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00:00
ok now the middle sum can be broken up into multiples of eulers totient function
because if gcd(a,b)=1, gcd(a,b+a)=1
sorry; it should be ka and kb
not just a and b
there's a constant
?
ok
(L+1-ka)(L+1-kb)
@NullOverNul $$\sum_{{\Large 1 \leq a,b \leq L} \atop {\Large \gcd(a,b)=1}}(L+1-ka)(L+1-kb)=(L+1)^2\sum_{{\Large 1 \leq a,b \leq L} \atop {\Large \gcd(a,b)=1}}1-(L+1)2k\sum_{{\Large 1 \leq a,b \leq L} \atop {\Large \gcd(a,b)=1}}a+k^2\sum_{{\Large 1 \leq a,b \leq L} \atop {\Large \gcd(a,b)=1}}ab$$
@NullOverNull can you write out explictly for me,
are these nested sums or multiplied sums together
00:07
$$\sum_{{\Large 1 \leq a,b \leq 6} \atop {\Large \gcd(a,b)=1}}ab$$
can you write that sum out for me? I want to be sure im not misinterpreting what your asking for.
@NullOverNull these are individual sums on the last equallity the factors in front of the sums are being multiplied by the sum
that would equal (1)(1) + (1)(2) + (1)(3) + (1)(4) + (1)(5) + (1)(6) + (2)(1) + (2)(3) + (2)(5) + (3)(1) + (3)(2) + (3)(4) + (3)(5) + (4)(1) + (4)(3) + (4)(5) + (5)(1) + (5)(2) + (5)(3) + (5)(4) + (5)(6) + (6)(1) + (6)(5)
does this clarify at all
@NullOverNull ok we can re-write this sum as,
@JacobBlack what do you think I should study on math before going to college?
user19161
@Khromonkey You might want to read up on proofs.
what do you mean?
00:12
@NullOverNull one sec, im trying to type
user19161
@Khromonkey Well, how proofs go in general.
@JacobBlack could you recommend a resource please?
@NullOverNull can we go in a private chat?
@NullOverNull how do we do that?
00:13
@JacobBlack Do you mean sort of like Poyia's how to solve it?
not sure, can't find a make chat option
How do I start a private chat?
user19161
@Khromonkey No. Actually I am not sure what to recommend you. Oh well, actually you don't need to read anything, haha.
@jacob hiiiii
00:16
so I should chill through high school?
I think it may be a privilege issue
hi @Charlie
not enough reputation on my end
$$\sum_{{\Large 1 \leq a,b \leq L} \atop {\Large \gcd(a,b)=1}}ab=\sum_{1\leq a \leq L} a(\sum_{1\leq b\leq L, \text{ (a,b)=1}}b)$$
user19161
@Khromonkey You should focus on your school work.
00:17
@Khromonkey Hola!
now the inner most sum can be re written mostly in terms of eulers totient
@NullOverNull
yes, the inner sum can be written in terms of totient
that was the nature of the original answer given on the math stackexchange
$$\sum_{1\leq b\leq a, (a,b)=1}b=a\phi(a)/2$$
right, but i have kb
and so k * a * phi(a) / 2
we factored out the k^@
00:18
but then when i simplify the rest of the expression, weird things happen
factored out the k
ok wait
@JacobBlack I hate school
It sucks
user19161
@Khromonkey Just get into a good university.
@NullOverNull I got it
$\sum_{b=2}^{N} b^2k^2\varphi(b)/2-3bkN\varphi(b)/2-3bk\varphi(b)/2+N^2\varphi(b)+2N\varphi(b)+\varphi(b)$ is what I get at the very end
user19161
00:22
Don't choose those with big nice buildings but poor curriculum.
