Mathematics - 2021-10-02 (page 1 of 2)

leslie townes

00:11

there goes my career of being the fastest answering gun in the west to score that sweet, sweet attention my ego needs

shintuku

yes i was quite surprised by the consummate alacrity of the answer

Semiclassical

you rep-monger you

user193319

Let $X$ be the space of all continuously differentiable functions on $[0,1]$. Define $\ell : X \to \Bbb{R}$ by $\ell (f) = f'(1/2)$. I am trying to determine whether $\ell$ is a bounded linear functional. My thought is that it isn't bounded, but I can find a sequence of elements showing that it isn't bounded. I tried $f_n(x) = x^n$ and $f_n(x)=nx$, but neither of these worked.

I could use some help.

Semiclassical

missed the $ in front of [0,1]

user193319

@Semiclassical Thanks for pointing that out.

Semiclassical

00:21

np

probably want to design a function which, for finite $n$, is well-behaved at $1/2$

but which becomes problematic as n->infinity

something like 1/((x-1/2)^2+1/n^2)

i think that particular example doesn't work but something in that vein

leslie townes

how are you norming X?

user193319

The maximum of $|f|$.

leslie townes

something like semi's idea would do it. e.g. arctan(n(x-1/2))

or f_n(x) = g(n(x-1/2)) where g is your favorite bounded function and g'(0) is nonzero

user193319

00:38

So, $f_n(x) = n(x-1/2)$ works too?

Because $g(x) = x$ is bounded and $g'(0) = 1 \neq 0$...?

Ted Shifrin

You should make clear what the norm on $X$ is?

user193319

The maximum of $|f|$ over $[0,1]$.

Ted Shifrin

Far from the one I expected.

Ah, leslie already asked.

Then for sure unbounded.

user193319

I don't see how that works...$||f_n|| = \frac{n}{2}$, and $$|\ell (f_n)| = |f_n'(1/2)| =n$$ which isn't greater than $n^2/2$.

Yeah, my intuition is that it shouldn't be bounded, because the derivative of all such functions at $1/2$ should be arbitrarily large; slopes can get as large as possible. But I am having trouble defining a sequence of functions showing that it isn't bounded.

Semiclassical

something as simple as 1/(x-1/2+1/n) should work

copper.hat

00:46

Try $f_n(x) = \sin n (x-{1 \over 2})$. Then $\|f_n\| \le 1$ but $l(f_n) = n$.

Semiclassical

oh, hmm

my example has a discontinuity at x=1/2-1/n

so that won't work

Amin Idelhaj

Hey guys

Haven't been here in a while!

leslie townes

user, i said bounded function

try sin(x) or arctan(x) in my recipe instead of x

i meant bounded on all of R, if that wasn't clear

copper.hat

@user193319 sometimes the best answer involves $\sin$.

leslie townes

there copper goes again

Semiclassical

00:51

it is a source of good examples

copper.hat

its that catholic upbringing.

user193319

@leslietownes lol thanks!

leslie townes

i instinctively went to arctan instead of sin to avoid sin

Semiclassical

i'd probably do arctanh

leslie townes

sickos

copper.hat

01:09

the wages of $\sin$ are good.

Ted Shifrin

I always prefer to cosin.

copper.hat

very shifty

Ted Shifrin

That was one of my nicknames in high school.

Joe Shmo

01:42

does anybody have a good guess what a reduced rv is? prof is probably translating his exercises from french

normalized?

leslie townes

looking on french wikipedia suggests maybe one rescaled so its standard deviation is 1

assuming rv is random variable

Joe Shmo

rv is a random variable

thanks leslie

right, so $X \sim N(0, 1)$ ?

leslie townes

i don't know if mean 0 is a part of it or not

i'm looking at fr.wikipedia.org/wiki/Variable_centr%C3%A9e_r%C3%A9duite

Joe Shmo

no theyre centered gaussians

centered, "reduced" gaussians

youre the man! thanks.

