Mathematics - 2018-01-22 (page 2 of 6)

Akiva Weinberger

1:00 AM

Oh, you're right

Sorry

orbit-stabilizer

2

Q: How can a probability density function (pdf) be greater than $1$?

The PDF describes the probability of a random variable to take on a given value: $f(x)=P(X=x)$ My question is whether this value can become greater than $1$? Quote from wikipedia: "Unlike a probability, a probability density function can take on values greater than one; for example, the unifo...

TIL

Akiva Weinberger

But it can't be negative, right?

Monolite

correct

Akiva Weinberger

Then I think your inequality is false

Let $f(x)=p(x)$ be $2$ on $[0,\frac12]$ and $0$ everywhere else

Then $\int f(x)=1$ but $\int p(x)f(x)=2$

Monolite

niice

orbit-stabilizer

1:05 AM

@MeowMix forgot to say, I liked your visualization.

@MeowMix did you just use HTML, CSS, and vanilla JS?

no css frameworks?

Xander Henderson

pmfs and pdfs are really the same thing at the end of the day

you just have to interpret them rightly

Semiclassical

1:23 AM

it's just like the difference between mass and mass density, really. The nucleus of an atom has a finite mass, but it's all packed into such a small volume that the density is very high.

Faust

how many densitys is an atom?

Semiclassical

don't know off the top of my head...hmm

I'll be boring and look it up :p

"Nuclear density is the density of the nucleus of an atom, averaging about 2.3e17 kg/m^3." (Wikipedia)

Faust

lol

thats a really freaking big number

Semiclassical

If I wanted to estimate it, I'd start from 1 nucleon having a mass of roughly 1 GeV/c^2 and a radius on the order of (I think?) 1 femtometer.

But then I'd need to multiply stuff out and blah blah blah

Mitch

If you cannot explain something in simple terms, you don't understand it.

Tweeted by ProfFeynman on January 21, 2018 at 2:14 PM

This really bothers me

Semiclassical

1:29 AM

By comparison, the density of water is 1000 kilograms per cubic meter. So the nucleus is about about 10^14 times as dense as water

Mitch

sure, you can't realy on your implicit inner definition to communicate that you understand it

but the whole point of scientific language is that you don't have to explain in a long paragraph every time you would normally want to use jargon

Didn't Voevodsky have a quote to the contrary of Feynman's?

I'm googling everything and can't find it.

Xander Henderson

@Semiclassical I was thinking more that it is the difference between Lebesgue measure and counting measure

Mitch

Maybe it was the Einstein quote about explaining to an 8 year old?

Xander Henderson

but, uh, to-MAY-to, to-MAH-to, I suppose

Semiclassical

@XanderHenderson eh, it's the same principle regardless

$\int_V p(x)\,dx=1$ doesn't imply $p(x)<1$

Xander Henderson

1:33 AM

$\int_A \,\mathrm{d}P$ :P

the measure of any set must be less than 1, but the Radon-Nikodym derivative needn't be so

which actually seems like a reasonable way of giving intuition to moment generating functions

if you are happy with the Radon-Nikodym derivative ;)

Semiclassical

My intuition for moment generating functions are "generating functions are awesome"

Xander Henderson

heh

Mitch

@Semiclassical They are!

DarkRunner

Hi all;

I'm doing a number theory problems and I'm having some trouble

P: A certain natural number is congruent to 4 (mod 9), congruent 1(mod 5), and congruent to 5(mod 8). a) Show that the number is congruent to 1 (mod 3)

So far, my thoughts have been:

Akiva Weinberger

@Mitch That doesn't mean that you can't use jargon, but it means that you should be able to explain what each bit of jargon means

On the other hand, when you're teaching someone something, you shouldn't hide the jargon from them, either. Explain the concept, and when that's done, give the concept a word ("jargon").

DarkRunner

1:48 AM

Since 9 is a multiple of 3, the residues of any natural number modulo 9 will have a onto relationship with the residues of any natural number modulo 3.

As, described by this diagram:

Akiva Weinberger

> “Triboluminescence. Triboluminescence is the light emitted when crystals are crushed…”

I said, “And there, have you got science? No! You have only told what a word means in terms of other words. You haven’t told anything about nature – what crystals produce light when you crush them, why they produce light. Did you see any student go home and try it? He can’t.

