Well, I drank at home bottled beer, each bottle is 500ml

Then I drank four pints (568ml)

Hm, not bad :)

Still headache?

Not from the hangover, no.

I'm finally over that.

@tb no, if I cap tomorrow and get an accept, I will be over 20K

@AsafKaragila spending time with me will do that to you ;-)

@robjohn Why? Are you an ebola-like virus carrier?

@robjohn oh. Tomorrow we can trust your integrals, not just believe them ;)

@AsafKaragila Who shots whisky?

@tb I don't think I could ever trust integrals..

@tb if I can make those goals.

Then you can help us delete that answer!

Right. I just need to write that last assignment on Whitehead and then I am all free for the thesis!

I thought trusting someone was a step before believing in them?

@Gigili Um, no.

@Gigili I don't lol that often here. But that was a good one :)

Bery nice :P

Well. I am off to bed for real this time. Be seeing you.

How do one find the last three digits in

Me too, good night.

$$ (2 \cdot 2011 -1)!! $$ ?

Wait... what is that about polynomials?!

I mean to um opp all the odd integers bellow 2011

$1 \cdot 3 \cdot 5 \, \dots \, 2009 \cdot 2011$

@Gigili good night, Gigili

@AsafKaragila nice, I missed that... :)

Good night!

:-)

@N3buchadnezzar With a computer, working modulo 1000?

Hi Henning

@tb Hi

@HenningMakholm Sorry to follow-up on what you said yesterday: what was that about well-known users that seem to think about reducing their contributions? (If you don't want to elaborate I understand, of course)

@tb I have nothing more definite than a feeling of deja vu. At the time I wrote it, it looked to me like you might be quitting in anger over something, and there has been a couple of those here recently. It may be more "active chat users" than "rep superstars on the main site", though.

E.g. Kannappan and QED.

@HenningMakholm Thank you. No, it's not in anger, more in a bit of disappointment in some trends that I seem to see: soft (often rather silly) and very elementary questions are the ones that receive attention. It was always that way, but I think it's getting more accentuated because of the steady increase in activity (which btw. is getting to a level that is simply too much for me). When I started participating there were a lot less questions and thus a lot more answers worth reading, I think.

Yes, I grasped that later.

I see the problem -- basically the exponential influx of mindless homework questions is at its core, I think -- but am not sure what could be done about that. Have a secret handshake between People Worth Reading and jump to a new site every few years as they become too popular? An explicit no-homework policy would easily turn a site into an elitist police state, with too high barriers of entry to keep the user base up as people find other interests...

I always liked that there was no entrance barrier except a little bit of coherence...

Hi Victor.

Bye Victor.

It is very hard to come up with a solution. The homework tag helps a bit, but you can't filter it out because there are some very good homework questions that I wouldn't want to miss. I would very much like the possibility of favoriting some people I would really like to follow up on (best in combinations with tags). Especially those in the middle field of reputation. There are some twenty to thirty users whom I'd like to follow closely and the possibility

of displaying the threads they have participated in would be very convenient and increase my interest. It's simply no longer possible for me to do that with bookmarks in combination with tags.

Maybe I should experiment a bit more with RSS at least on MO it works quite well.

Hi world!

Hi, leo

that's the first example in the books to learn Java

For any programming language, I would think.

Oh, I was just wondering what the heck Java had to do with RSS...

Isn't the classic Hello World! ?

@tb Indeed

in large enough time we will answer all the exercises and all the problems of all the textbooks

in this site

Fortunately, there are other questions, too...

@tb One could probably use the Stack API to scrape user profiles and create a personal RSS feed with just the posts of those users you want to stalk^H^H^H^H^Hfollow.

@HenningMakholm well, basically every interesting page has RSS, so choosing those pages judiciously, I think a solution not involving any kind of "programming" (which I would very much prefer) should be possible. I will try that for a while. What annoys me a bit about this is that many of the unexpected gems will surely be lost.

