@Danny have you an idea , i think about contradiction but i don't find any reselts

@Vrouvrou i dont think iam the right person to ask , but did you write down your question ? i cant see it

@Danny yes

iam actually buys eating gummy bears

@Danny :D

I WANT THAT

@Vrouvrou this is beyond my scope of knowledge , there are surly people here that could give you good answers, did you ask on the Forum? they usually answer very fast

oh u did ask,

yes i ask it on the forum

@Vrouvrou dont worry i bet someone will give you answers soon

i hope that

iam actually sure of that

@Charlie is there a shortage of gummy bears in brazil?

@Danny nope, as far as I know

then what are u wating for

:)

I have to go to a market

and walk

lazy

:(

:P

@robjohn let me know if you find a closed form.

@Chris'ssis for your sum which was the one I did but divided by $n$?

@robjohn Sorry, I don't get ...

@Chris'ssis $\sum\tan^{-1}(1/n)-1/n$

@robjohn Yes. Didn't you work on this version?

@robjohn also don't miss the question involving Fibonacci numbers --- it's more than amazing. :-)

@Chris'ssis $\gamma+\frac12\log\left(\frac{\pi}{\sinh(\pi)}\right)$

@Chris'ssis Not until this afternoon

$$\sum^\infty_{n=0}\frac{(n-1)^{1-n}\;\Gamma(n+1, n-1)}{n!}$$

@robjohn I admit I'm very tired today but I need to ask for which question is that answer?

@Chris'ssis Hang on, I'm checking

@robjohn ok

I am trying to simplify $-\gamma-\frac i2\log\left(\frac{\Gamma(1-i)}{\Gamma(1+i)}\right)$

which is the answer.

-0.27557534443399966272

@robjohn True.

@robjohn That answer you previously posted is very different from this one.

@Chris'ssis not so different. It is just the imaginary part of what I gave the real part to before..

@robjohn did you follow the Riemann zeta function way? I need no detail, just largely speaking.

@Chris'ssis I don't know what way you are talking about... I used the digamma function

@robjohn OK. This also has to work I suppose.

@robjohn It's not bad to have more tools at hand when facing such questions. (2, 3 solutions)

@Vrouvrou I will look at your question ...

@robjohn hi. Are you familiar with group rings ?

@mick not really

@robjohn Do you know what it is ?

What's the appropriate Fourier transform extension to use on the half-line $[0, \infty)$?

@robjohn Which of the 2 questions seemed more difficult to you? This one or that one you solved yesterday? How about a difficulty scale?

Is it true that " gone " will return ?

These questions are amazing to attend without using any computational software.

My master says he is one of the best !

forgive me robjohn for saying soo

@Vrouvrou Im sorry i dont know the answer :/

@mick what about them?

@anon this question : math.stackexchange.com/questions/583981/…

polysign numbers, I seem to recall those are a crank reinvention of the wheel

based on my skimming of the question, I'd say it would take me 20 times more effort to understand what your question is, translated into standard terms and concepts, than it would be to answer the question itself once understood!

@Chris'ssis I have not found a good closed form other than one that uses complex functions, i.e. $\Gamma(1+i)/\Gamma(1-i)$

@Chris'ssis Do you have a simpler looking answer?

Presently I have a calculus book where I want to come up with new solutions for each of the author's solutions. Maybe I send him my collection after I finish all his questions.

@robjohn No, I don't have any simpler answer.

@Chris'ssis So your answer also involved $\Gamma$?

@robjohn Yes.

@Chris'ssis Ah, then perhaps I will stop trying to simplify it :-)

@anon I provided the isomorphisms to e.g. R(X)/(1+x+x^2) so the " translation " is already done ...

@robjohn that form it's OK for me. Well, I don't know any other simpler form.

@robjohn Which of the 2 questions seemed more difficult to you? This one or that one you solved yesterday? How about a difficulty scale? (I think you missed this row I also wrote above)

@Chris'ssis The one yesterday took longer, but I used a lot of what I did yesterday to answer this one.

