Mathematics - 2012-03-10 (page 3 of 3)

Kannappan Sampath

15:00

@AsafKaragila Definitely yes. : )

Asaf Karagila

@KannappanSampath Excellent. I will ping you when I return.

Skullpatrol

@robjohn Have you ever read any of Donald Knuth's The Art of Computer Programming books Rob?

Kannappan Sampath

Sure. :-)

Skullpatrol

The Art of Computer Programming

The Art of Computer Programming (acronym: TAOCP) is a comprehensive monograph written by Donald Knuth that covers many kinds of programming algorithms and their analysis. Knuth began the project, originally conceived as a single book with twelve chapters, in 1962. The first three of what were then expected to be a seven-volume set were published in 1968, 1969, and 1973. The first installment of Volume 4 (a paperback fascicle) was published in 2005. The hardback volume 4A was published in 2011. Additional fascicle installments are planned for release approximately biannually. History ...

robjohn

@Skullpatrol I have them (at least the first 3)

Matt N.

15:07

@tb Good : )

You're quiet today.

Ah, must be the phone.

Will you disappear for long on Monday?

t.b.

Am I being too harsh by suggesting that this question is off-topic? I don't see mathematical content there.

Matt N.

Not sure. It does seem to have mathematical content. I'll leave the decision to others.

t.b.

@MattN Yeah, I'm constantly distracted... I was just saying that I have plenty of time until Monday morning, after that I won't be able to be around that much.

@MattN how is there mathematics? It's not used in a technical sense.

Daniil

hi

t.b.

hi Daniil

Skullpatrol

15:14

@robjohn I found the first one too challenging and gave up. The same thing happened to me with the Feynman Lectures, have you read those to?

The Feynman Lectures on Physics

The Feynman Lectures on Physics is a 1964 physics textbook by Richard P. Feynman, Robert B. Leighton and Matthew Sands, based upon the lectures given by Feynman to undergraduate students at the California Institute of Technology (Caltech) in 1961–1963. It includes lectures on mathematics, electromagnetism, Newtonian physics, quantum physics, and the relation of physics to other sciences. Six readily accessible chapters were later compiled into a book entitled Six Easy Pieces: Essentials of Physics Explained by Its Most Brilliant Teacher, and six more in Six Not So Easy Pieces: Einstein's ...

Jeff

hi again all. does mathoverflow.net not use our same universal stackeschange id? I want to favorite this: mathoverflow.net/questions/8846, but can't.

t.b.

@Jeff No, MathOverflow is not part of the stackexchange network. It just uses (an old version of) the same software. But you can create an account easily.

Jeff

well, a new userid won't let me save favorites to my universal Se id. oh well.

i thought there was an expert's math site on Se, too.

Matt N.

@tb I voted. Your second comment answers the question. I think though it's not off topic because the person is trying to understand a text about maths. Clearly, the guys on german.SE or whatever wouldn't be able to help him. So where else can he ask this?

@tb I don't think I like that.

Kannappan Sampath

Holy...., OP does not understand what conjugacy means and uses the terms conjugacy classes. My bad. I am gonna have tough time making him understand as I am more towards believing he won't know group actions as well. : (

t.b.

15:20

@MattN there will be plenty of overlap :) I'll just not be able to be around as much as I was this week.

Matt N.

@tb Ok. Nice : )

Jeff

hey, is it possible to get @mehdi to choose an answer: math.stackexchange.com/a/97279/22544 ?

Matt N.

@tb I like how you consider RL the distraction from chat and not the other way around : )

t.b.

:)

robjohn

15:42

@Skullpatrol I have those, too :-)

Kannappan Sampath

What should I do? OP does not know group actions but posted a question finding the conjugacy classes. Should I write about all of that?

robjohn

off to the park. bbl

Kannappan Sampath

@tb You many have some suggestions to my previous question. =)

Matt N.

@KannappanSampath I'm not sure. I think I might do that but I also think there are people who would give up.

Kannappan Sampath

@MattN I am one of those, largely because, I don't find enough time. :/

I'll write up later when I am dead for the day. Now that I'm alive I can do some CA.