I like big nice libraries
But the most important thing is definitely the teachers and curriculum
user19161
Of course, the library should be well stocked.
so then you don't think I should do anything?
user19161
I am talking about superficial things like glass windows or whatever.
$$\sum_{1\leq b\leq ka, (a,b)=1}b=\frac{(k-1)ka\phi(a)}{2}+k\frac{a\phi(a)}{2}$$
user19161
00:25
There is too much superficiality in this world.
@NullOverNull now we can chose a multiple of k, so that ka is close to L
The school where I go to has the worst math teachers in the world
however its called MIT
user19161
@Khromonkey You are an undergrad in MIT?
@Khromonkey I know of atleast 3 MIT's in India.
Luckily MIT has a horrible campus
00:26
@JacobBlack when did you start wearing glasses?
So you're covered there
No i am a high school student in MIT
Monterrey institute of technology
user19161
LOL
@NullOverNull which MIT has horrible campus?
00:26
Really? Why?
or in spanish Instituto Tecnológico de Monterrey
Just my opinion
user19161
@Charlie I think in high school.
Felt very dry/concrete
00:27
@NullOverNull Its really hard to talk to you here, and it doesn't sound like the problem is that important to you, id say you are going to end up with sums that cant be evaluated to be honest
not the sort of vibrant grassy academic sphere you envision for a stereotypical campus
@JacobBlack me too, when I was 16
@Ethan The sum I posted above is what I get when I reduce the whole thing to a single sum-loop
The problem is that when I then pull out the phi(N), the sum fails and that makes little sense to me
somehow something turned into a floor function in the translation
user19161
@Charlie Sometimes I wanted to go for Lasik, but not much point in it, I like my Spiderman glasses which are very sexy.
anyhow, i appreciate your help
it's a tough summation
00:29
@JacobBlack woooo ;)
Could somebody help me with algebra?
user19161
@Charlie I don't like normal glasses, and others don't like Spiderman glasses.
@NullOverNul it depends how you define tough, I don't think you are going to end up with a "nice" closed expression, you will most likely end up with easyier sums, that cant be evaluated directly
@JacobBlack I like spider man
@PeterTamaroff what is the problem?
00:31
@Ethan Right; but the pieces with the totient function, after pulling everything apart, act in weird ways that don't normally make sense
And I can't help you btw
@Khromonkey ???
what is the problem you need help with?
>"As an example we shall show that the dihedral group $D_n$ (...) is defined by the relations $$\tag {33} x^n,y^2,xyxy$$
>in the free group generated by $x,y$. It is clear that $D_n$ is generated by the rotation $R$ through an angle of $2\pi/n$ and the reflection $S$ in the $x$-axis. We have the relations $$\tag{34} R^n=1,S^2=1,SRS=R^{-1}$$

>Hence $D_n$ is a homomorphic image of $FG^{(2)}/K$ where $K$ is the normal subgroup generated by the elements $(33)$. We shall now show that $|FG^{(2)}/K| \leq 2n$ which will imply that $D_n\simeq FG^{(2)}/K$. Let $\bar x=xK$,$\bar y=yK$ (...). Then, s
what part don't you understand?
00:32
I cannot understand "Hence $D_n$ is..."
Why is that so?
What would the homomorphism be?
I don't know.
Oh, well.
what is a homomorphism?
and what is a free group?
@Khromonkey Are you trolling me?
lol
00:33
Really, I'm not in the mood.
user19161
@Ethan You omitted a word there.
I offered help before the question was asked
user19161
I just realised Firefox is at 19 now.
ohhhh I think I see why
All I know about algebra is the first two chapters of dummit's book and I forgot most of it by now
00:35
the totient translation only works for a>=3
where the totient is even
something else has to happen for the 1 and 2 cases
@PeterTamaroff I am not trolling you
do you guys play chess?
@Khromonkey Thanks, anyways,
No problem
Anything else you need just ask
user19161
I am sorry I cannot help you @peter. But when I am better, I will help you with 9000 questions a day lol.
@JacobBlack When will you be better?
user19161
00:40
@PeterTamaroff Maybe never, I don't know.