SmokenSieEinBitteChebaHitBits

👽💿

shintuku

02:34

any way to get \bbig( in mathjax?

leslie townes

$( \big( \Big( \bigg( \Bigg($ ?

shintuku

my god, I was doubling the b instead of the g

thanks

leslie townes

\left ( and \right ) often but not always do the job automatically. in my experience sometimes they are too big.

shintuku

\left( \right) failed me miserably because i had a single line

robjohn

04:18

@shintuku not sure what you mean by that.

shintuku

@robjohn $\forall x \forall B \left( (x \in A \land B \subseteq A \land \phi(B)) \implies \phi(B\cup\{x\}) \right) \implies \phi(A)$

\left( and \right) not getting bigger

leslie townes

05:37

makes sense. i think \left ( and \right ) attempt to auto-fit to the height of the expressions within them, and would not be affected solely by levels of nesting

copper.hat

06:01

gimme a pencil & paper. none of this perverted latex stuff.

geocalc33

@SmokenSieEinBitteChebaHitBits hey

It doesn't matter how you warp the product...it matters that you warp the product

geocalc33

06:22

Does anyone know how to assign a measure of curvature to the stellated octahedron?

SmokenSieEinBitteChebaHitBits

06:39

@geocalc33 what do you mean by assign to?

Take the stellar octahedron. Say $O$ and write $f(O) = $ some measure. Then $f$ is an assignment of measure.

geocalc33

@smoke

never mind ill look into it another time

BannedUser

Biology is harder than math cs and physics combined

SmokenSieEinBitteChebaHitBits

07:03

@geocalc33 look at this

post by another person in primes

I want to note that your conjecture is equivalent to $\lfloor \dfrac{p_{an}}{p_n} \rfloor = a$ for the case $a = 2$. I did that by expanding the remainder you have going on into $p_{2n} - (p_{2n} - \lfloor \dfrac{ p_{2n}}{p_n} \rfloor p_n) = p_n \iff $ the shorter form . This is quite an elegant theorem then! — SmokenSieEinBitteChebaHitBits 6 hours ago

Then general theorem is then that for all $a\in \Bbb{Z}, z\in \Bbb{Z}$ we have $\lfloor \dfrac{p(an)}{p(n)}\rfloor = a$.

Where $p(n) = n$th prime number

So probably not easy to prove but not too hard. In the post they prove at least case $a = 2$.

So I guess define $p(0) = 1$, $p(-1) = -1$, $p(-2) = -2$ and so on...

Actuallky you need $p(-1) = -2, p(-2) = -3$, etc.

@geocalc33 did you see it?

geocalc33

yes!

SmokenSieEinBitteChebaHitBits

Isn't it pretty?

Someone else (not me) made that post

geocalc33

link?

SmokenSieEinBitteChebaHitBits

7

Q: How to prove than $p_{2n}-(p_{2n}\mod p_{n}) = 2p_{n}$ ? where $p_{n}$ is the $_{n}$th prime number ? (for $n$ > 1)

Let the prime function $p_n$ be the $n$th prime number. For example $p_1$ = 2, $p_2$ = 3, $p_3$ = 5, $p_4$ = 7, $p_5$ = 11 etc. I noticed something with the prime function : it seems than $p_{2n}-(p_{2n}\mod p_{n}) = 2p_{n}$, for $n$ > 1 For example : $p_{4}-(p_{4}\mod p_{2})$ = 7 - (7 mod 3) = ...

The form they have in the title is $f(2n) \pmod {f(n)} = f(an) - af(n)$. We can think of the thing on the right as a measure of how far from satisfying scalar multiplication, $f$ is.

Then the set of all $g(n) = f(an) - af(n)$ such that $a \in \Bbb{Z}, a \neq 0$ forms an abelian group under pointwise addition. I.e. $(g' + g)(n) = g'(n) + g(n)$.

geocalc33

07:25

nice!

SmokenSieEinBitteChebaHitBits

I can't figure out the rest though

too dumb

LOL

geocalc33

I'm dumb too

I have a very low IQ

SmokenSieEinBitteChebaHitBits

bashes heads with geocalc

The reason for using functions as a group is the modulus grows $f(n)$

robjohn

@shintuku As leslie says, that is as it should be: \left( and \right) only grow the delimiters as much as needed for the content. Nesting does not require larger delimiters.

geocalc33

I have a very low IQ

I have an average IQ

actually I think I have a very high IQ

SmokenSieEinBitteChebaHitBits

07:30

I got bad dunna or D.N.A. as some peoples call it!

geocalc33

I got dat DNA haha

that double helix one

SmokenSieEinBitteChebaHitBits

@geocalc33 want to Zoom right now and screenshare?

geocalc33

screen share what?????

