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16:15
@Dodsy had to leave for TA
16:28
I was hoping to get a little help on bayesian inference, since my TA is being unresponsive. I got this HW problem mathb.in/154743, description and my thought process is in the link.
16:43
\begin{align}
\sum_{m=1}^n\left(\frac{H_m}{m}\right)^2 & =\sum_{m=1}^n \left(\frac{H_m}{m}\right)\sum_{k=1}^m\left(\frac{H_k}{k}\right)\\
& =\sum_{m=1}^n\left(\sum_{k=1}^n\frac{H_k}{k}-\sum_{k=1}^{m-1}\frac{H_k}{k}\right)\frac{H_m}{m}\\
& = \frac{1}{2}\sum_{m=1}^n (H_n^2-H_{n,2}-H_{m-1}^2+H_{m-1,2})\\
& = \frac{1}{2}(H_n^3-H_{n,2}H_n-nH_{n-1}^2+(2n-2)H_{n-1}-2n+2+nH_{n-1,2})
\end{align}
I am not very convinced about this, it seemed it will blow up while it should not
o wait nvm, careless mistake, I forgot to multiply Hm/m
will deal with these later...
@BalarkaSen You'll appreciate this
On my Facebook feed, some page posted an image of the Borwein integrals, above the line "Top 10 Anime Betrayals"
2
17:06
@Akiva loool
17:21
Field of scalar: $F$ or $K$?
17:32
12
Q: Can unprovability unprovable? Is there an $\omega$-fold unprovability?

Sunghyuk ParkI was just thinking about unprovability. I just wanted to know if it is possible to make a concrete boundary between provable problems and unprovable problems in a certain axiomatic system. We know that there is a statement that is true yet unprovable. Then is it possible that a statement is tr...

Mathematics, where even the unprovability of a statement is unprovably unprovable.
@AkivaWeinberger Mathematical Memes for Logarithmically Scaled Teens :^)
I am arclength parametrized.
17:54
Yo yo yo
Hmm. It turns out that the deadline for the next grant I want to apply for is Octiber 4th rather than sometime in April as I thought. Plus, there is also one further one with deadline October 1st. I hope I can recycle a large part of my proposal.
@Tobias is it a separate grant entirely?
@Daminark Yeah, these are three very different grants.
is poctdoctoral life entirely composed of granting miseries
Read PHDcomics to find out :P
17:57
It seems like I can use essentially the same project description for one of these new ones, but the one on October 1st has a 2 page limit which is bonkers.
@BalarkaSen Pretty much yes, until you secure a permanent position.
Yup, PHDcomics to the rescue :D
loooool
@TobiasKildetoft Rip
18:26
Hi. I am interested in the limit $L=\lim_{K\to\infty}\sum_{k=1}^Kk^{-\alpha}P_K(k)$ for some $\alpha>0$, and where $P_K(k)$ for $k=1,\dots,K$ are nonzero probabilities summing up to 1. Also, for each $k=1,2,\dots$ I have that $P_K(k)\to 0$ as $K\to\infty$. Am I right that since $L\leq 1$, I can use the dominated convergence theorem and in this way $L=0$?
18:38
how can i prove that the alternating group $A_{n}$ is simple iff $n \ne 4$
0
A: Understanding $\sqrt[i]{z}$ for $z\in\mathbb{C}, z\ne0$

Kenny Lau$$\begin{array}{rcl} z^{-i} &=& \exp(-i\ln z) \\ &=& \exp(-i\ln(re^{i\varphi})) \\ &=& \exp(-i[\ln_\Bbb R (r) + i\varphi + 2ni\pi]) \\ &=& \exp(-i\ln_\Bbb R (r) + \varphi - 2n\pi) \\ &=& \exp(\varphi - 2n\pi) [\cos(\ln r)-i\sin(\ln r)] \\ \end{array}$$ For an understanding, we must view number...

