Hello, need your help, What is the inverse of this function if $\Phi(t)=|t|^p$ ? $\int_{0}^{t}\frac{\Phi^{-1}(\tau)}{\tau^{1+\frac{1}{N}}}d\tau.$

What is $\Phi^{-1}(1)$? It seems to me that $t \mapsto |t|^p$ is not injective, and hence does not have a well defined inverse.

As such, that integral doesn't even make sense to me...

why ?

and for $\Phi(t)=\frac1p |t|^p$

I just explained why I am confused---the integrand doesn't even appear to be well-defined!

What is, for example, $\Phi^{-1}(1)$?

i want before find $\Phi{-1}$

Is that supposed to be $\Phi^{-1}$ as the inverse of $\Phi$, or as the reciprocal?

what is the inverse of $|t|^p$ ?, it is $|t|^{-p}$ no ? with $t\neq0$

yes @Semiclassical

...

headdesk

$\Phi^{-1}(t) \ne \frac{1}{\Phi(t)}$ in normal everyday notation...

@XanderHenderson youtube.com/watch?v=yqm5Dxl80B8

heh

just the first thirty seconds of that, really

So you want to find $$ \int_{0}^{t} \frac{1}{|\tau|^p \tau^{1+\frac{1}{N}}} \, \mathrm{d} \tau?$$

That's the goal? Evaluate the integral?

This reminds me of when a friend of mine was asked to prove $\sin^2(x) + \cos^2(x) = 1$, like I think it was for a high school diagnostic precalc test or something

But she was used to these powers denoting function composition instead of squaring and was like uh... are you sure about that?

heh

lol notation

$\sin^2(x) = \sin\circ\sin(x)$ forever and always

@XanderHenderson yes

I refuse to let physicists dictate to me that it should be $\sin(x)^2$

THAT IS JUST STOOPEED!

@Vrouvrou Do you know the fundamental theorem of calculus?

Also wait you're the second person I've seen who puts a comma before the $d\text{whatever}$

i like to do \int f(x)\,dx: $\int f(x)\,dx$

Just yesterday was the first time I saw this and I thought it was typo

That is $\int\,\mathrm{d}(whatever)$, you uncultured barbarian

just to make it look a bit nicer

ugh

@XanderHenderson Wait which physicists say that? I pretend to be one myself, and I (and my friends) always use $\sin^2x$

Yes, @Mr.Xcoder, you are part of the problem

I'm way too lazy to do \mathrm{d} all the time

@XanderHenderson Wat?

$f^n$ is the $n$-fold composition of $f$ with itself

not the $n$-fold product

If $\sin^2 x=(\sin x)^2$ is wrong then I don't want to be right

@Xander lol I just use $dx$, it's just that I didn't want to write $dwhatever$

$dx$ is wrong! $\mathrm{d}x$ is right!

ugh

too much effort

heh

I wouldn't write $\ln^2 x=(\ln x)^2$ though

cringes

there I would understand it as $\ln\ln x$

noöne should ever write that

$\ln^2(x) = \log(\log(x))$

i'd probably avoid it regardless

because we are grown ups, and $\ln$ is silly

lol

also, I don't speak French

we're not computer scientists and we're not engineers, so logs are base-e

Hi chat

in 'Merikan, we say "natural logarithm", so it should really be $\operatorname{nl}$

'MERIKA!

Btw was $\ln$ created just because someone was lazy to type $\log_e$ or just $\log$ all over again? :P

@Xander lol I usually just write $\log$ for base e

YES!

well, historically I think $\log_{10}$ really was the starting point

$\log(x) = \log_\mathrm{e}(x)$

@Semiclassical Yarp. Napier's bones and all that.

because the point of log tables was to facilitate decimal computations

Though sometimes in discrete stuff you care more about log base 2

same with slide rules

90% of the times the basis is irrelevant

e is basically a continuous 2

$\frac{\log(2)}{\log(3)}$ shows up in my work a lot

$\ln^{\omega}(x)$

@Daminark yeah, hence why I mentioned computer engineers

Ah tru

@XanderHenderson right, fractal dimension stuff

"fractal" dimension; heh :)

Oh that's like, Hausdorff dimension of the Cantor set or something, right?

pft! Hausdorff dimension is for plebes. :)