@tb They do? It's not very discoverable. As for losing unexpected gems, that sounds like a clear case of choosing between having your cake and eating it. Unless there's a feed for posts/comments with more than $n$ net upvotes...

You're probably right about the cake. Yes, they do have RSS on many pages here.

Ah, there's an RSS link in the HTML header. I was looking for some kind of feed link in the rendered page and forgot to check the address bar. On a profile page it seems to always lead to a feed for that user's posts. Apparently no way to get a feed for the comments tab.

Correction ... the same feed contains comments, but at the bottom of the feed.

Now I'm praying for three upvotes, a downvote and a screen shot...

@tb why a downvote?

I feel I've asked this before.

I see now: 33,333

I could just photoshop it ;-)

Hi

@robjohn that wouldn't count. In numerology cheating is frowned upon :) Hope you had a good dinner. Are you going to build some railroda tracks in India or Europe or elsewhere or is a different game up tonight?

@tb a badge, too :-)

@tb No, my friend has two tax meetings (for himself and for his aunt) next week, so he is preparing for those.

I had a sound sleep and woke up now!

@KannappanSampath sleep is better than coffee for theorems :-)

@robjohn This is a theorem in its own right! : )

still incomparable to food, showers, and transportation. I don't even like coffee.

@robjohn Oh, poor guy. Thanks for the badge and a wonderful one at that! I was sleepy for four hours and messed up my answer (which I liked but I'm sure I found David Mitra's solution somewhere along the way, but I'm too obsessed with Pelczynski recently...)

@tb downvote needed. I don't want to start now :-)

Which answer of tb are we talking about? Or are we at all talking about his answer? :-)

@KannappanSampath He wants a downvote so that his rep goes to 33333

@anon you don't like coffee and you don't like beer what kind of mathematician are you?

this would be a good one.

please please please please

@tb I don't like coffee, and I haven't had beer for a long time.

I like vodka and gin.

only one downvote!!

Take back the upvote!

no!!!

Let him settle for 44444! ;-)

The other way around! NOOOOOOOOOOOOO

Are we serious?

yes we are.

now you need 6 downvotes.

I give up. Thanks guys for trying.

almost: i.sstatic.net/OBf1W.png

I got it just after the upvote

@robjohn thank you. I will have to photoshop it then...

Well, @tb I think during my stay here, I'll keep upvoting some of your answers as I start understanding them in parts and bits!

Damn, sssticky keysss.

@KannappanSampath Thank you... :)

This link showed a nice picture!

I am now doing things veeerrrry slowly! Be it topology or CA! : (

faked: i.sstatic.net/asCZU.png

Holy!!!!!!!!!!!!!

yyyyayyyy!

I stand at 4770 for a long time!

@robjohn On the other hand, I prefer to have a clean slate downvote-wise... :) Thanks for the support!

I'll count the three badges I got as the missing three...

I had really liked some of my answers here I believe! Some Group theory answers in particular. (The most recent bitter experience excluded.)

@tb I have not downvoted anyone. I'd rather make suggestions.

@KannappanSampath bitter?

@robjohn sure, so much for keeping a clean slate... :)

@robjohn Ah, no. I wrote an answer but I wrote up a flawed answer and I had argued I was right foolishly. : ( I learnt something from there but still.... You know.

@KannappanSampath yeah, you feel foolish, but you learn something. You probably will remember it correctly in the future, too.

@KannappanSampath There were lots of good answers from you and it's best not to get too worked up about the votes. In the long run, the consistently good contributors stand out no matter how many downvotes they had...

@robjohn Most certainly yes. If ever I taught a course, this will be one of the illustrations, I would never forget to add!

Also, remember that you were under the distinct impression of being treated unfairly here, this makes you think less clearly. It was no more than a goof which you defended a bit too eagerly precisely because of that.

