Mathematics - 2017-03-10 (page 4 of 5)

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19:00

Forgot you raise the $-1$ to the power of the dimension :(

Mike Miller

❤️

Astyx

@CausingUnderflowsEverywhere Was that second part addressed to me ?

DHMO

@MikeMiller wait what... characters can have default color?!

user228700

19:39

Hi, everyone :-)

Simply Beautiful Art

o/

Astyx

Hi

user228700

I have a quick-ish question about a statement given in my book about the monotonicity of functions. This is the statement:

user228700

> "Let $I$ be an interval which is a subset of the domain of a function $f$. If $f^{'}(x)>0, \forall x \in I$, except for countably many points where $f^{'}(x)=0$, then $f(x)$ is strictly increasing in $I$"

user228700

If there are several countable points in $I$ at which $f^{'}(x)=0$, how can this function be classified as strictly increasing in $I$?

Semiclassical

19:43

Consider $f(x)=x^3$.

DHMO

@Kaumudi.H for example, consider $f(x)=x^3$

Semiclassical

loool

DHMO

and $I=[-1,1]$

Balarka Sen

I was going to give the same example but...

DHMO

$f'(x)=0$ when $x=0$

Balarka Sen

19:44

Too fast for me

Semiclassical

It is the obvious example, to be fair.

Astyx

Consider $f(x) = x^5$

DHMO

but the function is still strictly increasing

Balarka Sen

Anyhow, the point is there are only isolated number of points where $f' = 0$ so it doesn't matter

Daminark

Hey everyone!

DHMO

19:44

because countably many points can't form an interval

the definition of strictly increasing if $\forall a,b: a<b \implies f(a)<f(b)$

user228700

Wow, that was fast. Right, so "countable" means just one/two single points here and there?

DHMO

not really, that would be finite

Semiclassical

Not necessarily.

DHMO

countable means finite or bijects with the integers

Balarka Sen

Yes. It may not be one two: Q is countable.

Astyx

19:45

$\Bbb Q$ is countable and dense in $\Bbb R$

Semiclassical

It's just that the easiest example here is with only one.

(There's probably a good example with countably-infinite points where $f'(x)=0$ and $f'(x)>0$ otherwise.)

DHMO

I wonder if we can construct an example with $\Bbb Q$

user228700

I'm failing to understand this. Are u saying that a function is still "strictly increasing" if $f^{'}(x)$ is zero for a whole interval?

Balarka Sen

Hmm. $f' = 0$ on every point on $\Bbb Q$ implies (C^1 assumed) $f'= 0$ throughout, right? So you're not strictly increasing for that tho.

DHMO

@Kaumudi.H i'm saying that it can't be a whole interval

and you can't make a whole interval with just countably many points

which is why "countably many points" works

user228700

19:47

Alright, so it seems that I don't understand what "countably many points" means...

Balarka Sen

@DHMO Right, so there's no C^1 example. There should be differentiable examples.

Daminark

It might be easier to find examples in the integers?

Balarka Sen

@Kaumudi Google.

DHMO

@Daminark just "copy-and-paste" the $f(x)=x^3$

Balarka Sen

"Countable set" is one which is in bijection with N

user228700

19:47

@BalarkaSen ::Googling::

Semiclassical

Yeah, an integer case should be easy enough.

Probably some variation on x+sin(x) should work.

Balarka Sen

Sure, N is easy.

Daminark

If you stitch together $x^3$ in some sense, as @DHMO said, that'd probably work

Doing things in dense sets is some messy business

Actually an interesting problem we had last quarter was to prove that there was no function continuous only on $\mathbb{Q}$

user228700

> In bijection with N

user228700

What?

Balarka Sen

19:49

Don't you know what a bijection is?

DHMO

@Kaumudi.H it means you can define a one-to-one mapping from the set to N

meaning it only maps one element from the set to one element in N

user228700

Right, yeah, OK, I'm an idiot.

Balarka Sen

@DHMO No one-to-one is injection.

One-to-one AND onto.

DHMO

@BalarkaSen it can mean both, depending on context

Daminark

You need it to be surjective as well

Balarka Sen

19:50

But I guess I am thinking of countably infinite.

So DHMO is right.