Matt N.

15:49

@KannappanSampath Sounds good! : )

Kannappan Sampath

@MattN Are you going to do it too?

Matt N.

@KannappanSampath No, I'm reading something unrelated and after that I am planning to work on my seminar thingy.

Kannappan Sampath

@MattN Oh, sure, there are some LHFs in AM which I'll TeX up and put it up, me thinks.

Matt N.

@KannappanSampath Sounds good. Maybe you can give it to me to read when you're done : )

Kannappan Sampath

@MattN Surely, yes. :-)

Matt N.

15:55

@tb Silly in the sense that it's outside my power but I would definitely do it if I could. : )

Asaf Karagila

Oh those assholes. It took me like 20 minutes to fall asleep and they had to shoot again.

@Kannappan: So, what can I do for you today?

t.b.

16:17

@MattN sorry, some scripts were bunching up my browser's memory and made it hang itself (I hate it when that happens)...

Matt N.

@tb I'd blame it on Firefox. Since I switched to Chrome my life is okay again. As for your disappearance: I noticed but this time I thought that you'd probably be back in a few moments : )

(as opposed to thinking that evil higher powers had taken you away)

@tb Have you ever tried Chrome at all? I'm not trying to flog it to you, if you think FF is awesome then you should probably stick to it, I'm just curious.

t.b.

16:34

@MattN No, I haven't. Google is only marginally above MS in my view of the world.

Asaf Karagila

:-D

Matt N.

@tb Why is that? Can you elaborate?

Asaf Karagila

@tb So you still use ICQ? :-P

Matt N.

@AsafKaragila Or a telegraph... : )

Asaf Karagila

@MattN Or a homing pigeon.

Jonas Teuwen

16:38

NICE! The Dutch railway company has send me a subscription that expired last October.

After three months, they finally send me something.

Matt N.

@AsafKaragila Or smoke signals.

Asaf Karagila

@JonasTeuwen It would have been nicer if they had sent you whisky.

@MattN Or cave paintings "Teddy was here! 33,176 BC"

2

Matt N.

LOL I was going to write "Or tattooed slaves." next but I guess I can't go any more backwards in the timeline : D

Asaf Karagila

I didn't see that, Jonas.

Matt N.

Me neither.

@AsafKaragila You're so funny! Either that or I'm very easily entertained.

: D

Asaf Karagila

16:42

@MattN Well, you can take the premise that Star Wars happened a long long time ago in a galaxy far away, and assume that means further in time than cave painting. Then you can suggest service droids holding secret video recordings...

"Help me Teddy-wan, you're our only hope..."

Matt N.

: D

t.b.

@MattN Short answer: no, I'm not in the mood. Slightly longer answer: the main difference I see is that they do a very good job at selling themselves as the good guys "don't be evil" instead of coming over as purely capitalistic pricks.

Matt N.

Well, well... : )

t.b.

I'll have to go for today. See you guys either late at night or tomorrow.

Matt N.

Have a nice evening!

t.b.

16:54

Same to you!

Asaf Karagila

Ciao!

Asaf Karagila

17:08

Hi @Kannappan.

robjohn

@AsafKaragila no nap yet?

Asaf Karagila

I tried to sleep, but the jerks from Gaza decided to shoot again...

robjohn

Oh, I see I missed tb. :-(

Asaf Karagila

I slept for like five minutes maybe.

robjohn

@AsafKaragila then explosions entered your dreams?

Asaf Karagila

17:12

@robjohn No, we live just by the sirens. So it's just really loud.

robjohn

@AsafKaragila okay. I saw "shoot" and thought of gunfire.

