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01:00
otherwise, yes
@Xander I will keep that in mind. It'll be tricky to not do that for this exam since I've got my classes and they're currently my highest priority, but I guess that's why I'm taking the test twice just in case
@MatheinBoulomenos the point of it is that there's a lot of variance for us schools so there should be some way to measure if everyone has a certain base of knowledge. Plus i think grad school here pretty much always includes classes that you would get at an earlier level in a european school
bc we specialize in stuff later i guess
i guess in the taxicab norm it wouldnt be too ugly right
@Meow: But this is the theory for why you integrate what you integrate.
The most diffcult thing I integrated is $\int \sqrt{\tan(x)} \mathrm{d}x$
01:00
@Daminark You should also note that a lot of schools don't require the subject GRE
I just did that today with my 10th graders in the class I substituted for, Mathein.
No I confused square and square root, lol
Oh, we did $\tan^2$. $\sqrt{\tan}$ is way tricky, but cool.
$u = \sqrt{\tan(x)}$?
You'll get to a bunch of challenging ones in a few chapters in Spivak, Meow.
That one's in there.
01:01
This is pretty:
I'm outta here for now. Bye, all.
And some integrals related to Fourier transforms and convolutions in my analysis course were pretty tricky, I guess
the past few days ive seen fourier transforms everywhere
source: wikipedia
01:02
Bye @Ted
baricentric subdivision of a triangle?
thanks for helping me think through that problem @Ted see ya !
@MeowMix mmm Furrier transforms
@Xander I haven't though about too many schools tbh, the dream school is Michigan and sadly they require a GRE
four-yay
01:03
Yeah, lots of schools do require it ;)
@XanderHenderson yeah pretty much, since every iteration, the triangles get subdivided into 3 parts using the centroid
@Xander there are places that dont require it? TIL
Anyway I'm just gonna continue pushing it off for now at least :P
@EricSilva Europe ;)
and UC Riverside didn't require it
Actually that might not be a bad idea to consider tbh
01:05
and thus each triplet of triangles resemble the original except look skewed and squashed at every scale
I'm sure that there are others
if i leave the US i would wanna go to brazil
which i think is similar in structure to european schools
Hey all Ive reditted my question ... https://math.stackexchange.com/questions/2625467/algorithm-to-tell-if-an-infinite-sequence-is-random-or-not

Hopefully now it makes more sense
but i was heavily advised against doing this by someone i respect a lot so idk
@Mathein are European folk chill with American math students? I get the vibe that we do much less in undergrad than you but... Does it ever happen?
01:07
@Daminark we have some international students, yeah. I don't personally know American math students, but why not? I would recommend you do a masters, first though.
@Daminark I get the impression that folk start to specialize a lot earlier in Europe
In mainland Europe, the masters is like the first years of grad school
thus a European bachelor's degree in mathematics is somewhere between a US bachelor's and master's
with more math than a typical US bachelor's, but less in the way of general education courses
my second undergrad algebra course could probably pass as a grad course
*re-edited my question