Because I think I might slain a piglet right now, and drown a kitten with its blood.
user19161
But, there can be miracles when you believe.
@JacobBlack "Oh, I believe that I can solve this problem."
BAM. Books smashes me in the face. "TAKE THAT MOTHERFUCKER."
user19161
@PeterTamaroff Yes, that is when you try hard and think for the next week over it, and you might solve it.
user19161
That is how math works Pedro, one thinks for days and days, not a few seconds.
00:41
@JacobBlack Then it is not by believing, but with hard-friggin-work.
@JacobBlack are you sick?
@JacobBlack I know.
@Eugen Holy mother of Troy.
user19161
@PeterTamaroff I told you, you don't understand the definition of "miracle" and "believe" there.
@PeterTamaroff still doing anal?
@JacobBlack Ok, illustrate me, man.
00:42
what?
@Eugene Nah. I'm doing algebra now.
@PeterTamaroff You are now doing a Zebra? Gross.
@Khromonkey HAHAH your face must have been priceless.
user19161
A miracle is something hard to happen but can happen if you believe in yourself and try hard enough, QED @peter
00:43
lost interest already?
@JacobBlack Stop QEDing mundane stuff.
You're becoming a heretic.
@Eugene Nope, I kinda finished Spivak.
So now I'm doing some Algebra.
@PeterTamaroff i see
Next month I officially start college.
user19161
@PeterTamaroff So you see, everyone misunderstands Mariah Carey.
@JacobBlack She's a good singer, and that's about it.
I don't really care about most people, you should know that.
00:44
the lineup here has changed a lot
@Eugene So you think you can help me with free groups and relations?
@Eugene Ah?
user19161
@PeterTamaroff You must call him Dr Eugene now lol
@JacobBlack I mean, I admire Jim Carey as a comedian, but I don't give a rat's ass if the news is making a special section about him getting married, or splitting up, or having a kid, or getting arrrested, or into a fight, or debating child labor...
Can someone explain this to me? every graph with three vertices and degree at least n/2 contains a hamiltonian cycle
@JacobBlack I thought he was doing an MSc
00:47
what does it mean with degree? Isnt the degree for a vertex and not the whole graph?
user19161
@PeterTamaroff Yes, I was just trying to explain to you how I interpret that MC line.
@PeterTamaroff i was. and i don't know anything about free groups so I can't help you there.
@Eugene Uhh. Bummer. How did it go?
@PeterTamaroff how'd what go?
@JacobBlack I think that is your interpretation of what MC said, but I don't agree. She just said some corny stuff.
@Eugene The MSc
user19161
00:49
@PeterTamaroff Oh OK, now you know what I mean. Because, some people mock that line without knowing what I mean.
NBP
NBP
Ok guys this is ridiculous
I still got no answer Pete.
@NBP Please call me Peter.
@PeterTamaroff oh it went fine. finished 2 months before schedule.
Pete means blowjob in my language.
NBP
NBP
Really? Sorry
What language is that?
00:50
I actually write it phonetically as Pitt.
spanish
@NBP Castellano $\subset$ Spanish
wow blowjob and anal. interesting life there @PeterTamaroff
It is actually slang.
and i expect to be banned in 3... 2.... 1....
NBP
NBP
00:51
Ok Pitt, why haven't I got an answer yet
@Eugene Allow me to retort: Do you always finish so fast?
castellano is only what catalans and basques and galicians call it
user19161
@PeterTamaroff Tell that to Mr Sampras!
@Khromonkey OK: Rioplatense.
There you go.
@PeterTamaroff ? i've only done 1 masters so i don't know how to remark on that....
NBP
NBP
00:52
Why are you guys ignoring my plight? :/
we also call it that in mexico
@Eugene I'm being facetious.
@NBP what is the question
NBP
NBP
Khromonkey, probability.
user19161
00:52
@NBP Do you know that many of us here might know less than you?
