Mathematics - 2021-08-21

leslie townes

00:00

that's roughly how i meant it, shintuku. even f itself operates on both points and sets. some books will distinguish them.

shintuku

thanks a lot for the info!

SAJW

should newspaper(not scientific) articles use scientific notation of numbers that are way too big to grasp (or small)?

like what are a 3 trillion dollars, I couldn't grasp that from the word trillion

because such a number doesn't occur in normal life, if you don't count atoms of your body :D

Ted Shifrin

Worse for the common person.

leslie townes

interesting question. my dad was a journalist. it's sometimes helpful to use analogies. although it's hard to do it and maintain a reasonable notion of comparability because the analogies might be off.

SAJW

to be honest 20 is like the highest number I use right now. If something costs 100€ I just pay with a hundo...

leslie townes

00:13

it's difficult to compare spending that is used to finance something concrete and tangible than spending that just sends money out the door. accountants understand this.

per the mr show sketch (available on youtube), 24 is the highest number.

Ted Shifrin

How about 4.4 million deaths from COVID?

SAJW

stop right there

24 is the highest number

Ted Shifrin

So I will always be 24?

SAJW

not if you can age backwards :O

Ted Shifrin

No more induction.

All my disks in my spine think I’m infinitely old.

leslie townes

00:23

the boss did say that 24 is the highest number.

shintuku

01:12

$X$ is infinite, and let $C$ be the collection of all its infinite subsets. Suppose $C$ has a minimal element, $C^*$. Choose an element $c \in C^*$ and define $S = C^*\backslash \{c \}$.

do I need the axiom of choice to choose $c \in C^*$?

or can I give a non-arbitrary way to pick $c$?

(we don't know if $C^*$ has ordering)

oh: $C^*$ must have some sort ordering since it is a bijection with $\mathbb{N}$ by virtue of being infinite, no?

or maybe it doesn't have ordering itself, but we can pick a non-arbitrary $c$ using the bijection with $\mathbb{N}$, no?

shintuku

01:32

yes we can sunglasses

user76284

02:10

Is there a smooth function that is asymptotic to $\frac{1}{1-x}$ as $x \to -\infty$ and asymptotic to $x$ as $x \to \infty$? Like softplus, but with hyperbolic decay rather than exponential.

robjohn

02:36

@user76284 $\frac{x+\sqrt{x^2+4}}2$

user76284

02:56

@robjohn Perfect, thanks! I had a gut feeling I had to use $\sqrt{}$ somewhere. Your function is like $\frac{x + |x|}{2}$ with the soft absolute value $|x| \approx \sqrt{x^2+a^2}-a$.

robjohn

Hey, @Ted !

Ted Shifrin

Hi, @robjohn. Maybe my computer is sick, but I can't find the link to which you are responding.

robjohn

@TedShifrin which place am I responding?

Ted Shifrin

The one with the graph you just posted.

robjohn

The function I posted two lines up

to user76284

I posted the image for them and then paged you

two separate and unrelated actions, though they appear merged.

Ted Shifrin

03:08

No, no, no. I'm saying that the left arrow to user76284 leads me nowhere.

robjohn

I click on the arrow and the line above it is highlighted in yellow (just for a few seconds)

leslie townes

same for me.

Ted Shifrin

The line above it is from shintuku. I'm just confused.

robjohn

1 hour ago, by user76284

Is there a smooth function that is asymptotic to $\frac{1}{1-x}$ as $x \to -\infty$ and asymptotic to $x$ as $x \to \infty$? Like softplus, but with hyperbolic decay rather than exponential.

Ted Shifrin

I am phone-banking tomorrow for (against) the recall. I hope you guys will both do that.

robjohn

03:11

That was the line before my reply

Ted Shifrin

Those lines don't appear on my screen.

Weird.

robjohn

odd. refresh?

Ted Shifrin

Even after refreshing, nope.

robjohn

that is odd

Ted Shifrin

Well, I'm only 40% crazy.

leslie townes

03:12

i may not have the time but i've put my ballot in a dropbox already.

robjohn

@TedShifrin are you ignoring anyone?

leslie townes

thank you for doing that, ted.