lollll my answer got accepted
@NV-US it's probably better if you look it up in an algebra book, the proof I've seen was quite long for this chat
@AlessandroCodenotti they proved 2 lemma's before proving the theorom
@NV-US For $A_5$ (and potentially $A_n$), see this question.
i want to know, in the third line, how did they write (ab)(cd) = (acb)(acd)? and they wrote "Every element of An is a
product of terms of the form (ab)(cd) or (ab)(ac), where a,b,c,d are distinct elements
of { 1 ,2, . . . , n }", so where is the element (abc)=(ac)(ab)
18:48
Exercise 15.39 in the book I have uses guides to prove that $A_n$ is simple, and the guides fill up three quarters of the page.
@LeakyNun i will read the link, if u can please answer my questions
@LeakyNun which book?
@NV-US (a,b,c,d) after (acb) becomes (c,a,b,d); after (acd) becomes (b,a,d,c); which is the same as (ab)(cd).
@NV-US doesn't matter; the point is that it's very long for chat, as Alessandro suggested.
@NV-US (ac)(ab) is in the form (ab)(ac) [a,b,c,d are arbitrary distinct elements]
@LeakyNun what do u mean by " (a,b,c,d) after (acb) becomes (c,a,b,d)"
@NV-US do you know what (acb) means?
yes, a->c->b->a
18:55
so if you apply the permutation to (1,2,3,4) you would get (3,1,2,4) right?
 
2 hours later…
20:31
@Semi yea you need packages in sharelatex (sorry late response)
20:48
8
Q: Hahn-Banach From Systems of Linear Equations

bolbteppaIn this paper1 on the history of functional analysis, the author mentions the following example of an infinite system of linear equations in an infinite number of variables $c_i = A_{ij} x_j$: \begin{align*} \begin{array}{ccccccccc} 1 & = & x_1 & + & x_2 & + & x_3 & + & \dots \\ 1 & = & & &...

