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1:09 PM
hello..i am having problem in few questions of pigeon hole principle in discrete maths
 
yay, I did a Did :)
1 character(-ish) answer
 
can i give one question can you please expalin what to do in it
 
@ShivangiBhatnagar I can try
 
ok
show that every sequence of n^2+1 integer includes either an increasing sequence of n+1 numbers or decreasing sequence of n+1 numbers
 
@Chris'ssis Oh I see, thanks
@Chris'ssis What's $\sum_{k=1}^n\dfrac{1}{2^k-1}$ ?
 
1:14 PM
@Hippalectryon Ramanujan answered these questions.
 
@Chris'ssis But he isn't here to tell me :/ I need that for a probability exercise
 
@ShivangiBhatnagar You should try asking it on the main website, math.stackexchange.com instead of in this chat ... looks like a cool question, I will upvote it when you ask it :)
 
ok
we can not ask such question ..then i will not ask
 
Why not?
 
i mean here
 
1:15 PM
@Hippalectryon I answered it somewhere on main.
 
@Chris'ssis Do you have the link ?
 
actually i have searched there only first but i didnt get it
thats why i m asking here..
 
You can ask it yourself ... ?
On the main site
 
4
A: Find the infinite sum $\sum_{n=1}^{\infty}\frac{1}{2^n-1}$

Chris's sisI think you wanna see this: Ramanujan’s Notebooks Part I Click me and try Entry $14$ (ii) / pag 146 where you set $x=\ln2$ Chris.

 
1:16 PM
thanks
i am sorry this is not the question i am asking for
my question is of pigeon hole principle not of calculus
 
@Chris'ssis uh :/ thanks, but that's gonna be too hard for my probability exercise xD i'll try another way
 
@Hippalectryon I doubt you find something better than Ramanujan's work ...
 
@Chris'ssis Do you know a bit about probabilities ?
 
yes i know
 
@Chris'ssis I think I just took the wrong approach
 
1:19 PM
@Hippalectryon Out of practice. I was good years ago ...
 
Hi @ShivangiBhatnagar @hippa
 
hi sawarnik
 
where are you from?
 
@Chris'ssis Here's the exercise : $\Omega=\mathbb{N}^*,P(\omega=n)=\dfrac{1}{2^n}$, let $A_k$ be the event $k\mid\omega$. 1) Find $P(A_k)$ 2) Let B be the event "$\omega$ is prime", show that $\frac{13}{32}<P(B)<\frac{209}{504}$
@Chris'ssis I have found that $P(A_k)=\dfrac{1}{2^k-1}$
 
I never thought I'd make an answer like Did's one character one
but I've done it :)
 
1:23 PM
@Hippalectryon I'll keep this exercise and retake it when I turn back to the probability area. It seems interesting.
 
@Chris'ssis It's 5 dollars :P
 
@ShivangiBhatnagar You live in Delhi?
 
@Hippalectryon What do you mean?
 
@Chris'ssis 5 dollars if you wanna keep the exercise :DDD
 
@Hippalectryon lollllllll :-))))))))))))))))))) Nice!!!! :D
$$\LARGE{\text{I HAVE AWESOME NEWS!!!}}$$
 
1:26 PM
Tell us :D
The awesome news is that... you have awesome news
:D
@Sawarnik Can you help me on a proba exercise ?
 
@Hippa Proba-bly not.
 
Can we have a vector space that does not have any basis?
 
@Hippalectryon lol
@Sush @Sush!
 
@Sawarnik :c
 
@Sawarnik, hi!
 