Asaf Karagila

Well, there was gunfire involved... I think. :-)

robjohn

@AsafKaragila that can ruin your REM pattern

N3buchadnezzar

Hey =)

Very short question,how do I split a sum into odd and even terms? I want to show more explicit that all the odd tems vanish, but I do not know how to properly write it in mathematical termsn

Jeff

hey! you guys all still here, huh? :D

N3buchadnezzar

17:25

The sum is $$ \sum_{k=1}^{\infty} \frac{1}{n}x^n \left[(-1)^{n+1} + 1\right] $$

Jeff

@N3buchadnezzar i think every other term in that sequence is $0$, right? (and I think you meant to put an $n$ under the sigma)

Asaf Karagila

Well. I am going to cook and whatnot. See you guys later.

Jeff

whenever $n$ is even, then $n+1$ is odd and $(-1^{odd}+1)=0$, so I think this sequence only includes terms when $n$ is odd

N3buchadnezzar

@Jeff Yeah, I know

Jeff

...in which case it reduces to $\sum_n \frac 2n x^n$

N3buchadnezzar

17:30

if we call the series f(x), I was wondering how I could split it into f(2x)+f(2x+1)

Jeff

hmmm.....

N3buchadnezzar

And then show that f(2x)=0

Jeff

i haven't any clue :(

N3buchadnezzar

@Jeff ki

Jeff

i'm curious how to do it, though

N3buchadnezzar

17:40

I was thinking that it was just to mess with the indexes, but I am kinda in a hurry. So I just wrote a few lines explaining why the even terms vanish.

Kannappan Sampath

@Asaf Had to be away. Sorry. I am now here.

Jonas Teuwen

@robjohn The reason we can see distributions are normal functions is just because we have approximating sequences which do have our nice representation? :-).

Matt N.

@JonasTeuwen Is this also true for non-tempered distributions?

Jonas Teuwen

Why not?

Matt N.

17:55

Because the function space might not be dense in its dual.

Or might not even be isomorphic to a subset of its dual, no?

Jonas Teuwen

Hmm, might be possible. But for normal distributions it would work.

Matt N.

Does "normal" here have any mathematical definition? What is a normal distribution?

Jonas Teuwen

The test functions are the compactly supported $C^\infty$ ones.

Matt N.

I see.

Asaf Karagila

@KannappanSampath Are you still here?

Kannappan Sampath

18:05

@AsafKaragila Yes. : )

So the doubt I have had is:

Asaf Karagila

I have 20 minutes until I need to put the chicken in the pot, so you have me for this time.

Kannappan Sampath

In the most recent answer you had written on AC, you wrote that: The axiom of choice asserts that if we have a collection $X$ of nonempty sets, then we can choose from each one. What does "choose" mean? It means that there is a function whose domain is $X$ and for every $a \in X$ we have that $f(a) \in a$.

Asaf Karagila

Yes.

Kannappan Sampath

But, once I had asked a question on MO about AC which I subsequently deleted. There was a comment from Ryan Reich that said this is a misunderstanding. We can always choose elements, even infinitely many of them from a set. What is guaranteed by AC is, we can form a set from them. So, I don't see if what you write and this are the same.

If the set is nonempty, it is always possible to pick one or even finitely many elements from it, and this does not invoke AC but is merely a consequence of the rules of logic. AC says that given a set, you can pick elements from each of its members, and from those elements form another set. It's not the picking that's a problem, but the forming. – Ryan Reich Sep 19 at 22:49

Asaf Karagila

What? Link me to that question of yours.

Wait.

You said infinitely, but quoted finitely.

Kannappan Sampath

18:12

@AsafKaragila Oh, sorry the quote is right.

Asaf Karagila

He's wrong there.

Consider $X$ as a collection of nonempty sets. Now consider $\bigcup X$, by the axiom of union it is also a set. Every choice function is a function into $\bigcup X$ and its range defines a subset. By replacement axiom we have that this is a subset of $\bigcup X$.

You also have subsets of $\bigcup X$ which are not the range of a choice function (or transversal sets as Brian calls them).

The axiom of choice (or its fragments) tell you how many choice functions exist and thus how many (partially) transversal sets exist.

I feel, however, that I'm not answering your real question.

Kannappan Sampath

Can we do a bit slowly? I haven't yet digested your third post

Asaf Karagila

Well, you have about ten minutes. :-)

Kannappan Sampath

What are "partially" transversal sets?

