Mathematics - 2024-04-12

01:40

@onepotatotwopotato but btw, during the proof of the knot complement is a K(G,1) space, why can I assume PL embedding for $S^2\hookrightarrow M$ not just a topological embedding?

Thorgott

02:17

I mean, it doesn't really matter whether you take it PL or not

but also, PL and top embedding is basically the same in dim <=3, is it not?

I don't know what the precise statement is, but morally there should be no difference

one potato two potato

Doesn't the Alexander horned sphere matter?

And for the surface group question, can't I just do like this?: $\Gamma<\pi_1(M)$. Since $\Gamma$ is a surface group, it's finitely generated. Choosing a base point $\ast$, I represent each generator $S^1\to M$ as a based map. I consider an annulus neighborhood on each generator and they well-glued on the nbd of the base point. I write the resulting surface $N$. I add $2$-disks if $\partial N$ bounds a $2$-disk.

oh I need to add $2$-cells for the relation...

Thorgott

oh yeah ignore what I said

it's too late, I'm not thinking right

Silly Goose

02:50

do you guys think $e_{ij} = \delta_{ij}\delta_{kl}$ is a reasonable notation for the matrix with a $1$ in the $i,j$ entry and $0$s elsewhere

oh wait

nevermind

user 85795

03:07

honk

Obliv

04:27

i had a weird dream someone from the math chat room @'d me and they were a mod but pretending to be someone else because on their profile they had a photo reel of themselves

and a name that didn't match the people in the photos (i think I knew one irl)

WEIRD

one potato two potato

04:39

When I evaluate the complex limit $\lim_{z\to 2}{\sqrt{z}-2\over z-2}$, I should consider the branch cut and consider limits in both directions right?

one potato two potato

05:08

ignore it's just stupid

one potato two potato

05:55

Tarski monster group

In the area of modern algebra known as group theory, a Tarski monster group, named for Alfred Tarski, is an infinite group G, such that every proper subgroup H of G, other than the identity subgroup, is a cyclic group of order a fixed prime number p. A Tarski monster group is necessarily simple. It was shown by Alexander Yu. Olshanskii in 1979 that Tarski groups exist, and that there is a Tarski p-group for every prime p > 1075. They are a source of counterexamples to conjectures in group theory, most importantly to Burnside's problem and the von Neumann conjecture. == Definition == Let p...

iiiiiiinteresting

one potato two potato

06:07

it says that for given prime $p$, there exists an infinite group such that every nontrivial proper subgroup has order $p$.

Pizza

07:45

hi

Soumik Mukherjee

Hi

Pizza

i have a question

$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$$

why does this limit give $e$ as a result?
I know it's a notable limit , but I don't understand why this thing is $= e$

leslie townes

how you verify it formally would depend very sensitively on what your formal definition of e was (which would be true no matter what number you're talking about, but in some textbooks, e is literally defined by that limit)

a "calculus" explanation of it might be obtained by using l'hopital's rule to compute the limit of the logarithm of that sequence (which is [ln(1 + 1/n)]/(1/n)). you pretty clearly get 1 for the limit of the logarithm of that sequence, so you'd get e^1 = e for the limit of the original sequence

Pizza

08:07

mm wait

one potato two potato

It's not that intuitive for me that a fiber bundle can be orientable even if the base space is nonorientable

Pizza

@leslietownes yes

okok so

leslie townes

i don't put scare quotes there because the calculation is illegitimate or anything, it's just that if you put yourself in the business of deriving things from simple principles, things like "what e is" often come before things like what the natural logarithm is or l'hopital's rule

so depending on what you take for granted, using the latter to understand the former might be putting things in the "wrong" logical order

Pizza

@leslietownes so to prove that
$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$$ I can use hopital / taylor etc, while to prove that the sequence admits finite limit I have to prove that $a_n$ is monotone increasing and bounded?

