Mathematics - 2018-10-15 (page 4 of 4)

Semiclassical

6:00 PM

bah, $\leq 9$ in my last expression

Additionally, I have $E[X_k^2]=1$. Exploiting that, the above inequality can be rewritten $E[XY]+E[YZ]+E[XZ]\in [-1,3]$

Mike Miller

@Ted Cuz I think about this for nontrivial bundles.

Semiclassical

So that's a linear bound on the mutual covariances. By replacing one element of $u$ with $-1$ in three different ways, one can write down three other such inequalities

Mike Miller

But wedge just means 'stick sigma out front'.

Semiclassical

@MikeMiller I momentarily read that as Ted Cruz. I'm so very sorry for that.

Mike Miller

@TedCruz You are not competent enough in any amount of mathematics to be welcome in this chat

Ted Shifrin

6:02 PM

LOL

Oh oh, troublemaker @Alessandro is here

Alessandro Codenotti

I am

Hi everyone

MatheinBoulomenos

Hi @Alessandro

Alessandro Codenotti

Is there a geometric explanation of the suspension isomorphism in homology?

Ted Shifrin

Draw pictures!

Mike Miller

@AlessandroCodenotti yeah

Alessandro Codenotti

6:06 PM

@TedShifrin the only suspension I can draw is that of $S^1$!

Mike Miller

Well that's no good

Since the definition tells you how to 'draw' it

Ted Shifrin

Make your picture schematic in terms of a simplicial chain.

Mike Miller

Gluing two cones together

Alessandro Codenotti

Yeah sure, I was exaggerating

I know the glued cones picture

Ultradark

@TedShifrin what can one say about the natural logarithm of an algebraic number?

Mike Miller

6:09 PM

What is homology geometrically?

@Ted: I'll write up your question about the boundary operator when I'm at the airport.

Ted Shifrin

Airport? Where're you going this time?! :)

Ultradark

ohh, it's always transcendental.

by lindeman wierstrass theorem

Alessandro Codenotti

@MikeMiller the problem with the approach we're following is that in this generality I don't have geometric intuition. I guess I should think about singular homology to see the geometry though?

CaptainAmerica16

My drawings were accurate! I don't know where I'm going from here though...

Ted Shifrin

@CaptainAmerica: Just drawings won't do it. You actually have to do a calculation. If you put two vertices down, where must the third be to get the origin inside the triangle?

Mike Miller

6:12 PM

@Ted Home. I'd talk elsewhere about it.

@AlessandroCodenotti let me get you a blog post

Ted Shifrin

Ohh .. never mind :P

Mike Miller

assorteddetails.wordpress.com/2016/04/27/what-is-homology

Ted Shifrin

@Alessandro: Think simplicially rather than singularly.

Mike Miller

@Ted It's what you're dealing with soon. I won't say more.

Ted Shifrin

Ah, got it. My sympathies.

CaptainAmerica16

6:14 PM

@TedShifrin Yeah, I just wanted to see what I could get out of random attempts. (seeing if it was impossible or 100% of the time etc.) Now I have to try and do the actual probability part. lol

Mike Miller

Thank you.

Ted Shifrin

I guess you're right about our convergence to one another :(

Alessandro Codenotti

@MikeMiller Thanks! I know what to read this evening now!

Ted Shifrin

The math is crazy in here today. :)

I guess @Balarka went off to write down Riemann-Roch.

I have to give an MIT-sponsored talk to middle schoolers on Sunday. So I have to prepare a handout for that. That should take a few hours.

Ultradark

bye ted

CaptainAmerica16

6:16 PM

See ya Ted. Thanks for the suggestions and the problem :D

Alessandro Codenotti

@TedShifrin Hmm so if I have a simplicial structure on $X$ ($\Delta$-complex? I forgot how those are called) I get one for free on $\Sigma X$

Ted Shifrin

I haven't left yet :P

Ultradark

lol :)

Alessandro Codenotti

I need to get back to the dorm and write this down, can't be worked out on the bus... But I'll think about it later, thanks for the suggestion!

Ted Shifrin

A $\Delta$-complex is a bit weaker than a triangulation, but that's fine. OK, @Alessandro. Talk later.