1:29 PM
@Sush That isn't impossible
 
8
Q: Evaluating $\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx$

Chris's sisWhat starting point would you recommend me for the one below? $$\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx $$ EDIT Thanks to Felix Marin, we know the integral evaluates to $$\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$$

 
@Sush Are you out of that college yet? :P
 
@Sawarnik If you don't take the axiom of choice for granted, there are some v spaces for whom we do not know if they have a basis
 
@Hippalectryon @robjohn see above :-)
 
Doesn't mean it doesn't exist, though
@Chris'ssis :O WUT
@Chris'ssis Why is that awesome though ? (awesome, as in AWESOME)
It's great, but I wouldn't say AWESOME
user image
3
 
1:31 PM
@Hippalectryon Did you see thet no one knew the value of the integral so far?
@Hippalectryon That is a very clever piece of advice.
 
@Chris'ssis Oh indeed
@Chris'ssis It's on his profile
 
@Hippalectryon, I think it is possible for infinite-dimensional vector spaces only, right? Because finite dimensional itself means that there are basis to define its dimension.
 
@Hippalectryon Proba-bly the wrong ping?
 
@Hippalectryon Yeah, I know, I saw it.
 
@Sawarnik Oh indeed
@Sush Indeed
 
1:32 PM
@Sawarnik, No, I enrolled one.
 
@Sush In finite dimension, it is easy to prove that there always exists a basis
 
no in chennai
 
Ok :)
 
Anyone here can help me with some probas ?
 
@Sush You still didn't tell me its name :O :D
 
1:34 PM
@Hippalectryon, I think I read a proof.
@Sawarnik, :)
 
@Sush Tell me!
Plz :D
 
@Sawarnik, why?
@Sawarnik, i'm at DSE.
 
@Sush :O
You said you were in some random Gujrat college? :O
 
@Sawarnik, I did my undergrad from there.
 
@Sush But you said you didn't change yet? :O
 
1:40 PM
@Sawarnik, are you preparing for ISI B.Stat or something?
 
@Sush Not thinking of it now :D
Nothing sure.
 
"But you said you didn't change yet?" Pardon but i can't understand what u wanna say@Sawarnik
 
@Sush Nuthing.
@Sush So when did you get into DSE?
 
@Sawarnik, In July.
This year.
 
@Sush That is why you were not here those days?
 
1:43 PM
@Sawarnik, oh yes. i missed this site a lot.
 
$$P(V)=\displaystyle\sum_{\omega\in\mathbb{N}^*}P(\omega)P\left(\bigcap_{k=1}^{\‌​lfloor\sqrt{\omega}\rfloor}\overline{A_k}\right)=\sum_{\omega\in\mathbb{N}^*}\dfr‌​ac{1}{2^\omega}\left(1-P\left(\bigcup_{k=1}^{\lfloor\sqrt{\omega}\rfloor}{A_k}\ri‌​ght)\right)\le\sum_{\omega\in\mathbb{N}^*}\dfrac{1}{2^\omega}\left(1-\sum_{k=1}^{‌​\lfloor\sqrt{\omega}\rfloor}P\left(A_k\right)\right)=\sum_{\omega\in\mathbb{N}^*}‌​\dfrac{1}{2^\omega}\left(1-\sum_{k=1}^{\lfloor\sqrt{\omega}\rfloor}\dfrac{1}{2^k-‌​1}\right)$$
@Chris'ssis ^
 
^Eww.
 
Urm why does it error -__-
It displays fine on hostmath.com
 
@Sush Was getting into DSE a breeze for you? :D
 
$$P(B)=\displaystyle\sum_{\omega\in\mathbb{N}^*}P(\omega)P\left(\bigcap_{k=1}^{\‌​lfloor\sqrt{\omega}\rfloor}\overline{A_k}\right)$$ $$=\sum_{\omega\in\mathbb{N}^*}\dfrac{1}{2^\omega}\left(1-P\left(\bigcup_{k=1}^{‌​\lfloor\sqrt{\omega}\rfloor}{A_k}\right)\right)\le\sum_{\omega\in\mathbb{N}^*}\df‌​rac{1}{2^\omega}\left(1-\sum_{k=1}^{\lfloor\sqrt{\omega}\rfloor}P\left(A_k\right)‌​\right)$$
$$=\sum_{\omega\in\mathbb{N}^*}\dfrac{1}{2^\omega}\left(1-\sum_{k=1}^{\lfloor\sq‌​rt{\omega}\rfloor}\dfrac{1}{2^k-1}\right)$$
@Chris'ssis ^
Maybe you see a way to simplify the last expression ?
 