Ted Shifrin

I have been ignoring people ages ago, but not anyone recently that I remember.

robjohn

they are on your list. That is why you don't see their question

Ted Shifrin

@leslie Any recommendations on what to do about the disastrous part 2 of the ballot? Friends here are voting for Faulconer, but he's turned into a complete Trompie.

leslie townes

03:14

i left it blank because one of my more politically active friends told me it was a good idea. i got a text message from the campaign of another democrat but i'd sent it off by then.

Ted Shifrin

Interesting. I wonder what this person did to render me irate in the past.

I don't think it's a good idea. I think we'd better to do our best to minimize the damage if we lose.

I've forgotten how to see my "ignored" list.

leslie townes

i can't stand this stuff. they can't win elections anymore and they know it so we get all of this horses--t.

Ted Shifrin

Yes, this was all Tromped up with Tromp money, for sure.

robjohn

Trump was mainly going after republicans that didn't agree with him, but he couldn't resist going after Newsom

leslie townes

the last recall election did have the beneficial effect of putting arnold schwarzenegger's signature on my phd diploma. he wasn't too bad. but whoever we get this time will be.

Ted Shifrin

03:17

Way bad.

leslie townes

the annoying thing about all of this is i don't like gavin newsom at all. i just can't stand these end runs around the democratic process.

Ted Shifrin

I don't have a strong opinion against him, for sure, but he sure was an idiot re French Laundry. That's what precipitated all this.

@robjohn, how do I find my ignore list?

robjohn

go to your profile and look into the preferences

Ted Shifrin

I give up.

robjohn

I've added a link to the line above

Ted Shifrin

03:24

I couldn't get there, I dunno. I have no recollection who any of those people are.

robjohn

you should be able to click on your name to the left of any of your chat messages. click on "user profile". once there, click on the "prefs" tab

Ted Shifrin

I did that directly and it didn't get me where you took me.

robjohn

odd

its even more than odd

shintuku

this must be some sort of sign

maybe we're not meant to click the link

Ted Shifrin

Yeah, a sign for me to resign completely.

robjohn

03:29

did you click on "user profile" or "user profile on math.stackexchange.com" by accident?

I don't think that should be different for mods, other than we can remove people from the ignore list for others.

problem with being a mod is you can't see what the usual user experience is.

I've often wished for a suspend my mod privileges mode so that you can see what things look like

or a what do things look like when I have 150 rep mode

Ted Shifrin

Yes, we mortals are lowly creatures.

Nano Miratus

03:58

@PM2Ring thanks!

@PM2Ring this seems like something a computer in 10 years could just calculate in like 10 minutes...nice!

LearningCHelpMe

04:15

Hello all. Is there a chat room equivalent to this where I can ask classical mechanics/physics questions

user76284

@LearningCHelpMe chat.stackexchange.com/…

LearningCHelpMe

@user76284 thanks!

user178758

05:58

@robjohn may I suggest creating a new separate account, using a different email address, to find out what things look like?

robjohn

@user178758 There is nothing wrong with having two accounts with the same email address as long as they don't interact in any way. The problem is that you cannot simply set the reputation to what you need to see what you want to see.

user178758

I guess that's where asking different kinds of questions with those alternate accounts comes into play.

;-)

robjohn

@TedShifrin this says that you should be able to manage your ignores from the prefs tab in your chat profile.

leslie townes

our IT people say this is a big problem with a lot of software. there's a huge amount of personalization and you can't see what is or isn't happening for an individual user without a lot of work on the back end. and of course user error, not system design, is 80% of tech issues.

but if you don't have the visibility, you can't know.

robjohn

what do you mean? or are you talking about something else?

