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14:06
@PedroTamaroff K-I-S-S-I-N-G??? >8(
@skullpatrol ?
@PedroTamaroff You know what I'm talking about pal.
@Pedro good morning
14:22
Hello Professor @TedShifrin
morning, mr skull
how are you?
doing just fine ... planning out what I have to do to make it through the next week or so ...
morning, mr @Pedro
hi @Paul
@TedShifrin Hello.
Hi @TedShifrin I have an English question for you
14:36
Hi Ted, How was dinner yesterday?
@Ted is the right person for English questions. @Ted, I think you said you had some type of connection with the Georgia review
LOL @Paul ... nothing spectacular :P Planning one for guests for next Saturday, though :)
Yes? @Ian
no, no, @Paul: I merely said I used to know the editor.
@Ted Do you take a serious interest in cuisine, and own lots of cookbooks etc?
You should ask @Paul your English questions and ask me your American questions :P
yes, @Paul, a whole wall full of cookbooks. Don't know what'll happen with them when I downsize soon.
@Ted, I think I am extremely knowledgeable about usage matters in all forms of English, when compared to other maths types, even the native speakers.
@Ted, can't you scan them into your PC?
I wasn't suggesting otherwise, @Paul. I do not have the patience to scan dozens and dozens of books. It would take a day to do one of 'em.
14:40
math does have a lot to do with language...
There may be fancy machines for this, but we don't have one at school ...
@TedShifrin I wrote an answer firstly defining a "connectable number" and then changed it to "connectible number". Do you think they are ok? Is there a better name for suggesting that a sequence can be connected in another one?
@Ted Yes, and I wasn't suggesting that you were suggesting otherwise. Just trying to volunteer as a good person to ask. Of course, that doesn't preclude the possibility that my ramblings on this forum may be error-prone.
@TedShifrin $\min\{n>0:g^n\in N\}$ is usually called the indicator of $g$ in $N$?
Well, @Ian, it seems you're inventing a word and a concept. I have no idea what it should mean. But if I had to spell it, I'd use the second.
14:42
Or was it index?
I dunno, @Pedro. What is your context?
@TedShifrin Group theory.
@TedShifrin hm, ok. The "connectable" came from my native brain (conectável)
Indicator rings a number theory bell. But I dunno.
If we have a subgroup $N$ and an elt $g\in G$; the least $n$ for which $g^n\in N$ is called...?
14:43
@Ian, beware of coining your own definitions. You should not introduce new terminology if standard terminology is already available.
I think it is called the index of $g$ in $N$.
Index is ordinarily used for a subgroup, so I would think index would apply when $g\in N$. Hmm.
For example, let's call an operation good-sequence-behaved-three-term-wise if for all a, b, c we get a* (b*c) = (a*b) * c. That's not good -- just say "associative".
LOL @Paul.
So what is your notion, @Ian? I.e., what is the definition of this term?
@Ted, I'm sure you've seen lots of stuff like that as an experienced math educator?
14:45
@IanMateus able to connect?
not so often as you'd think, @Paul ... I've had issues a few times with my own research :)
So, @Pedro, if $\langle g\rangle = H$, you're looking at $[H:H\cap N]$?
@PaulEpstein you could also call it "grouping"
@TedShifrin Don't know.
Well, does my formula agree with your definition?
I haven't checked.
14:50
I'm giving you an index :) I think it's what you're talking about.
@Paul: Do you have any idea what @Ian's notion of "connecting" might be?
yo @Charlie
@JessyCat you like the Cheshire cat?
I've always been a cat person, but not a Cheshire cat.
how's my favourite cyber cat @Charlie?
@Ted No, I don't but alarm bells should be raised when someone inexperienced in maths wants to coin a new term. It's a bit analogous to a software developer wanting to write their own square-root function instead of using a standard implementation.
14:54
@Pedro & @Ted, I do like the Cheshire Cat, but on general, I'm more of a dog person
It seems more people are dog persons ...
Erp, in general.
Dogs show feelings, cats... not so much.
Hm, about to hit 50k.
My cats always did ... maybe it's the way I treat them. But, sadly, they're all gone now.
@Pedro, exactly!
14:56
I've basically quit answering stuff on main, so my rep is languishing.
burnt out ...
so soon?
grading 35 papers every 10 days will do it to me
@TedShifrin this cames from this question, I wanted to say that I can find a "nice sequence" (such that the sum of adjacent numbers is always prime) for two numbers and "connect" them to form a third one. For it to happen, I need the last number ($n$) to be either in the start or in the end of the sequence (like a LEGO piece)
14:57
Amen to that!
@PaulEpstein yeah, but I unfortunately don't know one :/
sort of like dominos, @Ian. Can you give me examples?