Asaf Karagila

Sets that would correspond to choice functions on a proper subset of $X$.

For example singletons, which chose from only one member of $X$.

Kannappan Sampath

18:22

So, Partially transversal sets are images of Choice functions restricted to some proper subset of $X$?

Asaf Karagila

It's an ad-hoc definition. But yes, this is what I meant.

Kannappan Sampath

OK. I think I got this. So, how does this help?

Asaf Karagila

I'm not sure anymore what your question is, I think I got lost in my own words.

What was that deleted question of yours on AC, by the way? (if you have link I could read it there...

Kannappan Sampath

@AsafKaragila Do deleted questions exist on MO?

Asaf Karagila

Oh. I don't have sufficient reputation on MO yet. ;-)

Kannappan Sampath

18:26

I have a .txt file that contains the question, an answer and the comment. Shall I link it here?

Asaf Karagila

Sure. Is that your question to me now as well?

Hm, it seems my time is up. I will be back in a bit, I need to put the chicken on the stove.

Kannappan Sampath

@AsafKaragila Sure. : )

@AsafKaragila No. My question now is: Does picking elements from a collection of sets require AC or not? As you wrote in the recent answer, seems yes. But, Ryan seems to say no.

(Of course, collection is non-empty.)

Asaf Karagila

How many do you want to pick?

I wrote this and that about finitely many choices without AC.

Kannappan Sampath

I want to pick finitely many, say.

Looks like Ryan talks about non-empty sets (collection), we can pick finitely many elements (sets) from it but we cannot pick elements from these sets without AC, generally. Am I right in making this statement?

@Asaf Sorry, here is the ping and my question is just above this. ^

Please ping me, I am another tab. : )

Asaf Karagila

18:45

@KannappanSampath No, you're not right. You can always pick finitely many elements from finitely many sets. The sets only need to be nonempty, and they could be finite or infinite. If you want to choose from infinitely many sets at once then you need the axiom of choice.

Kannappan Sampath

@AsafKaragila I think I am understanding. : )

Let me think about this and get back. Thanks for your time.

Will the user whose Comments I moved to a room on system's prompt also be notified of such a thing having happened?

Asaf Karagila

@KannappanSampath Sure. I discuss this in both the answers I linked you. Read those as well.

robjohn

19:59

@JonasTeuwen there are functions which converge weakly to the distribution

Jonas Teuwen

@robjohn Yes, that's what I mean.

So that is the reason?

Jeff

@Jonas: =0 Have you been sitting there this whole time waiting for an answer or somethin'?

Jonas Teuwen

@robjohn When we write a distribution as a 'normal' function we actually mean such a limit?

@Jeff Yes.

robjohn

@JonasTeuwen well, such functions can represent distributions as well because they are dense in the functionals. But essentially, yes.

Jeff

@JonasTeuwen lol

Jonas Teuwen

20:04

@robjohn Working with them as functionals is quite a pain.

I'll just pretend I'm a physicist.

Matt N.

20:48

Accepted. Yahoo.

Kannappan Sampath

@MattN ??? =)

Matt N.

Victor accepted my answer : )

How is CA going?

Kannappan Sampath

I'm doing good. I have written out some answers and all that. =)

@Matt I came in to actually ask, if I say $R$ is homomorphic image of $S$, then, I mean, there is a surjective homomorphism from $S$ to $R$. Is this right?

Matt N.

Yes.

Jeff

@MattN yaaay

Matt N.

20:56

Yes : )

Kannappan Sampath

@MattN Thank you. : )

Matt N.

@KannappanSampath Pleasure : )

Jonas Teuwen

21:38

@robjohn What do you think about the uglyness of my integral here: math.stackexchange.com/questions/118253/… ? :-). (at the bottom)

In my original integral I can take the series expansion of the $\exp$, but then I get arbitrary powers of $\cos$ and that makes it really hard.

Jeff

@JonasTeuwen that was the one from yesterday!

Jonas Teuwen

Yes, I have slept for four hours and then I continued.