@robjohn Might we say that they are very difficult? The first one didn't have a solution for 6 months.

@anon crank is a bit disrespectfull , amateur is better imho ... ( crank suggests (inconsistant) nonsense )

(a) crank needn't suggest inconsistent, (b) it is inconsistent

@anon it is not inconsistant. they are group rings. Like R(X)/(1+x+x^2)

@robjohn Personally I think they are not difficult, it's about training with such questions. With a bit of clever manipulation we're immediately done.

If we are talking about polysign numbers, I am with anon

@mick for example, the geometric representation and discussion indicates 2.3-0.5 is different from 1.8

@OldJohn Im sorry but you guys are wrong , it is consistant. It might be " confusing " but not inconsistant.

@Chris'ssis I wonder whether the person who gave the unjustified solution actually had a solution.

@mick also, that is a quotient of a polynomial ring (you meant to write R[x]/(1+x+x^2)); a group ring for example would be R[x]/(x^3-1), which is R[C3]

@anon aha but then - is not the usual - . he uses - in two signed and three signed

@OldJohn hi John!

@robjohn Ah, yesterday you weren't here at the moment I saw his comment on meta. I was so angry ... he didn't have! One that loves calculus would be so so proud to post such a solution!

@mick Do you have a proof of consistency ?(not consistancy)

@Charlie Hey!

@OldJohn how are you?

@Charlie Fine here, thanks - just helping my wife with her new laptop (Windows 8 - yuk!)

@Chris'ssis I commented on his meta question as politically as I could.

@anon Are you sure R[C3] is not R[x]/(1+x+x^2) ? Afterall (x^3-1)/(x-1) = 1 + x + x^2 ?

@Charlie How are things with you?

@OldJohn oh! Hehe

@robjohn that solution is good to be added to a very nice solutions collection.

@OldJohn I'm fine, it's not so hot now:)

@mick yes I am sure. R[x]/(x^3-1) is the direct sum of R and R[x]/(1+x+x^2) by CRT

@Charlie Excellent! - almost freezing here these days

@robjohn I also love to find proofs, but when one is simply keener than me and finds a nicer way, I recognize that and want to learn. That comment on meta was so ridiculous ...

@anon hmm I read ( by a real mathematician !? ) that it was R[x]/(1+x+x^2) .... What is CRT ?

chinese remainder theorem (the general abstract algebra version). I find your real mathematician comment amusing.

@OldJohn whats the coldest you saw there?

@Chris'ssis but he said his was the same as mine.

@robjohn Jesus ... :-). I better say nothing. :-) Maybe in his nicest dreams ....

@anon But then the complex numbers are isomorphic to R[x]/(x^3 - 1) ?

@Charlie a couple of years ago we had something like 18 degrees C - but that is very rare

@mick absolutely not

@Chris'ssis I don't see any reason for him to complain, he still has the most reputation from that question.

@robjohn that problem requires a very experienced person in analysis ...

C is isomorphic to R[x]/(x^2+1). the ring R[x]/(x^3-1) has zero divisors and idempotents (which C does not), and is dimension 3 as an R-vector space (as opposed to C, which has real dimension 2)

@OldJohn 18 ? 18 bellow zero or 18,18?

@robjohn Yes, but your answer was chosen. :D

@Charlie sorry, yes - 18 below zero (centigrade)

@OldJohn yeag, it's very cold :P

@Charlie Coldest I have ever experienced -anywhere

@robjohn I mean this question -- math.stackexchange.com/questions/393013/… -- some here might miss the masterpiece

:-)

@robjohn and I downvoted his answer some time ago ...

@mick yes, C is isomorphic to R[x]/(1+x+x^2) (as well as R[x]/(x^2+1)), but the second thing is not a group ring, and is not isomorphic to R[x]/(x^3-1) (which is a group ring, it is R[C_3]).

@Chris'ssis you downvoted vladimir ? think he stole the answer or used a computer ?