leslie townes

08:42

i really don't know what tools you can use for what purpose in a classroom setting

i was trying to explain it more in the relative sense of, if you believe in things like the natural logarithm and l'hopital's rule, then you must also believe that that limit is e

if you want to work it out from "first principles," you need to specify what those principles are, including what "e" is defined to be, and unfortunately there are a lot of different ways of doing that

55

Q: About $\lim \left(1+\frac {x}{n}\right)^n$

I was wondering if it is possible to get a link to a rigorous proof that $$\displaystyle \lim_{n\to\infty} \left(1+\frac {x}{n}\right)^n=\exp x$$

49

Q: Limit of $(1+ x/n)^n$ when $n$ tends to infinity

Does anyone know the exact proof of this limit result? $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n = e^x$$

limits proof-writing

none of those answers are models of exposition, but i link to them because they illustrate the variety of ways in which you might approach the problem, depending on what you are comfortable starting with

0

Q: How do we know that the limit of $(1+1/n)^n$ exists as $n$ increases without bound?

I am starting to learn Calculus now and my learning pathway for finding the derivatives of exponentials and logarithms goes like this: I assume that $\lim_{n\to+\infty}(1+ \frac{1}{n})^n=e$. I accept this by definition. Then, I do all the necessary computations to prove other stuffs from it. He...

calculus limits limits-without-lhopital

Pizza

wait wait

$e = \lim_{n \to \infty} a_n$ with $a_n = \left(1 + \frac{1}{n}\right)^n$

this is the definition from my book

leslie townes

okay, so the reason that limit is e is "by definition." if you're asking why the limit ought to exist in the first place, some of that is addressed in the answers above, particularly the last answer

Pizza

it is written that the definition is justified by the fact that: 1) the sequence $a_n$ is monotone increasing 2) the sequence $a_n$ is limited

leslie townes

OK, so in your book, you are apparently at a point where you know that monotone increasing sequences that are bounded above (we wouldn't say "limited" in English) have limits

if your question is about why that sequence is monotonic, again, this is covered from various points of view on the site, depending on what you are comfortable using

46

Q: Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing

Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence. I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$ I am trying to go ahead of this step.

sequences-and-series

13

Q: Prove $(1 + \frac{1}{n})^n$ is bounded above

I've checked similar questions on the site but couldn't find satisfactory solutions or hints. Also, is there a more general approach to proving whether a given sequence is bounded below or above?

sequences-and-series exponential-function

Pizza

yes but these 2 things only tell me that the limit exists and is finite, to say that the limit tends to $e$, can I verify it as you said previously?

leslie townes

08:53

i can't tell if it's a language issue or something else. if your book defines that limit to be e, there is nothing to "verify"

you don't need my permission (or anyone else's) to see how that limit being e is also consistent with facts about things like the continuity of the natural logarithm and l'hopital's rule

but that's back in the realm of explaining that fact relative to other facts, not "verifying" anything

Pizza

ah ok ok, clear

thanks!

Jakobian

@Obliv it was me

Jakobian

09:26

@onepotatotwopotato reminds me of Prufer group, but the one you're talking about is a little different in that the Prufer group has subgroups of order $p^n$

I assume there can't exist such abelian group either (?)

@AlessandroCodenotti you said one should use Samuel compactifications because its more natural. But I am skeptical. You see, you come from the background of topological groups and topological dynamics, so its natural you might advocate for something like Samuel compactification since its relevant there. So, what justification do you have to call it more natural?

one potato two potato

The existence of Prufer group is not strange. The existence of that monster group is strange.

Jakobian

Oh wait. I read "Monster group" and ignored the link because I thought its talking about the finite simple group

sorry

@onepotatotwopotato the existence is for primes $p > 10^{75}$, so quite large primes. I wonder if its for all primes as well

It doesn't exist for $p = 3$

Thorgott

09:46

15

Q: Tarski monster groups: for which primes they don't exist?