@Daminark yeah but they usually care about stuff being $O(\log n)$ which doesn't depend on the base

Tbh spaces with non-integer Hausdorff dimension should be banned waits for storm

(though that happens to be the Hausdorff dimension of the usual ternary Cantor set)

You also have base-2 stuff implicitly when you talk about half-life and such

@Alessandro yeah for algo purposes that's def true

But in practice you'd never bother to talk about $\log_2 x$

I do like the notation $\log_2x=\lg x$

new rules: $\log = \log_{\mathrm{e}}$, $\operatorname{lg} =\log_{10}$, $\operatorname{lame} = \log_{2}$

just to annoy everyone

lol

oh, and $\ln$ for natural logs, like this one: kmart.com.au/wcsstore/Kmart/images/ncatalog/tf/4/…

:eyes:

(I wish we had emojis like Discord)

we don't need them

nah. log_10 = common log, so $\text{clog}\equiv \log_{10}$

YES!

But thonk tho

$$\ln^{\omega}(x) = \ln\ln\ln\ln\ln\ln\ln\ln\ln\ln\cdots (x)$$

$\text{wog }x$

But that is a constant!

um... no.. wait... it is either a constant, or $-\infty$...

no... not that, either

nevermind

I don't want to deal with the interval $(0,1]$

but for $x > 1$, $\log$ is contractive, no? so there is a fixed point

or maybe not...

ugh... I need to go make more coffee

Well I mean $\log^2(x)$ is already bounded above by 5

ah... no! I'm an idiot. $\log^n(x)$ is eventually negative, at which point the wheels fall off

$\log^{\omega}$ is bullshit

and I just spent way to much time thinking about that (and thinking wrongly, too) :(

MOAR COFFEE

@Daminark really

let me check this

I don't believe it

ln(ln(10000000000000000000000000000000000000000000000000000000000000000000000000‌​000000000000000000000000)) = 5.4

(and both functions are increasing and tend to infinity, so)

That number doesn't actually exist though

oh norman

Does anyone want to think about a boring question?

no, I only want to think about nonboring questions

i just need x so that log(logx)>5, so x>exp(exp(5)), so why is it bounded by 5? @Daminark

oops

logx.logx

@Shobhit it isn't bounded by 5

he's trolling

oh ok

I wouldn't call it "trolling" so much as "being disingenuous with the intention of evoking laughter"

maybe I am old, but trolling requires some intention to distress or anger the target of the trolling

the chat's trash today

UR TRASH TODAY!

cries

i cant hear you over the sound of my REEEEE

Xander has been slain

@XanderHenderson has been pretty emotional lately

This is a safe space Xand, we're here for you

Safe spaces are the worst. X<

ugh... I managed to type up an entire lemma yesterday which wasn't utter crap

GO ME!

@XanderHenderson In content or in how it was actually typed up?

@XanderHenderson I typed up a lemma too

it's 2:00 already and I had hoped to have written like 5 pages already

3 guesses how much I've done

@0celo7 A lemma?

the lemma was yesterday

1.5 pages? @0celo7

it rhymes with "hero"

The inverse of the mass of a neutrino? Thats a lot of pages!

measuring page length in inverse kilos

hmm

actually in geometric units that might make sense

My day has been similarly unproductive @0celo7

debating whether I should prove the solvability condition for the Poisson on a closed manifold in my thesis

@TobiasKildetoft The lemma is something that I proved a year or two ago, but it needed to be written up nicely; the proof that I wrote a year or two ago was nearly incomprehensible. :\

In total content, it amounted to about six lines.

YAY ME!

Well, that is progress

I also wrote a footnote

Technically, two footnotes. (1) As a historical note, the notion that is now commonly referred to as the Assouad dimension was originally introduced by Bouligand \cite{bouligand1928}, but received little attention at the time. It was reintroduced much later by Assouad \cite{assouad1979} in his 1979 thesis, where it was used to study certain embedding problems.

and (2): The Minkowski dimension was also introduced by Bouligand \cite{bouligand1928}.