1:45 PM
@Sawarnik, i was and am a very lazy and careless guy, so i could have done better but yes it was a breeze.
@Sawarnik, omg hope u didn't notice lady-lazy mistake!
 
@Sush haha i didn't, but now i know
lol :D
 
@Sawarnik, hihi!
 
@Sush How is DSE like? Hectic?:
 
@Sawarnik, thanks for increasing my vocab. I think it is hectic, but the atmosphere can be improved.
 
Ok :)
 
1:52 PM
What's DSE ?
The Dinausaurs Section of Education ? :D
 
:D
 
:D
 
Oh that makes sense
>.>
 
<.<
 
@skullpatrol, hey!
 
1:54 PM
hi pal
 
that still doesn't tell me what DSE is xD
@skullpatrol Can you help me with some probas ?
 
askaway
 
@Hippalectryon, hmm, that's Delhi School of Economics.
 
@Sush oh ok
@skullpatrol see below
34 mins ago, by Hippalectryon
@Chris'ssis Here's the exercise : $\Omega=\mathbb{N}^*,P(\omega=n)=\dfrac{1}{2^n}$, let $A_k$ be the event $k\mid\omega$. 1) Find $P(A_k)$ 2) Let B be the event "$\omega$ is prime", show that $\frac{13}{32}<P(B)<\frac{209}{504}$
34 mins ago, by Hippalectryon
@Chris'ssis I have found that $P(A_k)=\dfrac{1}{2^k-1}$
I have tried :
11 mins ago, by Hippalectryon
$$P(B)=\displaystyle\sum_{\omega\in\mathbb{N}^*}P(\omega)P\left(\bigcap_{k=1}^{\‌​lfloor\sqrt{\omega}\rfloor}\overline{A_k}\right)$$ $$=\sum_{\omega\in\mathbb{N}^*}\dfrac{1}{2^\omega}\left(1-P\left(\bigcup_{k=1}^{‌​\lfloor\sqrt{\omega}\rfloor}{A_k}\right)\right)\le\sum_{\omega\in\mathbb{N}^*}\df‌​rac{1}{2^\omega}\left(1-\sum_{k=1}^{\lfloor\sqrt{\omega}\rfloor}P\left(A_k\right)‌​\right)$$
11 mins ago, by Hippalectryon
$$=\sum_{\omega\in\mathbb{N}^*}\dfrac{1}{2^\omega}\left(1-\sum_{k=1}^{\lfloor\sq‌​rt{\omega}\rfloor}\dfrac{1}{2^k-1}\right)$$
 
Ewwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww.
 
1:56 PM
But that's still pretty ugly
 
@Sush But you still didn't tell me your first name :O
:D
 
2:13 PM
I guess I'll ask on main
 
Bye.
 
@MikeMiller $\mathbf Z_p$ acts on the solenoid (the inverse limit of inverse system of $S^1$s). Question is, whether the action is free and properly discontinuous. Fixed point free part is obvious, but it's not clear if the action is prop. disc. as the solenoid $X$ not only has the profinite topology but is also inherited with the topologies from the $S^1$s. In fact I guess it's less likely for the action to be prop. disc.
Although if it is, then you have the striking (to me?) identity $\mathbf{Z}_p \cong \pi_1(X/\mathbf Z_p)$
 
No you don't, @Balarka. You only have the theorems about $\pi_1$ in the case that that your spaces are path connected and locally path connected. Solenoids are neither.
 