Feynman was known for coming up with his own terminology and notion...
well, Feynman earned the right
he started young
15:00
So did Atiyah, Bott, Singer ... in math. But we're not those people :P
true dat^
Someone might just have the talent or knack for coming up with better, more efficient, or more descriptive notations. We all have unique talents, so we shouldn't discount ourselves @Ted
@TedShifrin sure. Consider a nice sequence (we can always find one for a prime twin by defining a sequence $S_p$, wait a bit) of length, say, $11$. I can construct it as $(1,10,3,8,5,6,7,4,9,2,11)$ (note the pattern) and take another one of length $k\leq(p+1)/2 $, say 4. The four one can be simply $(1,2,3,4)$ (connectible, the $4$ must be in the end). Now I construct another one of length $p-k+1=11-4+1=8$ by the following algorithm
@Ian: I looked at the link. So I see only one sequence there. I'm still not sure what you intend. Given sequence $\{a_n\}$ you want a new sequence $\{b_m\}$ with the property that $b_m = a_{f(m)}$ for some function $f\colon\Bbb N\to\Bbb N$?
This discrete math stuff is just so not my way of thinking .... AGH.
@TedShifrin I want a sequence such that the sum $a_k +a_{k+1}$ is always prime, right? Ok. I can always make it for a prime twin $p$ in $(p, p+2)$. Just take $(a_n)_1^p$ with $a_{2k-1}=2k-1$ and $a_{2k}=p+1-2k$
15:06
Before you worry about a word, give a precise definition, please! :)
Isn't it just what I wrote?
I don't see what you mean about connecting only at the end. You need to be able to do the connecting at each element of your second sequence.
@TedShifrin A connectible sequence is a sequence of length $n$ such that the sum of any adjacent numbers is prime and $n$ is at the end or at the start of the sequence. A number $n$ is connectible if there is a connectible sequence of length $n$.
and unfortunately definitions are written in words
Your original question was phrased in terms of arbitrary sequences. So now you're talking only prime sums and finite sequences.
@skullpatrol Unfortunately?
Length is a terrible word there.
15:10
@PedroTamaroff see^
Oh, no, you mean length, and the $n$'s have to agree.
I now don't see why the word connectible at all. You're not trying to connect two (infinite) sequences.
@TedShifrin sorry, I've been thinking about this for some time, this is already implicit in my mind :\ which word can I use?
Why not just call a number $n$ interesting if it has that property?
No, I mean the length one. Ok, let's call $n$ interesting
Is $(\sqrt{2}+\sqrt{8})$ irrational?
15:12
Simplify $\sqrt8$, @cyril.
@cyril Yes.
well ok
Great ...
Sophie Germain does not approve it
15:14
Nor do I, @Ian . :D
@cyril: Is $\root3\of 2$ irrational?
that's really wrong to say x² rational => x rational, sorry
So what are you asking us?
nothing actually, just realized this
@TedShifrin why do you burden yourself with such a heavy workload?
so close to retirement
15:19
Because I'm no longer doing research and I contribute by doing the best job I can of teaching. The sad thing is that so many of the students don't reciprocate by trying to do their best. If I were going to quit caring and doing my best, I would already have retired.
@TedShifrin ok, let's redefine some things. A finite sequence $S$ is interesting if it is a permutation of $\{1,2,\ldots, n\}$ and it either starts or ends with $n$. A number $n$ is interesting if there is an interesting sequence $S$ with $|S|=n$. Is that definition ok? I know $S$ is not a set, but $|S|$ seems to fit nicely
@TedShifrin Why don't you research anymore?
I'm ok with it, @Ian, but I would skip the first step. Just define it for the natural number $n$. $n$ is interesting if I can re-order the set $1,2,\dots,n$ so that ...
I still talk math with colleagues and help them with their research, @Balarka, and same with grad students, but my heart has always been in teaching undergraduates, and I've concentrated on that for the last ten years or so.
@TedShifrin have you seen this?
@TedShifrin I see. I would be much more interested in researching than teaching peoples, if I ever get the chance. By the way, are you a frog or a bird?(ams.org/notices/200902/rtx090200212p.pdf)
15:24
Yeah, @skull, @Mike was posting about it a week ago.
@TedShifrin Ok, now the claim is that given a prime pair $(p, p+2)$, $p$ and $p+1$ are always interesting. Moreover, if $2k-1\leq p$ and $k$ is interesting, then $p+1-k$ is interesting. That is the main theorem in the answer
I have no idea, @Balarka.
I guess I need to start worrying about clarity. Hehehe
A standard classification of birds and frogs are given in the article, you can look at that yourself.
OK, @Ian, this seems cleaner to me.
15:25
@TedShifrin not interested in "following" it?
Nah, @skull. Not at the moment.
As I said, I'm burning out on this place.
I'm out.