Poor Bouligand. :(

the problem is that you need to be a bit careful with the lagrange multipliers

Hmm, not sure I have yet to include a footnote in any of my papers

And now I need to write up this generalization of the Harvey-Polking Estimate

But the proof is six handwritten pages. cries

on the bright side, I have nominally started drafting my thesis, so it has been a good week overall

@XanderHenderson did I tell you I got my tex working

You did. But I still think that paying for TeX is gross

I paid nothing...

Why would you ever have to pay for TeX?

@0celo7 You paid in your dignity. Proprietary TeX packages are gross.

it looks so much better than computer modern though

But there are free (as in speech) alternatives to computer modern

even if I don't understand your beef with it

I never had any issues with computer modern either. But I really care very little about fonts at all

I don't really have a beef with it

I just think this font is a better fit for my thesis

computer modern is too...algebraic

@0celo7 Ahh, you need something closer to comic sans?

ah shit I know I have this proof on a paper somewhere

@TobiasKildetoft funny

@TobiasKildetoft do you know Deligne cohomology ?

@Adeek not by that name at least

@TobiasKildetoft not comic sans i.gyazo.com/f240fbe83b03770d1fdc0c2ffa88f92c.png

Ugh... you are one of those mathematicians that likes to use phrases like "it is clear"

UGH!

@XanderHenderson I'm actually rewriting the proof right now

it made way more sense when I wrote it the first time

it was clear then...

oh, the first sentence is indeed clear

I am not making any judgement at all about whether or not the proof is clear; I am complaining about the phrase "it is clear".

Phrases and words like "it is clear", "trivially...", and "it is obvious" set my teeth on edge

@XanderHenderson "By standard arguments we may work in diagonal basis for A"

better

@XanderHenderson I find that phrase fine, as long as it really is clear to anyone in the intended readership

@0celo7 I like that phrasing!

"without loss of generality we may assume A is diagonal"

it communicates to me that I should not be looking for deep reasons why the statement holds

@TobiasKildetoft It may be a matter of taste, but "it is clear" is contentless

matrix norms do not depend on the basis chosen

@XanderHenderson As I just said, it really is not

this is clear to people

(well, good matrix norms)

@TobiasKildetoft If everyone who ever used the phrase used it exactly as you claim that do, perhaps not; but it often means "I'm too lazy to provide a proof."

or "My intuition tells me that this is true, but I haven't bothered to check"

I don't want to write the proof, sure

it would involve writing down what a change of basis means and then doing computations involving the definitions

@XanderHenderson In those cases it is bad, sure. But when I read stuff, I usually see it used properly (did a quick check through the last paper I refereed, and indeed, every instance of "clear" was in the good usage)

but it really should be clear because the norms are defined on the level of linear transformations, not their matrix representations

And I don't see how asserting that "it is clear that X" is any more indicative of "there is nothing deep going on" that "X holds"

so to me it is clear that one can compute them in any representation and get the same result

@0celo7 Again, "By standard arguments..." is better phrasing.

I also think that "It is clear" is rather condescending.

Obviously, it is a matter of taste, but those kind of phrases strike me as sloppy and in poor style

if I remove "it is clear" are you happy

not changing anything else?

@XanderHenderson I find it essentially the opposite. The authors assume that you are familiar enough with the prerequisite theory to not need it spelled out

@0celo7 Indeed.

@TobiasKildetoft Then why bother with stating that it is clear? Why not just state the result?

@XanderHenderson just stating the result leaves it open for interpretation why the author thinks it's true

@XanderHenderson As I said, I know what sort of ease I should have in proving the result when it is stated like that

So does "it is clear"

maybe it follows from something before, maybe not, who knows

@TobiasKildetoft do you agree that my claim is "clear"

Either a result really is clear, in which case there is no need to add additional fluff; or it is not clear, and should be explained.

@0celo7 I didn't look at the claim. I was too distracted by the font :)

@TobiasKildetoft good or bad?

The fact that the lemme is higher than a kite is kind of distracting, too.

@XanderHenderson wot

oh, 4.20

It is clear what I meant. :P

that chapter is like 10% done

it'll be 4.69 in a bit

hur hur

@0celo7 Well, at least you are numbering inside the sections

I just refereed a paper that just had consecutive numbering

and here I was, worried that definition 2.18 was too many things for one chapter

@XanderHenderson the chapter currently has 51 items

sadly there's no natural way to break it into two chapters

example 4.7 is five pages long

with three figures

The final result of that paper is Lemma 65

and that is single spaced; it is eventually going to have to be doublespaced

yuck

(yes, the last result is a lemma)

@Tobias: I decided in my books to number everything within sections, using C.S.x for theorems, etc.