SmokenSieEinBitteChebaHitBits

website

geocalc33

screen share screens?

SmokenSieEinBitteChebaHitBits

07:36

That works on desktop or mobile device

in Zoom

geocalc33

it's 3:36 AM I can't

SmokenSieEinBitteChebaHitBits

K cool

Web stuff is difficult

Might be harder than math itself

geocalc33

my intuition tells me that you may be right on the money

you may be situated directly on the money itself

SmokenSieEinBitteChebaHitBits

I'm using a Python library: Django - that's the HTTPS server app

I'm using a CSS library: Bootstrap5

I'm using an already built CD editor: q.uiver.app

But modding out the UI into a django-bootstrap5 template

So that it supports iPhone 5 even

Currently that quiver site doesn't work very well from an iphone

geocalc33

@SmokenSieEinBitteChebaHitBits you know what a color plot is?

domain coloring I think

SmokenSieEinBitteChebaHitBits

07:41

Nope

geocalc33

consider a surface $S$ that intersects itself but the colors at the intersection are different

google.com/…

something like that

love_sodam

09:33

7

Q: Help in finding $\lim_{n\to\infty}\Bigl( \sum_{k=1}^{n} \frac{1}{({n \atop k})} \Bigr)^n$.

I am not able to get a solution for this problem . Of finding the limit $$\lim_{n\to\infty} \left( \sum_{k=1}^{n} \frac{1}{\binom{n}{k} } \right)^n$$ I have tried using Mathematica and that numerically evaluates it to $7.3890560989 \cdots$ Which motivates me to think it is $e^2$ . Thanks for help

combinatorics limits

Is there any solution without big O?

geocalc33

$\nabla^2_1 f = \nabla^2_2 f = \cdot\cdot\cdot = \nabla^2_n f$

the worst partial differential equation

robjohn

10:25

@love_sodam $$\begin{align}1+\frac2n\le\sum_{k=1}^n\frac1{\binom{n}{k}}&\le1+\frac2n+\frac4{n(n-1)}+\overbrace{\frac{6(n-5)}{n(n-1)(n-2)}}^{\substack{\text{$n-5$ terms}\\\text{each at most $\binom{n}{3}^{-1}$}}}\\&=1+\frac2n+\frac{10n-38}{n(n-1)(n-2)}\end{align}$$

robjohn

10:40

It is big-O in spirit, but no big-O

love_sodam

11:00

@robjohn After taking $n$ power, how can I conclude the RHS goes to $e^2$?

robjohn

Use this and the Squeeze Theorem

love_sodam

Cool! thanks

BannedUser

11:23

Transcription, the rate of translation $\alpha_p$ depends on the number of ribosomes and the concentration of the protein building blocks. But also, translation can't take place without transcription, the first step of the protein production process. There needs to already be mRNA for the ribosomes to translate.

Rate of protein production is given by the translation rate, multiplied by the total number of mRNA molecules $m$:
$\frac{dp}{dt} = \alpha_p m$

I don't understand why $\alpha_p m$?

love_sodam

@robjohn I always wondered how all those ideas come up to you

BannedUser

I don't understand why mathematically they use this differential equation model...

BannedUser

11:38

If translation depends on depends on amount of mRNA then why do you multiply $\alpha_p$ by $m$?

robjohn

@BannedUser if you double the density of the mRNA molecules, does it not make sense that the reaction rate would double?

is it $\frac{\mathrm{d}p}{\mathrm{d}t}$ or $\frac{\mathrm{d}a_p}{\mathrm{d}t}$?

BannedUser

dp/dt

what do you mean by reaction rate?

robjohn

isn't the translation a reaction?

BannedUser

Ah I got biology background of kindergarten so I don't know anything about it ;_;

I was thinking in terms of if a man can kick a ball at a time and if you give him 100 ball then still there will be no change in his rate in this case man is ribosome and ball is mRNA but now I think I need to know how ribisome works

I think above will be true if it works like tap water. The more pressure the more flow...

I guess I will just assume it true since I don't want it to be a biology class.

robjohn

11:56

Think if you have 100 men and one ball, they can only kick one ball at a time but up to 100 balls, if you give them $n$ balls, they can kick $n$ balls at a time.