@tb Thank you for the compliment. :-)

@tb I think, yes! But, well, I learnt something there. I probably must remember the way we prove that quotient group actually is a group under the given structure. I reflected on some of these recently. I went through several times this part in two days.

@KannappanSampath In any case, try to follow Matt's advice and sleep over it before you think about leaving again. That would be to no one's benefit, I believe.

Okay, the birds are singing for more than 20 minutes now and I wanted to go to bed 4 hours ago. So, I wish you all a good start into the week, good night y'all!

@tb Good Night! Sleep well!

I will head out for the breakfast and may be see you all in the evening! Later folks! :-)

I'm off to bed (and then work) too. Good week, whoever is still up.

Hi @rob @kanna

Hey @Bull

Hi @RajeshD

Hi @RajeshD :)

@AsafKaragila twimg0-a.akamaihd.net/profile_images/1897160184/Omega_Shoe.jpg

A shoe for set theorist :]

$S: \begin{cases}x^{3}-xyz^{3}+yz^{5}=1 \\ xy^2+yz+2xz^3=0\end{cases}$

Now what is the tangent to this hyper thing? If it was in two terms, $\bar{n}=\frac{\partial \bar{r}}{\partial u} \times \frac{\partial \bar{r}}{\partial{v}}$.

I meant that like this:

$\bar{r}=\bar{r}(u,v) \text{s.t.} \begin{cases}x=x(u,v)\\ y=y(u,v) \\ z = z(u,v) \end{cases}$

My first idea was to partial-differentiate S -functions but cannot yet see how to proceed after that.

Say the parameter is $t$: implicitly differentiate wrt that and you have an underdetermined linear system in $(x_t,y_t,z_t)$ with coefficients as polynomials in $x,y,z$. Since the actual magnitude of the tangent is arbitrary, set one of them equal to $\alpha\ne0$ and solve for the other two; if the only solution for the other two is trivial i.e. $(0,0)$ then set the first one $=0$ instead.

What do need $z$ for? What is the point of this?

@anon I tried to express the vars with less vars trying to eliminate things to express the original thing in a more nicer way. Now I have $y=f(x,z)$ and $x=t(y,z)$ but cannot see how it helps with the tangent (where $f$ and $t$ are some functions).

I don't understand your perception of "nicer" here. Your $f$ and $t$ functions are complicated.

@anon yes but I cannot understand "Say the parameter is $t$", there is no such par in the original?

Each of the two original equations gives a surface, and their intersection is a union of curves, no? Locally the curves can be parametrized with $x(t),y(t),z(t)$, no?

You were comfortable with inventing parameters $u$ and $v$ on the fly, in the hypothetical case that you were working with surfaces instead of curves.

Actually in my comment you starred it should be "if no solution exists" rather than "only the trivial solution exists," sorry about that.

@anon I read p.798 on my book, I am not yet comfortable what it means.

p.802 has some similar example, thinking...

That's about the normal vector of implicitly defined surfaces, not the tangent vector of implicitly defined curves.

but I still cannot see how I can get from 2 curves to something one, their intersection point and tangent there? p.804 problem 15 that I am thinking.

@anon Is this the theorem here (Wikipedia) which I need to use for this problem "Implicit function theory"?

Why are you differentiating S1 and S2 wrt x, y, and z?

What do "Maarita avaruuskayran" and "tangentin suuntavektori pisteessa" translate to?

I think "determine the spatial curve" and "the direction of the tangent at the point". You'd have to ask Mahnax to be sure.

@hhh When I implicitly differentiate $S_1$ I get $$\frac{d}{dt}\big(x^3-xyz^3+yz^5\big)$$ $$x^2\dot{x}-(\dot{x}yz^3+x\dot{y}z^3+3xyz^2\dot{z})+(\dot{y}z^5+5yz^4\dot{z}$$ $$=(3x^2-yz)\dot{x}+(y^5-xz^3)\dot{y}+(5yz^4-3xyz^2)\dot{z})$$ Did you follow that, and can you do that for $S_2$?