Kannappan Sampath

@Jonas Now I understand how GEdgar trolled you. =)

Jonas Teuwen

Do you happen to know the power series of $\cos^m \alpha$ for $m$ an integer?

Jeff

@JonasTeuwen i know the power series for $\cos \alpha$. and i think i have a method for finding the first few terms of raising the series to a given power (does that make sense)?

Jonas Teuwen

21:46

I need the whole bloody series.

Jeff

@JonasTeuwen it might take you a while (the series is infinitely long)

Jonas Teuwen

They have invented this "$\sum$-notation some couple of hundreds year ago @Jeff.

Jeff

that's true. but the only way i know to take that to a power is to write out the first few terms and apply the method i'm thinking about. I'm sure I could do it for $m=2$. not sure for $m>2$.

Jonas Teuwen

I need it for general $m$. Give me the power series. Plz.

I have put up a recursion but I can't seem to solve it using 1) Mathematica 2) Generating functions.

Kannappan Sampath

Is that a second order recursion?

Or what is it?

Jeff

21:55

power series of $\cos a $ centered around $a=0$ is $1-\frac 12 a^2 + \frac{1}{4!}a^4 - \frac {1}{6!} a^6 +... = \sum_{n=0}^{\infty} \frac {{-1}^n}{2n}a^{2n}$

Kannappan Sampath

Edit you LaTeX soon. @Jeff It does not compile/.

Jeff

I don't know why the mathjax won't render. i think i did the sigma part correctly, but use the left side of the equal as my definitive answer. @Kan

I don't know what's wrong w/it

Kannappan Sampath

One of the frac's for 6! is messing it up.

Jeff

@Kan K. I got it now

Kannappan Sampath

@Jeff Good Job. =)

Jeff

21:57

@Kan $\cos$ is even, so all the powers are even. the terms alternate sign

Kannappan Sampath

@Jeff OK. But why did you ping me?

Jeff

not sure what you mean?

Jonas Teuwen

@KannappanSampath math.stackexchange.com/questions/118677/…

Kannappan Sampath

@Jeff Why is this comment aimed at me?

@JonasTeuwen Holy crap monkey, It is an mth order recursion, I know not I can solve it. :/

Jeff

Did you not request it?

Kannappan Sampath

22:01

I, when?

Jonas Teuwen

Don't be afraid @KannappanSampath 8-).

Jeff

@Kan wrong person. sorry. @Jonas, see 16:55 for the series.

Jonas Teuwen

I know what the power series of trigonometric functions are. Srsly.

Now, take it to the $m$-th power and give me the solution.

Jeff

@Jonas ok. hold on. first i want to determine what it is to the 2nd power, and work up from there.

Jonas Teuwen

I know what it is for $2$-th power.

anon

22:04

@Jonas: you could say $$\frac{(-1)^n}{(2n)!}\sum_{|\lambda|= n}\binom{2n}{2\lambda_1,\cdots,2\lambda_m}$$ is the coefficient of $a^{2n}$ in $\cos^m a$.

Jonas Teuwen

I know what it is for the $3$th, $4$th, $5$th and even $6$th but not more.

Jeff

@jonas well you are way ahead ofme

Jonas Teuwen

@anon Aha! Could you elaborate a bit on that? :-).

Where did you get this? Out of thin air? I suppose not.

anon

No, simple generating function stuff, let me type!

Jonas Teuwen

Nice.

If you type it you can also type it in the box under my question.

Then you can get ++++.

Ilya

22:09

@Jonas: Michael is so Hardy

Jonas Teuwen

@Ilya Yes.

Jeff

3774451 if $m=2$, does @anon's eq=$1-a^2+\frac {3}{4}a^4-(\frac{2}{6!}+\frac{1}{4!})a^6+(\frac{2}{8!}+\frac{1}{6!}+\frac{1}{4!})a^8-...$

Ilya

@Jonas I would talk more to you, but I wanna sleep. The hangover takes me over

Jonas Teuwen

@Ilya Tomorrow then 8-).