@leslietownes yes, when trying to help someone over the phone and you can't see their screen, you have no idea what they are talking about.

leslie townes

06:13

if you're asking me i'm just commenting on software in general.

robjohn

people don't always use the same terminology.

leslie townes

i had a very difficult interaction with a superior. she wanted something to be a certain way for a client. i didn't understand what they were talking about. it turned out to be a microsoft word setting and not something you could embed into a document. they simply did not believe me and had me bother half of our IT staff about it.

probably 3 hours of completely wasted time.

split across several people.

user178758

yup, a screenshot may help :-)

leslie townes

i said "this is someone's word settings, not a document setting." three hours later, technical staff confirmed. it was a friday night.

exasperating.

user178758

as they say, a picture tells a thousand words...

shintuku

06:22

i mostly panic when pictures talk

user178758

:D

SAJW

10:55

why does the taylor series expansion of arctan(x) converge faster for smaller x? where can I read about that?

PM 2Ring

12:49

@SAJW Well, with smaller x, the powers of x get smaller faster. And it's an alternating series, so the error from truncating the series at some finite point is <= the first omitted term. Eg, if our last term is $x^{41}/41$, the following term is $x^{43}/43$, and our error must be <= to that.

PM 2Ring

13:17

To continue what I was saying in the Python room, we can rearrange the standard tan sum formula to verify these arccot relations. To make things more compact, I'll write a(x) for arccot(x).

Now $$\tan(U+V) = \frac{\tan U + \tan V}{1 - \tan U \tan V}$$
Taking reciprocals,
$$\cot(U+V) = \frac{\cot U \cot V - 1}{\cot U + \cot V}$$
Let $u = \cot U, v = \cot V$
So $U = a(u), V = a(v)$
Then
$$\cot(U+V) = \frac{uv - 1}{u + v}$$
and
$$a(u) + a(v) = a\left(\frac{uv-1}{u + v}\right)$$
Eg,
$$a(3) + a(7) = a\left(\frac{3\cdot7-1}{3 + 7}\right) = a(2)$$

So if $a(w) = a(u)+a(v)$, then $w = \left(\frac{uv-1}{u+v}\right)$. But also, $v = \left(\frac{uw+1}{u-w}\right)$.

That gives us an easy way to generate some integer $(u,v,w)$ tuples. Let $w=u-1$. Then $v=u^2-u+1$

PM 2Ring

14:10

Let $u=w+d, v=w+f$
Then
$$w+f = \frac{w(w+d)+1}{w+d-w}$$
$$w+f = \frac{w^2+1+wd}{d}$$
$$fd = w^2+1$$
Now we can generate all the $(u,v)$ for any $w$. We just have to find the factor pairs $fd$ of $w^2+1$.

shintuku

14:48

any sort of interesting or noteworthy ways to characterize the degree to which a function fills an interval of real numbers? what I mean is, e.g., considering functions whose image is a subset of $[0,1]$, we could have $Im \ f = [0, 0.5]\cup[0.9,1]$, $Im \ g = \{\{0\}, \{1\}, \{2\}, \{6\}\}$, $h$ is non-surjective with codomain $[0.4, 0.5]$ and has the image of a countably infinite sequence converging to 0.5, $Im j$ is the irrational numbers on the interval, $Im k$ is the algebraic reals, etc.,

I'm looking for some way to categorize that isn't cardinality (since most of the time it is either aleph null or aleph 1 and nothing else) and could give an idea of how close we are to continuity

PM 2Ring

With a bit of algebra, we can find various interesting relations. A very useful one is
$$a(u) = 2a(2u) - a(4u^3+3u)$$
Eg,
$$a(5) = 2a(10) - a(515)$$
In 1706, John Machin found
$$a(1) = 4a(5) - a(239)$$
This historically important formula has its own [Wikipedia page](https://en.wikipedia.org/wiki/Machin-like_formula). And there's a huge collection of Machin-like relations at the old http://www.machination.eclipse.co.uk/ site

Sha Vuklia

yo @LeakyNun, you were good at logic, right?

Leaky Nun

decent i guess

Sha Vuklia

I'm not very good/familiar with inference rules involving quantifiers

say I have the following hypotheses:

for all x, (S(x) -> A(x))
for all x, (S(x) -> [A(x) <->B(x)])

Which inference rule(s) can I use to conclude:

for all x, (S(x) -> B(x)) ?