What can be said about the set of primes $p$ for which it is proven that an infinite group with all non-trivial proper subgroups cyclic of order $p$ doesn't exist? Specifically, what is the largest such $p$ (say $p_0$)? All I could find in the literature is $p_0\le 10^{75}$, but I admit I didn't ...

gr.group-theory

Jakobian

$10^{75}$ is an old bound. The best current is 1003 (odd) proved in Adyan, S. I.; Lysënok, I. G. Groups, all of whose proper subgroups are finite cyclic. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 55 (1991), no. 5, 933–990; translation in Math. USSR-Izv. 39 (1992), no. 2, 905–957. Note that for some exponents, say, powers of 2 Tarski monsters do not exist for simple reasons. — user6976 Dec 1, 2019 at 1:44

Finitely generated groups of exponents 2,3,4 and 6 have been proved finite. Exponent 5 (together with lots of other small exponents) is unknown. — Derek Holt Jul 23, 2013 at 10:17

according to this comment there is no such group for $p = 2$ and $p = 3$

but $p = 5$ seems to be unknown, and for all primes $p > 1003$ such group exists

with $997$ being unknown again

seems like there might be some prime $p_0$ such that such group doesn't exist for $p\leq p_0$ and exists for $p > p_0$?

I wonder if thats known

probably not

they say its quite hard to show that such group doesn't exist for a prime

Burnside problem

The Burnside problem asks whether a finitely generated group in which every element has finite order must necessarily be a finite group. It was posed by William Burnside in 1902, making it one of the oldest questions in group theory, and was influential in the development of combinatorial group theory. It is known to have a negative answer in general, as Evgeny Golod and Igor Shafarevich provided a counter-example in 1964. The problem has many refinements and variants that differ in the additional conditions imposed on the orders of the group elements (see bounded and restricted below). Some of...

this problem for $p = 5$

research about Burnside problem seems to died down a little bit

Alessandro Codenotti

10:11

@Jakobian the Stone-Cech compactification is a topological construction. It allows extending continuous maps to compact spaces, but it completely ignores the uniform structure. The universal property of the Samuel compactification instead allows extending uniformly continuous maps toward compact uniform spaces

Jakobian

@AlessandroCodenotti I agree that ignores the uniform structure, but is it really so weird to use it if your uniformities are defined in terms of pseudometrics and your goal is to extend those pseudometrics?

Alessandro Codenotti

It comes up in dynamics because for a topological group $G$, the samuel compactification $S(G)$ is the biggest compactification on which the natural action $G\curvearrowright S(G)$ is continuous, but even in the context of uniform spaces by themselves, in my opinion, uniformly continuous functions are the natural notion of morphisms and being able to extend them is the natural requirement

Thorgott

spoken like a true category theorist

Jakobian

@AlessandroCodenotti This is all assuming $G$ has left uniform structure or however its called, right?

or right depending on which side you prefer

I mean you'd still have to construct Samuel compactification which is just extra work and if you already have Stone-Cech constructued then might as well use that

Alessandro Codenotti

@Jakobian Yes, you need to choose the correct one depending on whether you think about G as acting on itself by left or right translations

Jakobian

10:22

There's probably a way to obtain Samuel compactification from Stone-Cech compactification here

Alessandro Codenotti

Stone-Cech is the spectrum of the algebra of continuous functions, Samuel of the uniformly continuous ones, so that duality gives a surjective map $\beta X\to S(X)$. But I'm not sure whether I'd call this obtaining the Samuel compactification from the Stone-Cech one

Jakobian

@AlessandroCodenotti If you use uniformly continuous functions on a uniform space to induce a uniformity, then does the completion give you the Samuel compactification?

maybe you need bounded uniformly continuous functions

Alessandro Codenotti

Yes, bounded (also for Stone-Cech I skipped a bounded)