“But if, instead, you were to write, ‘When you take a lump of sugar and crush it with a pair of pliers in the dark, you can see a bluish flash. Some other crystals do that too. Nobody knows why. The phenomenon is called “tribolumi…

(Source)

DarkRunner

I looked at the hint on the back of the book, and it just says, "Use what you know about the number modulo 9"

Anyone know if I'm on the right track, or if anyone could provide some hints, that would be greatly appreciated

Akiva Weinberger

@DarkRunner Well, you said it was 4 mod 9, right?

(So it's one of 4, 13, 22, 31, etc.,)

DarkRunner

@AkivaWeinberger, right so it's any number in the form of 9k+4

Akiva Weinberger

Mhm, so now take that mod 3

DarkRunner

1:55 AM

Oh! wow

I feel so dumb

Faust

was just about to say that

the first line was all you needed

DarkRunner

Wow

k thanks people!

Faust

if you know what a number is say mod 15 you know what it is mod 5 and mod 3

the opposite of the CRT

that happens to be easier

if i have a topology group and want to make a room on the site for that can i?

William Oliver

Lets say that we have the hyperreal number h: 2, 3, 4, 2, 3, 4, ... (2, 3, 4 repeated), is a h necessarily an integer no matter what ultra filter we choose on N? (I.e. equal to either 2, 2, 2, ... or 3, 3, 3, ... or 4, 4, 4, ...)

If we choose an ultrafilter such that {n: h(n) = 4} is quasi big, then we can't choose {n: h(n) = 2} and {n: h(n) = 3} can't be quasi big right? So if i is 4, 4, 4, 4, ... then {n: h(n) = i(n)} would be quasi big and therefore h = i, but we could do the same thing if we choose 2 or 3 to be quasi big in our ultra filter. My question is, is it always the case that …

Akiva Weinberger

So technically I finished my CS project but it's ugly

so I have until midnight to un-uglify it

Faust

2:03 AM

@AkivaWeinberger un-uglify

Akiva Weinberger

@WilliamOliver Yes, $h$ is necessarily an integer.

Faust

I like this statement

Akiva Weinberger

It will either be $2$, $3$, or $4$.

The intersection of two quasibig sets is quasibig, right? So it can't be more than one of those

DarkRunner

Sorry to disturb again, but I understand that if you know what a number is mod x, you know what it's residue is for all mods that are factors of x. But what if a number is a multiple of x?

Akiva Weinberger

Since the union of finitely many quasismall things is again quasismall, it can't be none of them

@WilliamOliver Another way to think about it is to notice that it satisfies h = floor*(h)

DarkRunner

2:08 AM

Like if we know x= 1(mod 5), how can I show it's also congruent to 1 (mod 60)?

Akiva Weinberger

Like, its floor is itself, which means it's a (hyper)integer

(And it's not infinitely large, so it's just an integer)

William Oliver

@AkivaWeinberger Thanks! Was my reasoning correct? I guess we would have to choose either {n: h(n) = 4}, {n: h(n) = 2}, or {n: h(n) = 3} to be quasi big because for every set $S$ either $S$ or $S^c$ has to be quasi big. So once we choose one, the other two would have to be not quasi big. Since for every quasi big set $S$, if $S$ is a subset of a set $B$ then $B$ must be quasi big, therefore no subset of $2, 3$ for example, can be quasi big.

Semiclassical

@DarkRunner you can't, because that's false without further conditions

William Oliver

What is floor* ?

Akiva Weinberger

@DarkRunner You know how even numbers end in 0, 2, 4, 6, or 8?

Semiclassical

2:09 AM

one can quite easily come up with numbers which aren't congruent to 1 mod 60 but are congruent to 1 mod 5.

Cookie Toast

Suppose $\textbf{x},\textbf{y}$ $\in$ $\mathbb{R}^{n}$, $\textbf{x}$ and $\textbf{y}$ nonparallel. Prove that if $s\textbf{x} + t\textbf{y} = 0$, then $s=t=0$. My attempt:
If $s\textbf{x} + t\textbf{y} = 0$, then $\textbf{x} = -\frac{t}{s}\textbf{y}$ This is a contradiction as $\textbf{x}$ and $\textbf{y}$ are not parallel. Therefore the only solution is $s$ and $t$ are $0$.
Did I miss anything and/or is more rigor required?