math.stackexchange.com/…
01:08
@XanderHenderson i think part of the weirdness is that "European bachelors" means more than "US bachelors" bc there's waaaay more variation for US schools
@EricSilva I'm not sure that "more" is quite right
different is definitely true
i mean it like has a more concrete meaning
I heard there are some colleges were you start with proofs in the third year or something?
Yeah that's apparently a thing
@MatheinBoulomenos it is not atypical for a US institution to not offer a proofs-based course until the third year
not every student starts college having taken AP calc
01:11
In us anyone can become a mathematician , in europe they do a natural selection in the first 2 years
in fact it's more normal for that to happen than what happens at my and @Daminark's institution
It's very atypical that math majors and engineering majors take the same courses here
which means that a lot of students have to get through a year of calculus, then a year of linear algebra and differential equations, before they are considered "ready" for upper division math
differential equations is not even a required course here
@MatheinBoulomenos You could transpose ANT with representation theory or K-theory tbh
01:12
I mean, you usually prove Picard-Lindelöf in first year analysis
of course, most of the R1 schools get into real math much sooner
since they can typically assume a stronger background upon admission
depending on the school you might see very very little proof based math in a math major
at a state university near my hometown it was normal to take 2 proof based classes in the entire 4 years
@EricSilva Indeed. At my current institution, most math majors can get away with only a couple of proofs based courses, and neither analysis nor algebra is required
yeah at the school i mentioned their first year phd courses were a dummit and foote course and a baby rudin course and since they mostly just accepted their own students it wouldve been most of the students first analysis and algebra courses
@Secret To backtrack a bit, my advisor would say that a fractal is "an object [whatever that means] to which we can associate an appropriate zeta function [whatever that means], and that this zeta function can be analytically continued to a meromorphic[?] function on a domain where it has poles [or perhaps some kind of natural boundary of singularities, or something like that] that are strictly complex [and therefore not real]"
01:22
That's a strange characteristic, as superficially I don't see how zeta functions (assuming your advisor is referring to something similar to riemann zeta) are related to how a geometry behaves. But I guess I would had to be in your field to understand better...
@Secret There is a relation to Riemann's zeta, but that is not exactly what is meant
though it turns out that Riemann's zeta does occur as the geometric zeta function associated to some ordinary fractal string
and that the zeros of the Riemann zeta function are in some sense related to the spectrum of that string
which is nifty
@XanderHenderson there are a lot of things with zeta functions
yeah, and I never knew nor aware of all of this before today
@Xander is there any relation between the spectral zeta functions arising from riemannian geo
like varities over finite fields
01:28
hey can you please help me how to calculate e(i)s with matlab in this question math.stackexchange.com/questions/2625761/…
@EricSilva I think that is where the idea came from, but I don't know that piece of the theory well enough to give a definitive answer
I was able to calculate by hand up until e4
but then it gets super crazy
interesting
4
Q: Mandelbrot set and riemann hypothesis

PMayHas anyone tried to make a connection between the Mandelbrot set and the non-trivial zeros the zeta function? Looking at the Mandelbrot set, it appears that all points are to the left of the line 0.5 + t*I, and to the right of the point -2. So the set is bounded on the left by the trivial zeros...

some random googling lead me to this
@MatheinBoulomenos Indeed; a friend of mine wrote his doctoral thesis on zeta functions associated to elliptic curves
01:29
that's considered fractal geometry?
not really, but my advisor has wide tastes
i need to learn about those spectral zeta boiz
@XanderHenderson umm ... sorry if this is a dumb question but could this pattern be considered a fractal ?
enter image description here
then maybe we could speak and find where the connection might be @Xander id be interested to know
01:31
3
Q: Algorithm to tell if an infinite sequence is random or not?

More AnonymousBackground Imagine a computer that has finite memory but produces an infinite output. Our aim is to find if it using a deterministic algorithm or a non-deterministic(in layman's term having an element of randomness in it's) algorithm. The below algorithm I have proposed seems to do exactly that....

