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1:55 AM
@user1618033 How is the book?
 
 
1 hour later…
2:58 AM
Is $\ln\omega$ defined?
 
$\omega=...?$
 
ordinal
 
ah. no idea.
 
I think I am OCD
I have rewritten my introduction 4 times in the past 3 days
 
Hello!
@Semiclassical iterated more?
 
3:11 AM
it's running overnight on campus
 
Hope it goes better this time!
 
el hostpital's rule
what is wrong with the students cant they speak french
 
You just combined the Spanish "el" with a non-existent word "hostpital"
it's either l'hôpital or l'hospital
 
I think Forever Mozart was quoting a student?
 
3:24 AM
yes
 
Oh, sorry
@ForeverMozart You're French?
 
no but I can speak some
 
I see
 
i never knew it would be useful
until I had to read some math papers from the early 1900s
 
3:48 AM
i'm more ambivalent about this answer than I expected. on the one hand, as an answer to the question it's clearly nonsense
on the other hand, there's an infinite product being appealed to which I haven't seen before
namely $\displaystyle \prod_{p=-1\text{ mod }4}\frac{p^n+1}{p^n-1}$
where the product is over all primes congruent to -1 mod 4
and that's a beast i know nothing about. so i'm curious if there's a valid identity the author is appealing to, albeit one which doesn't establish what they imagine it does.
 
@LeakyNun In the surreal numbers, you mean?
 
I think so?
 
Note that all finite ordinals are integers, so $\ln\alpha$ isn't an ordinal even if $\alpha$ is finite
 
What about $\lfloor\ln\omega\rfloor$?
 
$\omega-1$ doesn't even make sense in the ordinals, as $\omega$ is the smallest infinite ordinal. It does make sense in the surreals, though.
I don't know if $\ln\omega$ makes sense in the surreals, but my guess is "yes".
(But I'm really not sure.)
The surreals, if you don't know, are kind of like the ordinals, except expanded into a field. So, you can always subtract and divide, and $1+\omega=\omega+1$ (which isn't true in the ordinals). It takes some care to define them properly.
It's also too large to be a set, like the ordinals. It has the property that any (set-sized) field is isomorphic to a subset of the surreals. So it essentially contains every possible field.
 
4:05 AM
no one has been able to convince me that surreal numbers are useful
 
People have argued that about math as a whole, though.
Or at least the really pure bits
 
How do perform subtraction in a cyclic group?
 
@Semiclassical let $\chi$ be the nontrivial Dirichlet character mod 4, and $L(s,\chi)$ the corresponding Dirichlet character. By looking at Euler products, that infinite product is $\zeta(n)L(n,\chi)/\zeta(2n)$.
@JesterTran subtraction only makes sense if the group operation in the cyclic group is written additively. and in that case, what you're asking is not much different from "how do I do a group operation in a group" which is pretty broad.
 
@JesterTran Like, for example, in $\Bbb Z/7\Bbb Z$, you'd want to know how to find $\bar 3-\bar 5$?
 
4:08 AM
i know something that nobody else in the world knows
 
@Semiclassical actually I forgot to look at the 2-factor in the Euler product, which might be where the 2^n-1 in the denom is from.
 
very good feeling
 
I see
 
@ForeverMozart Is it a secret?
(I love secrets. They're so fun to spill)
 
right. which means that the result being quoted is plausible, albeit incapable of doing what they think it can
 
4:12 AM
depends what alan thinks he's doing.
as you say, it doesn't really give a closed form, just an interesting formula (if true)
 
right.
 
@AkivaWeinberger yes it is a math secret
but soon it will be revealed :)
 
And it has a wonderful proof but this chat is too short to contain it?
 
yes unfortunately it takes 8 pages
 
4:39 AM
hi
 
lol hi
 
0
Q: Calculating the period of a modular congruence

user19405892How do you in general calculate the period of a number $x^m$ modulo $n$ using the Chinese Remainder Theorem if $n$ is squarefree? We see that it is sufficient to find the residues modulo every prime factor of $n = a_1 a_2 a_3 \cdots$ where $a_1,a_2,\ldots$ are the prime factors of $n$ listed in a...