Ultradark

6:20 PM

can somebody verify my proof

it's not really a proof it's just a question

Semiclassical

I'm always disappointed when an interesting problem turns out to only have a solution in terms of a transcendental equation

Mike Miller

@Ted You know I think about this as singular submanifolds.

Ultradark

why?

Mike Miller

Which is why I think this is quite clear, from that perspective

Ultradark

oh because transcendental equations can't really be solved analytically oftentimes

Leaky Nun

6:24 PM

@MikeMiller hi

Semiclassical

because it means that you're going to have to resort to numerics to get the answer, and usually you might as well have done so from the get-go

Secret

I wish we have a more direct grasp on what exactly make something transcendental

Mike Miller

Hi

Ted Shifrin

@MikeM: Who are singular submanifolds? I'm losted.

Secret

almost all proof by contradiction never make use of the transcendental properties of the assumed number of function

Ted Shifrin

6:25 PM

hi @Leaky

Secret

Thsy just try to bound it so that it becomes a contradiction, so showing it is not algebraic, and hence by law of excluded middle, must be transcendental

Semiclassical

In my case, it's $\sin\theta-\theta \cos\theta=\pi/2$ with $\theta\in (\pi/2,\pi)$

no real hope of getting an exact answer for that

Ultradark

secret what would you like to get?

Secret

A list of properties every transcendental has to obey

Ultradark

hm

Secret

6:27 PM

The only known universal property of transcendentals is that they don't solve polynomials and their repeated fraction does not repeat nor terminate

but almost none is known about other properties

Semiclassical

Or, equivalently, find $\theta$ in that interval such that $\int_0^\theta t\sin t\,dt=\pi/2$

Mike Miller

@TedShifrin Homology classes. :)

Modulo bordism.

Ted Shifrin

Oh, sure.

Mike Miller

Poincare style.

Ted Shifrin

But for Alessandro's purposes, I think suspending a bunch of simplices is pretty clear.

Secret

6:29 PM

The proof of lindlemann weistrass seemed to suggest transcendentals may have a universal property that when compared to rationals, it somehow deviates faster than if it is a rational for their nth term approximation

but I have not checked whether it is really the case

Mike Miller

He started non-singular, but Heegaard pointed out that led to errors. So he moved to simplicial (basically the same idea)

@Ted Agreed.

Ultradark

@secret @Semiclassical don't know if this helps but: $e^{e/\ln(x)}=e$ is always transcendental when $x \in \Bbb Q(0,1)$

by lindeman wierstrass

in this case $x=e^e$

Secret

It is also interesting to note that some transcendentals such as $\pi$ and $e$ has geometric relations that are basically "here's a circle", "here's some oblique line such that blahblahblah is satisfied". That is, when expressed algebraically, all of this is a countably long series expression, but somehow mapping that to curves $\Bbb{C}$, we get only a finite sentence

Ultradark

@Secret I can prove the transcendence of $e^e$

Secret

Do we have some theorem that governs which countably long schema of proofs can be mapped to a finite proof using analytic geometry/high school geometry like arguments, that I might need to dig into the literature to find out

@Ultradark show

Ultradark

6:36 PM

okay it's probs wrong but i'll go for it

CaptainAmerica16

@TedShifrin I don't know if it matters, but the triangles that have the center inside are scalene and the ones I've drawn that don't are isosceles. I'm going to work on the vectors things now that I have measurements.

Ted Shifrin

That doesn't seem right, @CaptainAmerica. Certainly an equilateral triangle has the center inside.

CaptainAmerica16

Whoops, I haven't drawn an equilateral facepalm (I got ahead of myself.)

Secret

@Ultradark ?

Ted Shifrin

For isosceles, you can get both, @CaptainAmerica.

CaptainAmerica16

6:42 PM

Hm. Maybe I should just forget the drawings. Should I have started purely from computation?