@Chris'ssis I just upvoted your comment on Mhenni Benghorbal's answer. I would like to see how he got that.

@OldJohn wow

@OldJohn there now? ouch!

@mick Well, I think a lot of people would have liked to see his answer ... but nothing (if there was one)

@robjohn No - currently here it is about +3 or +5 ... -18 is incredibly rare here

@anon But P3 is isomorphic to C so P3 must also be isomorphic to R[x]/(1+x+x^2) ( since C is ! ? )

@mick I don't know what P3 is

@OldJohn Ah, we've had rain here, but I don't think it has dipped into the 30's F here recently.

@anon Then what are arguing about ? I thought it was about that ? You called it inconsistant too ??

@robjohn That's good - but in some ways I prefer cold compared to rain - I much prefer hill-walking on frozen ground rather than soggy ground :)

@OldJohn For traversing, snow is good for walking, but driving on frozen roads is like playing air hockey.

@mick I thought we were talking about R[x]/(1+x+x^2). if you mean the polysign numbers P3 as in the link, the discussion of those is rather confusing and I can't make heads or tails of it. I am not arguing with you; I am trying to teach you math!

@mick no, I said R[C_3] is iso to R[x]/(x^3-1) (not the brackets, not the parentheses), although I guess it is also isomorphic to R[x]/(x^3+1) (as it happens)

@OldJohn However, sitting inside on a rainy day with a fire blazing, listening to the rain can't be beat.

@anon yes sorry typo of me

@robjohn absolutely! - I hate driving on ice with a passion. I remember once driving (at 5 mph) towards another car and having absolutely no control ...

... luckily, he moved away before I hit him :)

@OldJohn Yep. I've seen video of a bus, parked at the side of the road, spontaneously start sliding down the road into parked cars.

@robjohn add hot chocolate, marshmallows and internet

@anon Indeed!

err, *note the brackets

@robjohn yep - nothing you can do about it

@anon A good supply of videos also helps :-)

@robjohn I guess the bus driver lost his $\mu$

@Alizter I don't know if the driver was even in the bus.

Has anyone ever looked into the function defined by the series $z+z^2+z^4+z^8+z^{16}+\dots$?

@OldJohn yes , its a classic

@robjohn and also a hot series ... :-)

@OldJohn f(z^2) = f(z) - z :)

@mick and how exactly does it behave as $z$ moves towards the unit circle in $\mathbb{C}$?

@mick (3) R[C_3] is a group ring, (4) no R[x]/(1+x+x^2) is not a group ring (it is not presented as one, nor is it in any way isomorphic to a group ring over the reals), (5) the ring R[x]/(1+x+x^2) is a quotient ring of a polynomial ring by a principal ideal; it is isomorphic to C by sending the coset x+(1+x+x^2) to either primitive cube root of unity

@OldJohn look up "lacunary series"

@anon I have - but nothing I have found fully describes the behaviour as $z$ approaches the boundary

@OldJohn Consider the possibility that the behaviour is not yet fully understood.

@anon but from what I have found, it seems to behave more hideously than any holomorphic function I have ever come across

@DanielFischer Yep - I reckon it has no radial limits anywhere, no non-tangential limits anywhere, and probably no (finite) asymptotic values ... really nasty (but a fantastic counter-example)

@anon (5) what is a coset ? So I should retitle my OP and talk about quotiënt rings ( and define R[x] ) ?

@mick you do not need to define R[x], it is a polynomial ring. how in the world are you even talking about quotient rings such as R[x]/(1+x+x^2) when you don't know what a coset is!?

@OldJohn Yes, it is pretty nasty. I don't know much about its behaviour, though.

@anon Also I think x + (1+x+x^2) must be send to 1^(1/3) and not the other root of unity ...

@mick what is 1^(1/3)?

do you mean exp(2pi*i/3)?

it doesn't matter which root of unity you send it to, as long as it's primitive

@anon im not familiar with quotiënt rings ... I thought I was talking about group rings ...

@mick What does your "couple" notation indicate?