the pic in it?
In mathematics, particularly linear algebra and numerical analysis, the Gram–Schmidt process is a method for orthonormalising a set of vectors in an inner product space, most commonly the Euclidean space Rn equipped with the standard inner product. The Gram–Schmidt process takes a finite, linearly independent set S = {v1, ..., vk} for k ≤ n and generates an orthogonal set S′ = {u1, ..., uk} that spans the same k-dimensional subspace of Rn as S. The method is named after Jørgen Pedersen Gram and Erhard Schmidt, but Pierre-Simon Laplace had been familiar with it before Gram and Schmidt. In the theory...
@MoreAnonymous Anything can be considered fractal, depending on what you mean by fractal; my advisor's answer, I suspect, would be "What is the associated zeta function?"
you basically wrote a for loop to iterate the process
@MatheinBoulomenos My advisor seems to find interesting connections all over the place; he started in QM and wrote a couple of books on Feynman's operational calculus
@Xander that's cool
01:33
which turned into fractal geometry
which led him to analytic number theory and the study of zeta functions
I know less about analytic number theory than I'd like, but the formula for the residue of the Dedekind zeta function is pretty amazing
zeta functions does have a lot of relation to quantum mechanics and quantum field theory for some reason, probably because of how regularisation works
(of which I don't have the background for QFT yet, and mostly heard from my peers talking about it)
the theory of fractal strings is related to the wave equation, which requires a Laplacian, which quickly gets you into the analysis of unbounded operators, which is basically where QM lives ;)
@MatheinBoulomenos It was easy to do using Newton's method. It gives you after three iterative 1.4857... while the exact value is 1.47..... Thank you for your help!
@user777 glad to help
01:39
Mona: Actually I recall the algorithm I use for gran schmidt in my PhD, is basically reading the section in wikipedia which about implementing the numerical stable version of gran schmidt, and then just manually translate that text into python code
(plus coverting all instance of addition into kahan summation to increase the numerical stablity)
The resulting code works very fast and done 360 deg of 3x3 rotation matrix multiplications in roughly 2 seconds
02:19
@MoreAnonymous If youve ever taken a basic calculus class youd know that sequences are defined to be real valued functions on the domain of the natural numbers. There is nothing that special about that fact. The fact that you dont know that tells me you need to first brush up on whatever branch of math is termed as "the study of sequences". At least the elementary stuff.
I have a set Z/36Z. I am asked to find the order of $\bar 2$.
@Yashas ask on main
I haven't completed yet :p
The series which I got after multiplying is: 2, 4, 8, 16, 32, 28, 20, 4, 8, ...
The definition of order an element is the smallest positive power to which it must be raised to give $e$.
$2^8 = 2^2$
$2^6 = e$
but $2^6$ is 28
28 isn't actually the identity element, right?
@Yashas the operation in this group is addition
If the operation was multiplication, it wouldn't be a group, right?
02:26
right
So that's why I was ending up with weird contradiction?
right
02:42
@TheGreatDuck yea ... I've always paid little heed to definitions and always used an intuitive feel instead for the subject. But you are correct this I should pay more attention to definitions so as to avoid miscommunication
How do I go about showing that Z/nZ is not a group under multiplication? I am able to show that 1) it is closed under multiplication, 2) has an identity $\bar {1}$, 3) associative
@Yashas Does $\bar0$ have an inverse
Ok. I was silly.
@Yashas it is a field* if n is prime
02:59
@Jacksoja how come I can't view ur math.stackexhcnage profile?
@MoreAnonymous am protected
ah ... k... I was curious
I enjoy my privacy
are u into pure math or number theory?
Lol, I love how it sounds like "Which are you? Math or number theory?"
03:02
hahaha ....
anyway If u were into either I was gonna ask u this question that has kept me up ... @Jacksoja math.stackexchange.com/questions/2625467/…
Sorry can't help you with that
ohk ... its really been buggin me
I feel like I may have accidently found something super important and cool
(cool) obviously being of more value ;)
03:18
2
Q: how to hide my profile on stackexchange sites?

DaveI am planning to hide my profile on one of my SE site. I am not asking to hide my network profile. While that would be my main goal, but I know that is not possible. What I want is no one can click on my username to see my profile page. That way I have effectively disabled my network profile. I ...