There will be some extinction when $m$ gets very large, but for normal densities, I assume it follows the product $a_pm$

love_sodam

If $a_n\to 0$ as $n\to\infty$, is it obvious that $\lim_{n\to\infty}(1+a_n)^n = \lim_{n\to\infty}(1+a_n)^{\frac{1}{a_n}na_n} = e^{na_n}$?

BannedUser

Oh now it makes sense $\alpha_p$ has already included amount of ribosome and other stuff so multiplying by amount of mRNA makes.

Thanks for football analogy!

love_sodam

ah it's obvious

BannedUser

I do remember I have derive this e^x definition using that... wait let me see notes

you can go in reverse direction

obvious is relative which and 90% of time only works on geniuses ;_;

I got IQ of most genius dog or chimpanzee on the Earth at least

sequence and a_n works same as x to 0 for more additional info

@love_sodam

geocalc33

12:34

2

Q: conformal compactification $\overline G$

Construct a conformal compactification, $\overline G$ of $G:=\Bbb R^{1,1}_*$ and/or provide a diagram of the conformal compactification of $G?$ conformal compactification Let $G$ have the metric tensor (not necessarily positive definite): $ds^2=\frac{dxdt}{xt}.$ This link, Penrose diagram, (und...

geometry differential-topology conformal-geometry semi-riemannian-geometry compactification

anyone see how to do this?

Been stuck on this for some time

BannedUser

12:52

I don't even remember compactness.

I don't remember topology. May be because I haven't seen it's application yet.

robjohn

13:07

@BannedUser It lets you use certain parking spots...

2

user1115542

14:14

This is part of the prep work for Spivak. Find all numbers $x$. Problem: $$|x-1|\cdot |x+2| = 3$$ I am running into a problem when taking the absolute value of $x+2$. Here is my work. $$|x+2| = \begin{cases}
x+2 , x + 2 \geq 0 = x\geq -2 \\
-(x+2) , x + 2 \leq 0 = x \leq -2
\end{cases}$$ Does anything obvious jump out?

Semiclassical

14:24

ugh, casework

also, writing $x+2\geq 0 = x\geq -2$ really is not a good notation

$x+2\geq 0\implies x\geq -2$, maybe

beyond that though i don't see anything strange. $|x+2|=\begin{cases} x+2, & x\geq -2 \\ -(x+2), & x\leq -2\end{cases}$

user1115542

Yes... I did wonder about the notation. Could you perhaps explain why your notation is preferred.

Semiclassical

the main annoying point is that you'll need to do the same for $|x-1|$, and then combine then for the product. so you need to figure out the intervals on which both, one, or neither are positive respectively

you're presumably trying to say that $x+2\geq 0$ and $x\geq-2$ are equivalent

but equals signs don't serve that purpose

user1115542

Correct... that is what I am trying to say.

Semiclassical

i mean, suppose you wanted to say that 2=2 and 3=3 are equivalent

using = sign would make that 2=2=3=3

which is obviously a bad idea

i'd probably just show it as steps:
\begin{align}
|x+2|&=\begin{cases} x+2, & x+2\geq0 \\ -(x+2), & x+2\leq 0 \end{cases} \\
&=\begin{cases} x+2, & x\geq -2 \\ -(x+2), & x\leq -2 \end{cases}
\end{align}

though tbh the second equality is sufficiently obvious in my brain that i'd just omit the first piecewise entirely

anyways, this is just presentation stuff

user1115542

May I ask you to explain the notation $\implies $. This is something that is clearly widely used, certainly on this board.

Semiclassical

14:34

sure. it's literally just \implies

user1115542

Yes... I like your presentation...it's much neater and as you say, less clutter for the brain

Semiclassical

i.e., "X, therefore Y is true"

user1115542

OK... thanks. But, when I look at the presentation, it's much easier to use.

Semiclassical

yeah. takes some experience for all this, anyways

beyond the presentation, though, i don't see anything wrong with your |x+2|

user1115542

So, now I can at least tackle the underlying problem with a little more certainty. I will try and make it a question, as Spivak and others seem to use different approaches, which is the part I like.

