Mathematics

12:23 AM

Is there a trick to checking whether or not I can isolate $y$ from an equation? For example, I have something like
$$y + 2\ln |y+1| = x + 5\ln |x-3| + C$$ Aside from spending 30 mins playing around with my algebra, is there anything I can look for?

Jacksoja

12:35 AM

@AkivaWeinberger Hello Akiva

@AkivaWeinberger Did you read Rudin book ? analysis?

Semiclassical

@CookieToast I wouldn’t hope for an explicit solution for $y(x)$ in that case, no

It’s not always obvious though

Cookie Toast

@Semiclassical is there a general method of telling that a function has no closed form solution without having to mess around with algebra for a while?

Semiclassical

General method, probably not

But if you’ve got both powers and exponentially of y, that doesn’t bode well

For instance, there’s no elementary way to solve $x=ye^y$ for $y$

Prototank

so can we just get wild embeddings of any surface

Balarka Sen

yup

Prototank

12:49 AM

by cutting the horns off of alexander's horned sphere, cutting out a disk from the desired surface, and gluing the horn-cap in?

Balarka Sen

take connected sum with alexander horned sphere

Cookie Toast

Gotcha @Semiclassical

Prototank

sweet

Cookie Toast

Lol the guy across the street from my house has gotten locked out of his car with the car horn stuck

Jacksoja

Sweet

@CookieToast Will this be on the Youtubes?

Semiclassical

12:57 AM

meant to say *exponentials not exponentially above

What's going on here right now is snow and a good load of it

Jacksoja

How does one operates with Big O notation

it is very comfusing soemtimes

Cookie Toast

@Jacksoja I would if I didn't have so much homework. Our first exam is in a week and a half, and our prof hasn't even started lecturing :O

Semiclassical

it's up to about 9 inches now

Jacksoja

@CookieToast what subject?

Cookie Toast

@Semiclassical do you live somewhere that snows a lot, or is this a new thing?
@Jacksoja Diff. Eq. :P

Semiclassical

1:00 AM

Minnesota

Jacksoja

@CookieToast no way your prof will run thru everything in 1 week

Semiclassical

so it's not unusual to have a storm like this in the winter, but it's definitely not what we get on a day-to-day basis.

Cookie Toast

Ik, I don't know what he's thinking @Jacksoja. Luckily I got ahead over winter break.

@Semiclassical My roommate is from MN and she says it can get pretty hairy there

Semiclassical

yeah

it's not cold right now at least

by which I mean it's 30 F outside

so it's freezing, but it's not too cold :P

Cookie Toast

1:16 AM

An irrational number is a number not expressible by a ratio of integers, which means that its decimal expansion goes on forever, and never repeats in any sort of pattern. Does that mean that it is impossible to theoretically come up with a function $f(n)$ that will return the $n$th digit of an irrational number?

Semiclassical

Bailey–Borwein–Plouffe formula

The Bailey–Borwein–Plouffe formula (BBP formula) is a spigot algorithm for computing the nth binary digit of pi (symbol: π) using base 16 math. The formula can directly calculate the value of any given digit of π without calculating the preceding digits. The BBP is a summation-style formula that was discovered in 1995 by Simon Plouffe and was named after the authors of the paper in which the formula was published, David H. Bailey, Peter Borwein, and Simon Plouffe. Before that paper, it had been published by Plouffe on his own site. The formula is π = ...

Cookie Toast

For example, $f_{\pi}(1) = 1, f_{\pi}(2) = 4, f_{\pi}(n) = digit(\pi, n)$

Semiclassical

So at least in binary, there actually is an algorithm

Cookie Toast

Does this concept extend potentially to all irrationals?

Semiclassical

No clue. I'd guess not.

Cookie Toast

1:18 AM

Also, what's the difference between an algorithm and a function?

Semiclassical

the number of manipulations involved, really

or, well

Cookie Toast

I ask because I'm curious about the function $f:\mathbb{R} \to \mathbb{R}$ where $f(x)$ returns the average of the digits of $x$. I was thinking for irrationals you could just take the limit of the Bailey-Borwein-Plouffe-esque formula for each irrational as $n \to \infty$.

Semiclassical

the function is the mapping (in this case, from n to the nth digit)

the algorithm is the process which cranks out this mapping.

Cookie Toast

Ah, so an algorithm is basically a function involving "steps".

And even if not, $f:\mathbb{Q} \to \mathbb{Q}$ would be interesting

Semiclassical

There's a remark to the following effect: "D. J. Broadhurst provides a generalization of the BBP algorithm that may be used to compute a number of other constants...Explicit results are given for Catalan's constant, $\pi ^{3}$, $ \log ^{3}2$, Apéry's constant $\zeta (3)$ (where $\zeta (x)$ is the Riemann zeta function), $\pi ^{4}$, $\log ^{4}2$, $\log ^{5}2$, $\zeta (5)$, and various products of powers of $\pi$ and $\log 2$."

So there's some crazy stuff out there

That said, there's a big difference between pi and an arbitrary irrational number.

We know a looot of formulas for pi.

Xander Henderson

1:25 AM

$\pi$ is stupid, $\tau$ is life!

Cookie Toast

I prefer to eat at least $3$ pi, thank you very much :P

Xander Henderson

I am confused... what do you mean by "average number of digits" of an irrational number?