hey can you please help me with this? stackoverflow.com/questions/48493913/…
a = sin(t);
plot(t, a, 'r', 'DisplayName', 'a');
fhat = (21./(8*pi.^10))*(33*pi.^4-3465*pi.^2+31185)*t.^2 +(3750*pi.^4 -30*pi.^6 -34650*pi.^2)*t.^3 +(5*pi^8-765*pi.^6+7425*pi.^4)*t;
plot(t, fhat, 'c', 'DisplayName', 'fhat');
hold on;
Missing a "hold on" before plot (t, fhat ....))
Matlab saids you hold on after every instance of plot
03:36
@Secret still the same problem paste.linux.community/view/1d9ff7e8
can you help please?
imgur.com/a/z8FGE @Secret this is what I get
04:05
@MonaJalal Note the y axis, which is times *10^4. Your fhat function gives a very huge value between (-3,3). Thus your sin t is actually there, just the scale is too large to see it. Try restricting the y axis range with ylim
04:24
can some one explain the last step in this i don't get how they get that fraction i.imgur.com/JLBS3se.png
wouldnt that make sinh^2 x = cosh^2 x - 1 ?
because the fraction cancels down to just -sinh^2 x
oh wait a moment
nope, you get $1- \tanh^2 x$
im cancelling the fraction down wrong aren't i
Update: Elon Musk has raised another two million dollars for his tunnel boring company by selling flamethrowers
lol
@WDUK $\frac{\cosh^2 x}{\cosh^2 x}\neq 0$
04:28
yeh i got it now
1 - tanh^2 x = sech^2 x = 1/cosh^2x
unrelated tangent: If $x/x = 0$, we will get weird Wheel business
i always cancel fractions down wrong drives me crazy
04:39
anyone familiar with linear programs with degenerate basic feasible solutions?
quick question, im trying to explain to some one who is new to u substitution with integrals, and am trying to show how it cancels out a trig function.

they understood it but i am wondering, have i actually written my integrals syntactically incorrect here:
https://i.imgur.com/ZSZy0ZX.png