Semiclassical

14:37

sure

ultimately you will need to look at the cases $x\geq 1$, $x\leq 1$ for $|x-1|$

user1115542

@Semiclassical Thank you so much for your time and input

Semiclassical

and figure out how those cases overlap with $x\geq -2$, $x\leq -2$

np

Ryan G

15:00

@user11155 "May I ask you to explain the notation $\implies.$" Maybe take a look at the second bullet point of this Answer that I wrote 2 days ago: math.stackexchange.com/a/4264196/21813

user1115542

@RyanG Very nice, thank you.

Ryan G

@user1115542 No worries

user1115542

I am beginning to think that there might be an erratum in the answer book for this question. (Chapter 1, Question 11(viii) I will try and and set it out here in a few moments.

user1115542

15:18

So, the problem is, as stated above, $$|x-1| \cdot |x+2| = 3$$ And, removing extraneous work, we have $$|x-1| = \begin{cases}
x - 1, \quad x \geq 1 \\
-(x-1), \quad x \leq 1
\end{cases}$$ and $$|x+2| = \begin{cases}
x +2, \quad x \geq -2 \\
-(x+2), \quad x \leq -2
\end{cases}$$ Now Spivak writes this in the answer book. " If $x > 1$ or $x < -2$ then the condition becomes $$(x-1)(x+2) = 3$$" Should not the second inequality be $x > -2$?

BannedUser

16:06

studypug.com and brilliant.org ads are haunting meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

and what is with these grammarly and masterclasss

skillshare

I remember when I was learning topology I was just drawing brunch of circle :(

user1115542

16:50

oops... I think I just realized my error, or as I am sure they say in the parlance.... Spivak: $1$, Me: $0$. My error was failing to realize that those inequalities ie $x > 1$ or $x < -2$ produce factors that are either both positive or both negative. Duh!

robjohn

@Ted: good morning

Ted Shifrin

Good morning @robjohn

How're you two doing?

robjohn

I'm pretty good, still working on Dolly

Ted Shifrin

Well, OK. Add a persistent floater in my left eye to our list of casualties.

Soon we'll need a hospital room for our chat.

robjohn

@TedShifrin is it fixed or does it move around?

@BannedUser does a brunch of circle mean pancakes?

leslie townes

16:59

i wonder which horse medication would work on this. there's only so many possibilities to check.

Ted Shifrin

@leslie Or maybe, in my case, cow.

@robjohn Moves around lower left.

user1115542

@leslie I think bleach should cover it all... with a bit of sun!!!

Ted Shifrin

I have an appointment with my ophthamologist in a month. Damn, that word is impossible to spell.

leslie townes

yeah, definitely the sun should be involved. maybe crack open one of those old thermometers and dumb its contents on the eye.

user1115542

Yep...mercury, bleach and sun. What could possibly go wrong :-)

leslie townes

17:02

the healing properties of beams of light are well known.

i wanted to know why it was so confusing to figure out where words came from. the entomologist was no help whatsoever.

Ted Shifrin

You're getting the gnat of this, @leslie.

leslie townes

i first experienced floaters on a visit to denver when i was about 30. i couldn't tell if i'd always had them and just never noticed because i'd never been anywhere where the sun gets as bright as it does in denver.

anyway, i haven't been back.

Ted Shifrin

But have they?

leslie townes

who knows?

Ted Shifrin

Yeah, I first noticed it yesterday in bright sunlight, but now it persists.

copper.hat

17:10

Call your ophthalmologist and tell them.

Ted Shifrin

I'll call Monday if it's still here.

Add that to my bad, bad disks in my back. The joys of old age.

Semiclassical

I noticed floaters as a kid but they were a source of fascination not frustratation

And not persistent

leslie townes

17:31

i just heard from downstairs, "don't kick the phone out of my hand!" i think the munchkin is ready to have that cast taken off.

Ted Shifrin

Well, her parents sure are.

robjohn

@TedShifrin I get them all the time, but they float around and I don't notice them for the most part. In bright light, they are easier to notice.

@leslietownes did you say "cast" or "head"?

leslie townes

i dunno, i'm not a doctor. we'll see what the orthopedist says.

robjohn

"I'm a lawyer, not a doctor!"

Ted Shifrin

Most lawyers think they know everything, even mathematics.

robjohn

17:47

@TedShifrin yeah, but it's when they try to sue someone for breaking a lemma that things get sticky.

or the Law of Large Numbers...

leslie townes

there are lemma laws in effect in most states to protect the consumer.

robjohn

@leslietownes and orange you glad.