Mazltov, @Pedro :D
15:26
@TedShifrin slow down a bit pal
Maybe you could call the number $n$ attainable or something, @Ian.
What's the best result to the asymptotics on the number of integers $n$ s.t. $(n, \varphi(n)) = 1$
LOL@Pedro. You silly goose.
Now I could see your joy if you'd just beaten me 6-0 6-0 in tennis. (Just as likely.) :D
Quick sifting through Selberg gives somewhat like $e^{-\gamma} x / \log \log \log x$
15:29
@TedShifrin yeah, but at least it wasn't a cranky ianish one :D
Are anyone here familiar with Roy's approach to Schanuel's conjecture? I just finished Zilber fields, and am trying to move onto Roy conjecture and differential operator theory.
Is it me or am I wandering through many things at once?
16:27
is it proven that there is no finite and trivial ($+x-x$) linear combinations of irrationals giving a rational?
o well it's stupid, it's more or less by definition I guess
@cyril What do you mean by 'trivial'?
@skullpatrol: Almost considering to work on a presentation for a seminar that I'm going to do.
Dang, time travel
@BalarkaSen: Hi Ballu :D
16:43
@Nick I much prefer Balarka.
@BalarkaSen where irrational terms appear at the most once
@cyril How about $\left ( \sqrt{2} \right)^2$ then?
@BalarkaSen: What about $\text{BaSe}$?
@BalarkaSen linear combination with rational coefficients
16:50
@cyril $\sqrt{2} - \sqrt{1 + 1}$.
@cyril: I don't think anyone has proven that
@BalarkaSen: trivial!
You need to rigorously define "term appears only once"
@Nick Which one would be non-trivials, then?
Greetings!
@BalarkaSen: ... nobody knows. it hasn't been proven that it can or can't
Today I've created a new class of integrals.
16:52
@BalarkaSen yes this will be hard :)
@BalarkaSen you genius, I need your help
@Chris'ssis Fire it.
Are you really 14?
@BalarkaSen Tell me a nice alternative form for $$\psi(1-i)+\psi(1+i)$$
@GabrielR. Yes, and it's tiresome to answer that to everyone. So assume that the "y" as gone off to get some rest and "e" is dead. S.
@Chris'ssis Are those digammas?
16:55
@BalarkaSen yes
That's awesome then
@Chris'ssis Then try out $\psi(1 + x) = \psi(x) + 1/x$ and $\psi(1 - x) = \psi(x) + \pi\cot(\pi x)$
But that's not nearly nice. let me see what else I can do.
I want something nice.
Ah, I seem to have done something.
@BalarkaSen $\sum_{i=1}^n r_i a_i$ where $\forall i, j \neq i : a_i \neq a_j$
where $a_i$ are irrationals, $r_i$ rationals, can such a term be rational
17:09
Nope, not yet.
right, even with n=2
It was directed towards chris'sis, not you =D
I will look at it though, @cyril.
@BalarkaSen ok en.wikipedia.org/wiki/Irrational_number#Open_questions even if wikipedia isn't the best source
@cyril ? I don't see the article referring to anything relevant to your question.
but the question was about existence, I messed up nvm
17:13
? existence ?
@TedShifrin You could just set your preferred tags on the front page to, say, only DiffGeo or similar and ignore the mass of homework questions. Or just show up to chat and nowhere else like me :P
(does $(a_i)$ and $(r_i)$ exist)
@Chri'ssis I conjecture that the minimal extension in which $\psi(1+x)+\psi(1-x)$ can be expressed contains $\psi(x)$.
On the other hand, the same is not true for $\text{Li}_2(1+x)+\text{Li}_2(1-x)$, as it can be expressed in $\mathcal{EL}$ for $x = i$, namely, $\frac{-1}{2}\log(-i)^2$
Interesting, isn't it?
@cyril Yes, then it's not always known. What you refer to is known to us transcendental number theorists as 'linear independence'
Probably the best result known to us is that at least one of $\pi + e$ and $e^{\pi^2}$ is transcendental.
(The former can be extended to algebraic independece)
@BalarkaSen I'm sure there is a way to make things beautiful ...
This is what my intuition tells me ...
@Chris'ssis Not always. But I'd rather not be skeptical and search for some identities.
I'll let you know if I find any.
17:22
@BalarkaSen I appreciate!
@Chris'ssis Are you expecting some surprising thing as foxtrot?
Let's not laugh everybody. mathworld.wolfram.com/FoxTrotSeries.html
@BalarkaSen My intuition just bothers me ... (and never lies to me)
@BalarkaSen did it (without pen and paper)
Interesting. I found myself working with something titled "An Example By Littlewood" which seemingly started with ring theory and ends up with presentation theory. Not only that, but an interesting reference is given 'I. Davis and Balarka Sen. Hardy-Ramanujan, stochastic systems and maximality methods. Journal of Singular Lie Theory'
@Chris'ssis Did what?