I am wondering if my formula $t = \text{lcm}(a_1-1,a_2-1,\ldots)$ is correct
 
5:00 AM
Hello all!
 
user147690
5:20 AM
@EricStucky Do you know basic things about Coxeter groups?
 
user147690
95% sure I remember you writing something about them on your blog
 
user147690
Actually yes, I found some mention. But what ever happened to Cluster algebras for you btw?
 
Hey Saw, how's it going?
 
jeje
I still don't know how to introduce cluster algebras
I know something about coxeter groups, @Alex
 
user147690
@EricStucky I have to calculate polytopes for 14 clusters, RIP
 
5:26 AM
O.O
wai
 
user147690
I think there are 16 length 6 ways of using transposes s_1,s_2,s_3 to get from 1234 to 4321
 
user147690
(Looking at the permutohedron)
 
user147690
@EricStucky Trying to prove a theorem and then a conjecture ;P
 
user147690
But I would be happy if there were 6 reduced words in the coxeter group that get from 1234->4321 if that makes sense
 
user147690
Please tell me if that makes no sense haha
 
user147690
5:31 AM
@EricStucky Yes, they seem hard to motivate without lots of even harder to motivate background hahaha
 
I only see 14
12*
 
user147690
@EricStucky 12?
 
Actually no I don't see even that many, let me try again
 
user147690
123121 121321 123212
Symmetry
321323 323123 321232

132312 132132
Symmetry
312132 312312

213231 213213 231231 231213 212321 232123
 
user147690
16 that I see
 
5:35 AM
Okay I found 16
 
user147690
Are these all reduced words?
 
user147690
If that makes sense by itself
 
user147690
I guess I need (W,S) to state the real problem don't I
 
I don't see how they couldn't be reduced words
 
user147690
Damn thats what I thought
 
5:37 AM
in this context the only words that aren't reduced have two same numbers next to each other, yes?
 
user147690
That's what I thought too
 
user147690
rekt
 
user147690
There are 36 sides right?
 
user147690
36 edges*
 
user147690
5:38 AM
36e,24v,14f
 
user147690
fits euler formula
 
in the permutohedron?
 
user147690
Yep
 
user147690
Sorry, forgot only I am looking at the picture hahaha
 
yeah 36
oh yeah Euler :/
wait f=14 am I being trolled hard rn
 
user147690
5:42 AM
Why?
 
user147690
8hexa
 
the S3 permuto has 5 edges right?
 
user147690
6rect
 
No it's a hexagon why does a pentagon come to mind
idk point is I'm not being trolled
 
user147690
Oh good haha
 
user147690
5:44 AM
yeah and there are 2 reduced for the hexagon
 
user147690
s1s2s1 s2s1s2
 
user147690
So there are i reduced
 
user147690
And I am meant to get an $i\times k$ matrix, where there are $k$ terms in those reduced words
 
user147690
So I got a $2\times 3$ matrix for the hexagon, that described each of the side lengths
 
user147690
Now I want a $n\times m$ matrix where $n*m=36$, for the $36$ edges, and I am getting shrekt, since $16\not|36$
 
5:45 AM
oic
 
user147690
If there were 6 reduced words I would have $6\times 6$ gusta mucho
 
But yeah, I have Official™ confirmation that 16 is correct
 
user147690
Kind of confused though, since there are 2 reduced words for S3, but s1s2s1=s2s1s2
 
oh wait nope this isn't the long cycle :/
 
user147690
5:50 AM
My friend is working with hook characters soon though :D (in linked paper)
 
user147690
I was really hoping I could get that 16 up to 18 haha
 
user147690
But I checked it three differentish ways
 
Yeah :/
 
6:08 AM
The OEIS gives this formula: $$\frac{\binom n2 !}{1^{n-1} \cdot 3^{n-2} \cdot\cdots\cdot (2n-3)^1 }$$
for ways to do this in Sn.
 
