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12:00 PM
Just ignore
 
Or just call a supermoderator.
Oh boy. 3.
 
what like batman?
 
confused
 
@Vibhav So, any number theory lately?
 
so since $f(0)=0$ and $f(1)=1$ and it is injective, the function is strictly increasing?
 
12:02 PM
@VibhavPant Have you seen the solution I've posted to your problem?
 
@Hakim Yep
thanks!
@BalarkaSen there are 7 months left for CMI's entrance exam, and I have other topics to complete too :P
 
@VibhavPant You're welcome! ;)
 
oh right. that exam.
 
@BalarkaSen I promise to do as much as Number Theory as you want when I get into CMI :P
 
heh. no hurry.
 
12:04 PM
But then I want to purse Complex Analysis ;)
 
CA is friends with number theory.
 
yeah
that too
 
Yeah after complex anal its numb
 
IIRC the prime number theorem does involve some complex analysis
 
@VibhavPant Yes.
My favorite proof is through CA.
But there are other "elementary" proofs through (real) analysis too. See [Erdos-Selberg proof].
 
12:06 PM
Also, does how important really is geometry in the Argand Plane?
 
Why do some people not want there integrals solved using CA? They act like it's witchcraft or something.
 
@sarah Yeah.
 
@VibhavPant Well seeing as complex numbers are vectors...
 
CA is a tool and makes life easier. Why not use it?
That Feynman thing is just stupid.
 
@BalarkaSen I think people are scared of the unintuitive nature
 
12:07 PM
@sarah Maybe.
 
all closed completely analytic contour integrals are 0
that startled me at first
 
aren't complex numbers a form of witchcraft?
9
 
@cirpis No only sedions
 
@sarah that is explained by a little physics.
smacks himself
 
Im not really fond of proofs that make use of physics
ducks
 
12:08 PM
Nor me.
But that particular fact is well-explained by physics.
 
I am unaware of physics. I used to think string theory was the theory of symbolic strings. And that formal language was a part of it..
 
I like physics. Not experimental physics though.
mechanics
I never really enjoyed the "scientific method" of trial and error.
especially because we were required to use trial and error in solving third degree polynomials
I would have just done newtons and guessing.
 
@sarah No no. Think of the Cauchy theorem like this : You are applying force and the corresponding displacement is with you. But you are doing this at infinitesimal level and in a closed curve if you add up the displacements of a point traversing through the path then it's 0 (it comes back to where it started looping).
This is just a naive explanation.
 
@Alizter My textbook has questions like "Find the circumcentre of the triangle whose vertices are given by the complex numbers $z_1, z_2, z_3$". Arent these problems better suited to coordinate geometry?
 
@Alizter Nor me. Yuck.
 
12:11 PM
Ohh yes @BalarkaSen I know how to think of it now ;)
 
@sarah The thing is that contour integration is related to vector calculus and everything there can be explained through physical means.
 
@VibhavPant You will find that complex numbers (not necesserily analysis) tie in with coordianate geo
It is just vectors with multiplication such that it adds the angles around the unit circle
 
@VibhavPant Complex numbers are more or less the coordinate space but with a kick of the ideal $(x^2 + 1)$
 
What @BalarkaSen said
 
@Alizter The generalization is called Hilbert's Nullstelansatz, btw. That's what I am reading right now.
The path to algebraic geometry.
 
12:14 PM
@BalarkaSen yeah, but dont you think they would be better in texts on coordinate geometry?
 
@VibhavPant Well, $i$ is not quite a transcendental over $\Bbb R$ =)
 
@BalarkaSen this book doesnt introduce the complex-number-as-an-ordered-pair idea
 
@BalarkaSen hmm. I need to focus on studying well for these two years ahead. Especially cambridge entrance exams. The math questions are so difficult that nobody has ever gotten full marks
 
@Alizter Olympiad questions again, I guess?
 
@VibhavPant $x^2+1$
@BalarkaSen err no
 
12:16 PM
@Alizter hm?
 
Calculus, Coord Geo, Complex numbers...
I just have to be really good at them
I am taking 3 a levels in math
 
I hate coord geo.
 
oh and mechanics + stats
 
smacks @Alizter
 
@BalarkaSen why?
 
12:17 PM
@cirpis I am more of an algebra guy. It depends on taste.
 
I never liked statistics, mainly because of the $\text{horrible}$ way it was taught to me in school
 
Data charts and stuffs. Yuck.
 