Ultradark

@secret Assume by contradiction that $e^e$ is not transcendental. We can construct: $e^{e/\ln{x}}=e; x\in \Bbb Q(0,1)$ The solution is $x=e^e$. By L.W., $x$ must be transcendental. So we have reached a contradiction. Hence, $e^e$ must be transcendental for $x\in \Bbb Q(0,1)$

Secret

This is wrong. L.W. said nothing about $e^{transcendental}$

L.W. only said $e^{algebraic}$ is transcendental

Semiclassical

In that vein, there's no reason to expect that $x=e^e$ is a rational number between 0 and 1

(unless there's some other meaning intended to $\mathbb{Q}(0,1)$.)

Secret

Some random idea probably because I am too tired by the chemistry stuff:
Is it possible to express $e$ as an infinite sum of ln expressions?

Semiclassical

maybe, but that wouldn't help much here. L.W. is for a finite number of $e^\alpha$ with $\alpha$ algebraic

Secret

6:47 PM

No, I am thinking about:

Suppose $e$ can be written as a convergent sum of the form:

$$e = \sum_{n=0}^{\infty} \ln(a_n)$$

Semiclassical

So, write $e^e$ as an infinite product

Secret

Take $e$ both sides, since $e$ is a holomorphic function, and the sum is convergent, we can move the $e$ inside:

Semiclassical

uh

"e is a holomorphic function"?

("exponentiation is a holomorphic function"?)

Secret

uh, $e^z$ is entire, right?

Semiclassical

right. but $e$ is a number, not a function :P

Ted Shifrin

6:50 PM

@CaptainAmerica: Drawings are important. I asked you earlier to think about where the third vertex should go relative to the first two ...

Secret

ah, bad terminology, sorry about that

Anyway:

$$e^e = e^{\sum_{n=0}^{\infty} \ln (a_n)} = \prod_{n=0}^{\infty} e^{\ln (a_n)} = \prod_{n=0}^{\infty} a_n$$

Semiclassical

well, note that $e=\sum_{n=0}^\infty \frac{1}{n!}$. therefore $e^e = \prod_{n=0}^\infty e^{1/n!}\implies a_n=e^{1/n!}$

Secret

ok, that's very bad, all the $a_n$ are transcendental, meaning anything can happen, nvm then

Semiclassical

that's one identification of $a_n$, at least

Secret

acutally, I am not sure if $e^{\frac{1}{n!}}$ and $e^{\frac{1}{m!}}$ for integers $n\neq m$ are algerbriacally independent

Semiclassical

6:55 PM

there's an infinite number of such identifications, since it amounts to a series with terms $c_n=\ln a_n$ converging to $e$

Ultradark

@Semiclassical okay I see what you mean

Secret

If there exist an identification such that $a_n$ are all algebraic, then we can bound that product more easily since we knew more about algebraic numbers

Semiclassical

In general, if something isn't known to be transcendental, there's probably a good reason for it. new results in transcendental numbers are few and far between

Secret

29 mins ago, by Secret

The only known universal property of transcendentals is that they don't solve polynomials and their repeated fraction does not repeat nor terminate

All the proofs I have seen so far relies on proving something that is not algebraic, instead of trying to directly probe transcendence

Ultradark

@Semiclassical but isn't what I said correct?

wait no it's not correct

lol

MatheinBoulomenos

6:59 PM

@Secret the irrationals with repeating continued fractions are precisely the quadratic algebraic numbers, e.g. $\sqrt[3]{2}$ doesn't have a terminating or repeating continued fraction expansion

Semiclassical

@Secret reminds me of nothing so much as this: scp-wiki.net/scp-055

Secret

@MatheinBoulomenos ooops, forgot the non qudratic algebraic numbers

But the point is still maintain: We know almost no direct results on what exactly make a number transcendental

I am not sure if we have tools to directly probe transcendence. I knew there are two main methods in transcendental number theory, such as diophataine approximation (establishing bounds by approximating with rationals), and auxillary functions (certain carefully chosen functions that has many zeros at the number we want to probe, but otherwise I don't fully understood how they work yet)

Even less is known about composition of transcendental functions

Semiclassical

We're much better at saying what it means to fail to be transcendental

Ultradark

@Secret I can solve infinitely many of those

correction: I can solve an infinite number of transcendental equations

Secret

Actually, I think that proposal above is not entirely useless, but I don't know how one can prove or disprove that:

Basically the proposition is:

Semiclassical

7:06 PM

of course, this revolves around the notion that arguing about transcendent numbers is 'not entirely useless' in the first place :)

Secret

Proposition: Let $t$ be transcendental. Does there always exist sequence $(a_n)$ such that $a_n \in \Bbb{A}$ and:

$$t = \prod_{n=0}^{\infty} a_n$$

I am not sure what will a contradiction of this could look like if there is one

the issue is that when operations goes to infinity, a string of algebraic objects can become transcendental

Ultradark

$e^{1/\ln(x)}=x^n$

that's an analytically solvable transcendental equation

user602338

Hi guyz!

Secret

This is how transcendental functions like $\sin $ can give transcendental values for rational arguments, because an infinite sum is involved

user602338

Anyone here to answer a general question?

MatheinBoulomenos

7:09 PM

@user602338 just ask; don't ask to ask

Secret

Semiclassical

@Secret I think that boils down to: Can every transcendental number be written as the limit of an algebraic sequence?

user602338

How many books (not school or university ones) do ypu read every year?

Semiclassical

(a sequence of partial sums is just a special case of a sequence, after all)

Ultradark

1 @user602338

Secret

7:11 PM

@Semiclassical Well it has to be possible, because the rationals are dense in the reals, thus any reals (even indefinable ones) will be expressible as a convergent sequence of rationals, let alone algebraics

But infinite sums and products, I am not as sure

Semiclassical

Eh, there's no practical difference. If there's such a sum, then the sequence $a_1,a_1+a_2,a_1+a_2+a_3,\cdots $ converges to $t$

And if such a sequence exists, you can get a sum by taking differences of successive terms

user602338

@Ultradark where are yoh living?

Ultradark

@user602338 what's your math level

user602338

High school

Secret

hmm.. ok

Ultradark

7:13 PM

okay I live in the US

user602338

Where are you from ?

Ok

Semiclassical

to be sure, there are two versions of this

1) Does such a sequence exist for any transcendental $t$?
2) Given a particular such $t$, can I find such a sequence in a useful way?

Secret

I guess I might try to bound that product in wikipedia tomorrow, as it is 6:13 now and I must go to sleep
Leave any note and I will read it when I woke up

user602338

@Ultradark I just want to knosw that how do they teach you there!?

Semiclassical

The former is true, but the latter need not be.

Secret

7:14 PM

but yeah, 2) will be a useful point of investigation

Semiclassical

Night

user602338

How many years have you studied ?@Ultradark

Ultradark

@user602338 I've been studying all my life

CaptainAmerica16

@TedShifrin I think I need to take a linner break to clear my head. Be back later.

user602338

How old are you about?

Abve 30 years old? @Ultradark

Ultradark

7:17 PM

I am done with the press for today

user602338

Sorry @Ultradark but what is press?!

What do you mean?

I am asking that because i amn not an antive speaker

Anyone online here guys?

CaptainAmerica16

I'm here, but I'm not sure what he means by press.

Semiclassical

@MikeMiller wut if i only know a proof by induction

Leaky Nun

7:33 PM

@Semiclassical then prove it by induction

@CaptainAmerica16 hi

CaptainAmerica16

@LeakyNun Hey

Semiclassical

but induction is laaaaame

Leaky Nun

@Semiclassical no it isn't

it's the foundation of maths

Semiclassical

In truth, I do tend to find induction to be a rather disappointing form of proof at times.

CaptainAmerica16

I'm not good at induction yet.

Semiclassical

7:35 PM

It certainly validates a proposition. But I feel like it doesn't always give an insight into why it's true.

Mike Miller

Remind me what the ping was in response to

Semiclassical

your starred remark

For instance, it's easy to prove 1+2+3+...+n = n(n+1)/2 by induction. But the simple Gauss argument is both more memorable and more illuminating

To the extent that the proof by induction illuminates, it's because the way you might infer the formula is by looking for a formula $An^2+Bn+C$, imposing the base cases, and figuring out what A,B,C will need to be to make the inductive step work

Mike Miller

Ah yeah

Inducrion feels like brute Force

Semiclassical

the case of recursion relations is a bit special of course

to me the insight behind them is basically the other way around: If someone gives me a polynomial function $p(x)$ of degree $n$, then $p(x+1)-p(x)$ is a polynomial of degree $n-1$

And then it's not hard to work out which p(x) you'll need in order to get p(x+1)-p(x)=x, etc

Leaky Nun

7:51 PM

@Semiclassical but how would you formalize Gauss' argument?