user147690
Oh that's nice
 
This grows much faster than n!, so yeah the dream is pretty dead.
 
user147690
I was hoping to do for the next dimension
 
user147690
And then after that the problem has infinite bases of canonical type, and I don't worry anymore haha
 
Yeah, I'm gonna think and to see if the formula is straightforward because it seems nice :)
 
user147690
6:11 AM
Nice :D
 
I mean, I'm going to do that when it's not 1am :P
 
user147690
Sure :P
 
user147690
Like when it's 1:30am haha
 
^ where is the lie tho
 
user147690
wat
 
6:13 AM
What are you two up to?
 
user147690
Lie as in Lie algebra
 
user147690
@SAWblade He was helping me out because he is a top lad
 
Helping you out with what? :0
 
user147690
I am trying to construct polytopes from clusters
 
user147690
Clusters attached to the simple roots of the lie algebra sl(4) (dual canonical type basis for dual to $U(n_+(4))$ and then map this to the shuffle algebra, and then cluster mutate, and then draw polytopes :P)
 
6:16 AM
Ah, shame, I do not know what those are. xD
 
oh dear goodness I am so hopelessly a combinatorialist
those words actually made sense to me
 
user147690
Hahahaha nice
 
what the hell man
 
Ah, I aim to become a combinatorialist. :0
 
Well, the only advice I have for you is to read Proofs and Confirmations but I'm not sure that's actually good advice.
The book is about as gentle to algebraic combinatorics as exists in written form, but, like, maybe take a class instead of trying to get it from a book :/
Anywho I'm in 'giving random strangers advice'-stage sleep dep so g'night :P
 
user147690
6:22 AM
Good night, thanks for your help :D
 
npnp
 
user147690
I'll tell you if I work it out :P
 
please do :)
 
6:59 AM
I'm trying to understand Linear Feedback Shift Register (LFSR) behavior in terms of finite fields and primitive polynomials. I don't have any knowledge of finite fields. Where should I start?
 
user147690
7:10 AM
@0x1000001 I don't have any knowledge on LFSR so I don't know how much theory you need, also are you a mathematics major ?
 
user147690
I am fairly sure cryptography needs galois theory around here (in your education [assuming you are a math major])?
 
EE sophomore, I know nothing. But I know that LFSRs have (2^n)-1 states, where n is the number of bits. In part because of the (2^n) - 1 states, we should be able to model the LFSR with primitive polynomials.
 
user147690
What is a sophomore again
 
user147690
Freshman->Sophomore->Junior->Senior?
 
Yep
 
user147690
7:20 AM
Can you explain LFSR's to me with an example?
 
Absolutely.
An LFSR uses a shift register and XOR logic gates to create a psuedorandom pattern.
If we take a 3 - bit register:
1 1 1 - here n = 3, as there are 3 bits
 
user147690
Yep
 
"tap" the two right most registers, lead values (voltage high) to XOR
 
user147690
So you check (111) by XOR on the last two entries, and obtain 1?
 
user147690
Sorry 0
 
user147690
7:24 AM
since XOR
 
yep. feed that back to the left side of the register, shift everything over - now its 011
 
user147690
Ahhh sure
 
user147690
So (111)->(011)->(001)->(100)->(010)->(101)->(110)->(111)
 
user147690
So it is cyclic
 
The galois field arithmetic allows us to determine which registers to tap to ensure a maximal length sequence, ie so it doesn't die and just shift zeroes
How it does, I don't know. It can be done with primitive polynoms, though from what Ive read
 
user147690
7:28 AM
Sure, I'm thinking on it
 
Thanks!
 
user147690
If we had n=4, with tap being the two rightmost registers still, how do we 'start' it? Do we get to choose arbitrarily a nonzero start?
 
We could preset it to all 1's. Let me make certain that is best answer, though.
Ok. Just as long as we have a non-zero bit, we'll avoid the trivial solution.
 
user147690
Sure
 
Ex: 100 -> 010 ->101 -> 110 -> 111 ...
 
user147690
7:43 AM
Are you working with Fibonacci LFSRs and/or Galois LFSRs, or something else?
 