Leaky Nun

oh no

what are your rules?

Sha Vuklia

14:59

ehh, the ones that is usually used in metatheory discussing formal first-order thy?

I was trying to give a meta-argument formally

Leaky Nun

do you have like a list of rules you can use

Sha Vuklia

not explicitly, because I was trying to formalise a meta-argument

I'll show the statement

So I wanted to prove that for a well-formed formula $\mathfrak B$ we have: $\mathfrak B$ is false for an interpretation $M$ iff $\neg\mathfrak B$ is true for $M$

shintuku

instantiate your universal variable

all of them

and then do hypotheticals

Sha Vuklia

$\neg\mathfrak B$ being true for $M$ means that for each sequence $s$ we have that $s$ satisfies $\neg\mathfrak B$

PM 2Ring

@shintuku Mandelbrot was investigating that sort of thing in the complex plane when he discovered his eponymous set. He was investigating quadratic Julia sets. Sometimes a Julia converges to a connected blob, but sometimes it falls apart into a Cantor dust. The border of the Mandelbrot set is the border of those two behaviours: Julia sets of $z \Leftarrow z^2 + c$ with $c$ inside the M set are connected.

Sha Vuklia

15:03

while $\mathfrak B$ being false for $M$ means that for each sequence $s$ we have that $s$ does not satisfy $\mathfrak B$

so I wrote: $S(x)$ means that $x$ is a sequence

and for all x, S(x)->A(x) means that every sequence x satisfies $\mathfrak B$

and for all x, S(x)->B(x) means that every sequence x does not satisfy $\mathfrak B$

so $B(x)=\neg A(x)$

and I wanted to formalise the argument a bit

shintuku

@PM2Ring i was thinking about the cantor set too, I'll search for mandelbrot set and julia set, thanks!

@ShaVuklia instantiate your universal variables and do hypotheticals

Sha Vuklia

The argument is: a sequence $s$ satisfies $\neg \mathfrak B$ iff $s$ does not satisfy $\mathfrak B$. Hence, all sequences satisfy $\neg\mathfrak B$ iff no sequence satisfies $\mathfrak B$; that is, $\neg\mathfrak B$ is true iff $\mathfrak B$ is false

@shintuku thx, imma look up what that means

shintuku

"Given a: Sa -> Aa"; "Given a: Sa -> [Aa <-> Ba]"

PM 2Ring

@shintuku No worries. IMHO, it makes more sense to look at this stuff on the complex plane, rather than restricting it to the real line.

Just looking at the real part of a function is like trying to study a scene by looking at it through the gap between two fence palings. You get tantalizing glimpses, but it's hard to make out what's really going on. ;)

shintuku

@ShaVuklia you might find the rules I mention under the names "-> introduction" and "∀ elimination"

Koro

15:12

0

Q: Showing that $\int_0^{\infty}\sin t^2\,dt$ converges

I want to show that $\int_0^{\infty}\sin t^2\,dt$ exists. I know that we define the above integral as $\lim_{b\to \infty}\int_0^b\sin t^2\,dt$, if it exists finitely. Let $I:=\lim_{b\to \infty}\int_0^b\sin t^2\,dt=\lim_{b\to \infty}(\int_0^1\sin t^2\,dt+\int_1^b\sin t^2\,dt)$ We can do a variable...

is my proof correct? thanks.

Sha Vuklia

@shintuku thx, let me see

shintuku

example (i'm using fitch-style deductive system):

Given b:
	Sb -> Ab
	Sb -> [Ab <-> Bb]
	Suppose Sb:
		Ab ^ [Ab <-> Bb]
		Ab
		Ab <-> Bb
		Bb (from the previous two lines)
	Sb -> Bb (we've supposed Sb, and we've acquired Bb through this supposition)
∀x Sx -> Bx

the premises being those you mentioned

last line is from the fact $b$ was arbitrary

Sha Vuklia

Thanks, that is amazing. I will tell you if it made sense as soon as I've found a good list with the rules I need

shintuku

@PM2Ring hm, would you go as far as to say that the continuity of the real numbers depends on the complex numbers?