I refer to the the fact that i have integrals of x but i have du at the end, is that still valid since i showed what dx equals? I'm under the impression i would need dx technically but couldn't figure out a neater way to show it.
with a standard for problem Ax = b and x >= 0 and A mxn, I am struggling to find an example that has an x in the feasible region with only m positive components that isn't a basic feasible solution.
hmm
hi
if one subgroup acting on a set and fixes and element
what can we say on its conjugate?
does it also fix the same element?
If $gx=x$ then $hgh^{-1}(hx)=hx$, it seems to me
so if $g$ fixes $x$ then $hgh^{-1}$ fixes $hx$
@AkivaWeinberger thanks
Any game theorists hanging around?
05:01
So cyclic notation for symmetric groups on finite sets, say S_3, are three distinct group elements written the same as (123)?
@CookieToast me
Yay! Have you ever heard of the card game War?
sure what about it?
i just realized what you meant by game theorist.
facepalm
@Corellian There's one group element that has three distinct ways of writing it
(123), (231), and (312) are all the same group element
(the one that sends 1 to 2, 2 to 3, and 3 to 1)
So you're not game for game theory? D:
@TheGreatDuck
05:06
i dont know it
i was thinking you meant this
@AkivaWeinberger Right, I was thinking backwards
if there's a complex number like $z^3-8i$ couldn't this be factored the same way as difference of cubes?
$z^3-8i = (z-2i)(z^2+2zi+4i^2)$
Ahhh lol @TheGreatDuck :P
that kind of game theory id have no problem talking about for hours
not the channel itself
but theories
however i did promise my birds to put them to bed 8 minutes ago so brb
05:09
S_3 = { () , (123), (132), (13), (23), (12) }
(they kinda know in general how to "read the clock")
both 3-cycles can be written three ways, the three transpositions two ways, and the identity is... identity
no im dead serious
@Corellian Yup
05:10
Ty!
they understand at least that the hands move and its for telling time
XD
In S_4, we have things like (12)(34)
even if they cant read the numbers they can tell what it means, lol
so that's not a cycle, but rather two 2-cycles put together
but in S_3 everything is a cycle
@Corellian
birds are smart
05:16
@AkivaWeinberger I see. naturally <4 symbols allows cyclic permutations only
05:43
im back
06:26
I feel so conflicted omg
I think the sign is throwing me off cuz http://prntscr.com/i70eci
or maybe I'm messing myself up because the numerator is that bar version and then regular z is z^3+8i OMG
throws a baseball at the problem I know taking the complex conjugate could be a bad idea
wolfram why are your signs backwards? prntscr.com/i70f8w
if I use that though the numerator and denominator cancel and I also get -12 as the limit
but I don't get why the signs are switched I mean
$z^3-8 = (z-2)(z^2+2z+4)$ that's probably in the real plane
prntscr.com/i70g4z and prntscr.com/i70f8w signs are switched. Does this have to do with being in the complex plane?
06:42
What is the LCM of $0$ and $n$?
least common multiple
I'm just factoring by difference of cubes but when I'm dealing with a complex number the signs are switched? how and whyyyy?
Oops, my question was not relevant to yours :p It was a separate question.
@Yashas It is $0$
06:45
@Yas what is the definition of the LCM?
I always thought that LCM of two numbers will be at least big as the larger of the two.
What is the definition of the LCM?
@orbit-stabilizer if multiples of $n$ and $m$ are listed, the smallest common element in the lists is the LCM
@Yashas It will be, with the correct notion of "big"
@TobiasKildetoft ?
n (not equal to zero) is always bigger than 0?
06:48
in the case of lcm and gcd, one should order based on divisibility, rather than by the usual order
which makes 0 the largest natural number
In arithmetic and number theory, the least common multiple, lowest common multiple, or smallest common multiple of two integers a and b, usually denoted by LCM(a, b), is the smallest positive integer that is divisible by both a and b. Since division of integers by zero is undefined, this definition has meaning only if a and b are both different from zero. However, some authors define lcm(a,0) as 0 for all a, which is the result of taking the lcm to be the least upper bound in the lattice of divisibility. The LCM is the "lowest common denominator" (LCD) that can be used before fractions can be added...
That should clear it up
@TobiasKildetoft so that list is 1, 2, 3, 4, 5, ..... up to infinity and then 0?
@Yashas except that they are no longer arranged in a line, as they cannot all be compared
(also, not up to infinity, as that is not a number)
@Tob extended reals :P
@orbit-stabilizer extended in which way? Also, we are not working in the reals here anyway
06:55
In mathematics, the affinely extended real number system is obtained from the real number system ℝ by adding two elements: + ∞ and – ∞ (read as positive infinity and negative infinity respectively). These new elements are not real numbers. It is useful in describing various limiting behaviors in calculus and mathematical analysis, especially in the theory of measure and integration. The affinely extended real number system is denoted R ¯ {\displaystyle {\overline {\mathbb...
07:07
@TheGreatDuck sorry I had to step away from my computer dude!
Hey @orbit-stabilizer, how's the homework coming? :P
@usukidoll (2i)^3 = -8i so the top is z^3-8i = z^3+(2i)^3
ARGH
so then by difference of cubes I should get something like
$(z+2i)(z^2-2iz+4i^2)$ and since $i^2=-1$
$(z+2i)(z^2-2iz-4)$
can't believe I got tricked on that -_-! thanks for clearing it up
@anon Hey anon :D how to show that if a subgroup fixes an element so does its conjugate?
that needn't be true
that is the last piece I need to put ._.
oh you remember our example?
07:13
yes
okay :) ehm
and I got -12 just what I needed
but I didn't say the conjugate subgroup fixes the same element as the original subgroup
If $H$ fixes $x$ then $\sigma H\sigma^{-1}$ fixes $\sigma x$
how to show this?
just check it
07:14
oups ><
that was trivial and somehow I did not figure it out -_-'
thanks anon :D
mmhmm
and so the argument is
since K did not fix anything
this would lead to a contradiction
I wonder if there is a stronger statment
as in the number of elements that H fixes, in relation to its cónjugate
$x\mapsto \sigma x$ is a bijection between their fixed points
so they do have the same order :D
because if both fixed something, and had different number of elements being fixed
one needs to use this argument
in the case of both of them fixing same number of elements , those arguments wont work
@anon anyways thanks anon :D i hope we can talk more tomorrow =P I need to sleep its 8 am and i have lecture in 6 hours >< hope i can get 5 hours sleep
Good night ! :)
night
07:21
if we take the limit of Arg(-1) is that -pi?
Arg(z)
since z = x+iy then
Arg(x+iy)
but then there's only x = -1 and y =0 because -1 is the real part and 0 is the imaginary part so
Arg(-1+0i) = Arg(-1)
Hi; To solve "The perpendicular dropped from the vertex of the right angle upon the hypotenuse divides it into two segments of 9 and 16 feet respectively. Find the lengths of the perpendicular, and the two legs of the triangle.", what I did was set up the equations ${ a }^{ 2 }+{ b }^{ 2 }={ 5 }^{ 2 }\\ { 9 }^{ 2 }+p^{ 2 }=a^{ 2 }\\ { 16 }^{ 2 }+p^{ 2 }={ b }^{ 2 }$. But after I plugged in, I got a negative value for p. Does anyone know why?
@DarkRunner wouldn't it be $a^2+b^2=25^2$?
oh lol @anon thanks
08:08
Hello!