Akiva Weinberger

@WilliamOliver If $f$ is a function from $\Bbb R$ to $\Bbb R$, then $f^*$ is its extension to the hyperreals, defined by $f((a,b,c,\dots))=(f(a),f(b),f(c),\dots)$. I don't know if it's standard notation

But it's just extending a function $f:\Bbb R\to\Bbb R$ to a function $f^*:\Bbb R^*\to\Bbb R^*$ (where $\Bbb R^*$ is the hyperreals)

William Oliver

Ah gotcha

Thanks a lot!

Akiva Weinberger

@WilliamOliver Right

@WilliamOliver The fact that hyperreal extensions of functions exist is essentially the reason why hyperreals are so useful.

A function $f:\Bbb R\to\Bbb R$ is continuous at $x$ iff, for every hyperreal $y$ infinitely close to $x$, we have that $f^*(y)$ is infinitely close to $f^*(x)$.

William Oliver

@AkivaWeinberger Yeah that floor function proof is super useful and elegant, I like it.

Akiva Weinberger

2:15 AM

(I'm considering the real number $x$ to be the same as the hyperreal number $(x,x,x,\dots)$. I guess theoretically you could call that $x^*$, but that feels like too many stars)

@WilliamOliver Here's something interesting, by the way.

Define $N$ to be $(1,2,3,\dots)$.

Semiclassical

@CookieToast That's a good idea, but note that in writing $s/t$ you're implicitly assuming $t\neq 0$

Akiva Weinberger

So $N>5$, for example, because $5=(5,5,5,\dots)$, and a quasibig subset of $(1>5,2>5,\dots,5>5,6>5,7>5,\dots)$ is true, right?

Cookie Toast

And would simply specifying as such be ok, since I'm showing a contradiction only if $s,t$ $\neq$ $0$? @Semiclassical

Akiva Weinberger

@WilliamOliver And similarly, $N$ is bigger than any real, so it's infinitely large

Now, let $f(x)$ be $x^2$.

Semiclassical

Not quite. You need to prove that both of them must be zero.

William Oliver

2:18 AM

@AkivaWeinberger right

Akiva Weinberger

$N$ and $N+\frac1N$ are infinitely close together, right? (That is, their difference is infinitely small, or smaller than any positive real)

But $f^*(N)=N^2$ and $f^*(N+\frac1N)=(N+\frac1N)^2=N^2+2+\frac1{N^2}$.

That is, $f^*(N)$ and $f^*(N+\frac1N)$ are not infinitely close together, since their difference is $2+\frac1{N^2}$.

But $f(x)=x^2$ is continuous! Doesn't this contradict what I said above about continuous functions?

William Oliver

Oh, wow, interesting

Akiva Weinberger

The resolution is that $f$ is called continuous if it's continuous at every real point $x$.

In other words,

Cookie Toast

Hmm... Well if we first assume $t$ $\neq$ $0$, then we have $\textbf{x} = -\frac{s}{t}\textbf{y}$. $s = 0$, $t \neq 0$ doesn't work because then you'll have $\textbf{x} = 0$. If we then assume $s$ $\neq$ $0$, then we have $\textbf{y} = -\frac{t}{s}\textbf{x}$. $t = 0$, $s \neq 0$ doesn't work either because then we'd have $\textbf{y} = 0$ And we already know we cant have $s,t$ both $\neq 0$ because then $\textbf{x}$ and $\textbf{y}$ are parallel.
Does that cover it @Semiclassical?

Akiva Weinberger

$f$ is continuous if, for every real $x$ and every hyperreal $y$ infinitely close to $x$, we have that $f^*(x)$ is infinitely close to $f^*(y)$.

One of them needs to be real.

Still, that leaves us with a question

Semiclassical

2:23 AM

Where did you assume $\mathbf{x},\mathbf{y}$ are nonzero?

William Oliver

Oh thats really interesting, so it is technically allowed for two infinite numbers to be both infinitely close and not infinitely close?

Semiclassical

It's true that $\mathbf{x}=0$ would be a problem, but why? The problem as you wrote it never specifies that.