Every element of Sn can be generated by the (n-1) transpositions (12)(13)(14)...(1n). I want to ask what was the intuition behind this? Or was it just observation that (1i)(1j)(1i)=(ij). I hope there was some intuition, because most proofs rely on some tricky non-trivial identity, that i cannot "observe".
@TedShifrin Hello Mr. Shifrin! I am not sure, have you got my message here?
21:05
@Kirill Seems that he is gone again, lol.
@ShaVuklia np
@Semiclassical Hi I am kind of bored today, lol.
@Semiclassical can u help me with the above problem
@NV-US my intuition is to note that any permutation can be represented by a string diagram
by that I mean: Suppose you want to represent the permutation (124)
to represent that with a diagram, I write the numbers 1-4 on two different lines
and draw lines (not necessarily straight) from top to bottom, with each such line showing the destination of each element 1-4
so for (124) i'd get
I'm drawing as u type
21:14
same
i'm doing a quick paint doodle
i have a question, if an object was thrown down below from a 70 foot cliff with an initial velocity of 70 m/s , why would the end velocity be zero?
and would the acceleration be 9.8 or -9.8 since its trying to pull the object down?
@NV-US there we go. it's actually easier to draw by hand, since squiggling in paint is a bit annoying
so (124) sends 1 to 2, 2 to 4, 3 to 3, and 4 to 1
What you can notice then is that you can't draw the above diagram without lines crossing
with that in mind, we can deform the above picture a bit without changing its content (drawing):
The only thing I know how to do in Paint is resize and crop, lol.
I know how to draw lines!
...yeah
Wow, I must learn from you some day.
Wait, there is one thing I would like to know though.
How do you make a photo say black and white in Paint? Or give it a grey overtone or other overtone or something like that?
21:20
not sure what you mean by that. start with a color photo and make it black-white?
Yes.
I clicked on a few buttons but that did not achieve what I wanted.
not sure, tbh
I have one more doubt after this @Semiclassical
something something greyscale I think
Are u still drawing? @Semiclassical, if its that much of a hassle, just type, i'll draw on my notebook
21:23
there we go.
so now it's ordered with one intersection in each layer
Layer?
Oops, that looks ugly, lol.
@Jasper yep
Whats layer?
@NV-US sure. there's four layers. in the first layer: 1->1, 2->2, 3->4, 4->3
(it's not drawn perfectly)
in the second layer, 1->1, 2->3, 3->2, 4->4
21:27
@Semiclassical i dont get it
let p(x) = (piecewise function) { 3x-2 where x cannot equal 1/3 and 4 where x = 1/3} find the limit as x goes to 1/3
how would i find a limit for this
@MATHASKER The limit is independent of the value of the function at that point itself.
...? i dont quit get it
one more try
if you follow that diagram from top to bottom, you have (for instance) 4->3 ->2->2->1
21:30
@MATHASKER As x gets closer and closer to 1/3, what does 3x-2 get closer and closer to? Note that x doesn't take the value of 1/3 in this process.
now compare that with my original diagram (the one that actually looked nice)
the first crossing you see as you trace the lines down is between line 3 and line 4.
the next one down is between 2 and what was line 4, and so forth
@MATHASKER Yes, I already understand your question perfectly.
basically i've manipulated those strings in such a way as to isolate each crossing.
21:32
@jasper -1 at 1/3 but it can't be 1/3rds so
the punchline is this: each such crossing, when you arrange them in this way, is just a transposition
@MATHASKER First, do you understand the problem?
so from there I read off that (124)=(12)(34)(23)(34)
i don't know somewhere near the negative 1
yea kind of @jasper
So this way of reading string diagrams lets me read off the product of transpositions immediately
21:34
@MATHASKER If it helps you should draw a graph to see what is going on. You can see then that as x gets closer and closer to 1/3, y gets closer and closer to -1, so the required limit is -1. It's not important what y is when x is 1/3 itself in this case.
@NV-US despite the poor pictures, does that help at all
Yes it does, thank you
But why did u go from bottom to top to write (124) as product of transpositions
oh so the limit is -1 what about the 4 doe do we not have to write anything for that
21:36
@MATHASKER The limit is -1 and you have answered the question. End of story.
because I have in mind that the bottom is the output, and that permutations act from left to right. so the leftmost transposition would be the last one performed
that may not be the usual convention, though.
but it asks for limit of p(x) which includes 4 too doesn't it.? or do we not have to worry about that...@jasper
back later
Ty @Semiclassical
@MATHASKER The question asks for the limit of p(x) as x approaches 1/3, which we have just found out.
@MATHASKER I suggest you read your textbook or notes again as it seems your concept of limit is not very clear at the moment.
21:41
@jasper oh ok ya the teacher just taught it to use yesterday and im still kinda understanding it
@MATHASKER Once you understand something, everything will follow. The problem with many students now is that they practise endelessly without understanding, wasting all their time.
@Jasper thats partly a teaching issue, I think: a good teacher designs their course so as to make it hard for students to blindly pattern-match
There is pattern matching in math, of course, but of a different sort: a novice tried to match problems to patterns of symbols, whereas an expert matches problems to patterns of reasoning
Reminds me of a distinction I made in an undergrad paper which I was always proud of: the difference between being literal, literate, and literary
(Ramble ramble ramble)
22:04
@Semiclassical Yup, and bad teachers are more common than good ones, lol.
22:18
hi chat
23:04
is it incorrect when my peers state $5^3$ five to the power of tree?
"five to the power of three" is a valid way to say 5^3.
In idiomatic English it's usually "five to the third power"
What if someone asks what the power is in this expression $5^3$ and they say it's 3. Would that be incorrect as the entire expression is actually the power?
@NV-US: What you'll want to understand at some point is what happens when you do what's called conjugation. When you do $\sigma\tau\sigma^{-1}$ it's like a change of coordinates. So if $\tau = (12)$ and $\sigma$ sends $1 to i$ and $2 to j$, then $\sigma\tau\sigma^{-1} = (ij)$.
@CausingUnderflowsEverywhere: Imprecise language, again. $3$ is the exponent.
