic trek.

Akiva Weinberger

Still feel like it's 1am rather than 8, but yep, awake

who is dogatemy?

^^^^

Akiva Weinberger

Or, rather, 2 rather than 9.

I think I just demonstated my point.

6:42 AM

Close, anyhow, DogAteMy.

You remonstrated your point.

i dont understand.

Akiva Weinberger

@TheGreatDuck 'Tis I

raises eyebrow

Only one?

you're akiva.

Akiva Weinberger

6:43 AM

Yes.

@TedShifrin yes. Don't you do that?

Akiva Weinberger

@TedShifrin Of course

and you're also dogatemy?

Akiva Weinberger

According to Ted at least

okey dokey.

6:44 AM

0

Q: Find three types of function .

If be a continuous function at $x = 1$, we have to find the value of $4g (1) + 2 f (1) – h (1)$. Assuming that $f (x)$ and $h (x)$ are continuous at $x = 1$.� I try alot but not getting any start . Could anybody help me in this.

limits continuity

Can anybody help me in this

@TedShifrin i would imagine you are good a phrasing things clearly. Do you see what's confusing in this post? math.stackexchange.com/questions/2045828/… I don't really see it. I guess I'm just blind to it. Y'know? It's hard to tell when you already know what you're asking.

Now, does @TedShifrin have many eyebrows?

@TheGreatDuck: I roll numbers of eyes (up to 9, I think), but I don't raise an eyebrow.

...

Akiva Weinberger

He means exactly what he says he means.

6:45 AM

are those the eyes in the back of your head for watching misbehaving students? (just kidding)

@AkivaWeinberger I didn't need clarification, thank you.

Akiva Weinberger

Sure

But if someone hands you $C(x)g(x)$ how do you recognize which part is $C(x)$? That's totally impossible. BTW, I don't think the abstract algebra tag is remotely plausible.

Ted must have a full 360$^\circ$ vision

or rather, a full $4\pi$ steradians

For example, @TheGreatDuck, I write down $\dfrac{x^2}{\sqrt{1-x^3}}$. What is $C(x)$?

Yes, @Brody, Ted obviously sees in solid angle :P

(including through his own neck, apparently)

I finally (inadvertently) got three (secret) hats. But they're not showing up yet.

@TedShifrin read the question fully. C(x) is a variable I am solving for. It's a question involving solving for unknown functions. f and C are literally the unknowns.

so in that case, C(x) has the solution set of all functions

cause it's not there. XD

6:50 AM

I understand that, @TheGreatDuck. But if I gave you that function and told you it was $C(x)g(x)$, unless I know which part is $g(x)$, I can't possibly tell you.

You have a penchant for posing ill-posed questions.

no you miss the point

@TedShifrin How to prove that $\forall K,a>0: \exists M: \forall t > M: \Gamma (t) > K e^{at}$?

Well, elucidate.

C(x) would never be written as anything other than C(x)

as I'm solving for C(x)

@DHMO, $\Gamma(t)$ is the actual Gamma function?

6:51 AM

yes

@TedShifrin for instance

f(x) = C(x)cos(x) + u(x - 3) where f is continuous and C is piecewise constant

find the set of all C that work.

@TedShifrin you could also prove the (weaker) alternative that involves switching the second and third quantifier

What is $u$, @TheGreatDuck?

the particular question I'm asking is how to pull C(x) to the side so that it's being added to a function rather than multiplied by a function.

@TedShifrin Heavyside

From your original question, I would never have understand anything remotely like this. So

6:54 AM

"f and C are "variables" that I am attempting to solve for."

there's no universal way to approach this. Right, you're basing it on the jump discontinuity of $u$ and needing to pick $C$ appropriately.

I'm not asking how to solve it

I'm just asking how to rip C off to the side

I don't know why you're so obsessed with these kinds of questions, but clearly you are.

after that I already know how to solve it.

You cannot rip $C$ off to the side.

It's all about understanding the discontinuities. It's not about equalities of functions or modified functions.

Anyhow ... back to @DHMO.

6:56 AM

-_- I mean take a function on both sides that results in something where C is being added to something else rather than multiplied. It's just an algebra issue really. I just don't know of a function that has such identities.

@DHMO: Do you understand why it should be true? Do you understand the growth of $n!$?

I believe you're wrong, @TheGreatDuck, but I know better than to try to fight.

@ted yes i understand

So if you know Stirling's formula for $n!$, you see why it's true.

@TedShifrin I haven't even claimed anything so I don't see how I can be wrong by merely saying what my question asks.

Akiva Weinberger

@DHMO Well $Ke^{at}=e^{\ln K}e^{at}<e^{t\ln K+at}=e^{(\ln K+a)t}$ for $t>1$, so you only need to prove it for $e^{at}$ rather than $Ke^{at}$

6:58 AM

@ted do i really need to invoke Stirling?

@AkivaWeinberger nice perspective

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