Jakobian

@AlessandroCodenotti no, both work

you skipped the word "maximal" in spectrum if anything

okay but now I see some people define Samuel compactification in terms of completion so it would be backwards for me

since I was looking for existence of the completion, so I can't use Samuel compactification for that in that case

now that I know the completions exists it seems more natural to me to define Samuel compactification as the completion of another uniform structure coming from the bounded uniformly continuous maps

unless you have an idea how to define it from scratch

Alessandro Codenotti

10:37

"Spectrum of the C*-algebra of uniformly continuous bounded functions" is a definition from scratch (not the only possible one though)

Jakobian

yeah. Probably doesn't matter if we use bounded functions for this one either, if we define it as spectrum (since taking spectrum already makes this thing compact)

I mean okay suppose I have Samuel compactification defined and all of its properties

@AlessandroCodenotti well what exactly does the property say?

wouldn't I have to juggle between two uniformities here?

if I wanted to go with the same proof then why would I need uniform continuity preserved

what did you have in mind?

Alessandro Codenotti

You have a uniform space $(X,\mathcal U)$. The Samuel compactification $S(X)$ has the property that any uniformly continuous function $X\to K$, where $K$ is compact (so uniformly continuous wrt to $\mathcal U$ and the only compatible uniformity on $K$) extends to a uniformly continuous function $S(X)\to K$

Jakobian

I'd still need to define uniform structure on the completion somehow and Samuel compactification changes uniform structure

I don't see how its better

maybe if I were to use Samuel realcompactification?

or did you have a proof of the type "embedding into a product" in mind?

yeah I don't get it, for the proof I'm going with it doesn't seem to matter if I use Samuel compactification or Stone-Cech because the proof seems to boil down to the same thing. Unless you have a different proof in mind

Alessandro Codenotti

10:56

You can also do an "embedding into a product" construction of the Samuel compactification. But I guess it doesn't matter after all to show completions exist, my main point was that it is a more natural notion of compactification for a uniform space

Jakobian

and it probably is, just probably not for such basic constructions

well, thanks for the conversation I've learned some stuff about Samuel compactification

psie

11:11

I don't understand a basic claim made in this answer. If $F$ is continuous and is the distribution of $X$, then $P(X=x)=0$. The argument made is that $P(X\in (x-\delta,x+\delta))$ can be made as small as we want by shrinking $\delta$ from the definition of continuity of $F$. I don't see how.

$F$ being continuous at $x$ means $|y-x|<\delta\implies |F(y)-F(x)|<\epsilon$. Assume $y>x$, so that $F(y)-F(x)=P(X\in (x,y])$, but how did the answerer get $P(X\in (x-\delta,x+\delta))$?

Jakobian

@psie $P(X\in (x-\delta, x+\delta))\leq F(x+\delta)-F(x-\delta)$

if $\delta\to 0$, right side $\to 0$

from continuity of measure $\lim_{\delta\to 0} P(X\in (x-\delta, x+\delta)) = P(X = 0) = 0$

Or like, you know, $P(X=0)\leq P(X\in (x-\delta, x+\delta))$

However you want to justify those things

psie

ok, thanks 👍

psie

11:52

@Jakobian basic question; to use the continuity of measure, we need to have a countable collection of sets, right? Are the sets $X\in (x-\delta,x+\delta)$, "indexed" by $\delta$ (which is continuous), countable?

Maybe they are not indexed by $\delta$...?

Jakobian

$\lim_{\delta\to 0^+} f(\delta)$ exists iff $\lim_{n\to\infty} f(\delta_n)$ exists for all sequences $\delta_n\to 0$ such that $\delta_n > 0$ and we can make it so that $\delta_n$ are decreasing

it doesn't matter that its indexed by $\delta$ which comes from uncountable set

Xander Henderson

@Jakobian Since you like to be pedantic about silly topological counterexamples, this is true as long as the space is, uhhh... first countable? (Which the reals are, as is every space that anyone actually cares about, but...) :P

Jakobian

Right, I forgot to mention that $\mathbb{R}$ is first countable, silly me

Xander Henderson

If I ever find the time to write a calculus text, I want to do everything in terms of sequences. Convergence of sequences is, I think, a lot more intuitive for most students. Continuity is a little tricky, because you quantify over all sequences converging to some point, which is hard, but I don't think that it is really any harder than $\varepsilon$-$\delta$, and students tend to think of continuity sequentially, anyway.