Semiclassical

I think the 'disguised' version I wrote earlier works just fine

Let $j>0$ be some integer or half-integer number. Then $j+(j-1)+...+(-j+1)+(-j)=0$ by inspection.

Now add $j$ to each of the $2j+1$ terms. Bam, $2j+(2j-1)+\cdots+1+0=j(2j+1)$.

Semiclassical

8:07 PM

One thing I will say in favor of induction over Gauss’s symmetry argument, though, is that it’s the former that generalizes in a handy way

Mike Miller

9:02 PM

Flight delayed an hour so gonna miss my connection

Love that

Eric Silva

@MikeMiller a geometer missing his connection

what a tragedy

Mike Miller

Lmao

Leaky Nun

9:24 PM

@MikeMiller so what are you going to do?

Mike Miller

Not my choice

Waiting in line to find out what they'll do for me

Alessandro Codenotti

@MikeMiller you're not having a lot of luck with planes lately...

You got stuck in Münich too iirc

Mike Miller

Yes

But that was a 10 hour delay and I was sick so that was about as bad as possible

Eric Silva

@MikeMiller if i were u id prob just lay down and die

Alessandro Codenotti

Don't say that out loud, it can always get worse if there's an airline involved

Mike Miller

9:30 PM

@EricSilva I don't want to live but I'm too afraid to die

Eric Silva

the temporary struggle

Mike Miller

9:54 PM

@Ted This is the gorgeous note of Michael A I was thinking of.

Also @Eric though you won't parse all of it yet

Fargle

Howdy everyone

Ultradark

hi

fargle

Daminark

Aw man, nerds

How's it going?

Leaky Nun

@Daminark I learnt BCT for T2 locally compact spaces

and managed to prove by myself that open susbet of T2 locally compact space is locally compact

Fargle

Pretty good. Just finished helping a friend do DC circuit stuff in preparation for his final today.

ÍgjøgnumMeg

10:07 PM

@Dami wagwan ma nurd

Fargle

I think he'll do fine, but he has a fair helping of test anxiety, which I hope he's able to push through

Daminark

Nice! I think that version of BCT has come up in topological dynamics at least once

@Igjo not too much, how about you?

Mike Miller

@LeakyNun Wow Nerds really is accurate

Also I got on a later flight so I'll be home just an hour late

Daminark

Well that's at least something

ÍgjøgnumMeg

@Daminark not muuuch, just arrived home and came here to observe

as always

Daminark

10:09 PM

Nice

Got psets due in AG and AT so tonight is gonna be a bit busy

But after that things are gonna cool down a bit

ÍgjøgnumMeg

Nice, I can't wait to start back up

Just reading and not attending courses is a bit lame

pilotmath

hey guys

Ultradark

hey pilot

I found a terrible way to count primes

Mike Miller

One by one

1,2,3,4,5...

Ultradark

So $\pi(x)$ counts primes. That is to say, it counts the number of primes up to a given integer $x$. This function is very important in number theory. I was wondering how well the following counts primes: $\phi(x)-c(x) = \int_2^x e^{1/\ln(x)}dx-\int_2^x 1 dx .$ I tried making a table of values for $\pi(x)$ and comparing them to $\phi(x)-c(x).$

Rithaniel

10:21 PM

Hmmmm, I wonder if you could have a worse way of counting primes than doing so one-by-one.

Ultradark

it definitely over counts the primes. for example looking at $\pi(x)$ and $x=10^5$

gives you $9592$ primes

and $\phi(x)-c(x)$ gives $10119$ primes

so it's definitely over counting

so yeah, that's what I had to say

pilotmath

10:47 PM

$\sum_{n=1}^{\infty} 2^{-n}$ is absolutely convergent, right?

Fargle

Any convergent series with only positive terms is automatically absolutely convergent, @pilotmath.