Astyx

Yeah, terminology is not very strict here AFAIK

Daminark

Lol we need to establish in this chat what the terminology will be and declare it binding on all mathematicians

DHMO

@Kaumudi.H are there more integers or more even integers?

Astyx

Depends how you define "more"

user228700

@DHMO Are u trying to explain to me what "countably infinite" means?

Akiva Weinberger

19:53

https://en.wikipedia.org/wiki/Minkowski's_question_mark_function

@DHMO @BalarkaSen

Zero derivative on the rationals.

DHMO

@AkivaWeinberger ?

Balarka Sen

@AkivaWeinberger Crazy.

Akiva Weinberger

@DHMO Yes

DHMO

@Kaumudi.H yes

@AkivaWeinberger I was trying to make a joke response from the name of the function

user228700

Is it really important to understand the statement given in my book?

DHMO

19:54

@Kaumudi.H heh

Balarka Sen

Not if you don't want to.

Akiva Weinberger

@DHMO Yes, I know

user228700

I'm running low on time so I think I'll put that off for another day.

Akiva Weinberger

I wonder if one could build a contraption that lets them steal from vending machines

Semiclassical

What surprises me a bit is that the set for which $f'(x)=0$ only has to be countable.

Balarka Sen

19:55

"Come back another day to another person"

Akiva Weinberger

without smashing the glass or anything

Balarka Sen

@Semiclassical I don't think this is true without any assumption on $f$.

DHMO

@BalarkaSen can it have zero derivatives on the cantor set instead?

Astyx

What's an example of a strictly increasing differentiable function of which the derivative is zero at every rationnal ?

Akiva Weinberger

@Semiclassical I think it might be true for any null set

Semiclassical

19:56

Hmm.

Akiva Weinberger

@Astyx I gave it above

Astyx

Oh thanks

Akiva Weinberger

https://en.wikipedia.org/wiki/Minkowski's_question_mark_function

Semiclassical

What bothers me is that said countable set could still have a limit point.

Balarka Sen

@DHMO I think so. Take the bump function on each step in the Cantor set, and arrange so that it has strictly decreasing height. Uniform limit should do it.

DHMO

19:57

@BalarkaSen so it's not necessary to be countable @Kaumudi.H

Semiclassical

But intuition doesn't count for a lot in real analysis.

user228700

I still don't really get the statement in my book though. I'm asking again because I'm confused--does "countably many points" just mean that all these points can't be in a single interval within the subset of domain we're considering?

DHMO

@BalarkaSen is "measure 0" a necessary condition?

Akiva Weinberger

@Astyx I don't think it's differentiable actually

user228700

'Cause my textbook follows up with this:

DHMO

19:57

@Kaumudi.H "countably many points" just means "biject with N"

Balarka Sen

@DHMO I don't see why you can't do the same thing for Smith-Volterra Cantor sets.

Semiclassical

Well, there are two easy applications of said result and a hard one.

DHMO

@BalarkaSen because I've never heard of it

Balarka Sen

They are Cantor sets of positive measure.

DHMO

I figured it out

Balarka Sen

19:58

I am not really sure of the construction I have in mind anyway. Whatever.

user228700

> "$f^{'}(x)>0$ at countably many points implies that $f^{'}(x)=0$ does not occur on an interval which is a subset of $I$"

DHMO

that is crazy

Semiclassical

The first easy one is $f(x)=x^3$ as above. There you have exactly one point where $f'(x)>0$ fails, and the above indicates that $f(x)$ is increasing.

DHMO

@Kaumudi.H because an interval must have uncountably many points

Semiclassical

Another would amount to: Take the graph of $f(x)=x^3$, and insert a flat segment at the origin.

Balarka Sen

20:00

Isolated is good intuition but countable sets need not have isolated points. Rationals are countable!

DHMO

:35957064 that's a handwavey intuition but I think that's good enough for you

Semiclassical

Then it would fail to be increasing since the interval for which $f'(x)>0$ is not countably infinite (it's just an interval of real numbers).

Balarka Sen

$\{0, 1, 1/2, 1/3, 1/4, ...\}$ is countable too - and you can convince yourself that $0$ is not isolated.

Astyx

@Kaumudi.H As a matter of fact for the statement you only need to know that there is no interval on which $f'=0$

DHMO

because you're only studying for JEE right? @Kaumudi.H

Semiclassical

20:01

What's hard is: Can you have an example for which there are an infinite number of points where $f'(x)>0$ but for which $f(x)$ is still increasing?