2:28 PM
oh noes
you're right
hrmph
 
@Chris'ssis That's a funny one
 
@MikeMiller suggestions on how to patch that up?
 
@Hippalectryon Yeah, indeed.
 
@MikeMiller Isn't it true that complement of non path connect spaces inside a path connected space is path connected? Then how about thinking about $\prod S^1 - X$?
 
1
Q: Probability of an integer being a prime

Hippalectryon$\Omega=\mathbb{N}^*,P(\omega=n)=\dfrac{1}{2^n}$, let $A_k$ be the event $k\mid\omega$. 1) Find $P(A_k)$ 2) Let B be the event "$\omega$ is prime", show that $\frac{13}{32}<P(B)<\frac{209}{504}$ One can see easily that $P(A_k)=\dfrac{1}{2^k-1}$ To find $P(B)$, I did the following : $$\begi...

Shameless advertising
 
2:40 PM
test
hi
can i ask something about homolocial algebra?
 
@Teddy It's $10 :D
 
I don't have suggestions to patch it up, @Balarka. For screwed up spaces, like solenoids, the fundamental group is just not the right tool to study them.
 
haha
i am broke
 
Personally, my taste is for things that are locally contractible, so I can't say I'm passionate about solenoids.
 
@Teddy Ask anyway, we'll take your liver
 
2:43 PM
my question is related to this math.stackexchange.com/questions/1038286/…
 
@MikeM my interest is purely number theoretic
 
i dont have enough reputation to comment -.-
in this task, you have a long exact sequence ..->C_k->C_k-1->C-k-2->...
 
@Balarka Then you will get bored of solenoids quickly.
 
Can anyone confirm the answer to this question is correct?
 
where (C*,d*) is a chain complex of free modules over a ring R
 
2:46 PM
When do the kids here have their exams?
 
@JasperLoy Where ?
 
@Hippalectryon Anywhere in this chat, lol.
 
sure, i am just looking at a few examples, preparing for doing this in general for gal(\bar q/q) @Mike
 
@JasperLoy What kids ? What exams ?
It greatly changes
 
and the problem is, they use the splitting lemma. but i only know this for short exact sequences
 
2:47 PM
@Hippalectryon Kids like you. Any school exams, lol.
 
lol
 
but they have a long exact sequence
 
@JasperLoy For instance, I have competitive exams at the end of the school year, and that's all
 
lol at what, @Mike?
 
@Hippalectryon I see. My last exam was about ten years ago, lol.
 
2:48 PM
@JasperLoy e__e
 
i know you can a long exact sequence split into a short exact sequence
but then it doesnt fit
with the chain-homotopy
 
I lol frequently, @Balarka.
 
@mike I am now deciding which dictionary to get for French and German...
 
jasper do you lern german?
hallo wie gehts dir?:)
 
you have said that before, @JasperLoy
 
2:52 PM
@Teddy I hope to soon, but I have not started, lol. Hi! I remember you are German!
@BalarkaSen Yes. Sometimes I take years to decide on a book. For example, I took years to decide on my 12 holy math books.
 
I'm not going to engage in dictionary chat because I think it's unhealthy.
 
LOL
 
@Teddy I am going to use Assimil German With Ease to learn German. Have you heard of Assimil products?
 
no i dont know assimil
 
Ah, I thought Assimil is very popular in Europe.
 
2:57 PM
hallo, wie gehts dir? = hello, how are you ?
;)
 
@Hippalectryon Now I'm curious too see who are the real integration gurus ... :-))) $$\LARGE{\text{500 Points Bounty}}$$
9
Q: Evaluating $\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx$

Chris's sisWhat starting point would you recommend me for the one below? $$\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx $$ EDIT Thanks to Felix Marin, we know the integral evaluates to $$\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$$

2
 
Anyway @teddy Assimil is a French company started in 1929.
 
i dont know. i learned english and spanish in school
 
@Chris'ssis :DDD
 
@Jasper ah ok=)
 
3:03 PM
@M.N.C.E. I just granted 500 points bounty for the question above. You might like the question since it involves integrals.
I also added this message that is important
The bounty will be granted for a simple solution that only uses methods of real analysis. Good luck and have fun!
@Hippalectryon btw, from yesterday's work, even if things were wrong, there is a cute question there tthat we shouldn't let go.
 