@BalarkaSen I do it for the grades
 
@Alizter You want to go to cambridge?
 
@Alizter Fortunately I don't have to do it. Unfortunately I can't do A levels at maths either.
 
12:19 PM
@BalarkaSen Yuck? You can do a lot with statistics
 
@Hakim ducks
 
r9m
@Alizter so is the functional eqn resolved ? :)
 
@r9m Oh yeah that tinh
@sarah Yes
 
@r9m I just reduced it to $f(f(f(z)))) = f(z)$
What are the solutions to this?
 
r9m
@Alizter what solutions did ya get ?
 
12:20 PM
sub $f(z)=t$
 
r9m
oh! right :)
 
the function is its ownincerse
god typing and eating is hard
 
r9m
@Alizter :P I am doing the same thing .. :P over that I have a malfunctioning touch pad :P
 
@r9m Use a mouse like a real man ;)
Actually I use a track ball but what ever
My wrists will start hurting and I cant play piano without wrist bands then everything sucks
 
r9m
@Alizter my mouse lost its tail a long time ago :P ..
 
12:23 PM
@BalarkaSen $f(f(f(z)))) = f(z)$ isn't enough to determine $f$
 
@Hakim A function that is it's own inverse
$f(x)=-x$
 
r9m
@Balarka we need $f(f(x)) = x$ ;)
 
Therefore its bijective
 
@Alizter Hmm I thought that there were more such functions than $f(x)=-x$?
 
there are
 
12:25 PM
@Hakim I know.
 
$f(x) = x$ would work too in this case
 
$f(x)=1/x$
 
yup
 
$f(x) = -1/x$
 
Though I dont know the complete question
 
12:25 PM
$f(x^2+yf(z))=xf(x)+zf(y)$
 
$f(x)=\frac1{2x}$
$f(x)=\frac1{rx}\,r\in\Bbb R$
 
note that $f(0)=0$
 
Such functions are known under the name involution
 
@Hakim Yes.
Anything of order 2 are called involutions.
$1 \in \Bbb Z/2\Bbb Z$ is an involution too.
 
But then there are infinitely many solutions as shown here
 
12:29 PM
@cirpis Why do you think complex numbers are witchcraft?
 
@Hakim I just showed there are
 
@BalarkaSen twas a joke
anywho
 
@BalarkaSen I think it is a reference to pre Gauss mathematicians attitude.
 
@Alizter $1/rx$ aren't the only ones
 
@Hakim But show there are infinitely many
 
12:30 PM
$x=1,z=1$ gives uz $f(1+y)=1+f(y)$
 
@Hakim Are there countably infinitely many?
 
r9m
@Alizter restricting $f$ as $\mathbb{R}^{+} \to \mathbb{R}^{+}$, with $f(xf(y)) = yf(x)$ and a $f \to 0$ for large $x$ should do ;)
 
thus a recurrence is given
 
@cirpis and if $f(1)=1$ then we have distribution over addition
 
@cirpis we don't know if f(1) = 1
 
12:31 PM
@Alizter Here they are: $x\mapsto c-x$, $x\mapsto 1/cx$, $x\mapsto \tfrac{c_1-x}{c_2x+1}$ I guess there are more
 
we know
 
@cirpis eh?
 
$f(f(z))=zf(1)$
 
oh right.
 
if $x=0,y=1$
and you proved that it is its own inverse
 
12:32 PM
What are you guys trying to solve?
 
thus $f(1)$ can only be $1$
 
@sarah our miseries.
 
hehe =)
 
r9m
CFing the chat with FE :P
 
$f(x2+yf(z))=xf(x)+zf(y)$ was the question
 
12:33 PM
7 mins ago, by cirpis
$f(x^2+yf(z))=xf(x)+zf(y)$
JINX, @cirpis
 
@Alizter Here's another type: $x\mapsto \sqrt[n]{c-x^n}$
 
find all real valued real defined functions that satisfy that
the only one if $f(x)=x$
as we just found out
also i have nothing against complex numbers
 
wait a minute. $f(f(z)) = zf(1)$. How does it follow that $f(1) = 1$? I am not sure.
 