(I'm kind of sad that Ted isn't here to tell me that I'm an idiot.) :(

Jakobian

I certainly used nets in the past just because I liked the mentality of sequences being easier to use than filters

(which a limit as $x\to y$ of real numbers is a particular case of if you take the filter generated by $\{((y-\delta, y+\delta) : \delta > 0\}$ or uh, $\{((y-\delta, y+\delta)\setminus \{y\} : \delta > 0\}$)

Xander Henderson

12:09

@Jakobian Indeed. I can grok nets. I've never needed to think about ultrafilters.

psie

@Jakobian ok, so you're saying that we can treat the limit $\lim_{\delta\to 0}P(X\in (x-\delta, x+\delta))$ as a limit as $n\to\infty$. What confuses me here is that $P$ is a set function, so do we just replace $\{X\in (x-\delta,x+\delta)\}$ by some sequence of sets, e.g. $\{X\in (x-1/n,x+1/n)\}$? I'm not sure I did this correct, simply exchanging $\delta$ for $1/n$...

Jakobian

12:22

@psie you can do that

any decreasing sequence $\delta_n$ converging to $0$ will work

since you're dealing with a decreasing sequence of sets so its a monotone limit anyway

psie

ok 👍

Jakobian

you don't have to use that argument if it confuses you this much

psie

yeah

Jakobian

just saying $P(X\leq x)\leq F(x+\delta)-F(x-\delta)$ and taking $\delta \to 0$ is enough

the reason why I'm so comfortable with using real limits is the fact that this is a monotone family of sets

indexed by real numbers $> 0$

so if you take any coinitial sequence $x_n$ converging to $0$, say $x_n = \frac{1}{n}$

the limit will be the same, really

where coinitial means for every $r > 0$ there is $n$ with $r \geq x_n > 0$

psie

interesting

John Zimmerman

12:35

Guess who independently discovered and posed a reformulation of an open problem in the field of differential topology. Me. I did. Let that sink in.

CowperKettle

13:00

3

Xander Henderson

@CowperKettle MY CULTURE IS NOT YOUR COSTUME! :P

(Little do I know, Mr Binden is the son of a cultural anthropologist, and was raised in southern Africa.)

user 85795

13:14

how do you make Algebra "interesting"...

(in that given middle school environment)

or should one even try

Xander Henderson

13:28

In the US, someone who wants to teach middle school mathematics typically spends two or three years taking courses in methods and pedagogy which are meant to give that person the skills to make mathematical topics interesting (or, at least, engaging) to their future students. A quick description of the material covered in those classes is likely not possible in chat.

The general idea is to tap into the fact that humans are kind of innately pattern seeking, and mathematics is, in large part, about finding and describing patters. So you are trying to hook students through exploration.

Give them interesting problems to solve which don't initially look like algebra problems.

Moreover, humans are social animals, so make students work together to solve problems (or compete with each other, perhaps, in a low-stakes kind of way).

Jakobian

@XanderHenderson I very much don't like working together with other people

Xander Henderson

@Jakobian I've noticed.

Jakobian

That's not the point

The point is that there exist people who find working together to be stressful and/or unpleasant

user 85795

how can you encourage students to work through the class notes on their own at that age

Xander Henderson

Most people will eventually end up in a position where they have to work with other people. There are very few positions in society where you can be a misanthrope and get away with it. Part of middle school education is teaching students how to get along, and work with, others. Yes, some students find it "stressful and/or unpleasant". But this is something that they need to learn to either get over or tolerate.

@user85795 You don't? Middle schoolers are not, by and large, independent learners. They still need a fair amount of hand-holding.