SAJW

@PM2Ring why are we using arccot(x)? Isn't that using Pi, so we can't really use it for calculating Pi? I mean the equation: $arccot(x)=\pi/2-\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{2k+1}$

PM 2Ring

15:23

@shintuku I'd say that the real line is just a 1D slice of the full behaviour of the function, and the patterns on that slice make more sense when you can see what's happening on the plane.

@SAJW No, we don't use that equation with pi in it. :) We just use the reciprocal of x in the arctan series.

$$arccot(x) = 1/x - 1/3x^3 + 1/5x^5 - 1/7x^7 + \ldots$$

SAJW

ah ok, now I understand. I think

geocalc33

can all of math be done in a multiplicative manner?

PM 2Ring

So it's really just the arctan series that I used in that Python program.

@geocalc33 No, you need both addition & multiplication to make things interesting.

If you have a group with one operation, you're free to call it addition or multiplication. But if you have a field, you need both.

Sha Vuklia

@shintuku Thanks, I figured it out now (how do to it with the axioms and inference rules in my book) by using the structure of your proof. That was very helpful!

shintuku

np!

geocalc33

15:38

@PM2Ring okay gotcha

Sha Vuklia

@LeakyNun sry, there were rules after all, but I didn't realise it

Leaky Nun

ok

PM 2Ring

@SAJW BTW, if you want to see a pi program that can calculate many thousands of digits quickly, take a look at stackoverflow.com/a/26478803/4014959 I kind of understand why that algorithm works, but not well enough to explain it. But it's hard to resist a pi program that doubles the number of correct digits on every loop. :)

geocalc33

@PM2Ring what if you had isomorphic binary structures to addition and multiplication?

SAJW

@PM2Ring it was in general not about the actual formula, but rather the implementation. Still looking at it ;)

PM 2Ring

15:49

@SAJW I've written lots of different pi programs over the years. It's a strange hobby, but oddly satisfying. About 30 years ago, I used a(10), a(239) & a(515) to compute 12 digits of pi by hand, with no computer or calculator. :)

@geocalc33 Well, that's ultimately just addition & multiplication, isn't it.

SAJW

@PM2Ring I worked myself on a geometric solution. it divides the circle in 4, then in 8, then in 16 ect. and always takes the new circumference for pi=C/d. It's also easy to show that Pi must be close to 3 with that. (make a hexagon with sidelength 1 which is close to a circle)

my algroithm was probably slow, can't remember

and it got to the numberlimit of the pc fast

so after a point it multiplies with 0

PM 2Ring

@SAJW That's basically what Archimedes did, although he might've started with hexagons rather than squares.

SAJW

@PM2Ring I'm not that arrogant to think noone has done it before, but I can say I didn't cheat :) (because internet that days sucked)

PM 2Ring

Most pi algorithms ultimately come down to estimating either the circumference or the area of a circle, in various ways. But sometimes the circle underlying the mathematics isn't easy to see.

One historical reason for the popularity of using arccot p

SAJW

@PM2Ring but where can I look up this arccot(x)=arccot(y)+arccot(z)?

or arctan for that matter

PM 2Ring

16:04

One historical reason for the popularity of using arccot is that it avoids square roots. Doing stuff with polygons, you tend to run into square roots a lot. And before computers, square roots were annoying to calculate.

@SAJW I'm not sure what you mean. Do you understand that stuff I posted earlier? Are you familiar with that formula I used for tan(U+V) ?

SAJW

@PM2Ring it's magic to me, is that part of the definition of tan? never came across it

PM 2Ring

16:19

@SAJW Oh, ok. No, it's not part of the definition of tan. You can prove it using geometry. Wikipedia has some proofs. en.wikipedia.org/wiki/Proofs_of_trigonometric_identities But Wikipedia can be a bit overwhelming when you're trying to learn new mathematics. There are formulas for sin(A+B) and cos(A+B). They're a little bit easier to derive from geometry. And using those formulas you can easily make the formula for tan(A+B)

When I was at school, we learned those angle sum formulas, and how to derive them, in the 3rd or 4th year of high school.