How many homomorphism $f:\mathbb{Z}_4\rightarrow S_4$ are there? How could we count them?
08:20
@MaryStar remember that such a homomorphism is uniquely determined by where it sends a generator
Also, please at least mention that you asked on the main site also, so we don't all tell you the same things nd waste our time
@TobiasKildetoft We have that $f(a)=f(a\cdot 1)=f(1)^a$ where $a\in \mathbb{Z}_4$, right?
Does this mean that there are $4$ such homomorphism? @TobiasKildetoft
@MaryStar No
08:38
So is it wrong that $f(a)=f(a\cdot 1)=f(1)^a$ ? @TobiasKildetoft
@MaryStar das ist richtig
aber es sage dir nicht, das gibt es 4 Homomorphismus
Ok! The order of $f(a)$ must divide the order of a, which is 4, right? @LeakyNun
So, we have to findhow many permutations of S4 have order that divides 4, or not? @LeakyNun
We have 1 identity (order 1), 6 transpositions (order 2), 3 products of two disjoint transpositions (order 2), 6 4-cycles (order 4). So in total we have 1 + 6 + 3 + 6 = 16 elements of S4 that have order that divides 4, right? @LeakyNun
Does this mean that there are 16 homomorphisms?
Are my thoughts correct? @TobiasKildetoft
 
1 hour later…
09:57
If f is an injective function N >N, and the sequence $(a_n)$ converges, why does $(a_{f(n)})$ converge too?
*from N->N
@MaryStar ja
@JDizzle just use epsilon definition
@LeakyNun Ah ok! Thanks!
would it be right if I say at some point the sequence must go past the element with the biggest distance to the the limit and after this point apply the definition?
Maybe simple question, if $f(x)$ is monic and $p_i(x)$ is monic, $i = 1,\dots,n$, then from $P(x)Q(x)f(x) = \Pi_{i=1}^n p_i(x)$, why must $P(x)Q(x) = 1$?
It was simple, degree's...
10:18
@JDizzle just apply the definition
I don't know what you mean by that informal statement
@bolbteppa it may be any invertible element in your ring, I think
11:02
how do we calculate Arg(-1)?
the end result is pi if it's computation but the limit is -pi ? :/
I think it's similar to what beep-boop did math.stackexchange.com/questions/837978/…
How to get the limit $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) = \frac{1}{2}$ ?