@Chris'ssis Which one ?
 
@Hippalectryon $$\sum_{k=0}^{\infty} B(a/2,k/2+1)$$
 
3:18 PM
Ah yeah
 
@Hippalectryon actualy, we may write it as $$\sum_{k=0}^{\infty} B(a,k/2+1), a>1, a\in \mathbb{Z}$$
 
Indeed
That's what I wrote on my paper
@Chris'ssis $a>1,a\in\mathbb{Z}$ ??
@Chris'ssis Did you mean $A\in\mathbb{Z}\cup]1,\infty[$ ?
 
@Hippalectryon lol, I don't know why you play like that. :-) It's exactly what I meant above your last message.
 
@Chris'ssis I'm not playing :/ for me, $P(A),P(B)$ means $P(A)$ AND $P(B)$
When you write $a>1,b>a$ it doesn't mean $a>1$ or $b>a$
 
@Hippalectryon Now let's see how you recommend me to finish the series ... I saw the theoretical part ... :D
@Hippalectryon The nicest thing would be to finish that series without using integrals, but that sounds crazy, right?
 
3:29 PM
@Chris'ssis :O
Kind of.
 
3:53 PM
@Chris'ssis Thank you for the offer. I have an approach in mind and will try to answer it if I can.
 
4:04 PM
Say I have $n$ urns and $m$ balls. The distribution of the balls in the urns is equiprobable. What's the proba that one given urn has exactly $k$ balls ? @skullpatrol @MikeMiller
 
@M.N.C.E. OK, great.
 
r9m
4:21 PM
@Chris'ssis very Nice !! :D
 
@r9m hehe, yeah, it is! Let's see now the most brilliant solution that one ... :-)
 
@Hippalectryon Since you're looking at one given urn, isn't this just the Bernoulli distribution with $p=\frac1n$?
 
@DanielR :O me stupid
 
r9m
@Chris'ssis okay !! :-) The integration gurus must be working on that :-)
 
The bounty will be granted for a simple solution that only uses methods of real analysis. Good luck and have fun!
 
4:23 PM
 
@r9m ^^^ (simple solution)
 
r9m
:D 'kay !! 'kay !! 'kay !!
 
@Hippalectryon lol
 
hi @Chris'ssis
 
@user2179021 Hi
 
4:34 PM
I am hoping my latest question is at last in your area of interest! math.stackexchange.com/questions/1039829/…
please tell me it is so :)
 
r9m
@M.N.C.E. Nice solution !! :-)
 
@r9m Thank you.
 
Anyone here uses Tox ?
 
what is it?
(no)
 
4:40 PM
sfw?
 
Of course
 
It's a math chat after all
22 hours ago, by Nick
@Twink A pony went into a coma, ate some fondue and got pregnant?
 
you know that these things are exactly the worst things to use if you want to avoid government surveillance?
 
@user2179021 I've heard some good things about this one
 
4:42 PM
@Hippalectryon that was aimed at you
 
Anyway, it can't be worse than Skype xD
 
@Hippalectryon well it can
 
the point is that only a small subset of people who have something to hide use something that is marketed as being for people who want privacy
everyone uses skype
so you are hidden in plain sight
 
Still, I don't do anything weird with it so I don't really mind :-)
 
4:43 PM
you are basically doing their job for them by using a niche product.Now they know you are interesting :)
 
But at least it seems to be 100% add free, fast, and updated often
 
why is it free?
what is their motivation?
 
that doesn't answer the question :)
this has cost a lot of time and money
why are they giving it away?
 
r9m
BBL .. I need some sleep .. ^^ good night !!
 