@mick no, you are talking about quotient rings (although group rings can be characterized as certain types of quotient rings). what did you think R[x]/(p(x)), where p(x) is a polynomial, meant?

@DanielFischer I find it fascinating - generally holomorphic functions are really nicely behaved, but as you move towards the boundary, they can become incredibly badly behaved - and as you say, it is likely that the last word has not yet been written about the situation

@KarlKronenfeld (a,b) = (b,a). additions are like (x,y) +(a,b) = (x+a,x+b) and multiply is like (x,y) * (a,b) = (xa,yb)

@mick And this is an element of something?

(You're splitting some ring up into a direct product, not sure exactly what and why)

@KarlKronenfeld what do you mean ?

@robjohn Are you sure that you think supporting asciimath in chat is a bad idea ? I agree that in the beginning there may be some confusion, but if it is promoted a little bit that chatjax has an update, then sooner or later most people will have the new version. And it wouldn't be a disaster is some people don't have the new version. The good thing about asciimath, is that it looks readable, even if it isn't rendered.

@90intuition I think it is best to use the same thing here as on the main site. It makes it easier to port expressions.

@90intuition We can't support both at once.

@robjohn agreed

@OldJohn By the way, do you know much about the boundary behaviour of $$\sum_{n=1}^\infty z^{n!}\,?$$ That should be pretty bad too.

hm.. escaping backtick works not anymore ? ` oh it does

@90intuition Does asciimath support everything that MathJax does? (a large portion of LaTeX)

@anon thanks. wish me luck with my new question :) take care

@DanielFischer That has the unit circle as a natural boundary, does it not?

@DanielFischer Not investigated that one - but it is also (very!) lacunary, so its behaviour should be very bad indeed

@robjohn Yep - I think every lacunary series has

@robjohn Yes, like $\sum z^{2^n}$.

Well technically asciimath is a part of mathjax.

but you can't write anything in asciimath syntax that you could write in latex like syntax

using mathjax

But with the script I wrote, you can use both. So it is not that you can't use latex anymore, it is that you can also use asciimath.

@mick I am writing a general-purpose response to your question anyways

@anon you are sure that the term group semi-ring or such is not better / does that even exist ?

Testing: ` 1^(1/3) `

@90intuition how does the script differentiate asciimath and interpret it properly?

@anon ah you can solve the question anyway , despite wrong terminology ?

Well asciimath is between backticks, not dollars. But as we use also markdown, you need to escape the backticks.

@mick in my answer I even say that group semiring would be a better term (for polysign numbers, anyway - I don't talk any more about those). I can give you the necessary background information to answer almost all possible versions of questions you could actually have in mind.

@anon wow thanks that would be awesome.

I might become a supersaiyan 2 mathematician afterall :)

well, you should follow along the standard educational track before that happens :)

@anon Thanks for the patience and answer .... I hope I will be able to understand the answer ...

@anon well sorry , but Im only 14 yo. This place might be the best for me :D

Another test: ` exp(2pi*i/3) `

there are no shortcuts to supersaiyanhood a la trunks and goten

@90intuition what on earth are you doing ? :p

@anon yes but education for 14yo is a bit under my level ... its too easy for me :/

@90intuition we already use the ` symbols around latex in order to prevent latex rendering by our chatjax code (in order to teach latex for example); having another usage that actually renders could potentially create issues

@mick I never said you have to do the track according to your age

@anon just saying the " standard educational track " is too slow for me at the moment ...

@mick I meant for example get a basic "modern algebra" or "abstract algebra" text and learn the standard concepts and definitions in the correct order, not be satisfied with the classes you're offered

@90intuition Once you are familiar with LaTeX it becomes a lot easier to read. There is no reason to have another typesetting format, as it would cause confusion (not to mention take the fun and control away from writing formulas). Sometimes \frac12 wins over 1/2

@anon I know you mean that , but that is for when im 17 or so I think ...

ahh you mean books ...