$\begin{align}
\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) &=
\lim_{n \to \infty}\left( \frac{(\sqrt{n+\sqrt{n}}-\sqrt{n})(\sqrt{n+\sqrt{n}}+\sqrt{n})}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}+\sqrt{n}\sqrt{n+\sqrt{n}}-\sqrt{n}\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
11:31
@jublikon You forgot to remove the first square root in the nominator in the second step. It should be $(\sqrt{n+\sqrt{n}})^2$.
I'm trying to understand robjohns proof here
3
Q: Proving Abel-Dirichlet's test for convergence of improper integrals using Integration by parts

vondipI'm struggling with the following calculus question. Let there be two functions $f,g : [a, \infty) \to \mathbb R$ such that: $g$ is monotonic, differentiable and has a limit at zero $f$ is continues such that $\int_a^b (f) < M \in \mathbb R$. Prove that integral $\int_a^{\infty} f(x) * g(x)$...

I've gone into the abyss
Basically I don't understand why he can justify the existence of the limit by Cauchy (i.e. $(4)$) but then needs an additional proof that the limit exists (i.e. $(2),(3)$)... Why is it not enough to just use Cauchy?
Can anybody help me out?
11:58
It is no wonder h bar is a lot more unstable. They lockdown a lot more frequently
Now even Kaumudi H joined the ranks. We are still not clear how to hack behind the scenes to confirm that
@philmcole thanks man :)
no problem
hi all
let $Q \subset \mathbb R^3$ be the subset where all coordinates are positive. let $f: Q \to Q$ be continous. how do i show that there exists a $v\in S^2 \cap Q : f(v)=\lambda v$ ?
12:17
@MatheinBoulomenos If I conjugate $(a,b)$ by $(1,2, \ldots, n)$ $k$ times, then I get $(a+k,b+k)$, so how do i get from that $(ka,ka+1)$?
Can someone please help with the above?
12:34
hello
@blat Positive, or nonnegative? I.e., is $Q$ open or closed?
i fond this : by the convexity of $f$ $f(s) +f'(s)(t-s)\leq f(t)$ is it correct ?
in The h Bar, 56 secs ago, by Slereah
Feel free to set the example
Let $set = \{example\}$
done
anyway, back to business...
Let G be a cyclic group of order 30. We consider all the subgroups H of G. Let k be the number of H.

Isn't k equal to the number of the divisors of 30, i.e., $30-\phi (30) =22$ ?
@MaryStar Why do you think you get the number of divisors by subtracting the totient function?
12:41
PLease is it true that if $f$ is convexe then $f'(s)\leq \dfrac{f(t)-f(s)}{t-s}$ ?
Isn't $\phi(30)$ the number of elements that are coprime to 30? @TobiasKildetoft
@MaryStar must a number be either a divisor of 30 or coprime to 30?
Ah yes. But how can we calculate then the number of divisors?
@Vrouvrou Yes. Draw a picture. To prove: show that $f(t)-f'(s)(t-s)$ is increasing for $t\in[s,\infty)$ and decreasing for $t\in(-\infty,s]$. (Remember that [$f$ is convex iff $f'$ is increasing.)
@Vrouvrou This one is not true.
12:47
@MaryStar you can just count them one by one
Ah ok!
It's not the same thing—if you divide by $t-s$, you might have to flip the sign in case $t-s$ is negative!
You'll end up with $\ge$ for $t>s$ and $\le$ for $t<s$.
Please someone take a look at this:
35 mins ago, by Silent
@MatheinBoulomenos If I conjugate $(a,b)$ by $(1,2, \ldots, n)$ $k$ times, then I get $(a+k,b+k)$, so how do i get from that $(ka,ka+1)$?
@Silent Yeah, I think he was mistaken in that step
@AkivaWeinberger, so how do we get 'simple transposition' if $\gcd(b-a,n)=1$?
12:57
How about this. Consider $(12\dotsb n)^{b-a}$. That should be $(b-a,2(b-a),3(b-a),\dots)$.
You also get that conjugating $(a,b)$ by $(12\dotsb n)$ $b-2a$ times gives you $(b-a,2(b-a))$.
Gotta run
ok!

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