@DavidWallace yes that translation seems correct.

But I don't know what a "spatial curve" is, really. Could be just a curve through space.

Or it could mean a curved surface.

That's what the two equations give, so yes.

@anon $x^2\dot{x}$, this term came from?

is the restraint $0\leq y \leq 1 \ \wedge \ 0 \leq x \leq 1$ the same as $0 \leq xy \leq 1?$

Ah, that should be $3x^2\dot{x}$.

@anon yes, then I start to understand ---second have to read it fully.

@N3bu: No, e.g. x=y=-1.

@N3buchadnezzar No, it's a much tighter constraint.

@DavidWallace How do I express it in a single restraint then ? =)

Why?

Why would you? How you expressed it is just fine.

BTW, "restraint" is the wrong word. You want "constraint".

Those $x$'s and $y$'s are violent offenders that need forcible restraints.

$xxx$

If you insist on having just one inequality, it could be $\max \{ (2x-1)^2, (2y-1)^2 \} \leq 1 $

but that would just be silly.

@anon $\dot{x}(y^2+2z^3)+\dot{y}(2xy+z)+\dot{z}(y+6xz^2)=0$ for $S_{2}$

Morning.

@HenningMakholm : D : D

@hhh Now plug in $x=1,y=2,z=-1$ to each.

Morning.

Evening.

@anon $\dot{S_{1}}(1,2,-1)=(7,0,4)(\dot{x},\dot{y},\dot{z})^{T}$ and $\dot{S_{2}}(1,2,-1)=(2,3,8)(\dot{x},\dot{y},\dot{z})^{T}$

@hhh: Where did our $\dot{x},\dot{y},\dot{z}$'s go?

and how are you getting vectors?

What do $\dot{S}_1$ and $\dot{S}_2$ stand for, I guess I should be asking.

oh, okay

@anon they are some sort of directions, directions at point $(1,2,-1)$ and they should somehow specify the tangent. Is the tangent just their sum or some combination? If it was a normal vector asked, it was just cross-product but here with tangent -- it is something different.

I'm checking your computations. Are you sure that 7 is correct?

@anon should be $5$.

Don't write $\dot{S}$. Instead, write the equations $(5,0,4)\cdot \vec{T}=0$ and $(2,3,8)\cdot\vec{T}=0$, where $\vec{T}=(x_t,y_t,z_t)$ is the tangent vector. This tells you the tangent vector is perpendicular to $(5,0,4)$ and $(2,3,8)$; geometrically, the cross product is also perpendicular to these two vectors, so the tangent vector has the same direction as their cross product.

Whence we can choose to set $\vec{T}=(5,0,4)\times(2,3,8)$.

$\vec{T}=(5,0,4)\times(2,3,8)=(-12,-32,15)\cdot (\dot{x},\dot{y},\dot{z})$

Why is the $\cdot(\dot{x},\dot{y},\dot{z})$ there?

...meant $-12\bar{i}-32\bar{j}+15\bar{k}$

okay, yes.

I see, I should use some unit vector signs...thinking..

spanned in the base $(\bar{i},\bar{j},\bar{k})$

@anon I am uncertain about cross product here -- look if we have had 3 surfaces, what about then?

Geometrically, what is the intersection of three (sets of) surfaces?

Depends on the surfaces.

@anon it can be a surface, line, point or nothing.

Making some generic assumptions, the intersection should be simply points.

Or finite collections of points.

But I don't like to make generic assumptions, so it could be anything at all, finite or otherwise.

Unless by "surfaces", you meant "planes", in which case hhh's comment is correct.

@DavidWallace What is the difference between surface and plane? I think I was thinking about simple planes -- surfaces can be of different form?

A surface can be bendy. A plane is flat.

So, $z=x^2+y^2$ is an example of a surface, but it's not a plane.

But $x+2y+3z=4$ is a plane.

And, obviously, all planes are surfaces.