(Same if all the terms are nonnegative.)

pilotmath

@Fargle I did not know that, thank you! Just to double check, that sum equals 1 right lol

Fargle

Indeed.

pilotmath

The infinite sum that is.

@Fargie do you have time for another question?

Leaky Nun

just ask

pilotmath

10:55 PM

question refers to this page: math.stackexchange.com/questions/2955683/…

But I'm trying to find the distribution/density function without convolution, as convolution hasn't been covered yet

Startec

Can somebody explain to me why `-log[H+] = -log[H+] + 4` reduces to `[H+] = [H+] * 10^-4` (There are supposed to be subscripts 1 and 2 for the [H+].

I understand multiplying through by -1 to get:
`log[H+] = log[H+] - 4`
But using exponentiation I thought you would get `[H+] = [H+] - 10^4` not `[H+] = [H+] * 10^-4`

I do not understand where that multiplication comes from

$-log[H+]_1 = -log[H+]_2 + 4$ reduces to $[H+]_1 = [H+] * 10^-4$

Ted Shifrin

11:14 PM

The subscripts are super important here, @Startec.

So you have $\log[H^+]_1 = \log[H^+]_2 - 4$, so taking $10$ to both sides we get $$[H^+]_1 = [H^+]_2\cdot 10^{-4}.$$

Remember that $10^{a+b} = 10^a\cdot 10^b$.

re @Fargle @Leaky

Leaky Nun

hi @chemist

Startec

@TedShifrin where did you get the $10^{a+b} = 10^a\cdot 10^b$

Ted Shifrin

That's the basic rule for exponents.

Try it out with $a=2$ and $b=3$. Write it out.

Startec

oh yes, but where does it apply in this equation?

Ted Shifrin

Oh, $a=\log[H^+]_2$ and $b=-4$.

Startec

11:18 PM

oh i see. Thanks!

Ted Shifrin

You're welcome.

user193319

Does $f_n : [0,1) \to \Bbb{R}$ given by $f_n(x) = x^n$ converge uniformly to $0$ on $[0,1)$ (notice that the interval does not include $1$)?

Ted Shifrin

Draw the graphs.

If you drawn an $\varepsilon$-fence around $y=0$, do all the graphs stay inside for large enough $n$?

user193319

I think so, because $f_n$ is decreasing.

Ted Shifrin

Draw it.

CaptainAmerica16

11:22 PM

Ooh, drawing.

user193319

But $f_n(0)=0$ for all $n$.

So I'm not sure what you're suggesting.

Ted Shifrin

I don't care about what's happening at $0$, @user193319. Draw the entire graph on $[0,1)$.

@CaptainAmerica: Are you poking fun at me?

Visualizing stuff is a huge part of mathematics ;)

user193319

Okay, I drew them. It looks like they are all shrinking to $0$, which is what I said above.

CaptainAmerica16

Lol

Startec

@TedShifrin wait I still do not understand why $-4$ turns into $10^-4$ instead of $-10^4$

Ted Shifrin

11:25 PM

They're all inside the $\varepsilon$-fence for the whole interval $[0,1)$?

The exponent rule, @Startec.

user193319

Looks like it: wolframalpha.com/input/…)

Ted Shifrin

Study it carefully.

It doesn't look like it to me, @user193319. They get all the way up to (but not including) $y=1$.

Draw the darned fence in there.

user193319

But none will ever equal $1$, because the functions have domain $[0,1)$.

Ted Shifrin

@user193319, so what? Don't they all cross $y=1/2$? $y=3/4$? My $\varepsilon$ is pretty small.

Ultradark

@LeakyNun can I ask a question

Ted Shifrin

11:28 PM

re @MikeM ...

Mike Miller

@Ted Did you get the ping about Michael:s note?

Ted Shifrin

Yes, I downloaded it, thanks.

Mike Miller

:)

My plane is FINALLy taking off.

Ted Shifrin

Haven't seen him in a year. Is he doing OK?

Safe travels.

This is the second plane?

Mike Miller

No.

Ted Shifrin

11:29 PM

Oh.

Mike Miller

Long delay.