Casually, people might call that the power.
Personally I've always enjoyed saying exponent, but my teacher stopped a student that said n to the power of x.
23:15
Is this all happening in English?
advanced mathematics.
No, no, language.
Oh boy, got me there for a second. Pure english.
Two possibilities. I'm used to USA and who knows where you are. Or ... the teacher is not competent.
Hello Ted ! :D
Hi @Kasmir!
UK English — so how did the teacher want it to be said?
She says using power interchangeably for exponent is a common mistake everyone does.
prove that the elments (a,1) , and ( b,1) commute ,then decuce the order of (a,b)
So how does she insist it be said, @Causing?
I'm glad someone cares about the way things were intended to be just as much as I do.
23:20
@Kasmir: You can't just do that. You need context.
Five to the exponent of tree naturally.
order of a =n order of b = m, (a,b)^l = ( a^l , b^l) ,where l is the lcm(n,m)
@TedShifrin oups sorry, its a direct product
@Causing: I guarantee you that virtually no one in the US would ever say that.
The thing here that i have to deduce the order from that the elements (a,1) , (1,b) commute
I dont think about it that way
Note that you typed it differently this time, @Kasmir.
23:22
prove that the elments (a,1) , and ( 1,b) commute ,then decuce the order of (a,b)
sorry about that
You typed (b,1) the first time.
Yes noticed that just now =p
This is an easy problem. Of course they commute. Figure it out.
Okay I'll wait for the day I'll work with someone from the US and they will be pale when I say to the exponent and it doesn't make sense. Im sure it's not a big deal I'll just have a striking accent to do with how I pronounce (exponent vs power)
m divides lcm (m,n) and n divdes the lcm ( m,n)
23:23
@CausingUnderflowsEverywhere: It makes sense. It's just unnecessarily pedantic to me.
hence (a,b)^l = ( a^l , b^l) = (a^kn, b^tm )
I would, however, ask for the exponent if I were inquiring.
was that for me ?@TedShifrin
haha okay=p
I showed that they commute btw
did not type it here because its avious
but the part of deducing the order of (a,b) from that is not clear
23:25
Why is the lcm the smallest possible power, @Kasmir? You must always show that.
I mean I did not need that the elements commuting
oh
hmm let me keep thinking :D
And you do need that they commute.
I started working from (a,b) ^n directly
= (a^n , b^n)
that number n, must be divied by both order of a and b
in order to get the identity element for each group
Can I just argue by contradiction ?
assuming there is another number less than lcm that fullfill that?
23:27
Yes, you could do that.
all righty :D
Have you proved in class that if $a$ has order $n$, then it follows that if $a^k=e$, it MUST be that $n|k$? ... That's where the contradiction proof goes.
Yes =p I started the proof that way
assumed that p is less than the lcm (n,m)
m divides p and n divides p
Well, if you've proved that fact once and for all, you do not need to reprove it. By definition (?), the lcm is the smallest positive number that is divisible by both. So there's nothing more to do.
WAIT is it then incorrect for me to say , lets convert this exponent into a fraction? should I really be saying power?
23:30
yes exactly :D
I bet I'm making the opposite mistake people in the US are making.
@Causing: No, I want to call it an exponent. But when I refer to the entire expression, I say "x to the yth power" or "x to the power of y."
Halo
How are you all
23:32
Morning
morning? isn't it like 4:32 there, faust?
pm that is.
yeah...
Faust is doing what MikeM always used to do. He always entered with "Morning."
^^
ah, I see.
23:33
Anyhow, hi, Faust and Dodsy.
Hello, Ted.
Im confused
I'm good at confuzling people, Faust.
Hi, @PVAL.
Let G be a group and define $H= \{g \in G | g^2 =e \} $ prove that if G is abelian then H is a subgroup
23:35
wow faust haha.
so i belive that if G is abelian that H is a subgroup
mostly by the trivial
and i can think of a counter example
when G is not abelian
You need abelian for it to be true.
Good.
wait are we talking little g or big G
Big G
BTW, Faust, you and Kasmir should talk algebra together. You're both doing groups.
23:36
i cant rember how to prove this
Kasmir?
@KasmirKhaan
This sounds like a variant of the exercise if $g^2=e$ for all $g \in G$ then $G$ is abelian.
Yell at me nect time you on
He's here, Faust.
@Faust by definition H is a subset of G
also a subgroup of G
@PVAL: But this generalizes to any $k$ :)
23:38
so what is true for G is true for the smaller set Hh
if G is abelian , H must be abelian =p
@TedShifrin Let me have my moment -.- :D
Sure, @Kasmir, but you need $G$ abelian to argue that $H$ is closed under multiplication.
well i belive that
ah nvm i got it
from what I understood he asked if H< G
can they have different structure?
no
but i need to prove H is a subgroup only if G is abelian
answer is no , because a subgroup is also a group so all definitions are inherated
23:39
not assume H is a subgroup
OH
okay ><
buy you are correct =)
you guys are lucky, :( I have to review rules of differentiation.
Dodsy: I love the Spivak-type course. I taught it 14 times or so, and loved it every time.
23:41
One more stupid question:
No question is stupid !
LOL, @Kasmir :P Cute.
Lots of questions are stupid.
Haha
@PVAL-inactive they are stupid after knowning the answer or when allready the answer is known
@TedShifrin our prof said that spivak would be his preffered text but he didn't want us to have to buy two textbooks so only assigns readings from James Stewart :(
23:42
For any $ n \geq 3 $ find two elements a,b of the dihedral group $D_n $ s.t o(a)=o(b)=2 and o(ab)=n
its like a problem vs exercice idea
let me put that on latex so i can read it
let a = rj and b = j this is always true for all n
@Dodsy: That's like reading a trash novel when you have Hemingway to read as a possibility :P
as ab will have order n
how do i show that this is true for all n?
@TedShifrin yes :P, I suppose I could do some quick readings from Stewart and then read from spivak to really learn.
23:44
Stewart is the standard stuff that you're expected to know, Nate, but you'll learn to be a mathematician from Spivak.
( note any reflection must have order 2)
for the diheadral groups, order 2 means reflection or rotation by pi / n if n is even
and in a specific case its easy to prove that ab has order n im just not sure how to do it for all n
if n is odd only reflections has order 2
thats why i used tow reflections
they always have order 2
and there product has the same order as r
which by defn has order n
23:45
@TedShifrin