Fibonacci LFSRs
 
7:56 AM
Hi @AlexClark
 
user147690
Hey @BalarkaSen
 
Sorry, I never replied back to your reply to my blowup question. Did you ask anything particular? I didn't notice.
 
user147690
@BalarkaSen I think not
 
OK, great.
 
user147690
@BalarkaSen I just told you which blowups I had looked at :P
 
user147690
8:00 AM
I do want to work out some tangent cones and spaces though, but not tonight
 
user147690
I just want to work out this LFSR thing above and then do my boring classes assignment
 
Right. Do you have a mental picture for blowups?
 
user147690
@BalarkaSen Depends on the example
 
user147690
I have one for some simple planar projective curves with singularities
 
Say, blow up $\Bbb A^2$ at a point.
 
user147690
8:02 AM
I guess I am meant to have a 'chart' at the blowup point?
 
user147690
(Being $\Bbb P^1$)
 
I can only guess what you mean by chart there, but that's probably not how I think about it.
 
user147690
I haven't done manifold theory yet :P
 
To understand blowups, you don't need to though.
There's a nice picture in Hartshorne, you might want to have a look. It's close to how I think about it.
 
user147690
@BalarkaSen I think I should buy Hartshorne. It looks really good
 
8:07 AM
Not my kind of book, but yeah, people think it's quite good.
 
user147690
Fultons intersection looks really good too
 
user147690
@0x1000001 Are you happy with obtaining the feedback polynomial?
 
I borrowed it from my algeo prof and read some from there, but it's terse.
Very hard read.
 
user147690
The latter?
 
Yes.
 
user147690
8:08 AM
Definitely then
 
Sure, I'm curious why it works. My goal has been to find a way to prove that we must have an even number of taps.
for maximal length, that is.
 
Hello!!

I want to find all the matrices in Jordan form with characteristic polynomial the $(x+2)^2(x-5)^3$.

Let $\mathcal{X} (x)=(x+2)^2(x-5)^3$.
The possible minimal polynomials $m(x)$ are the ones that $m(x)\mid \mathcal{X} (x)$, so
- $(x+2)^2(x-5)^3$
- $(x+2)(x-5)^3$
- $(x+2)(x-5)^2$
- $(x+2)(x-5)$
- $(x+2)^2(x-5)^2$
- $(x+2)^2(x-5)$
right?

How could we continue to get all the matrices in Jordan form?
 
@AlexClark If you ever want to understand something about blowups, shoot me a ping and I can say something if you want.
 
user147690
@0x1000001 Aren't you just working in $\Bbb F_2(X)$?
 
user147690
@BalarkaSen I will soon :D
 
8:13 AM
uh, sorry, I didn't follow your notation
@AlexClark
 
It all originates from the fundamental question of studying isomorphism types of varieties inside a birational equivalence class.
Just sayin' because blowups might seem unmotivated at first (which I feel like how you are feeling).
E.g., an interesting question is given a very singular variety, can I produce a variety birational to it which is smooth? That is, can I remove singularities by preserving the birational type? Blowup is a useful tool which can be used to settle this
 
This morning I've read your comment, and now I understand your words, that mine was a simple identity,; I've proved it with your equations, very thanks much @DanielFischer
 
user147690
@0x1000001 I've run out of time to think about your problem sorry. I'll tell you if I work it out, but I probably won't think much on it for a few days. Were you trying to solve this by some deadline, or just out of curiosity?
 
Trying to solve it by Monday morning. Thanks for the time you spent on it.
 
user147690
@0x1000001 Ahhh damn
 
user147690
8:22 AM
@0x1000001 Well then I should try to pass on my thoughts :P
 
user147690
So finite fields only take order $p^k$ for some $k\in \Bbb Z^+$ and $p$ prime
 
user147690
You need to check your feedback poly is primitive
 
user147690
Looks like you are working with $p=2$, i.e. working mod 2
 
user147690
With $k=1$, which seems strange
 
user147690
So you are working with $\Bbb F_2[X]$
 
user147690
8:28 AM
I.e. polynomials with coefficients $0$ or $1$, like (1 or 0) x^0 + (1 or 0) x^1 + (1 or 0) x^2 + ....
 