Ilya

maybe :)

@Jonas: Yesterday I've started correcting the midterm exams - and even some Dutch answers :) I felt the first time this feeling, when the Red Pen is in my hand...

Jonas Teuwen

22:13

@Ilya You felt The Power?

Ilya

I feel happy when I see that they answer in the right way

really happy

and a little bit of a power :) that's when you understand, that you're not a student anymore... eh, these says are never coming back

good night

Jonas Teuwen

Yes.

anon

@Jeff: No. Not sure where you're getting that from.

ymar

Hi. Could somebody help me with polynomial rings?

Kannappan Sampath

@ymar Depends on how elementary you want of them. Let me give a try.

ymar

22:22

@KannappanSampath Thanks. Let me type the question.

Jeff

@anon i didn't think it was :(. I got that by determining the a^0 term by squaring the first term of the series....

then i determined the a^2 term by multiplying the first term of the sequence times the second term and multiplying by 2 (all the ways to make the a^2 term)...

then used the correct combinations of the first three terms for the a^6 result. make any sense? if not, nevermind. :D

robjohn

@JonasTeuwen I haven't had a chance to follow things through. It looks complicated.

anon

@Jeff My formula says that the coefficient of a^4 would be (1+1+6)/24=1/3. If you want to compute the power series of cos^2 up to 2n, then take the series for cos up to 2n and then square it as a polynomial, then ignore any term that has a power greater than 2n. So (1-x^2/2+x^4/24)^2 after truncation gives 1-x^2+x^4/3.

robjohn

@Ilya good night

ymar

@KannappanSampath Gosh. I solved my problem while typing the question. If you want I can still tell you what it was about. (I simply misread something.)

Kannappan Sampath

22:27

@ymar Good that you figured it out. What was it about, anyway? =)

anon

@Jonas, @robjohn: One method might be to write cos^m as a linear combination of cos(va)'s by some reversal process on the chebyshev polynomials, or something.

Jonas Teuwen

@Ilya Good night.

@robjohn If you say it's complicated, it is probably hopeless 8-).

@anon Hmm!

anon

Actually, I think my formula could shortcut that be using the multinomial theorem...

ymar

@KannappanSampath It was about this question by Arturo on MO. He says that if we have a field $F,$ then the multiplicative structures of both $F[X]$ and $F[X,Y]$ are both isomorphic to the direct product of $F$ and a direct sum of $\aleph_0|F|$ copies of $(\mathbb N,+).$ That's true which we can see by using the unique factorization property of those rings.

robjohn

@anon That is what I would do in the case of Bessel functions.

anon

22:31

well, the formula I posted. I wouldn't call it "mine."

robjohn

@JonasTeuwen don't go by me, I haven't done many Bessel integrals.

Jonas Teuwen

@robjohn You would do what? 8-). I'm now going to watch a film bit tired now. I will look at it later!

Jeff

@anon "compute the power series of cos^2 up to 2n, then take the series for cos up to 2n and then square it" is the process I tried to describe. I must have done the multiplation wrong.

Jonas Teuwen

@robjohn I have only done one.

robjohn

@JonasTeuwen write the powers of $\sin$ and $\cos$ by sums of $\sin(mx)$ and $\cos(mx)$

ymar

22:33

@KannappanSampath But I thought it was direct product instead of direct sum. And then it's not true.

Kannappan Sampath

@ymar I see. It's easy to misread. : )

ymar

OK, now I can edit my own question by adding a link to Arturo's question there.

Now that I understand why this is correct.

robjohn

@KannappanSampath "It's not easy bein' green" -Kermit

Kannappan Sampath

@robjohn : )

BTW, I wanted to ask, in connection with a recent question, if $f'$ is unbounded on a closed set of $\Bbb R$, does that mean that $f$ is not uniformly continuous.

@robjohn

Asaf Karagila

Yes.

What the... that user has a porn site linked from his profile.

ymar

22:43

@AsafKaragila Someone might have played a prank on him.

Asaf Karagila

On who?

Matt N.