I changed my first connection. I have to change my second when mine is late.

user193319

@TedShifrin Hmm...Well, the problem I am trying to solve is the following: $g : [0,1] \to \Bbb{R}$ continuous with $g(1) = 0$ implies that $x^ng(x)$ converges uniformly to $0$ on $[0,1]$...Since $g$ is continuous, it is bounded by $M$. I was going to define $f_n : [0,1] \to \Bbb{R}$ to be $f_n(x) = Mx^n$ for $x \neq 1$ and $f_n(1)=0$; and then use the Squeezee theorem...

Ted Shifrin

OMG, so many legs, @MikeM. Good luck.

user193319

If that won't work, what else is there do to?

Ted Shifrin

Well, @user193319, what does this have to do with the question you're being stubborn about?

You obviously need a better approach.

Mike Miller

11:31 PM

@Ted No, I misspoke, then. I have exactly one connecting flight. But I changed it once. And now I will have to change it again.

Ted Shifrin

YOu need to think more about graphs and visualize what's going on, @user193319.

Oh, I misunderstood. I get it, @MikeM.

Mike Miller

No, I used the same phrasing with someone else and they were confused too.

user193319

@TedShifrin Because if that sequence does in fact uniformly converge, then since $|x_ng(x)| \le f_n(x)$ on $[0,1]$, it would imply $x^n g(x) \to 0$ uniformly by the squeeze theorem.

Ted Shifrin

I think you should say "I changed my connection the first time. I'll have to change it a second time ..."

@user193319: But you're wrong. So you need a new picture and a new understanding.

user193319

I know...So how else might I approach that problem? I tried using uniform continuity of $g$, but I couldn't see how to use it.

Ted Shifrin

11:33 PM

What's the MOST important thing you know about $g$ (other than continuity)?

Leaky Nun

Gov’t official allowed to re-enter Hong Kong at new West Kowloon terminal despite losing travel ID

user193319

That $g(1) = 0$, which is why I immediately jumped to defining $f_n$ as I did.

Ted Shifrin

Now draw pictures.

Don't use computers. Use your brain and pencil and paper.

user193319

I have been using my brain. I just used wolfram to conveniently show you what I had drawn on paper, rather than taking a picture and uploading it.

Ted Shifrin

Fair enough :)

Now draw $x^ng(x)$ for various $n$.

Leaky Nun

11:35 PM

@MikeMiller safe flight

Ted Shifrin

Start by drawing $g$, of course.

user193319

All right. I'll give that a try.

Leaky Nun

I just feel like some theorem will do it

sure, I understand this is important

(don't get annoyed :P

Ultradark

0

Q: Prime counting function $\phi(x)-c(x)$ vs. $x/\ln(x)$

So $\pi(x)$ is the prime counting function. That is to say, it counts the number of primes below a given integer $x$. This function is very important in number theory. I was wondering how well the following counts primes: $\phi(x)-c(x) = \int_2^x e^{1/\ln(x)}dx-\int_2^x 1 dx .$ I tried making a t...

I told you I found a bad way to count primes!

Ted Shifrin

There is a theorem, in fact, @Leaky, but one can just do it bare-hands for this example.

Ultradark

11:39 PM

And what I learned from this is that it is very hard to count primes

user193319

Does uniform continuity figure in the proof in any way?

Ted Shifrin

I think all you really need is continuity of $g$ at $1$.

But I might be wrong.

user193319

And you use this to tame $x^n$ near $x=1$, right?

Ted Shifrin

Yup.

user193319

Okay. Let me try putting the pieces together.

Ted Shifrin

11:46 PM

What I wanted you to see by drawing the original $x^n$ and the $\varepsilon$-fence is that the places the graphs cross the fence march off toward $x=1$ as $n$ increases.

Startec

How does an equation of the form $y + 1 = Ay$ simplify to $1 / (1 - A)$?

Ted Shifrin

Startec, you need decent math skills to succeed in chemistry. Just saying.

Leaky Nun

it doesn't

Ted Shifrin

Always put all the $y$'s on one side of the equation.

And use parentheses, is what Leaky is bitchin' about.

Leaky Nun

no

Ted Shifrin