I remember when I was around 14, in addition to learning calculus myself from my dad's college text, I dabbled in stuff beyond my ability and a couple of borderline crankish things (in number theory). I would have been much better served by simply picking up a basic abstract algebra or elementary number theory book and doing it that way from the get-go.

@Alizter Just to be sure I'm clear, it's not that you couldn't use LaTeX anymore. You could also use LaTeX if you prefer. It is just that you could look intuitive notation like a*b or 1/2 look like a \cdot b or \frac12 just by putting it in escaped backticks.

But maybe something else then escaped backticks would be better.

no, I am an advanced undergraduate student

@anon how old are you ? and do you intend to become a professor ?

I am 22. I will probably be a teacher or professor.

i hope i will be like anon some day ..... :p

well at least I have much time left to become like him when im 22 :p

@mick my best advice is to move from stage (1) to stage (2) described by terry tao

@anon Did you finish your answer ?

Terry tao is to advanced for me :) but ill read the link :)

best mathematician is both subjective and hard to judge, because nobody is versed in enough fields to be sure anymore. there are the big names of history - Gauss, Ramanujan, Euler, Riemann, Archimedes, and so on. perhaps the best one in recent times is Grothendieck.

Next week I am to give a presentation to (oldish) amateur mathematicians on the topic of n-dimensional space - I have a basic talk organised (mostly from things I have used in the past to fill in bits of lessons) - does anyone have any favourite facts/curiosities about n-dimensions?

like, geometric?

@anon mainly - I can't assume they have very much technical background, unfortunately

People.

I plan to tell them about volume of spheres in n-dim tending to zero as $n\to\infty$

@OldJohn Heh, I had heard about that a while ago. Quite crazy.

and I have found one or two past questions on MSE with useful things

Oi, @PedroTamaroff Where hast thou been?

@DanielFischer (!) At university, Daniel!

I did read your last ping to me, though.

Nifty!

@PedroTamaroff I reckon it just means that most of the "volume" in an n-dim cube is near the edge, rather than near the centre

@OldJohn Ah. The volume is given by $\Gamma,\pi$ and some powers, right?

@PedroTamaroff yep - and the gamma in the denominator wins

@anon thanks - I had seen a couple of those but not the pde one

@PedroTamaroff yes we use repeated integrals ...

@DanielFischer I have a new task. To show a Hausdorff locally compact space is Baire. I have a proof plan, I think. =)

@anon this one is my favourite so far

@PedroTamaroff A good plan?

@DanielFischer I hope so.

How does it look?

@DanielFischer Well. First, since every point has a compact nbhd we may work locally in a compact Hausdorff space. But a compact Hausdorff space is normal. And then I can mimic the proof for complete metric spaces. Namely, construct a decreasing sequence of nested closed = compact sets that will have nonempty intersection and this point will be in $\bigcap O_i$ and I win. Left or right?

Yes, that will work, unless you do something stupid.

byeeee !!

@anon Where is your answer to my question ??? I do not see one ??

@DanielFischer Hehehehe.

Do you have a proof of your liking?

still 0 answers ?? @anon

impatience, yeesh

@anon so your still working on it ? I thought you said you finished ... ?

Sorry to be impatient :)

@mick where did I say that?

For locally compact is Baire, @PedroTamaroff? The standard one is fine, I don't know an impressively slick one.

@anon my mistake sorry.

@anon is it alot of work ? I cant wait to see it :)

I wonder how many mathematicians play chess. I know ppl who play chess and quite a few like math ... I wonder if it also vice versa ...

@DanielFischer The standard one is what I proposed? (My lin. alg. professor gave me Hausdorff+compact => normal, which was the killing blow)

faill...

Hello, I would like to do some mathematics

@AkshajKadaveru go ahead :)

Its pretty easy

my first reading was "I would like to do some mathematicians"

latex doesn't work

see the "LaTeX in chat" link pinned to the starboard --->

oh

@90intuition What did you do?

bye all

My script failed :P

Trying to fix it now.

guys solve

Asciimath chatjax script