This page may be helpful:

108

Q: How can I understand and prove the "sum and difference formulas" in trigonometry?

The "sum and difference" formulas often come in handy, but it's not immediately obvious that they would be true. \begin{align} \sin(\alpha \pm \beta) &= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\ \cos(\alpha \pm \beta) &= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta \end{align} S...

intuition trigonometry

SAJW

we learned only to calculate in 5th-10th grade, I can't recall any proves done

Jam

the polynomials that cut a line as a projective variety in the n-projective spane shouldnt be homogeneous of degree 1?

SAJW

16:37

ah so cos(angle) is a length?

PM 2Ring

@SAJW The cos, sin, and tan functions are ratios of lengths. But it's often convenient to work with a right triangle with a hypotenuse that has a length of 1 unit.

schn

16:56

@robjohn , since you wrote the limit with respect to $hu$, I assumed all occurrences of $h$ and $u$ would be as $hu$. However, I guess writing the limit with respect to $hu$ stems from the Taylor expansion, which only holds as $hu\to 0$. Could you clarify why $f(hu)$ instead of $f(h,u)$ eliminates the need for both $h$ and $u$?

That is, if all occurrences of $h$ and $u$ were as $hu$.

robjohn

@schn little-o is a one variable concept. $f(x)\in o(g(x))$ means $\lim\limits_{x\to0}\frac{f(x)}{g(x)}=0$. Any extension to two variables, must be viewed through that limit. Any extension to more variables is highly dependent on the context. Trying to remove it from that context is often troublesome.

SAJW

17:14

can alot of sums that start at n=1 be rewritten such that they start at n=0 and vice versa?

in $\sum_{n=1}^{\infty}$

robjohn

@schn In the question, where we seem to be stuck, we look at the error term in a Taylor series, which is $o\left((hu)^m\right)$. Unfortunately, this little-o estimate is only valid for sufficiently small values of $hu$ and it is being used in an integral over $\mathbb{R}$. To actually use it in that context requires a more global bound.

for some $g$ so that $\lim\limits_{x\to0}g(x)=0$,
$$
\begin{align}
\int_{\mathbb{R}}k(u)\,o\!\left((hu)^m\right)\,\mathrm{d}u
&=\int_{\mathbb{R}}k(u)(hu)^mg(hu)\,\mathrm{d}u\\
&=h^m\int_{\mathbb{R}}k(u)u^mg(hu)\,\mathrm{d}u\\
\end{align}
$$
but without knowing more about how $g$ behaves away from $0$, we cannot make any further estimates.

SAJW

17:47

is it ok to post questionlinks here from math SE?

or when is it ok...

Clarinetist

@SAJW Just post it

robjohn

@SAJW it would be better to post a link to it if you wish to discuss it, but to get more attention for your question, it is better to post a bounty.

SAJW

´@robjohn ah ok, fair enough

SAJW

18:16

Can you give an example of a used sum that doesn't start at 0 or 1?

(in physics or biology or some other science)

robjohn

@SAJW You can reindex a sum to make it start anywhere you want.

SAJW

Yeah I got that. But what is the point of doing so?

It seems to me, that starting a sum at anything other than 0 is just laziness.

Ted Shifrin

18:33

How about $\sum \frac 1{\sqrt{n^2+n-4}}$?

Koro

Is my proof here correct?

3 hours ago, by Koro

0

Q: Showing that $\int_0^{\infty}\sin t^2\,dt$ converges

I want to show that $\int_0^{\infty}\sin t^2\,dt$ exists. I know that we define the above integral as $\lim_{b\to \infty}\int_0^b\sin t^2\,dt$, if it exists finitely. Let $I:=\lim_{b\to \infty}\int_0^b\sin t^2\,dt=\lim_{b\to \infty}(\int_0^1\sin t^2\,dt+\int_1^b\sin t^2\,dt)$ We can do a variable...

real-analysis solution-verification improper-integrals cauchy-sequences

Ted Shifrin

Way too belabored. I would do river’s deleted argument.