Balarka Sen

@Semiclassical Huh? Like $f(x) = x$.

DHMO

@Semiclassical I don't understand this

Balarka Sen

Did you mean something else?

user228700

@BalarkaSen And if $f^{'}(x)=0$ at these given points, it still counts as a strictly increasing function?

Semiclassical

Yeah, meant to say "where $f'(x)=0$."

Woops.

DHMO

20:01

we just discussed this above

we gave two solutions

@Kaumudi.H yes

Balarka Sen

@Kaumudi If you can give a function such that $f' = 0$ only at those points and $f' > 0$ otherwise, then yes.

Semiclassical

I should also amend that to "infinite number of points within a finite interval."

user228700

Alright, I think I get it.

DHMO

@Semiclassical that solution still applies

user228700

Thanks so much, guys :-)

Xam

20:02

Hello :)

DHMO

there are infinitely many rationals in any given interval @Semiclassical

Alessandro Codenotti

What's the exact question being discussed about f'(x)=0?

Semiclassical

@DHMO Sure, but so would "f'(x)=0 on the integers".

Balarka Sen

Nobody knows

There are multiple questions going on at once

@Semiclassical Not within an interval

DHMO

@Semiclassical what?

Semiclassical

20:03

I meant something simple.

DHMO

10 mins ago, by Akiva Weinberger

https://en.wikipedia.org/wiki/Minkowski's_question_mark_function

Astyx

This conversation is becoming more and more confusing

Balarka Sen

I think there's a simple example with 0, 1, 1/2, 1/3, ...

DHMO

@BalarkaSen does it have to do with sin?

Alessandro Codenotti

@BalarkaSen great!

Semiclassical

20:04

"Can you have an example for which there are an infinite number of points where $f′(x)=0$ but for which $f(x)$ is still increasing?" My point was that that's the relevant (hard) problem.

DHMO

^ and he means simple example

not ?(x)

Semiclassical

Well, I'm fine with ?(x). I just wanted to make explicit what the challenge was.

I was just summing up the conversation :/

Balarka Sen

@DHMO How about $x\sin^3(1/x)$

Semiclassical

Works for me, if it's valid.

Balarka Sen

Make it $x^2\sin^3(1/x)$ if you want, just to be C^1 at 0.

Zach Hauk

20:07

Hi

Semiclassical

That's not an increasing function, though (either version).

DHMO

@BalarkaSen I don't think either works

Astyx

^

Balarka Sen

Take modulus.

$x^2|\sin(1/x)|^3$

Semiclassical

Won't help. That function has zeros.

DHMO

20:08

still doesn't work

Balarka Sen

Boo.

Astyx

You can add a linear function to make it increasing or something ?

Balarka Sen

Hmm, yes.

$x^2\sin(1/x)^3 + x$

Not quite.

Astyx

Why $^3$ ?

Balarka Sen

To have the zero derivative phenomenon.

Like in $f(x) = x^3$

Astyx

20:10

I don't get it

DHMO

I think the integral of $x\sin \left(\frac{\pi }{x}\right)+x$ would work

Astyx

$2x$ rather ?@Balarka

Oh no

Balarka Sen

Ok, that's the wrong counterexample. There's no interval around $0$ in which it's increasing.

DHMO

@Semiclassical $f\left(x\right)=2\pi x+\sin \left(2\pi x\right)$ although it gives half-integer instead

Semiclassical

$f(x)=\pi x+\sin(\pi x)$ gives an example.

Yeah.

DHMO

20:12

bye

Astyx

It doesn't, (it has only one zero or something) but you can add polynomials so that every local minimum of the derivative of $f(x) = x^2\sin({1\over x})$ is 0

Semiclassical

so many deleted messages...

Balarka Sen

Ah, great.

Semiclassical

Won't $f(x)=\pi/x+\sin(\pi/x)$ then give an example over [0,1] then?

Balarka Sen

But I wanted it inside [-1, 1]

Akiva Weinberger

20:13

$x^2\sin(1/x)+x$, ya dinguses

Balarka Sen

@Semiclassical I think so

Nah, bad at x = 0

Astyx

By polynomial, I mean a function $g$ that is differentiable and a polynomial ae

Akiva Weinberger

@BalarkaSen Mine?