4:45 PM
@user2179021 Not really
It's a git
 
@r9m bye!
 
As a git, it's collaborative work
It's not like a real job
 
sdf
can anyone tell me: is a complex projective variety always a complex manifold? Presumably we need the variety to be smooth also? It seems like Huybrechts book says that we dont though which confuses me...
 
Yes, it needs to be smooth, @sdf.
 
I have to go.. bye for now
 
4:57 PM
@MikeM Can every hyperbolic group be realized as \pi_1 of some hyperbolic manifold?
 
I don't know. What's a hyperbolic manifold?
 
OK, now I am in trouble. Er... erm... scratches head... I have no idea.
 
Me neither.
 
sdf
Thanks, I was misreading the book. I guess the converse is true: any complex manifold is a smooth projective manifold?
*smooth projective variety
 
Nope. One needs much more than that: Kodaira's embedding theorem
 
5:23 PM
I need help. Say we got 2 fixed points $\pi_1,\pi_2$ and denote their distances from a point $\sigma$ $r_1,r_2$ respectively. If they satisfy $|r_1-|r_2|=\lambda k, k\in\Bbb Z(1)$ for some known $\lambda$ and create a coordinate system such that the perpendicular bisector is the y-axis and $\pi_1,\pi_2$ is on the x-axis, do the points on the locuses formed by (1)(hyperbolas) form a group?
First of all how can I denote the set with set builder notation of the family of hyperbolas?
Secondly, if they form a group, what are some isomorphic groups to it?
Maybe I'll take it to the main site.
 
@MatsGranvik Hi, what are you working on these days?
 
What do you propose the group operation is, @UserX?
 
I suppose they will form a group under addition
Every point will have an inverse on the symmetrical hyperbola
 
@JasperLoy I am trying to reverse a function.
 
@MatsGranvik Did you recently change your username?
 
5:31 PM
Huh? Under addition, @UserX? Draw a picture.
 
@JasperLoy No, why do you think that?
 
@MatsGranvik I always thought it was Gravnik and not Granvik.
 
@MatsGranvik So, I have been wrong all this while?
 
5:34 PM
Can you draw two points on it and show me what you suggest their sum is, @UserX?
 
@JasperLoy Search for "Sacha Baron Cohen Letterman" and click on the youtube video.
 
$(0,-\lambda)$ and $(0,\lambda)$
 
The sum of those is...?
 
(0,0)?
 
But that's no longer on the hyperbola...
 
5:37 PM
It is. Not on that hyperbola. It is in one from the families(the one with k=0)
That's why I need the family. So we get an identity element.
 
But every point in $\Bbb R^2$ is on a (possibly degenerate) hyperbola. So all you're asking is if $\Bbb R^2$ is a group under addition.
 
Wait what
 
Oh, I see. You're restricting to those with integer whatever you call it.
Sorry, I misread.
 
Yea
Otherwise it would be the whole plane
 
I am reading amazon reviews of several dictionaries and many have missing pages, lol. WTF publishers!
 
5:39 PM
Is it worth taking it to the main MSE?
Well, I've prepared it already. Feel free to answer there :)
The worst that could happen is downvotes.
 
Anyway, no, it's not a group. Try adding, like, any two points together.
Your inverse two work. Most don't.
 
But every point has an inverse
Here's a random one
 
But the sum of two points doesn't have to be in it. That's step 0 to having a group.
You're thinking about inverses when you don't even have closure.
 
I thought I had
 
Try to show it...
 
5:45 PM
Holy shit I don't. If I want closure I have to drop the integer part, which makes it R^2
 
yeah
 
Damn. Nice experience anyway :D
 
 
So, in general, instead of asking whether it's a group, just try to prove the group axioms :P If they hold, you've got a group...
@anon tragically it wasn't actually published as it would cost $150
 
haha
 
5:47 PM
or really not tragically because fuck the people who run those things
 
@anon That was a sudden entry.
 