@AlexClark Ah, so if I can show that the primitive polynom will only have even numbers of x terms (which each represent a tap) then I can show that the maximal length feedback equation requires even number of taps?
 
user147690
@0x1000001 Yep
 
user147690
But it seems very strange...
 
@AlexClark I don't really understand the math behind it, but it will certainly do for an answer. Thanks.
 
user147690
But yes, in your case the only primitive polys are $x+1$ and ones with odd number of terms (i.e. even number of taps)
 
user147690
8:31 AM
Since any poly in your case, if it has an even number of terms, i.e. odd number of taps, is divisible by x+1 (since 1 is a root)
 
@AlexClark odd number of terms means you are counting the 1 term?
 
user147690
Yep
 
@AlexClark Can you recommend any source for understanding/manipulating primitive polynoms?
 
user147690
If there are an odd number of taps, we have x^4+x^3+x^2+1 for example, then this means that x=1 gives 4\equiv 0, hence not primitive (since it should be irreducible, but has factor (x-1))
 
user147690
@0x1000001 Really it should be covered in Galois theory, but that's normally very hard math
 
8:35 AM
@AlexClark I'll take your word for it then :D
@AlexClark and thanks for your help
 
user147690
@0x1000001 No problem, sorry it couldn't be cleaner sounding haha
 
@AliCaglayan Hey! Well, writing a book is the most difficult thing I ever attended so far, harder than any of the problems in it. All is fine so far, still having to finish tasks to it. It's a long journey until you see it finalized.
 
@MikeMiller you around? Any idea on this?
 
8:54 AM
@MikeMiller I finally arrived at the statement of mirror symmetry (in physics).
 
Jim
9:13 AM
0
Q: Finding similar permutation.

Jim$A$ is a nonsymmetric matrix. $\beta_k, \gamma_k$ are sets of permutations. Each permutation acts on column of $A$. $\beta_k, \gamma_k$ have same permutations but different labeling is used to label columns of $A$. $\beta_k$ used one kind of labeling, $ \gamma_k$ used a different labeling. Con...

 
Hey guys the singular points of $(1-x^2)y"-xy'+4y=0$ are 0 and -1 right? Am I correct?
 
 
1 hour later…
10:29 AM
@LeakyNun I have seen logarithms of cardinals. But I have not encountered logarithms of ordinals.
 
10:39 AM
This might be a silly question, but here goes. In the lie algebra $\mathfrak{so}(2)$ does an element not look like $$\begin{pmatrix} 0 & -a\\ a & 0 \end{pmatrix},\quad a\in\mathbb{R}$$
 
user147690
10:51 AM
@mortjt Check this guy out: $\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}$
 
@AlexClark which is equal to $$\text{exp}\left(x \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\right)$$
@AlexClark and well, then the conclusion does follows? I just feel uncertain
 
1
Q: The values of constants in the Equation.

Adesh Tamrakar$$ \frac{\int_0^{4\pi} e^t(\sin^6at+\cos^4at)\,dt}{\int_0^\pi e^t(\sin^6at+\cos^4at)\,dt}= L, $$ the question asks the value of $a$ and $L$. My friend solved it by differentiating, but i didn't understood a single thing. May be it is a very simple one, but i don't know how to solve it.

please try to solve this..
 
11:43 AM
@MartinSleziak Thanks
 
How does one pronounce Dirichlet?
 
@LeakyNun Maybe Cantor normal form could give you something like logarithm with the base $\omega$? I am not sure. (I did not encounter situation where such things would be needed.)
 
Thanks
I'm just asking out of curiosity
 
@iwriteonbananas Fun fact. The fundamental group of the base space does not act on the one of the fiber (although it acts on higher homotopy groups and (co)homology groups), but the one of the total space does act on the one of the fiber, which is something that is almost never discussed in textbooks.
An instructive example is BH->BG->B(G/H) coming from a group extension H->G->G/H. The respective fundamental groups are the groups you started with and G/H does not act on H via conjugation as long as H is not abelian, but G does.
 