Akiva Weinberger

It's allowed for two infinite numbers to be infinitely close (like $N$ and $N+\frac1N$) but have their images in the function not be infinitely close (like $N^2$ and $(N+\frac1N)^2$), yeah.

But there's a question, which is,

William Oliver

Ooh right of course

Akiva Weinberger

What do we call functions for which, for every hyperreal $x$ and hyperreal $y$, if $x\approx y$ then $f(x)\approx f(y)$? (Using $\approx$ to mean infinitely close)

Cookie Toast

2:24 AM

Ah crap. I guess I would say "Since $\textbf{x},\textbf{y} \in \mathbb{R}^{n}$, then they are not necessarily zero. Hm...but that means $s=0$ or $t=0$ is possible a solution if $\textbf{x}$ or $\textbf{y} = \textbf{0}$ right? @Semiclassical

Akiva Weinberger

Is there a word for such functions, where we can do it even if they're both infinitely large?

Turns out, there is: It's called "uniformly continuous".

And, you might remember from analysis, that $x^2$ is not uniformly continuous.

Semiclassical

Well, I think the question is to which vectors a zero vector is parallel.

William Oliver

Ah thats interesting!

Akiva Weinberger

So, continuity is equivalent to, $\forall x\in\Bbb R,\forall y\in\Bbb R^*,x\approx y\Rightarrow f(x)\approx f(y)$.

Cookie Toast

Wait...since $\textbf{x}$ and $\textbf{y}$ are assumed to be nonparallel, they cant be $\textbf{0}$ right? Because by the definition, any vector is parallel to the zero vector?

Akiva Weinberger

2:26 AM

And uniform continuity is equivalent to, $\forall x\in\Bbb R^*,\forall y\in\Bbb R^*,x\approx y\Rightarrow f(x)\approx f(y)$.

Semiclassical

That's what I'm wondering, yeah.

Cookie Toast

So that's the little hint where we can restrict $\textbf{x}$ and $\textbf{y}$? This is from Ted's book, so I don't know if that helps.

Akiva Weinberger

I mean, technically I still have to prove that that's equivalent to the usual epsilon–delta definition. But it's true.

Cookie Toast

The definition he gives for parallel is $\textbf{x}$ is parallel to $\textbf{y}$ if $\exists$ $c$ $\in$ $\mathbb{R}$ such that $\textbf{x} = c\textbf{y}$

Semiclassical

There you go, then.

Akiva Weinberger

2:28 AM

Er, I should have written $f^*$ in both cases. But I guess it's not really necessary

Cookie Toast

If $\textbf{x} = 0$, then $c = 0$ satisfies any $\textbf{y}$

Semiclassical

Right.

Cookie Toast

Hmm, yeah I didn't even see that little hint until you pointed it out. Gotta read the questions more carefully

William Oliver

@AkivaWeinberger Thats extremely interesting! I am definitely seeing non standard analysis as much more useful than standard analysis

Cookie Toast

Thanks @Semiclassical :)

Semiclassical

2:29 AM

np

Akiva Weinberger

@WilliamOliver Yeah, and it shows that when you're really rigorous with infinitesimals, unexpected things can happen

Cookie Toast

Ok, now help with part (b): Prove the Riemann Hypothesis :P

Faust

gl

Cookie Toast

Extra credit if you can get it in under a year!

Akiva Weinberger

Incidentally, I just gave you a nonstandard-analysis proof that $x^2$ is not uniformly continuous. It's instructive to try to turn that into a standard-analysis proof that $x^2$ is not uniformly continuous (with the epsilon–delta definition).

Faust

2:30 AM

cookie promises he will give u a phd

prove $x^2 $ is not uniformly continuous is not bad

Akiva Weinberger

It's not too hard; the two proofs end up looking almost identical, except for a change in wording. (At least, I remember them being almost identical... it's been a long time since I've done it.)

Cookie Toast

heck I'll give you 2

Semiclassical

life's too short to worry about RH

Faust

@AkivaWeinberger i belive you can do it rather easily by contradiction on an assumtion that a delta holds for all values

Akiva Weinberger

@WilliamOliver Also, do you know any topology? Specifically, do you know what "compact" means?