@MikeMiller I read it WAS published for 150$
 
> The scheme has earned Vamplew some online recognition, but sadly his main aim remains unfulfilled. “They still haven’t taken me off their mailing list,” he said.
 
@anon Where did that come from?
 
the guardian
 
5:49 PM
I will be studying calculus in 35 days from now.
 
@JasperLoy hope you're in $\Bbb Z_{35}$ because I find no reason to not start today.
 
Bart and Alizter don't come often to chat these days, sad.
@UserX Well, there is one reason for me, but it's too weird for you.
 
@JasperLoy what's the reason?
 
@UserX It's one of my obsessions. Full stop.
@anon Are your exams coming soon?
 
A rather convenient obsession :P
 
5:54 PM
To understand how this obsession came into existence would take too many pages of explanation. The margins of this chat are too narrow for that.
 
-______-
 
Quoting Fermat lol
 
6:22 PM
hi
 
Hello
 
are you an asymptotics lover?
 
6:59 PM
Hallo everybody
 
0
A: Calculus Question: $\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx$

Chris's sisHere is one line proof using beta function $$\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx=\lim_{a\to1}_{b\to-1^{+}}\frac{1}{2}\frac{\partial B}{\partial a}\left(\frac{1}{2}(a+1)\frac{1}{2}(b+1)\right)=-\frac{\pi^2}{24}$$ Q.E.D.

 
hhh
7:16 PM
What is the name of the law with polytopes that defines Planes, Dots and vertices relationship?
Or in graph theory.

I am trying to deduce Gibs' rule $F=N-P+2$ (thermodynamics where N is the number of components, $P$ is the number of phases and $F$ is the number of degrees of freedoms)
(it was some very basic rule, now cannot remember its name: Euler rule, too many of them...)
Found it: Euler's formule with Planars graphs, here. Yes it is a direct corollary from Graph theory :)
 
7:35 PM
If $AB$ is invertible, then either $A$ and $B$ are both invertible or both are not. Am I right @DanielFischer?
 
@JasperLoy I come in here often but I leave quickly if there isn't a conversation I'm qualified to participate in
 
@Sush square matrices?
 
@Studentmath, frankly IDK!
 
@Sush Yes. If $AB$ and $B$, say, are invertible, then $$A(B[AB]^{-1}) = 1$$ and $$(B[AB]^{-1})A = (B[AB]^{-1})ABB^{-1} = BB^{-1} = 1.$$
 
@DanielFischer, thank you very much, sir!
 
7:40 PM
So then $A$ is invertible too.
 
@Daniel isn't it true that if $AB$ is invertible than $A,B$ are invertible?
Because we can get from $AB$ being invertible that $B$ is invertible, and then the above argument, $A$ is invertible too - so the case of both being singular isn't possible.
They are both certainly invertible
 
@Studentmath That depends on the setting. If $A,B$ are square matrices, for example, then yes. If $A,B$ are non-square matrices, then neither of $A$ or $B$ can be invertible, but $A,B$ could be. Another example are the shifts on the sequence spaces $\ell^p(\mathbb{N})$, with $A$ the left shift, and $B$ the right shift. Then $AB = \operatorname{id}$, but neither $A$ nor $B$ is invertible.
 
Ahah, thanks!
 
If $AB$ and $BA$ are both invertible, then $A$ and $B$ are invertible.
 
For matrices its rather straighforward as you can conclude from both AB and BA being defined that they are square matrices, isn't it so?
But I don't see how it works on the sequence spaces $l^p(\Bbb N)$.. will try to think about it, good exercise :) thanks again for clarifying
 
7:50 PM
@Studentmath No, the product of an $K^{m\times n}$-matrix and a $K^{n\times m}$-matrix is defined in both orders.
@Studentmath Think of left-inverses and right-inverses.
 

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