@XanderHenderson I will be mad if they make me change my formatting on this.

I saw someone submit a dissertation in 10 point font with small margins. I was amazed he got that through

for some reasno my school insists on double space with like 5 inch margins

looks like a single column

I am required to number (Chapter).x

Oh, I take it back, Tobias. I just use S.x ... and refer in words to the chapter if I'm in another chapter.

@TedShifrin But do you at least state the number next to where the result appears?

But I am numbering my equations (Chapter).(Section).x

@TedShifrin I have a reference to Stein section VIII.3.1.2

That is one of the things I dislike in the way Humphreys does things. He has no numbers next to the statement itself, relying on which (sub)section it appears in

very precise numbering

I try never to number equations. It gets out of hand. If I have to refer to the occasional equation, I do notate it somehow.

@Tobias I don't understand your question?

@TedShifrin My advisor loves numbered equations.

I think most mathematicians totally overdo that, and it makes things very hard to read.

Honestly, most mathematicians' writing styles suck.

Physicists number every equation

@TedShifrin In Humphreys' books, Theorem 1.1 will be the theorem in section 1.1. But when stating the theorem, it will just say "Theorem" rather than "Theorem 1.1"

I have been accosted by physicists for not numbering

@TobiasKildetoft I really like this style

I'm not good enough at formatting the sections to get it to work nicely though

Oh, no, I number the theorems, lemmas, propositions consecutively within sections, @Tobias.

@0celo7 Have you had to look up results in books using that style?

it is a lot of extra work

Anyhow, it's lunchtime, so I'm disappearing.

I think I only have one book with that style

There is Scott's pseudodiff book that has no theorems at all

just sections and you kind of have to figure out what's going on

The book we use for algebra has chapters, sections and subsections. The various results/examples are numbered consecutively by section. So 2.10.2 is both a subsection and a definition for example

this tends to cause some confusion when looking stuff up

yeah I get that

@TobiasKildetoft the worst is when the theorems, lemmas, and propositions are not numbered consecutively

like Prop 4.3 follows Theorem 4.5

Suppose I have two polynomials $P_1$ and $P_2$. For the first $n$ degree they are equal. What can I say about $P_1^k$ and $P_2^k$ for $k \leq n$

@0celo7 Yeah, at least that has been done properly in this book

hi, if k(x,y) is a piecewise function with 1 if x>=y and 0 otherwise, what is x-1/2+k(x,y)?

@quallenjäger You mean how high a degree they will agree up to?

For $P^k_1$ and $P^k_2$ yes

one thing to do is k_(x,y)=0.5 if x>=y and -0.5 otherwise, then x-1/2+k(x,y)=x+k_(x,y)

@quallenjäger Same one. That is still the highest one not involving anything higher

writing random parts of the thesis out of order seems to be working disastrously

but the first chapter is the hardest because I have to commit to stuff

And vice versa: Suppose I know $P_1^k=P_2^k$, can I conclude that they are at least equal in the first $n$ degree.

@quallenjäger you mean if the $k$'th powers agree up to that degree?

Yes

then no, because you can't even conclude that from the $k$'th powers being equal

Thoughts on this as motivation for Hahn-Banach:

I see Thanks

and $p>1$ for some reason

that makes me chuckle

Also the 'linear functional b' should be a G in the last paragraph

@TobiasKildetoft Cant I even conclude for the lowest order?

i.e. for the $k$-th order.

I mean, $p>1$ amounts to $q>0$ under that assumption

Fixed

@quallenjäger if $x^2 = y^2$ you can't usually conclude that $x=y$

$1/p + 1/q = 1$ implies if $p > 1$ that $1/q = 1 - 1/p$ is like $1/2$ or some ****

I see,

They will differ from the sign

only?

$1/q = 1-1/p=(p-1)/p>0$ if $p>1$

@quallenjäger for polynomials over a field, yes.

@TobiasKildetoft You are from Aarhus?