I was asking William

William Oliver

2:33 AM

@AkivaWeinberger Yes I do

Faust

sorry

Akiva Weinberger

Although I guess I can share it with you if you're familiar with hyperreals, it's quite cool @Faust

@WilliamOliver Right, there's a nonstandard analysis definition of compact as well.

Faust

I know of them i dont understand them very well

Akiva Weinberger

So if $X$ is a subset of $\Bbb R$ (really you can do this with any topological space it turns out), you can define $X^*$ as the set of things of the form $(a,b,c,\dots)$ where each of those is contained in $X$. Right?

Faust

hai

Akiva Weinberger

2:35 AM

It turns out that, $X$ is compact if every element of $X^*$ is infinitely close to some element of $X$.

DarkRunner

Hi guys; I'm working on a mods problem; If anyone could help, that would be great!

Faust

What do you mean exactly by infinitly close?

William Oliver

@AkivaWeinberger Ah that makes perfect sense!

Akiva Weinberger

For example, $(0,1)$ is not compact, since $(0,1)^*$ contains infinitesimals such as $(1,\frac12,\frac13,\dots)$, which is infinitely close to $0$ but not infinitely close to anything in $(0,1)$.

Faust

i see

@DarkRunner just ask

Akiva Weinberger

2:36 AM

Similarly, $[0,\infty)$ is not compact, since $N$ from before is in it but it's not infinitely close to any real.

DarkRunner

@Faust OK, Thanks

$Find\quad { 10 }^{ 10 }+{ 10 }^{ 100 }+{ 10 }^{ 1000 }+...+{ 10 }^{ 10000000000 }(mod\quad 7)$

Akiva Weinberger

And $[0,1]$ is compact, since you can take the "standard part" of anything in $[0,1]^*$ and land in $[0,1]$.

DarkRunner

So my first intuition is

William Oliver

Beautiful!

Faust

intresting

Akiva Weinberger

2:38 AM

(The "standard part" of a hyperreal is the unique real infinitely close to it, if one exists (i.e. if it's not infinitely large).)

DarkRunner

Well, one thing you can do is reduce is to 3^10+3^100, etc.

But that doesn't really help

Akiva Weinberger

@WilliamOliver We can use this to prove the extreme value theorem.

But it might take me a minute to remember how.

Faust

@DarkRunner what is $10^{10} \mod 7 $

or $3^{10} \mod 7$

William Oliver

@AkivaWeinberger Well, I have to go now, but thanks for showing me all of these awesome definitions using hyperreals! Far more intuitive if you ask me

Akiva Weinberger

You're welcome! And thank you, it's fun sharing this stuff

DarkRunner

2:40 AM

@Faust I'm not sure; I'm trying to find some pattern in the residues;

Let's see

Faust

Yes very informative @AkivaWeinberger

Akiva Weinberger

@Faust Did you see the thing I wrote above about continuity and uniform continuity using the hyperreals?

Faust

i read it yes

Akiva Weinberger

Arright

Cool innit

Faust

intresting ^^

DarkRunner

2:42 AM

I don't think there's any pattern in the residues; If there was, the above problem could be reduced really easily

Let me continue, however

Faust

it looks like each term is congruent to 4

yeah im going with 4

DarkRunner

Wow! I found a pattern in the residues; Ok now it should be easy to simplify

Faust

so you have 4 -3 +4 -3

DarkRunner

@Faust Let me check

Faust

10000000000/2 mod 7 is the answer

maybe no 2

DarkRunner

2:46 AM

Wait, I'm confused

Faust

ok

DarkRunner

So the residues of modulo 7 have a period of 7, right?

Faust

$ 4\equiv -3 \mod 7 $

and each term is congruent to 4

DarkRunner

Because we have 10 (mod 7) = 3, 10^2(mod 7 = 2), 10^3 (mod 7 ) = 6, 10^4== 4, 10^5(mod 7) = 5, 10^6 (mod 7) = 1

Oh, wait! it's acutally a period of 6!

I calculated wrong

Faust

?

you start at 10^10

DarkRunner

2:47 AM

Yeah, ok! I got my method, one second let me calculate it

Faust

$10^{10} \equiv 4 \mod 7 $ and $4^{10}\equiv 4 \mod 7 $ so every term is just 4...