@DavidWallace so hyperplane is still flat? Consisting of flat planes?

If three surfaces combine and create a surface as a result, then all three share some 2-dimensional region. If three surfaces combine to make a curve, then only two of them were sufficient to generate that curve (locally anyway) and the third was superfluous. These are rather annoying things that get in the way and have no bearing on the exercise we just did in chat.

Hyperplanes are also flat.

Are you sure about the second sentence in that long comment?

(is the definition of "flatness" that it consists of flat planes? So flat hyperplane can look like ball with many-many interval-defined flat planes or?)

Um, when I said "flat", I was trying to paint a picture, instead of defining something precisely. I guess in $n$ dimensions, if you can express the surface as $a_1x_1+\ldots+a_nx_n=b$, then it's flat.

@David: On assumption of sufficient regularity, fairly sure. @hhh: A hyperplane in R^n is just an embedded copy of R^(n-1) translated and rotated inside of R^n.

Yeah, I guess "sufficient regularity" and "locally anyway" get rid of most of the funny cases.

@anon, I like your definition of hyperplane though.

Hi folks

I am now uncertain when I should use implicit differentiation and when not.

If I want to find the greatest growth i.e. to find the max $\bar{u}\cdot \nabla f$, can I use here the implicit differentiotation or partial differentials?

(I have earlier used partial differentiations to $\nabla f$ but I am now wondering whether implicit differention could become useful in some situations?)

No, just note that $|u\cdot\nabla f| = \|u \| \|\nabla f\| \cos\theta$, where $\theta$ is the angle between $u$ and $\nabla f$

@anon if it is $\mathbb R^{n}$, is $||u||=(n)^{1/n}$ (1-length unit vectors)?

No: unit vector means $\|u\|=1$.

You're thinking of $\|(\underbrace{1,1,\cdots,1}_n)\|_n=n^{1/n}$, maybe...

yes

The Euclidean norm is $\|\cdot\|_2$, not $\|\cdot\|_n$. And "unit" means the length (norm) is one, not that each component is one.

ok, thanks -- have to run and think -- back soon...

bah

An integral stumped me

haha, you said stump. log cabin joke.

$$ \int_a^b \arccos \left( \frac{x}{\sqrt{(a+b)x - ab\,}\,}\right) \, \mathrm{d}x$$

@anon ^^, any hints anon?

nothing comes to mind

k

I`think I have tried everything I know.

Will try by parts now, I started with using trig identities, then I tried differentiating under the integral sign.

although that denominator has $x^2-(x-a)(x-b)$ ......

Yeah...

@anon I know the answer to this problem, would that help ?

sure

$$I = \frac{\pi}{4}\frac{(a-b)^2}{a+b} $$

pi/4 looks like it comes from an arcsomething

Thats why I thought about using $arccos(x) = \arcsin(\sqrt{1-x^2})$

When I integrate by parts, I am left with a huge mess, but it did get rid of the arccos part. And the first part evaluates to zero, Hmmm

I am tempted to ask about this one on the main site

I'm thinking a linear substitution $u=\text{something}\cdot x$ will help, with something being some ratio involving a-b and a+b

Euler sub?

googles

Setting $u = \text{denominator}$, gets rid of the square root atleast

@anon HEy

I think I figured out something about that integral

I change my question after you have answered it and then claim that your example does not work.

Note that $$\arccos(x)=\text{arcsec}(1/x)$$

And so

Maybe $x=\frac{b-a}{2}u+\frac{a+b}{2}$ will work better

$$\text{arcsec}(x) = \text{arccsc}\sqrt{1+x^2}$$

so we get

$$\arccos(\text{mess}) = \text{arccsc}\left( \sqrt{\frac{(b-x)(a-x)}{x^2}}\right)$$

What's scs?

heh heh

So hard to type thesedays

Not sure what that buys us though :p

Yes, "thesedays" is a difficult thing to type.

An affine transformation should make the thing inside the square root a square...