Ted Shifrin

LOL ... Better than my gnat.

This looked like it might be more interesting than it turned out.

leslie townes

and now i want donuts.

Ted Shifrin

My torus was not sweetened.

robjohn

18:07

. o O ( spectral donuts )

Ted Shifrin

Now if you draw the Villarceau circles on those spectral donuts, I'll be more impressed. :P

shintuku

is there a way to get mathjax for $\in a$ without space between $\in$ and $a$?

leslie townes

the general vibe of tex/latex/etc. is precisely to avoid making this kind of thing easy with out of the box components. you might need a new command for whatever you have in mind

Avra

hello,

Is stable sorting a mathematical concept please?

Koro

Hi :). Is there any way to find $\lim \left(((n+1)!)^\frac 1{n+1}-((n)!)^\frac 1{n}\right)$ without using Sterling's formula?

shintuku

18:12

argh, thanks for the answer

leslie townes

shintuku there are commands for 'negative space' for example, kind of the opposite of \, for positive space. but i don't know how a kludge of the form \in [followed by negative space] would space well in all use cases. e.g. it might look good for some letters/symbols of some widths but not of others, or look goofy if justification requires more spacing between characters.

there are often 'junior' levels of stirlings formula that handle limits like that. things you can prove with one or two lines of limit calculation. you don't always need the full asymptotics.

shintuku

woah, \!\! does exactly what i needed

thanks!

Ted Shifrin

you mean whoa?

shintuku

who-ah? no definitely not

Koro

@leslietownes Is it possible to find the limit using $n!^\frac 1n\sim n/e$?

Ted Shifrin

18:18

@Koro No, I doubt there is. Even with Stirling I think you need to be quite careful.

leslie townes

woooah was always how they transliterated tintin's dog barks into english

Ted Shifrin

What does $\sim$ mean? You have to be very careful when you're doing a difference and not a quotient.

leslie townes

koro i agree with ted. i do not think it is an elementary limit although i do not know that it is a cookbook application of stirling either.

it may yield to simpler techniques that are not special cases of stirling but i don't see how at the moment.

Ted Shifrin

Damn, leslie agreed with Ted again.

Koro

Professor Ted, I know of that expression ($n!^\frac 1n\sim n/e$) as it's proven in an exercise problem in Spivak's. By $ f(n)\sim g(n)$, I meant to state that $\lim \frac {f(n)}{g(n}=1$

leslie townes

18:20

i suggested that someone could understand an inequality by graphing something the other day.

Ted Shifrin

No, that is false.

In fact, the quotient looks like $\sqrt{2\pi n}$.

Oh wait, I'm wrong. $n!\sim (\frac ne)^n\sqrt{2\pi n}$, but when you take $n$th roots it's OK.

But I still think you have to be very careful with your subtraction.

Koro

yeah, I can't use this approximation there due to subtraction. I'll have $\lim \left(((n+1)!)^\frac 1{n+1}\frac1{n+1}\frac {n+1}n-((n)!)^\frac 1{n}\frac 1{n}\right)n$. So this outside $n$ will create problem.

I may be wrong but I think that Sterling's formula was made exclusively to handle factorials

Ted Shifrin

StIrling

Koro

argh, I misspelled the name :(

Ted Shifrin

You may get it to work using the "correct" version I gave you with the $\sqrt{2\pi n}$. There are further terms, if needed, too.

So the factor of $n$ may be compensated when you compare $(n+1)^{1/2(n+1)}$ and $n^{1/2n}$

Koro

18:30

yes, I agree that it's possible using Stirling's formula (with the big series version). I was wondering if it could be solved without it.

I'll try it once with Cesaro Stolz after "rationalizing" the expression.

Ted Shifrin

This is going to get yucky very fast.

Koro

yeah, I think Stirling's the only way ...

27

Q: How to prove $\left(\frac{n}{n+1}\right)^{n+1}<\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}<\left(\frac{n}{n+1}\right)^n$

Show that: $$\left(\dfrac{n}{n+1}\right)^{n+1}<\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}<\left(\dfrac{n}{n+1}\right)^n$$ where $n\in \Bbb N^{+}.$ If this inequality can be proved, then we have $$\lim_{n\to\infty}\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}=\dfrac{1}{e}.$$ But I can't prove this inequality. Th...

inequality factorial

I found this post somehow. It answers my question :)

Ted Shifrin

Looks like robjohn is using the approach I suggested.