Cookie Toast

I'm not completely sure, but I was thinking that if a function exists to calculate the nth digit of an irrational, then maybe you could take the $\lim_{n \to \infty} \frac{f(n)}{n}$

Xander Henderson

That will go to zero

since $0 \le f(n) \le 9$

Cookie Toast

@xander I by no means have a well thought out cohesive definition here :P

Yeah, whoops. It should be something like $\lim_{n \to \infty} \frac{\sum_{k=0}^{n}f(k)}{n}$

Xander Henderson

1:31 AM

If $a_n$ represents the $n$-th digit of the decimal expansion of a number, it might be meaningful to ask about $$\lim_{n\to \infty} \frac{1}{n} \sum_{j=1}^{n} a_j, $$ but I suspect that this would be $\frac{1}{2}$ for almost every real number

Maneesh Narayanan

Please help me. chat.stackexchange.com/transcript/message/42422196#42422196

Xander Henderson

where I mean "almost every" in the sense that the set of numbers such that the above limit is not $\frac{1}{2}$ is of Lebesgue measure zero

Cookie Toast

Would you define lebesgue measure for me real quick?

Daminark

The set of natural numbers which aren't even primes is Lebesgue null, now that I think about it

Cookie Toast

Though I still understand your overall meaning

Xander Henderson

1:32 AM

and by $\frac{1}{2}$, I actually mean $4.5$

Daminark

:thinking:

Xander Henderson

or no

that's not what I mean

what do I mean?

yes! 4.5...

@CookieToast Define Lebesgue measure real quick... uh... okay

um...

Cookie Toast

:P

Xander Henderson

mathworld.wolfram.com/LebesgueMeasure.html

Cookie Toast

Just wave your hands, make some vague references, say "lebesgue" a few times, and say "let that sink in for a few hours"

Xander Henderson

1:34 AM

basically, you want to be able to measure the "sizes" of arbitrary sets of numbers (or whatever)

and you want your tool for measuring these sets to give the "expected" values to easily understood sets

Maneesh Narayanan

am I asking bad question? please tell atleast that. none is responding to my question :(

Daminark

>arbitrary

Xander Henderson

for example, the "measure" of an interval $[a,b]$ should be $b-a$

@Daminark I'm getting to that!

Daminark

>tfw no AoC

Lolol yeah fair I'm just messing :P

Xander Henderson

It turns out that you can't measure completely arbitrary sets in this way, but you can come close-ish

The tool that does the job is called Lebesgue measure

Daminark

1:36 AM

It's a pastime of mine to just interject inconveniently with technical stuff

Xander Henderson

heh

Cookie Toast

So its an aproximation of sorts?

Xander Henderson

it isn't an approximation

Cookie Toast

$\longleftarrow$ knows nothing about set theory :P

Xander Henderson

Lebesgue measure can be though of as a function from a subset of $\mathscr{P}(\mathbb{R})$ to $\mathbb{R}_+$ that assigns a "size" to sets

and this isn't set theory, it is measure theory

Daminark

1:37 AM

You define the length of a set to be the inf of the lengths of coverings by intervals. For some sets, you try that but it turns out it just doesn't make sense, so you cop out and say they aren't measurable

Xander Henderson

or you talk about the Hausdorff outer measure ;)

in dimension 1

which is the same as Lebesgue outer measure on the real line

BUT BETTER!

Daminark

Tru

Actually we should define the Lebesgue measure as the Haar measure on R

Xander Henderson

That is the typical approach, honestly

except without the word "Haar" thrown about, 'cause topological groups scare baby grad students

and advanced undergrads

Daminark

Lol I feel like I'm in the womb rn

Xander Henderson

but I prefer to think of Lebesgue measure on $\mathbb{R}^n$ as the $n$-dimensional Hausdorff measure, normalized so that the unit cube has measure 1

because I am an analyst, not an filthy algebraist

Balarka Sen

1:40 AM

You're still an analyst. Disgusting

Daminark

But yeah usually I just thought you do either Caratheodory directly, or if you're a hipster you use Riesz representation

Cookie Toast

Hey, if I have $\frac{dy}{dx} = \frac{f(x)}{f(y)}$ that must satisfy $y(b) = b$, then I end up with $\int f(y) dy = \int f(x) dx +C$ and $C$ just equals zero right?

Xander Henderson

you need Caratheodory to get a complete measure, and one typically gets Hausdorff measure by applying the Caratheodory construction to the Hausdorff measurable sets

Daminark

Ugh geometers, go back to those jumping jacks you do whenever you explain something

Xander Henderson

I think that using the RRT would be circular somewhere along the way, but I can't put my finger on why

Daminark

1:42 AM

Waving hands so much you borderline achieve lift

heh

@Balarka

I am not a geometer

then what ARE you?!

@XanderHenderson I actually think it's been done, you can define this Daniell integral or something and it just works out

Balarka Sen

1:43 AM

I am a topologist. I wave my hands so much I fly into the outer space

Xander Henderson

ha!

@Daminark There is probably a way of doing it, I'm just not at all comfortable with it

Daminark

I mean after that Lie bracket stuff you were doing earlier I think you're starting to change. Denial is the first stage

Xander Henderson

ha!

axiomatizing integration seems like a cop out

Also, I worry that if we abstract integration too much, we'll end up doing category theory.

and that sounds painful

Daminark

"ears perk up*

Actually one of the problems we had in functional last week was to show that if you have some normed space X and a closed subspace Y such that Y and X/Y are reflexive, then X is reflexive

Xander Henderson

oh, the wife and kid are home; time go make dinner

laters

Daminark

1:47 AM

The grad students found this functorial way to do it cleanly I think, it was quite painful directly

See you!

Cookie Toast

bye @XanderHenderson!

Transcript for

$Mathematics$ Mathematics