Jakobian

13:41

@XanderHenderson If someone is a misanthrope and is being forced to work with others, wouldn't that potentially just reinforce their feelings about being misanthrope?

user 85795

shouldn't that age be the transition to be coming an independent learner

Xander Henderson

@Jakobian Again, most people will eventually have to come to terms with working with other people. Creating an environment where they are made to work with others, and in which they are more likely to be successful by working with others, is important for their social development.

@user85795 What do you consider "middle school"? How old are those students?

user 85795

13 - 18

Xander Henderson

@user85795 Oh, yegads, no! Not in the US. That's high school.

user 85795

ok

Xander Henderson

13:44

Middle school is typically 11ish to 14ish (i.e. 6th through 8th grades, more or less).

user 85795

right

Jakobian

@XanderHenderson Say you assign people project and then they only communicate through facebook or other social media. Then there is no pressure to work together and you can end up with something like a sink where one person is working for benefit of others

user 85795

that is when Algebra is introduced, right

Xander Henderson

By the time a student is 18, they should be reasonably independent learners (not necessarily completely independent). But that transition to independent learning usually goes along with puberty and the general desire to become socially independent from one's parents and other authority figures.

user 85795

11-14

Jakobian

13:47

Am I missing the point because I'm talking about students or people older than we are supposed to talk about?

Xander Henderson

@Jakobian I am not going to engage with about this any more. You have a view of the world which is very particular to you. You may very possibly be one of the exceptions to the "most people" which I have articulated a couple of times. But you can't project your experiences onto everyone else. Most people need to learn to work with other people, live and in person. For most people, these social skills are going to be vital to their long-term success.

It might be hard or uncomfortable for them, but it is, for most people necessary.

Jakobian

I wasn't even arguing with you?

user 85795

can we get back to the cartoon

49 mins ago, by CowperKettle

Xander Henderson

@user85795 In the US, students typically get a first taste of symbolic mathematics (i.e. basic, basic algebra) around 7th or 8th grades (12--14 years old). A first real course in algebra is typically taught in the 9th grade (14--15 years old), with a second course in 11th grade (16--17 years old).

Jakobian

Maybe I wanted something explained about psychology and how people handle those things or if you even think about them. But you are pulling a handle on me

user 85795

13:51

the teacher is obviously trying too hard

34 mins ago, by user 85795

or should one even try

Xander Henderson

When I taught middle school, I assigned one of my classes a project to measure the size of the Earth. I worked with another teacher about 400 miles to the south to implement Eratosthenes' method. That was a lot of fun.

Pizza

$\lim_{n \to \infty} (1+\frac{1}{n})^n=e$

$\lim_{n \to \infty} (1+\frac{1}{n})^n=L$
$\ln\left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\right) = \ln(L)$
$\lim_{n\to\infty} \ln\left(\left(1+\frac{1}{n}\right)^n\right) = \ln(L)$
$\lim_{n\to\infty} n\ln\left(1+\frac{1}{n}\right) = \ln(L)$
now taylor the second term $= \ln\left(1 + \frac{1}{n}\right) = \frac{1}{n}$
$1 = \ln(L)$
$e = L$
can this be a valid demonstration to say that that sequence tends to e?

Xander Henderson

@Pizza What is your definition of $\mathrm{e}$?

user 85795

@XanderHenderson did you work with the physics department

Jakobian

@XanderHenderson read above, its $\lim_{n\to\infty} (1+1/n)^n$

Pizza

13:57

$e = \lim_{n \to \infty} a_n$ with $a_n = \left(1 + \frac{1}{n}\right)^n$

Xander Henderson

@user85795 There is no physics department in the vast majority of American middle schools.

Pizza

@XanderHenderson

Jakobian

5 hours ago, by Pizza

$e = \lim_{n \to \infty} a_n$ with $a_n = \left(1 + \frac{1}{n}\right)^n$

Xander Henderson

@Jakobian If that is the definition, then there is nothing to prove.