All I can say is, UGH. Not my cup of tea.

leslie townes

we agree. again.

river li's response performs some very detailed work with higher order approximations. great stuff, but i hope they have found other, happier ways to spend the limited time we all have on earth.

Ted Shifrin

18:47

Meanwhile, I was just having fond memories of the probability course I taught when I answered this one.

leslie townes

nice question, and nice answer.

Ted Shifrin

As I recall, one of my not-so-great students in the class raised this question early on.

leslie townes

elementary probability questions on SE tend to invite answers which often eventually do get to the point, but have a lot of essay discussion in the middle that doesn't help very much. it's nice to see that not happen.

Ted Shifrin

Sometimes people do miss the point that an easier version of the question might lead to better understanding.

Koro

6

A: How to prove $\left(\frac{n}{n+1}\right)^{n+1}<\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}<\left(\frac{n}{n+1}\right)^n$

Hint: Applying Stolz–Cesàro theorem $$L=\lim_{n\to \infty} \frac{\sqrt[n]{n!}}{n}=\lim_{n\to \infty} \frac{\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}}{(n+1)-n}=\lim_{n\to \infty} \left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)$$ We have $$\ln L=\lim_{n\to \infty} \left(\frac{1}{n}\sum_{i=0}^{n}\ln\left(\fra...

This answer finds the limit using using Cesaro Stolz :)

leslie townes

18:56

i feel sorry for asaf karagila. "set-theory" and its variants are one of the most overused, abused tags and he seems to be the only person who spends any time at all on trying to clean that up.

although ironically this does result in boosting the kinds of posts that abused the tag.

Ted Shifrin

Oh, you don't want to know how many tags of complex geometry I've had to remove. Virtually an uncountable number.

It's unfortunate that people think that the beginnings of complex numbers should be labeled that. Who knew it meant several complex variables and complex manifolds?

leslie townes

there should be a badge for that. maybe even workers compensation.

Ted Shifrin

Probably the only badge asaf and I would share :P

leslie townes

i'm guessing the platform provides autocomplete suggestions on tags and people are just clicking the first thing they see

Ted Shifrin

So many newbies don't even know what the tags are supposed to mean. They just throw them in there.

leslie townes

19:01

math.stackexchange.com/questions/4266036/… is an unpleasant throwback to teaching 'intro to proofs.' we had a textbook so poorly written that it even included tables of what words might mean what other words. i can't think of a worse way to teach that stuff.

Ted Shifrin

I don't see the point of sterilizing the English language to the extent that synonyms aren't allowed. But for struggling students these things are hugely an issue. Implied quantifiers at the top of the list.

leslie townes

people in K-8 education struggle with this a lot. students reading word problems want to know which english words will tell them to multiply the numbers they see, vs. divide them. math as classical conditioning.

shintuku

maybe we should change "set-theory" for "axiomatic-set-theory"

leslie townes

i also hate the appearance of forall and exists in that. the symbols aren't gonna help.

Ted Shifrin

Yup. Well, I remember having similar discussions with calculus students about word problems and cues.

@shin But set theory is an advanced area of mathematical logic, like model theory and recursion theory.

Munkres taught me in point-set topology sophomore year NOT to use symbols except for $\implies$ and the obvious $\in$, $\subset$, etc.

shintuku

19:04

yeah, isn't it always axiomatic? unlike elementary set theory

Ted Shifrin

No, I don't think of serious mathematics as axiomatic — unless it all is.

shintuku

in any case i think asaf karagila is very cool they've answered most of my "set-theory" tags

Ted Shifrin

He's one of the original people here, I think.

leslie townes

prepending "axiomatic-" to all tags except stuff specific to elementary subjects would reduce the use of tags considerably.

Ted Shifrin

well, axiomatic geometry is one of my least favorite things.

leslie townes

19:07

the math.SE answer box should have a rigid format. two columns, one labeled 'statement' and one labeled 'reason.'

shintuku

i'm looking for a decent axiomatic geometry book, i've never found a good one

Ted Shifrin

That'll get rid of us.

leslie townes

gonna pitch this on meta, see how many upvotes i get.