17:38
The coolest thing is that researchers actually submitted some of these abstruse papers for publication in online math journals which granted publications!!! Although these journals are freely available for reading, the publication is quite expensive...
.. actually, I never knew that my grandma knows that much commutative algebraic and measure theory.
Great, now I think everyone's trying out that. Good bye Zombies!
18:39
@Ted, do you like the New York Times? I think it covers maths better than most UK equivalents.
@Paul They have good crossword puzzles
18:57
Question for set-theorists in here, is there any trick to prove that if <A,<> is a well ordered subset of <B,<>, and both A and B are infinite, that <A,<> is alike to <B,<>?
define alike
They're alike if and only if there is a one-to-one from A to B (or vice versa), that keeps the well-order
ah
Oh. I think I can say that there is a one-to-one function, since if B is infinite yet countable, |B|=|A|.. so I just have to prove that it also keeps the order..
err
you don't have |B|=|A| unless you specified that |A| was countable too...
oh
nevermind
19:01
Forgot to mention that
But it is, yes
actually if |B| is countable, then so is |A|
It is, and both are infinite so |B|=|A| and I get that there is one-to-one function.. question is how do I prove that it keeps the order.. they do have the same order, so it should be easy. And yet I am stuck.
actually i don't get the question at all
A is a well-ordered subset of B
so just pick your map to be the inclusion map
unless you have a completely different order on A, in which case I have my doubts you can do what you want to do
No, exactly the same order. But it also have to be unto, and the inclusion map is only one-to-one (since A is not the same set as B)
you didn't say that before either... :(
19:05
I know, sorry about that..
so you have that |B|=|A|
?
Yes, I managed to prove that.
And <=< (same order on both).
then it's out of my league :P I was just looking at the easy counterexamples and such
goo luck tho
Which is a strict well-order too, can prove that as well..
Thanks!
19:34
Figured it out eventually
19:56
How do I find the ranks and nullities and bases of the image and Kernel of T:U->V where U=R^4, V=R^4, T(a,b,c,d)=(a+c,c+d,a+b,b+d)?
20:43
Hey @charlie!
Hey @mike don't be scared, lol.
21:08
Only zombies left in this chat.
@nayrb Hi, Why did you delete your answer ? (f(n)<f(n+1)
@user112495 apply definitions from course
L-L
L-L
@BalarkaSen @Shivam Patel I have posted a question here in this chat.chat.stackexchange.com/transcript/71. I would be glad if you could answer it.
21:25
Hi @jasper
Where is @robjohn?
@robjohn, do you think $\psi(1+i)+\psi(1-i)$ may possibly have a nice form?
Hi convergent @chris
@Charlie CAT!!! Hello!!!!
How are you doing?
@chris melting! today the temperature got 38 °C
@Charlie I cannot think when is very hot.
21:38
@chris D:
@Charlie :D
Plus it's not raining, we are facing a lack of water @chris
@Charlie that's even worse.
@chris oh yeah :/
22:02
@Charlie I'm a bit concerned I'm far too quick-tempered, irritable in the last period of time. I should get some rest, 4, 5 hours / night don't seem enough. High temperature has similar effects on me, but this time there is a different cause. :-)
You should take longer peeps of rest @chris
Periods
Relax a bit
Give yourself a time
Charlie, where are you from?
Brazil
I recall from calculus you could sometimes determine whether infinity/[number]^infinity is 0 or something else.. anyone here recalls the method?
22:22
@Charlie Indeed! You're absolutely right!
@Studentmath what method? it depends to what you have
but n/2^n goes to 0
2^(2^n)/2^n is also infinity/number^infinity but it goes to infinity
Hey all, I have embarrassing question. You know in division, when you divide, for example, 8 by 3, you get 2 with a remainder of something. If you drop the remainder, you just have 2. What is that called?
the fact that 3 goes into 8 two times, does that have a name?
quotient
thanks
22:48
@JohnMerlino not quite a good example, since in this case the remainder and quotient are identical, 2
Dividend/divisor = Quotient + remainder/divisor
$\frac{11}{4} = 2\frac{3}{4} = 2 +\frac{3}{4}$
23:17
@skullpatrol thanks , Didn't know about divident
don't we have divider too?
@Theta30 That's the "divisor".
@Theta30 Thanks mate, And I was speaking about that.. Something with deriative. When you are unsure. But I guess it's rather certain for that
23:50
Good evening, someone knows how can I prove : that given infinitely many points, there is always other one which has an irrational distance from all of them. Please
not true
hm okay
So Perhaps have you seen before this exercise : math.stackexchange.com/questions/659023/…
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