Semiclassical

Mine.

Balarka Sen

@AkivaWeinberger Sounds cool!

Semiclassical

20:14

To modify mine to work, I'd need to replace $\pi x+\sin(\pi x)$ with something that's bounded.

Astyx

@Akiva It has negative derivatives at some points

That's the problem, the $2x\sin({1\over x})$ makes every thing go wrong

Semiclassical

$f(x)=1/(1 + 1/x + \sin(1/x))$ works for some purposes.

Akiva Weinberger

Oh :(

Balarka Sen

we're all rekt

Akiva Weinberger

$\int_0^x(1-\cos(1/t))dt$?

Astyx

20:17

Oh yeah it looks quite cool @Semi

Did you come up with this ?

Semiclassical

Yeah.

Astyx

How ?

Balarka Sen

It sounds pretty easy to make Akiva's example work by visually modifying the bad parts.

Semiclassical

I took the x+sin(x) version, replaced x->1/x to map the integers to reciprocals, and then plugged it into 1/(1+x) so that it'd be bounded.

Astyx

Huh, I don't get the concept, but cool :p

Semiclassical

20:20

so 1/(1+f(1/x))

Akiva Weinberger

@Semiclassical Ah, that's the best way, I think

Balarka Sen

@Astyx x+sin(x) works for vanishing set of derivative being N\pi. So you just need to make it fit inside [-1, 1]

Astyx

And is it differentiable at 0 ?

Semiclassical

Should be.

Astyx

@BalarkaSen Yeah I figured

Balarka Sen

20:20

It should even be C^1 at 0

Astyx

Doesn't look like it

It looks a bit like $x^2\sin({1\over x})$

Semiclassical

The plot is fun if nothing else: wolframalpha.com/input/…

Balarka Sen

Which is C^1 at 0, right?

Astyx

No ..?

Semiclassical

Though the interesting action is really in [-1/pi,1/pi]

Balarka Sen

20:22

Oh, I guess you're right.

Akiva Weinberger

It's bounded below by $1/(2+\frac1x)$ and above by $x$.

Astyx

The $-\cos({1\over x})$ term makes everything go wrong

Akiva Weinberger

So it's differentiable, at least.

Astyx

It still satisfies the IVT though (necessarily)

Akiva Weinberger

Yeah, it's not C^1

Semiclassical

20:22

Hmm.

Oh well.

Eric

Can't you just take any zero set $F$ you want, take $g(x) = dist(F, x)$, then then take $\int_{0}^{x} g(t)dt$.

Astyx

And the derivative would be 1 @Akiva ?

Akiva Weinberger

@Astyx Yeah

Astyx

Right

Balarka Sen

@Eric Yeah.

Akiva Weinberger

20:24

@Eric Yeah, I think so. I think we want an elementary function, though

Eric

ah ok

Balarka Sen

Which was the merit of Akiva's example back there.

Semiclassical

As elementary as we can make it, anyways.

As a C^0 example 1/(1+1/x+sin(1/x)) works, though?

Akiva Weinberger

@BalarkaSen Well, Semi's

Balarka Sen

You can actually bootstrap Eric's example to be smooth. Any closed set is zero set of a smooth function in R.

So you just integrate that.

@Akiva I meant your integral example.

Akiva Weinberger

20:26

Mine failed

Astyx

He meant the ? function I think

Semiclassical

I can live with a C^0 example, I think :)

Mike Miller

what are you trying5o do

Balarka Sen

@AkivaWeinberger This.

Astyx

Find a strictly increasing function of which the derivative cancels at an infinite number of point in an interval @MikeMiller

An elementary function

Balarka Sen

20:28

@Eric I guess this won't be strictly increasing in general, would it?

Nah, of course it would.

Eric

no not in general

Don't disturb

@Secret then calculate $$\sum_{n=1}^{\infty} (-1)^{n-1} \left(\frac{H_n}{n}\right)^3$$ Do you think it is possible to do that without using further auxiliary series whose closed-forms are not known?

Eric

oh yeah no sorry

integrals

Balarka Sen

Distance away from the set would be positive and that's exactly the derivative so yeah.

Semiclassical

Actually, maybe the approach i gave above is still useful. There I did 1/(1+f(1/x)) with f(x)=x+sin(x).