DarkRunner

OK, so the answer is 1, right?

Because 36 (mod 7) =1 ?

Yeah; my method was to see that the period of the residues of modulo 7 is 6, so take the modulo 6 of the power of the first few terms, and you see that their residue mod 7 is 4, so by induction, it holds for all 9 terms. So, all the terms will have a residue of 4. Therefore, you have 4*9 (mod 7)

Faust

should be $4^9$

Cookie Toast

@Semiclassical Part 2 asks to show if $a\textbf{x} + b\textbf{y} = c\textbf{x} + d\textbf{y}$, show $a = c$ and $b=d$. My attempt:
We have
$$a\textbf{x} - c\textbf{x} = d\textbf{y} - b\textbf{y}$$
$$=(a-c)\textbf{x} = (d-b)\textbf{y}$$.
Let $s=(a-c)$ and $t = (d-b)$. Then we have the same case as in example (a), and the only solution is $s,t = 0$ which implies $(a-c)=0$ and $(d-b)=0$ which implies $a=c$ and $d=b$.
How'd I do?

Faust

but both are congruent to 1

i have no idea how your using induction

DarkRunner

2:51 AM

@Faust You mean 4*9, right?

Semiclassical

Eh, I think you're making life too hard.

DarkRunner

Well, anyway, I got 1(mod 7), is that right?

Faust

shit yes

Semiclassical

What do you get if you just move all the terms on the right (in the first equation) to the left and arrange by like terms?

Faust

gj

DarkRunner

2:52 AM

@faust cool thanks!

Cookie Toast

@Semiclassical I have this innate fear of my proofs being too weak, since I have no instruction on formal mathematics. You are undoubtedly right though

Stan Shunpike

@TedShifrin I feel like my communication skills need work.

I recently had a classmate challenge me to write clearer. He said my writing was overcomplicated.

Faust

how do i show that if F is compact there are $ x,y \in F $ so that $|x-y| = d(F) $

Stan Shunpike

@TedShifrin I'm not terrible. But I have been trying to communicate some of my medical ideas and, even the smartest people I know, they have trouble understanding me sometimes. So I need to find ways to improve. Any ideas?

Faust

@StanShunpike it may sound stupid but when i write up a math assignment my goal is to always fit it on one page. it usually doesnt happen but i think its made me a better mathimatician

Cookie Toast

2:54 AM

@Semiclassical you just get $a\textbf{x} - c\textbf{x} + b\textbf{y} - d\textbf{y} = 0$?

Semiclassical

if you arrange that in like terms, that's $(a-c)\mathbf{x}+(b-d)\mathbf{y}=0$.

Stan Shunpike

@Faust One thing i've learned recently is like, even if i don't like the notation a pset gives me, i still use it now. because i realized, when TAs read the psets, they expect the notation they gave, even if its shitty.

@Faust so there's definitely room for improvement in how i communicate my math too

the page limit might be a useful idea

Semiclassical

But what did you already prove about $s\mathbf{x}+t\mathbf{y}=0$?

Cookie Toast

Isn't that what I did above?
https://chat.stackexchange.com/transcript/message/42412080#42412080

Semiclassical

Sure, but my point is that you don't need to redo the argument.

Cookie Toast

2:56 AM

Or is there something different?

Oh, gotcha. I don't need to restate the whole first half :P

Semiclassical

Right.

Once you've written the above, you can immediately use part (a) to conclude that $a=c$ and $b=d$, full stop.

Cookie Toast

I guess with the amount of rewriting I was doing, I might as well prove $1+1 = 2$ while I'm at it :P

Faust

@StanShunpike i have autism so i think in a bit of a weird way to say the least. For awhile there it was incredibly difficult for me to communicate about mathematics with someone who didn't have a PhD cause they simply couldn't understand me. i did find the one page thing helped also getting very clear on the definitions and making sure i had memorized there exact meaning helped alot. (don't think of it as a page limit more of a goal that you hope to reach at some point)

Cookie Toast

Theorem: $s = t = 0$: Proof: Let Axiom be the 5-tuple of letters of the english alphabet denoting a ... (5000 lines later) therefore $s = t = 0$

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$Mathematics$ Mathematics