Surely we know the notion of comparison tests.

Koro

Ted, this is part of an exercise problem in Rudin's PMA

and till that chapter, I don't think that comparison test for integrals has been introduced.

Ted Shifrin

You do not need Cauchy sequences to please Rudin.

Fine. Tie both hands behind your back, but leave me out of it.

robjohn

@Koro The comparison test won't apply to that integral.

Koro

18:45

@TedShifrin I'm afraid I can't see river's deleted argument. I think you can see that and professor Rob too

@robjohn I have no idea about that but I think my approach is rigorous enough. Isn't it, Professor Rob?

as I haven't covered comparison test for integrals in much detail yet.

Ted Shifrin

You’re doing the same comparison in your argument, so don’t use that excuse.

You don’t need Cauchy sequences. You just need the fact that a bounded monotone sequence converges.

Koro

but here I don't have a monotonic sequence

@TedShifrin ??

Ted Shifrin

Why does the improper integral $\int_1^\infty x^pdx$ converge when $p<-1$?

Koro

because $\lim_{b\to \infty} \int_0^b x^p \,dx$ exists finitely when $p\lt -1$

but I see where you're going. Integral comparison test

iff the series $\sum \frac 1 {n^{-p}}$ converges

Ted Shifrin

Why does the limit really exist? Monotone sequences bounded above. That’s comparison. Forget Cauchy nonsense.

Koro

18:59

The limit exists because: $x^p$ is continuous on $[1,b]$ for all $b\gt 1$ so $x^p$ is Riemann integrable on $[1,b]$ and in fact $\int _1^b x^p\,dx=\frac 1{p+1} (b^{p+1}-1)$, which has a finite limit as $b\to \infty$ if $p\lt -1$.

And why don't you like Cauchy?

Ted Shifrin

I like Cauchy when it’s necessary and insightful. Not when you’re using more elementary means and clothing them as Cauchy.

robjohn

@Koro proving that $\lim\limits_{x\to\infty}f(x)$ exists by showing $f(x_n)$ is Cauchy for all sequences tending to infinity is messy because you need to show they all tend to the same limit. It can be done, but it is messy and unnecessarily difficult.

Koro

In this case, it's not. I have taken that in consideration in my proof

robjohn

If you're open to other approaches, I can post one, but it is based on comparison.

Koro

Why it's not difficult in this case is probably because integrand is bounded

@robjohn okay :)

robjohn

19:22

done

Koro

20:18

@robjohn thanks a lot :)

I think that there's a typo in 1b as you're making a substitution there so I think integral limits should change.

Avra

20:30

Hello, given that geometric series is $\sum_{i=0}^{\infty} (kx^k) = \frac{x}{(1-x)^2}$ after differentiating both sides and multiplying by $x$, then how we get that $\sum_{h=0}^{\infty}\frac{h}{2^h} = \frac{\frac{1}{2}}{(1-\frac{1}{2})^2}$

i.e. based on the formula, we should get $\frac{2}{(1-2)^2}$?

where $\frac{1}{2}$ came from above please?

Ted Shifrin

22:00

@Avra: Look carefully. You have $h(1/2)^h$, not $h2^h$.

shintuku

22:52

for $f$ injective, it makes sense that $u \in U \implies f(u) \in f(U)$, but how exactly would I justify this?

i can draw a neat little diagram but this wouldn't hold in court

i'll do it through contradiction, but if someone knows a direct way, you're cool and you should tell me how

Ted Shifrin

Why does this have anything to do with injectivity?

Do you perhaps intend the converse?

shintuku

23:07

wait you're right, that's just what the image means

haha, i'm tired

thanks

Avra

23:40

@TedShifrin. Thank you Professor

Ruochan Liu

23:58

Hi, I was wondering why any dense, non-compact subset of $\prod I$ (an uncountable product of unit segments, which I know to be not first-countable), is not first-countable? I'm not sure the non-compact condition is necessary. Thank you!

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$Mathematics$ Mathematics