Hi @archipelago.
Haven't seen you around in a while.
 
11:58 AM
@LeakyNun I simply tried to search in Google Books and I found this in Enderton's Set Theory.
Logarithm Theorem. Assume that $\alpha$ and $\beta$ are ordinal numbers with $\alpha\ne0$ and $\beta$ (the base) greater than $1$. Then there are unique ordinal numbers $\gamma$, $\delta$ nad $\rho$ (the logarithm, the coefficient, and the remainder) such that $$\alpha=\beta^\gamma\cdot\delta+\rho \land 0\ne\delta\in\beta \land \rho\in\beta^\gamma$$
 
user211812
Hi, In Atiyah-Macdonald's text on commutative algebra, Prop 1.1 says: There is one-to-one, order-preserving, correspondence between the ideals b of A containing a(a is an ideal of A) and the ideals b'' of A/a given by phi inverse (b'')=b. Can anyone state what the order preserving bit means? Does it mean for two ideals P and Q in A/a and P is a subset of Q, then phi inverse(P) is a subset of phi inverse (Q)?
 
@AdeshTamrakar Use complex integral to deal with it :D
@MartinSleziak But... you would generate a number smaller than $\omega$?
 
That depends on the base, doesn't it?
You get $\log_\omega\omega=1$ as expected.
Since for ordinal exponentiation you have $2^\omega=\omega$, you would get $\log_2\omega=\omega$.
And it seems that if $\beta>\omega$, then $\log_\beta\omega=0$.
 
I see
 
@user340001 I think that in situations like this order preserving is usually related to ordering by inclusion.
BTW this seems to be called lattice theorem.
 
12:26 PM
@Kari dih-rih-CLAY, I believe.
 
I pronounce it as dih-rich-lay.
 
I've only ever read it, so I don't know for sure
 
Wikipedia says: "Johann Peter Gustav Lejeune Dirichlet (German: [ləˈʒœn diʀiˈkleː] or [ləˈʒœn diʀiˈʃleː]; 13 February 1805 – 5 May 1859) was a German mathematician who made deep contributions to number theory...."
 
dee-ree-CLAY or dee-ree-SHLAY, then, translating the IPA
where the r is rolled in the back of your mouth and the ay isn't a diphthong
 
So Wikipedia mentions both German and French pronunciation, if I understand it correctly.
 
12:37 PM
I looked at two videos just now and they both do DIH-rih-shlay
 
It seems that there were also some questions like this on the main. But not everybody agrees that they are on-topic. Relevant discussion on meta: meta.math.stackexchange.com/questions/19861/…
 
that's how i heard people say it too
 
BTW if you can read Russian, the transcription into Cyrillic is usually quite a good approximation of how some name should be pronounced: Дирихле, Петер Густав Лежён
The Russian transcription seems close to diʀiˈkleː
 
Does anyone know the best mathematical college in India?
 
1:02 PM
There is absolutely no way that this function above is invertible, right?
 
Right, because it's not injective.
 
so an invertible function must be bijective?
@MartinSleziak I would just read the IPA in the English Wiki German: [ləˈʒœn diʀiˈkleː] or [ləˈʒœn diʀiˈʃleː];
 
@LeakyNun That's right.
 
Though I have no idea why the final t was not pronounced; strange to see it in German
Usually in German every letter is pronounced
diʀiˈkleː = something like di-"r"i-KLAY
diʀiˈʃleː = something like di-"r"i-SHLAY
"r" is a uvular trill
 
yes and a bijective function should be (even if unbeknowingly) invertible
a surjective/injective fts is invertibe on a specified range
 
1:31 PM
Heya
 
Tommorrow I'm in Bucharest to meet some mathematicians ... (f**king long trip to do)
(perhaps arriving there to the evening)
 
(removed)
 
Do you know about the Franca Leclair approximation?
Riemann zeta zeros.
It approximates the points t such that Re(zeta(1/2+it))=0 and such that Im(zeta(1/2+it)) = not zero.
My hope was that by convolving the zeta function with the Moebius function over the divisors, one could push the Franca Leclair approximation closer to the actual zeta zeros. But it would of course only be approximately accurate. The error I guess would be about 3*10^-3 for the first zeta zero.