NBP
NBP
@JaconBlack I doubt it
@NBP I have no clue on it, I told you already. And don't feel ignored, that's how it usually goes.
NBP
NBP
I'm an undergrad
i'm studying at MIT so I can probably help
user19161
@NBP Do you know that many of us here are school kids?
NBP
NBP
00:53
1
Q: Problem with coupling (basic probability)

NBPIf I have two probability spaces : $\\\Omega_1=\{w^1_1,w^1_2,w^1_3\}$ with $P_1$ defined to be $P_1(w^1_1)=P_1(w^1_2)=P_1(w^1_3) = 1/3$ and $\Omega_2=\{w^2_1,w^2_2,w^2_3\}$ with $P_2$ defined to be $P_2(w^2_1)=P_2(w^2_2)=1/5$ , $P_2(w^2_3)=3/5.$ I am asked to find the (or a) smallest probabilit...

at least jordan isn't here.
2
NBP
NBP
@Jacon probably not many
user19161
@NBP This is the third time you misspelled Jacob.
where is Jacon hiding?
user19161
For a start, get your spelling right.
user19161
00:54
Then you can work on your theorem.
NBP
NBP
@Jacon sorry, I can't pronounce Jacob with a straight face
Jacob&
I apologize
Jacob two two
NBP
NBP
Khormonkey, I sent you the link
yes, I am analysing the question meticulously
NBP
NBP
Khromoney, what's meticulously?
00:56
with great detail
@Eugene That was what I was thinking. I could not remember the name though.
@Eugene I remeber that.
nope sorry I don't know
NBP
NBP
I take it MIT is not that good on the mathematics department
MIT doesn't have a mathematics department
00:58
Its all shopfloor guys.
user19161
Do you even know which MIT we are discussing?
@JacobBlack YES! WE DO!
NBP
NBP
@OrangeHarvester, will you reconsider helping me with my exercise?
@OrangeHarvester, I still have no answer, it's been 32 hours now
@NBP I am not saying no, but to put it in reference I have one question that has not been answered for a year... so...
And I no squat about probability, so sorry no, I cannot help.
NBP
NBP
@OrangeHarvester, I'm sorry send me the link I'll try to answer
user19161
01:00
@OrangeHarvester Also, I have a question with two incorrect answers lol.
NBP
NBP
@OrangeHarevester
@JacobBlack My answer is not incorrect!!!!!!!!
NBP
NBP
please reconsider
dude we can't answer your question!
@NBP I said I am sorry. I am trying to cover analysis now.
01:02
I'm in high school at the Monterrey Institute of Technology
user19161
@OrangeHarvester It is, I am afraid, and I hope you see it one day.
user19161
@OrangeHarvester Alex has already seen the gap in his answer when I discussed it with him in chat. Yours is a different problem.
@JacobBlack Okay. By the way, do you have a problem with the idea or exact logical formulation, if it is the exact logical formulation, then I am yet to get there.
user19161
@OrangeHarvester You see, the problem is that the idea is based on flawed logic.
user19161
Like I said, handwaving is very dangerous.
user19161
01:05
Too much handwaving and the answer becomes wrong.
are you talking about this answer? math.stackexchange.com/a/203451/33907
user19161
@Khromonkey Yes, we have discussed the various errors in its various revisions.
LOL, I love the part "various errors" in the sentence
I hate that part by the way.
01:29
@peoplepower !!!!
You there??
@PeterTamaroff Indeed.
Thank goodness.
I need some help to grasp relations on free groups.