Well it is almost a square... just set a=b

actually, switch the signs on my suggestion!

err, switch the fractions, whatever

hmm

btw I think you mistyped I in your question

fixed

I think I will give up for now

And I thought I started getting the hang of integration =(

@JonasTeuwen I saw the edit and was slightly puzzled... Next time you should maybe include a link to the Wikipedia page on the Iverson brackets :)

Yes. Next time he will not get an answer from me 8-).

I meant next time you use them.

Yes, I got that.

Ballsy... :)

I notice my title is long, should I shorten it a tad?

And is it better with or without fractions in titles? (I know I should stay away from displaystyle)

it's good as is

=)

@tb Mmm...

I have some doubts if what Harlekin asks is really what he wanted to ask...

@tb I think so, probably: "I'm too lazy to really think about this" 8-).

@Jonas: thanks for the confirmation that I haven't misinterpreted the question...

Hey Kannappan. How is topology doing?

Morning, Matt

Oh dear. This student's presentation about ruzsa calculus is worse than my PDO one.

Hi teddy bear

But at least she didn't bring her grad student boyfriend and have him sit in the first row correcting her mistakes like the girl that had the PDO lecture after me.

You just volunteered to give all three lol

@MattN yeah, that's a nice one. A further instance where an emoticon saves the entire situation :)

@MattN I can't think of a non-obnoxious thing to say about this...

Yikes.

@JonasTeuwen apparently it was... I was posting the answer in the hope of facing the challenge of solving the real question ;)

Oh, well...

The non mobile view on iphone is broken but in mobile i cant see whos in the room grrr

I can see him controlling his temper

@MattN does this help?

Its ok thanks. Going home now

Aloha @Gigili

Martin's retagging of this question as (homework) is somewhat odd in view of Alex's answer... :)

I have no clue what this question is after but I'm pretty sure that it isn't measure-theory.

@tb Doing good! And, all of a sudden, flow of ideas in CA too! It has been a good day! :-)

@tb Luckily he does not change the question 8-).

@tb Maybe he doesn't know about Relativity.

Holy.... There is an user posting all his web home work problems in Linear Algebra!!!

@Skullpatrol what on earth does relativity have to do with this?

J. M. IS BACK

Since the achievements of Physics have shown that there is no absolute measure unit of length and an attempt would be meaningless, furthermore there is no absolute measure unit of time. Relativity has shown that both of these quantities are relative.

hello :))))

Hi @JM

This isn't my official return, yet. But hello to you, t.b.! :)

and also you Kannappan.

I have a spot of time just today, but I need to be off in a few hours.

A clarification about closing questions as Abstract Duplicates:

Here is a specific instance where OP has posted one of the subdivisions of a problem as a question and all subdivisions in another question.

Link to the full question...

@JM In any case, I'm glad to see you back here.

Hope all is well.

Link to a sub-division....

@tb Well, I did miss you guys...

The feelings are mutual, then :)

What do we do with this instance? (Flagging it for a moderator ...?)

I see Asaf has a "Generalist" now... :D

@KannappanSampath Have you tried pointing things out to the OP first?

@JM There is a comment by another user there, left for the OP 15 minutes ago. Right, this is not long enough, though!

@tb The question's title asks for an "Absolute measure unit of length" and he asks for a unit of length to be defined absolutely.

Leave it for the time being. If the user makes no move within the next day, I think a mod summoning is then justified.

@JM Thank You. :-)

It looks like there's so much to catch up on. Anything I should be looking at first?

@JM I'm very glad to meet you on the last day I have time to hang around here extensively. As a matter of fact, I checked just last night whether you've shown up on main recently...

Whoa, 'Ello @JM

@JM care to read or answer?

@tb I think I can write at least one answer today... :)

@Gigili hey hey

@tb Oh! So, you are not going to be here tomorrow and hence forth? </3

@JM Some reading first. Two nice integrals: first and second

@tb On the other hand he could be asking for a unit of length to be adopted universally such as the metric system, but asking for justification from Physics or attempts from mathematicians will lead to Relativity.