Which is why I asked.

Jakobian

Just Pizza being confused

Leslie already told him that much

@Pizza We know that $(1+1/n)^n$ has a limit. The limit is a unique real number. We call that number $e$.

Xander Henderson

14:02

@Pizza If your definition of $\mathrm{e}$ is that $\mathrm{e} = \lim_{n\to\infty} \left( 1+\frac{1}{n}\right)^n$, then you don't (and, indeed, cannot) prove that $\lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n = \mathrm{e}$. That is the definition. There is nothing to do.

Jakobian

$e$ is, by definition, limit of the sequence $(1+1/n)^n$, in other words, by definition, $(1+1/n)^n$ converges to $e$

Pizza

ok so I just need to justify the definition?

Jakobian

yes, you just have to justify that $(1+1/n)^n$ has a limit, as in your book, and then you call this limit $e$

Xander Henderson

The only question I might have about that definition is whether or not it is a good definition, in the sense that when you define an object via a limit, you should check that the limit actually exists. I this case, the usual approach is a bounded+monotone argument.

Jakobian

this is precisely, no more and no less, what $e$ is

user 85795

14:05

accept it

Pizza

@XanderHenderson but in this case I haven't proven that it exists? And that's e ?

Xander Henderson

@Pizza I have no idea what you have done. The argument you presented above, however, does not prove that the limit exists.

Jakobian

you said that the proof that it exists is in your book, and that they show $(1+1/n)^n$ is a bounded monotone sequence, so it has a limit by a theorem that every bounded monotone sequence has a limit

Xander Henderson

Again, the usual approach here is to show that the sequence is monotone and bounded, and then apply an appropriate theorem (i.e. the monotone convergence theorem for sequences).

Pizza

@XanderHenderson I calculated the limit

Xander Henderson

14:09

@Pizza No, you didn't.

Jakobian

calculated?

$e$ is an irrational number and we know $e \approx 2.71$

Thorgott

I'm getting a déjà vu, didn't this conversation already happen a month ago

Jakobian

there isn't really an another meaning in which you can calculate it than to write some of its digits

Pizza

sorry, but isn't that calculation I did an explanation that I get "e" as a result?

Xander Henderson

@Pizza NO!

Your book defines $\mathrm{e}$ to be the limit of a particular sequence, assuming that the limit exists.

Jakobian

14:14

you get $e$ as a result because $e$ is defined to be $\lim_{n\to\infty} (1+1/n)^n$. Your work doesn't show that. This limit is $e$ because its what we called $e$

Xander Henderson

If that limit exists, then $\mathrm{e}$ is precisely the limit of that sequence. Period. That is the definition. The only thing which needs to be shown is that the limit exists.

Jakobian

I don't know what to say because I don't know whats the point of confusion here

Xander Henderson

In other news, I am attending a conference this weekend, and staying in a hotel right by the London Bridge.

Jakobian

maybe Pizza doesn't understand the formal treatment of the number $e$ because he was only taught how to calculate limits, and not how to understand and define limits

Xander Henderson

London Bridge (Lake Havasu City)

London Bridge is a bridge in Lake Havasu City, Arizona, United States. When it was built in the 1830s, it spanned the River Thames in London, England. In 1968, the bridge was purchased from the City of London by Robert P. McCulloch. However, McCulloch only had the exterior granite blocks from the original bridge cut and transported to the United States for use in the construction of a new bridge in Lake Havasu City, a planned community he established in 1964 on the shore of Lake Havasu. The only parts of the “New London Bridge” that made it to Arizona were the exterior masonry. The Arizona bridge...

Jakobian

14:18

@XanderHenderson I thought you aren't active in terms of research

you're coming back to it? Keeping yourself in the field?

Pizza

ok so from what I understand, I have to prove that $\lim_{n\to\infty} (1+1/n)^n$ admits limit ?

so I have to justify the definition

Xander Henderson