Ted Shifrin

To me that's an oxymoron, @shin.

Just go read Euclid.

shintuku

yeah, that's actually what i've been doing

Ted Shifrin

19:08

Maybe Hartshorne.

leslie townes

hartshorne was my only exposure to it. pretty good, although definitely not my thing.

shintuku

the other ones i found would just not give me a very clear deduction tree for arguments

Ted Shifrin

Oh, just what mathematics should be. Formalism.

puts shin on permanent ignore

shintuku

:'(

all i wanted to do was understand the parts of needham's complex analysis that have axiomatic geometry arguments

Ted Shifrin

Absurd.

leslie townes

19:10

that's how it starts. the next thing you know, you're stripping copper out of the walls of abandoned houses and selling it for drug money.

turn back while you still can, shin.

Ted Shifrin

I do not know that book, but certainly in that setting geometry should be analytic/synthetic.

shintuku

yeah i meant synthetic. isn't it the same as axiomatic?

Ted Shifrin

NOOOOOOOOOOOOOOOOOOOOOOOOO.

shintuku

ok i fault leslie for a comment he made like 5 months ago

Ted Shifrin

In synthetic, you work with models of the geometry, not with the axioms.

Well, @leslie is at fault for most of the crap here.

shintuku

19:14

whats a model of the geometry

Ted Shifrin

Doing projective geometry on a sphere (with identifications), doing hyperbolic geometry in either the half-plane or the disk.

Doing Euclidean geometry in the plane.

leslie townes

the statute of limitations on fault-finding for stuff i say is shorter than 5 months. i get away scot free on this one.

Ted Shifrin

That was "like 5 months ago."

shintuku

Apr 21 at 3:37, by leslie townes

maybe axiomatic geometry. we could just call it geometry. if i'm being honest i don't see a lot of those posts. and they tend not to specify their axioms, which is a minus for me.

the defendant has the right to remain silent

Koro

you found it so quickly @shin :)

Ted Shifrin

19:20

This seems worse than the people who call differential geometry anything with a manifold in it.

shintuku

there's actually very few results for "axiomatic geometry" in this chat

leslie townes

i stand by my comment that "euclid stuff" could properly be called axiomatic geometry. the problem isn't so much the choice of label, as the fact that it's a tag used by people who are indifferent to labels.

some kind of air show going on in huntington beach right now.

copper.hat

maybe kamala is arriving

leslie townes

haha, it's performance stuff. they're dropping chemtrails on all of the sheeple.

copper.hat

i must say she has been a bit of a disappointment to me

leslie townes

19:25

on the plus side, my 5G signal strength is excellent

copper.hat

that's the vax

everyone's a mesh now

leslie townes

one side effect of living near an army airstrip is your toddler yelling "I SEE A FIGHTER JET! I SEE A FIGHTER JET" and it turns out she's not lying

copper.hat

the G's are modulo 10, so after we reach 9G the next will be Oh-G

leslie townes

i do wonder when they'll give up with G's

shintuku

what does it even stand for

copper.hat

19:27

i'm watching the mil aircraft fly by on globe.adsbexchange.com/?icao=~298694

shintuku

gigabytes? i'm certainly not getting that many

copper.hat

generation

shintuku

oh that makes sense

copper.hat

watching the adsb tracks is amazing but a little scary from a data perspective

leslie townes

helpfully, the marketing people always roll out the next generation well before there is any new deployment of technology

copper.hat

19:29

5G is more like 4.1G really, essentially more local cells & support for it.

marketing and false news seem very similar

on a positive note, i noticed that philip morris is buying vectura. interesting take on supply chain.

hard not to be just a teensy bit cynical

my god my mood is devastatingly bad. i need to get out.

leslie townes

they should fully vertically integrate. the funeral home industry is in dire need of consolidation. many efficiencies to be achieved.

shintuku

maybe they'll do the xbox thing

they'll call the next generation 1G

copper.hat

there's a e300 doing part of the show.

cool toy

cradle to grave, we take care of you

later folks, i need to dilute my joy.

leslie townes

seeya.

$\Omega$G

Transcript for

$Mathematics$ Mathematics