@tb Ah. :( Your sabbatical's over?

@JM So to speak. I need to do a lot of more serious writing and take care of family matters...

That's funny; I was off for those particular reasons, also...

I don't know if this suits your taste but George has uncorked another gem...

joriki's solution for the first bit was neat, but I prefer rob's.

And this is one you can answer almost effortlessly :)

@JM!!!!

How are you?!

And where have you been?!

?!?

@tb Hmm, I think that one's almost a dupe; I did mention there's a determinantal form in the comment to Bruno's answer in the linked thread.

@MattN Hey Matt. Could be better. This isn't my official return yet, though. I need to be away again in a few hours.

I'm hoping to be regular again in two or so weeks.

@tb :,(

The story seems to originate with George Dantzig, who went on to become inventor of the simplex algorithm:

“...during my first year at Berkeley I arrived late one day at one of [Jerzy] Neyman's classes. On the blackboard there were two problems that I assumed had been assigned for homework. I copied them down. A few days later I apologized to Neyman for taking so long to do the homework — the problems seemed to be a little harder than usual... About six weeks later... [my wife] Anne and I were awakened by someone banging on our front door. It was Neyman. He rushed in with papers in hand, a…

(A G+ post that some of us might be interested in...)

@KannappanSampath Ah, that one's been in Snopes for quite a while; it's my understanding that it was Knuth who helped spread that anecdote further.

@JM I saw that, but I thought it would be a bit harsh to close as a dupe because of that.

@JM Not sure about it. Sorry about that... But it is the first time I ever came across that! : (

@tb Hmm, I guess I could repackage Bruno's answer for his convenience...

:3875113 You could put in (removed) in silver color too. This time I did not miss that! : )

@JM I was about to suggest that. Moreover, Jonas T. had some Bessel-related questions recently. I'm sure he would appreciate if you had a look and maybe even a pointer, especially on this one

Okay, that made me feel lonely... are there really no other special functions guys here? :D Gee...

well, there's always the mean square...

Afk for a while. bbl

Hmm, I think Jonas's stuff aren't the sort that I can do in one sitting... a fair bit of Fourier series theory would be needed.

Waiting for Jonas to show up and say: I hear Fourier?!

Hi ?

Hi! @N3buchadnezzar

God, may anyone help me? I can not use this chat in firefox!

It tells me I have not enough rep to talk!

@tb Oh, right, figures... :D

Jonas, if you can read this: my initial suggestion would be to look up Anger and Weber functions. Their forms are quite similar to what you're dealing with. I'll deal with your question when I'm actually back and your nut still hasn't been cracked.

Can someone suggest a shorter title for a blog post that starts from the basics of ideals and builds upto the fact that nilradical and Jacobson radical of some rings coincide, please?

The interesting bit for you would probably be the fact that they degenerate to Bessel functions for integer orders.

@JM there was a bit of an amusing situation last night here. The very moment I was typing "I always liked that there was no entrance barrier except a little bit of coherence..." someone showed up :D

@tb Speak of the devil... :D

I'm amazed at how "3+.1+.04+.001+..." got a crazy amount of votes...

That was not a real question in my opinion!

Hmph.. facepalm and looks seriously into the screen for some reply

@KannappanSampath So far as I can tell the abstract algebraists who hang around here aren't here now, so you're in for a wait...

@JM :/ Anyway, I shall wait!

@tb: That talk you had with Henning is interesting. I find it a bit scary that the last entry on the front page was only updated three hours ago...

@KannappanSampath what is the goal of the question, IOW the actual question?

@tb I would like to know a short title for a blog post that is described there...

@ypercube I like your finite geometry answer. But, lot of it is open. sigh

I mean, the very elementary isomorphism problem is unsolved!

Yay!! @tb I found a relevant post!