So once you assume both $p,q$ are positive then at least one of them has to exceed $1$

@TobiasKildetoft Do you know Mark Podolskij from Aarhus?

Yeah, for whatever reason heh

@quallenjäger Name doesn't ring a bell

But the end is like a kind of undeniable motivation for caring about Hahn-Banach before either even thought about it if you believe that Riesz theorem it seems

Ah ok sorry.

If you were living in 1912 that is

@quallenjäger Ahh, just looked him up. I have seen him around (he is actually in the office next to mine currently while renovations are going on where I am usually located)

He seems to be in statistics (which fits with the office location), which explains why I have no iea who he is by name

Haha he was my tutor in Germany

I see

cool

Maybe it's only $p > 1$ because it came from the moment problem originally which is like a $\int x^n f(x) dx = c_n$ kind of thing

Oh wait, $L^p$ spaces always have $p > 1$ for conjugates/duals or something

I think they are not locally convex or something for $p<1$

But anyway Hahn-Banach only needs a vector space

@AlessandroCodenotti wait what?

@bolbteppa $L^p$ is reflexive for $p\in(1,\infty)$

@Daminark what's wrong?

I didn't realize this was a thing on any vector space, i expected that you needed local convexity

if $P_n$ a polynomial of degree n such that n is impair does it mean it has at least a root since the only irreducible polynomials are of degree 2; is that true? or is there a counterexample!

Nope, it's really powerful

Of course you get nicer consequences (separation theorems for example) if your space is normed and everything

Just watch in a few years we're gonna have "Given an algebraic structure we can extend it to a bounded linear functional on lp" or something

Like, first it was in a Banach space, then normed, then locally convex, now literally anything

Maybe the $p > 1$ can be seen from that Holder inequality above, the sum would probably go nuts and you wont be able to treat $g$ or $G$ as duals and ruin reflexivity, hmm, anyway, main point is it seems Hahn-Banach is not so random

Might be worth noting that 1/p+1/q=1 is equivalent to (p-1)(q-1)=1

Lies! What kind of dark voodoo magic is that?!

Wait wait whaaaa

$pq + 1 - p - q = 1$

That just works

Yep

heh

TIL

also $q = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}p$

Are you sure though? Because this formula works for $p = 8$ and $q = \infty$ but the above doesn't

hrm... you have a point

Eh. I’m fine with it being valid for finite values

@Semi nah I'm talking about Xander's thing

okay, for all $p \ne 8$, we have $q = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}p$

does that work?

Ohhh

Alright I'm happy

Lol

then $8$ is a special case: $\infty = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} 8$

I guess the R.H.S. of that inequality is a semi-norm because theoretically the $f_i(e_j)$ stuff could be zero even when $c \neq 0$

Hello chat

so, if I define $$f(z) := \int_{E} g(x)^z\,\mathrm{d}\mu(x),$$ where $g:E\to\mathbb{R}$ is nicely bounded above and below (i.e. there are constants $0 < C_1, C_2 < \infty$ such that $C_1 < g(x) < C_2$ for all $x\in E$) and $\mu(E) < \infty$, why is it obvious that $f$ is entire?

no takers? bah... I'll have to check Conway when I get into the office tomorrow...

I guess it's not that obvious

Hi yall

indeed

can someone clarify soemthing to me about vector spaces?

if we take the 3-tuple (x,y,z) s.t x,y,z are integers

this is not a sub space of R^3, and the reason is not closed under scalar multiplication

But if we had our field to be something like Z/pZ

where p is prime, wont it be a subspace?

seems like it would be the entire space, no?

if we let F to be R or C i get that 1/2 for example will cause the closure of fail

ehm does not matter if it is all or not

how does they define the field F

should be true for all FIELDS?

Wait what are you saying?

I clearly don't understand the question...

Like there are a few things you may be asking right now

I'm not sure which

well i noticed i was not very clear

okay let me ask in a good way

so you see the 3 tuples i have right?

Are you asking whether Z/pZ^3 is a subspace of R^3?

no no

if $p$ is some prime number, then $\{ (x,y,z) : x,y,z \in \mathbb{Z}/p\mathbb{Z} \} = (\mathbb{Z}/p\mathbb{Z})^3$

the 3 tuple, x,y,z, such that we only consider intergers as entries