Does anyone know about p-adic expansions?

anon

@Jonas: EUREKA. I have a closed form, after many an error.

Kannappan Sampath

How the hell will I prove that, I am in soup, I started answering without reading the question! :/

anon

@MattN: What about them?

Asaf Karagila

22:45

I know a bit, @Matt.

Matt N.

@anon I thought that the $2$-adic expansion is just the number in binary. Apparently this is wrong. The notes linked here claim that $$ \frac{21}{50} = \frac{1}{2} +2 + 2^2 + \dots $$

But $\frac{21}{50} = 0.42$.

ymar

@AsafKaragila On that user who has a porn site linked from his profile. He might have left the computer without logging off.

Asaf Karagila

Ah.

anon

No, the $p$-adic expansion is certainly not the base-$p$ expansion.

Matt N.

@anon With inverted exponents.

anon

22:48

Nope.

Matt N.

The thing I'm complaining about is that this supposedly starts with $\frac{1}{2}$.

I think this should start at $2^2$.

Since $\frac{1}{2} = 0.5 > 0.42$.

But I probably misunderstand p-adic expansions.

anon

You do, very much. You need notes on what they are, try these on for size.

Kannappan Sampath

@AsafKaragila Did you read Aguirre's answer to that question? He confirms my inntuition with an example to counter the claim.

robjohn

@KannappanSampath It sounds like it to me.

Asaf Karagila

Yeah, I was thinking on a bounded interval.

robjohn

22:51

@AsafKaragila eh?

Matt N.

@anon Thank you.

Kannappan Sampath

@AsafKaragila On a bounded closed interval most certainly true.

@robjohn The result is false.

robjohn

@KannappanSampath Oh, yeah. $x^{1/3}$

I was thinking Lipschitz

anon

@Jonas: Finished editing my answer.

Kannappan Sampath

@robjohn Can you add that as an answer, please?

robjohn

22:53

@KannappanSampath to what?

Kannappan Sampath

@robjohn To this. I have an answer for neither closed nor open set $(0,1]$. Aguirre gives a bizzare construction but one of the kind I was thinking of. You can add this simple example.

Matt N.

@anon Oh I see. To compute the digits $a_i$ of $x$ we want $|x - (a_0 + a_1 p + \dots + a_{i} p^{i})|_p < \frac{1}{p^i}$. Seems a bit of hard work to compute these expansions.

Sorry for the edits : /

anon

There, that's correct!

Matt N.

The sum I saw on Wikipedia looked just like the normal p-ary expansion so I thought it was the same.

robjohn

@KannappanSampath Is that sufficient?

Kannappan Sampath

22:59

@robjohn Derivative does not exist at $0$ in your function, but OP says it exists at all points in $[0, \infty)$ but it is unbounded in that interval.

Sorry, noticed it only now

Someone retracted the vote. My bad. :/ (Anyway, I made it CW as it does not answer OP's question but will be useful in some cases to other users.)

@robjohn Not writing up an answer?

robjohn

@KannappanSampath No, I was not. I am working on something else. Perhaps in a bit.

Kannappan Sampath

@robjohn OK. Then, I'll go to bed now. Upvote later. : )

robjohn

@KannappanSampath :-)

Matt N.

23:17

Good night folks. See you tomorrow.

robjohn

@MattN good night

I don't know how many people look at the complex analysis questions. It is probably too late for a lot of users here anyway.

Asaf Karagila

23:52

I am starting to dislike that question about $|\mathcal P(\mathbb N)|=|\mathbb R|$.

Jonas Teuwen

@anon Cool!!

So, Chebyshev polynomials are not needed? 8-).

anon

Apparently not.

Asaf Karagila

Jonas is moving to set theory?

Jonas Teuwen

What? Why?

Asaf Karagila

Chebyshev polynomials are not needed!

anon

23:55

Asettheoristsayswhat?

Jonas Teuwen

Oh.

They are certainly needed.

anon

we are discussing this

Jonas Teuwen

I just have to find where.

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$Mathematics$ Mathematics