Akiva Weinberger

20:29

Can we get the derivative to vanish at the Cantor set? Or did we rule that out?

No, it should be doable

Semiclassical

But I could've done another function and taken g(f(1/x)).

Eric

i mean unless you just took $F$ to be something too big

Balarka Sen

@AkivaWeinberger You should be able to cook up a variant of the Cantor function.

Semiclassical

Is there another g(x) which would give g(f(1/x)) being smooth at zero?

Astyx

@Akiva Doesn't Eric's answer answer that ?

Balarka Sen

20:30

I was thinking of doing bump function on the complement of each step in the construction of the Cantor set, of smaller and smaller height and then take uniform limit.

Semiclassical

I have in mind the usual bump function e^(-1/x^2).

Balarka Sen

@Astyx I think he means an easier example.

"easier", whatever that means.

Paradox 101

Hi, can anyone help me with a problem concerning game theory?

Astyx

Yeah, especially when talking about the Cantor set :p

Semiclassical

For a smooth example on [0,infty) I think $\exp(-1/(1/x+\sin(1/x))^2)$ works, in fact.

Don't disturb

20:31

@Secret here is also the art $$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n H_n^{(1)} H_n^{(2)}}{n}$$

Then

Semiclassical

Actually, I can be even simpler: $\exp((1/x+\sin(1/x))$.

Don't disturb

$$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n H_n^{(1)} H_n^{(2)}\cdots H_n^{(k)}}{n}$$

Mike Miller

You can't get the cantor set to be the zero set of any analytic function (so of course not as the derivative of such), and elementary functions are analytic on their domains except for at (I think) a discrete set of points

Semiclassical

...bah, no, ignore me. that doesn't fix things at zero.

Balarka Sen

Well, that depends on what elementary functions mean but yeah fair point.

Don't disturb

20:33

@Semiclassical you have above some silly series to take

Semiclassical

Not that interested, really.

Don't disturb

@Semiclassical aha

Semiclassical

?

Don't disturb

I'm out to continue my research

Balarka Sen

lolwut

Astyx

20:35

Ehe

Akiva Weinberger

There's a $C^\infty$ function that's flat on the Cantor set, using bump functions like @BalarkaSen said

So, not analytic or elementary, but something at least

Don't disturb

@BalarkaSen is it better now?

I'm out now.

Balarka Sen

i wasn't laughing at your english sentence. it sounded fine

Mike Miller

Analytic can only get you discrete closed zero set

Akiva Weinberger

Huh, interesting. That makes sense, I guess

Balarka Sen

20:37

It's the identity theorem.

Akiva Weinberger

I vaguely remember that

Balarka Sen

Well that was a long long conversation with a lot of garbage said and garbage fixed

Akiva Weinberger

So it can have arbitrarily high measure.

By switching to a fat Cantor set.

Balarka Sen

Yup.

Akiva Weinberger

But it can't be conull.

(I think.)

(Conull meaning complement of measure zero)

Zach Hauk

20:46

by that definition, the irrationals would be conull, right?

Balarka Sen

Ya.

@AkivaWeinberger Conull guys are dense, right? So, for one, you can't have C^1 examples - if f' = 0 on a dense set you could extend by continuity to f' = 0 everywhere. I am not sure how to prove for differentiable.

conull in [0, 1] I mean

Akiva Weinberger

Oh, duh, Cantor staircase. (If you don't require differentiability everywhere).

Balarka Sen

Can you modify it to be diff?

Akiva Weinberger

Honestly doubt it

I'll try to see if I can prove it impossible.

@BalarkaSen It would have to be the Riemann integral of its derivative, right?

Balarka Sen

Who's to say if the derivative is continuous on a measure zero set?

Akiva Weinberger

20:52

(Up to constants.) If we can show that the Riemann integral is zero, we're done.

@BalarkaSen Oh, we need C^1 for the FTC?

Balarka Sen

Ah, I guess that's just by assumption.

@AkivaWeinberger Well to make sense of Riemann integral you need continuity outside a measure zero set.

But I guess this is automatic here, because it's 0 everywhere except a measure zero set.

Akiva Weinberger

My phone's about to run out of battery. Continue this conversation later?

Balarka Sen

Sure.

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$Mathematics$ Mathematics