I am given $\Bbb Z^3$ with addition $(x_1,x_2,x_3)\oplus (y_1,y_2,y_3)=(x_1+y_1+x_2y_3,x_2+y_2,x_3+y_3)$
This forms a group with $e=(0,0,0)$
Now, consdier the free group $FG^{(3)}$ with generators $\{x_1,x_2,x_3\}$
I have to show $G$ defined by the relations $x_1x_2x_3=x_2x_1$, $x_2x_3=x_3x_2$, $x_1x_3=x_3x_1$ on $FG^{(3)}$ is iso to $(\Bbb Z^3,\oplus,(0,0,0))$
I have found the following:
@peoplepower do you follow me?
@PeterTamaroff Yes
NBP
NBP
@OrangeHarvester, Please reconsider.
OK.
SO, I have found that $(1,0,0);(0,1,0)$ and $(0,0,1)$ actually generate this group
With $(a,b,c)=ae_1\oplus ce_3\oplus be_2$
(you can check that)
Moreover, if we set $x_1=e_3$, $x_2=e_2$ and $x_3=e_1$, the relations are fullfilled.
How do I move on?
@peoplepower
01:37
So you have a mapping into $G$.
Yes, I have a unique homomorphism sending the generators like I siad.
So if I find the kernel, I should be done.
But that seems impossible.
This is what my book has on relations:
>"A group $G$ is said to be finitely generated if it contains a finite group of generators $\{a_1;1\leq a_i\leq r\}$. Then we have the homomorphism $\eta$ of $FG^{(r)}$ sending $x_i\to a_i$. Since the $a_i$ generate $G$, this is an epimorphism and $G\simeq FG^{(r)}/K$ where $K$ is the kernel of $\eta$. The normal subgroup $K$ is called the set of relations connecting the generators $a_i$. If $S$ is a subset of a group, we can define the normal subgroup generated by $S$ to be the intersection of all normal subgroups of the group containing $S$. If $S$ is a subset of $FG^{(r)}$ we say that $G
Honestly, that is all.
Reading what Zach wrote here helped quite a lot though.
You have the homomorphism $FG^{(3)}\to (Z^3,\oplus)$ as given there.
Yes.
I have that.
You have that backwards, if I read correctly
So it simply sends a word in the group to a sum of versors
I map $x_1\to e_3$, $x_2\to e_2$ and $x_3\to e_1$
01:43
Yes,
check that the kernel contains all relations.
That is immediate.
As you concluded above, indeed.
And it is normal.
So it contains the normal subgroup generated by the relations.
So $G\to FG^{(3)}/K\to (Z^3,\oplus)$ where $K$ is the kernel of my above homomorphism.
Notice that the first is just canonical.
What do you mean by "canonical"?
Also, does the fact that the $e_i$ fullfil the relations matter?
01:46
You start with $FG^{(3)}\to FG^{(3)}/K$ and factor by the relations defining $G$.
I think so. Anon said I should look for generators satisfying the relations.
But I didn't know why.
He usually does that =P
@PeterTamaroff Yes, and my above statement requires this fact.
@peoplepower Right, I see. I see why it is important.
@peoplepower What do you mean by "canonical"?
@PeterTamaroff It is based on the choice of map $FG^{(3)}\to FG^{(3)}/K$, usually called canonical; though I have not read enough Jacobson to know his terminology.
OK, I think you mean "natural" map, which is $a\to aK$, yes?
01:50
Yes, it is categorically natural too, and that might be his terminology.
As for the problem, we have set-theoretic inverses $G\to (Z^3,\oplus)$ and $(Z^3,\oplus)\to G$ where the second is the map you found initially.
But shouldn't we be consdiering $FG^{(3)}/K \to \Bbb Z^3$?
I think what I said above is fine; on the other hand notice that the map you gave is an isomorphism and if we want to look in that direction we should be proving $G\to FG^{(3)}/K$ is an isomorphism too.
@peoplepower Where does it follow that "...we have "set theoretic" inverses..."
?
Oh, duh.
The fact it is an iso follows from quoting by the kernel.
Since now the kernel is trivial.
Ah, you are getting used to the first isomorphism theorem. We could also use the definition of isomorphism here.
@peoplepower =)

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