My current reading list is:
Calculus of One Variable by Keith E. Hirst.
Calculus With Applications by Peter D. Lax and Maria Shea Terrell.
Calculus with Vectors Jay S. Treiman.
Spivak's Calculus
James Stewart Calc

For only one class :S
Yes I am not sure if you have to do that for n even and n odd or not=p
LOL, I don't know most of those, Nate.
hmm
I think Spivak and Stewart should suffice. Maybe this prof is a bit nuts.
He's pretty awesome.
23:46
maybe i should post it
I like him thus far.
Not yet fraust
Kasmir is on thinking mod
That's great, Nate.
lol
I'm ignoring you guys, @Faust @Kasmir.
23:47
i asked a good question though ted!
D_2n = {r,s | r^n = s^2 = 1 , rs = sr^-1 }
@TedShifrin OMG why ? :(
nono
dont pick that
I thought it was funny.
everyone was like ooooo
Why @Faust
pick the reflection rj and the reflection j
they both have order 2
there product is rj j = r
and r always has order n
23:49
Why what, @Kasmir? That's a definition.
Yes that was meant as a hint
hmm
all elements of D_2n can be generated by r and s
reflection and rotation
ohh
we use rho and sigma i dont know what you use
23:50
u use a wierd ewfn
defn
Play well together, children :P
i know it
mines half of yours
@Dodsy: I think humor (particularly in hard classes) is a useful teaching tool. At least, it was for me.
think my $D_4 $ is your $ D_8$
23:51
Right. I use 4 (4-gon) but lots of books use 8 (order 8).
yes some notation is D_n
Its not a problem
@Semiclassical Dang it, I tried to apply your Euler-Maclaurin approach to a seemingly well behaved integral, that is,$$\int_0^{2\pi} \ln\left(1+\frac{\sin(2x)}{2\sin(x)} +\frac{\sin(3x)}{3\sin(3)} \right)~\mathrm dx$$Though it doesn't seem to work. I assume its not analytic lol, which would suck.
one inficies order and other inficies vertices
yeah ^^
anyway s^2 = 1
23:52
ok
what is the product of 2 different s's ?
hmm, interesting
if it is a rotation then you are done
because order of all r is n
doing rotation n times you get back where you started
we saying the same thing in dif notation
oh okay then ><
23:53
=P
@SimplyBeautifulArt should that be 3 sin(x) on the bottom?
thanks +)
@Faust we can study together , like give each other tips and execercies from lectures :D
It also appears to cause problems for desmos to integrate over that mess
23:55
@Faust btw not all rotations has order n , it must be the rotation by 2pi / n
@KasmirKhaan yeah that would be great ^^ i love abstract algerbra.
^^ Good comment, Kasmir.
Thanks Ted _
yeah thats why i picked rj cause it will always have order n
err
not expaling that well
@SimplyBeautifulArt actually, it seems to be analytic in the nbhd of the real line
which is what one needs for the story I was telling to work
23:56
but if i picked r^2 j * j then in D_4 it divided the group and wouldnt yield the needed result
Nate, I guess some of my students might describe my classes as trying to stay afloat in the deep end. I've never heard that metaphor before.
whats up
Hi Meow.
@Semiclassical Then perhaps the $n$th derivative of that thing blows up to infinity too fast?
23:57
@Faust: Don't leave that up for long.
not allowed?
Not wise for you.
@Faust All rithty, I allready got 3 assigment , 15 good exercies
btw Ted is right , I copied the email
@Kasmir: Write down the addy so he can remove it.
that shouldn't be an issue, since it's explicitly $2\pi$-periodic
so the only place it can blow up is in the imaginary direction, and that's not an obstruction
23:58
Ill send you now the assigments that we got =p @Faust
kool ill pass on the stuff i ahve aswell the course notes are great with ltos of worked out explinations

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