In order for this to work one would need to invert functions like in the plot above.
Ok, this last sentence about inverting I am a bit unsure of.
And one thing. There is no simple relation for the complex argument Arg[(complex number 1) + (complex number 2)], is there?
 
1:55 PM
Does anyone know how to maximize $a^y * b^x \leq limit$ with respect to x, y?
 
@Danu So what is it? Sonething something Landau-Ginzburg models?
 
$$\left\{\left\{y\to \frac{\log \left(c a^{-x}\right)}{\log (b)}\right\}\right\}$$
:29822255
 
@MatsGranvik ?
 
c= limit
I don't know more than that.
Solve[a^x*b^y == c, y]
Do you need partial derivatives instead?
 
@MatsGranvik I did that see above
 
2:15 PM
[hello](www.google.co.uk)
[link](www.google.com)
Why it is not working?
 
add http:// and try again.
 
Yeah! Thanks
 
user211812
@MartinSleziak Thank you very much. Thanks again for editing my post for I did not know how to represent an ideal generated by an element of a ring.
 
3:20 PM
Why is this number so large? https://oeis.org/A114893
9146.698193171459265866197663466341252....
never mind,
 
Can anyone help with b ii). I think I need to find a parameterisation of that surface but I am not sure how to do it.
 
Can someone have a look at this
0
Q: Am I using sandwich theorem incorrectly?

ramsayI saw this question and wondered how OP of that question was able to do : $$0<\sin x+1<2$$ this $$\frac 0{|x|}<\frac{\sin x+1}{|x|}<\frac 2{|x|}$$ and when $x\to \infty$ he got the limit evaluated as zero. Why i wondered is because it is not working on this inequality $$2\leq x+ \frac 1x\leq 20...

 
3:39 PM
@ramsay are the comments not helpful?
 
4:11 PM
Hi anyone skilled on random graph?
 
What is the domain of $\sqrt{[x]-x}$ where $[x]$ is the greatest integer of x. The domain should be all integers right?
 
No, of course it's much larger than the integers.
 
How @BalarkaSen ?
It should not be. For the expression to be defined (if you consider the function from reals to reals) $[x]\geq x$
And that is true for only integers
 
What is the greatest integer function? Isn't it the greatest integer closest to $x$?
In that case $[0. 6] = 1 > 0.6$... and $0.6$ is not an integer...
 
No @BalarkaSen that is not the definition
 
4:25 PM
That's the floor function.
Both your notation and name is not standard.
 
I have been taught that as the greatest integer function
 
Look, they are the same.
the greatest integer function is the floor function.
 
I have never heard of the former terminology.
 
Yes . I agree to that@LeakyNun
That shows the power of notations
 
@Albas All non-negative integers
 
4:28 PM
@LeakyNun it is also good for negative integers $[-3]=-3$
 
Oh, mea culpa
 
So $-3-(-3)=0$
 
I mean, by definition $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. Then trivially that expression is defined only when $\lfloor x \rfloor = x$, i.e., when $x$ is an integer.
So what you said is trivially true, yes.
 
@BalarkaSen thank you. My book was showing the empty set for some reason so I wanted to check
 
4:43 PM
If I have $a_n + a_{n+1} = 4n + 2, a_1 = 4$, how can I prove that the terms in odd and pair places are arithmetic progressions each one?
I thought about finding $a_{n+2} - a_n$ but don't know how. Any help?
 
$a_n + a_{n+1} = 4n + 2$ and $a_{n + 1} + a_{n+2} = 4(n+1) + 2$. Subtract the former from the latter
 
Oh right, thanks
 
Did you know that the b-th root is edible? I just found out.
 
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