Surely only $f(x)=x$ and $f(x)=0$?
 
you proved that
$f(f(f(x)))=f(x)$
 
12:35 PM
@cirpis me? I proved that $f(f(f(z)))) = f(z)$.
Not that it's an involution.
 
substitute $f(x)$ with $z$
 
@cirpis That doesn't cover all of $\Bbb R$
 
what?
 
we don't know if it;s surjective
 
oh
 
12:36 PM
=|
 
gimme a second
 
r9m
@Balarka it is surjective :)
 
well
 
@r9m wat
 
r9m
@Balarka $f(f(x)) = xf(1)$ right ? :)
 
12:37 PM
If $f(1)=0$ then $f(x)=0$.
 
Much collaboration. Such math. Very progress. Wow.
@r9m yes. missed that.
 
$c \oplus x$ also satisfies $f(f(f(x)))) = f(x)$ (where $\oplus$ is the xor operator)
 
r9m
and if $f(1) \neq 0$ .. then you have a surjective baby
 
already proved that, @VibhavPant
 
12:38 PM
@r9m Then it must be $f(x)=x$ for $f(1)!=0$
 
so my proof isnt false?
 
yes.
$f(x^2) = xf(x)$
 
r9m
@sarah right :)
 
keep that in mind.
 
Two solutions. i have proof but I have to go. $f(x)=x, 0$
 
12:39 PM
what now?
 
Here's yet another function, this time from Mathematica: $x\mapsto\tfrac12\left(\sqrt{c_1^2+c_2-4x^2}+c_1x\right)$
 
I might aswell go too then
bye
 
@Hakim that doesn't satisfy $f(x^2) = xf(x)$
i should go too.
 
@BalarkaSen So $x\mapsto cx$ is the only solution then
 
yes.
indeed it is.
 
12:41 PM
jea
 
well no. $x \mapsto x$ is.
 
r9m
ya .. they could be anything that preserves reflection on $y=x$ .. :P so uncountable
 
$cx$ is involution iff $c = 1, 0$
@r9m i think they are uncountable.
11 mins ago, by Balarka Sen
@Hakim Are there countably infinitely many?
 
@BalarkaSen idk
 
me neither.
 
12:42 PM
@BalarkaSen What about $x \bmod c$ and $c \bmod x$?
 
heh.
not functions from reals to reals.
 
@BalarkaSen What if we introduce floors
 
@r9m @Alizter @Hakim @cirpis @sarah so it's settled everyone. the only solutions are $f(x) = x, 0$
wait, what?
 
yes
$\Delta$
 
how to make a plastic chasis? i know its not related but i dont where to ask
 
12:44 PM
@Hakim Erm. I dunno.
i gotta go now.
bye.
 
@BalarkaSen Bye
 
r9m
bbl .. :) I have a load of new manga volumes to catch on to :D
 
everyones leaving?
 
Im still here
although Im doing problems on continuity
 
1:20 PM
@ArthurFischer If you have time, can you have a look here in tagging chatroom? (I am not sure whether my ping from here reached you; since your comment is already deleted.)
 
1:49 PM
I have updated my question with some code:
http://math.stackexchange.com/questions/873899/successive-ratios-of-a-sequence-is-this-limit-true
It relates the error in approximation of this sequence:
http://math.stackexchange.com/questions/870989/number-of-compositions-does-this-sequence-have-a-closed-form
to the roots of the equations -x^3 + (y+1)*x^2 + 1 == 0 solving for x.
 
2:35 PM
@MatsGranvik Can you check some values of the formula I give here? I think it checks out, but it'd be fun to also see some values computed.
 
@PedroTamaroff I will have a look.
 
3:19 PM
@PedroTamaroff
The Mathematica line:
Table[Sum[(Floor[(n - 3*j)/4] + 1), {j, 0, Floor[n/3]}], {n, 1, 32}]
gives:
https://oeis.org/A025767
1, 1, 2, 3, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 15, 17, 18, 20, 22, 24, 26, 28, 30, 33, 35, 37, 40, 43, 45, 48, 51, 54
 
@MatsGranvik you missed the iverson bracket!
OH WAIT.
You have me an idea!
 
Table[Sum[(Floor[(n - 3*j)/4] + 1)*If[Mod[n - j, 2] == 0, 1, 0], {j,
0, Floor[n/3]}], {n, 1, 32}]

gives:
https://oeis.org/A005044
0, 1, 1, 2, 1, 3, 2, 4, 3, 5, 4, 7, 5, 8, 7, 10, 8, 12, 10, 14, 12, 16, 14, 19, 16, 21, 19, 24, 21, 27, 24, 30
 
3:36 PM
cool
 
4:11 PM
so uhhhh.. anyone want a tricky maths problem of sorts?
 
4:23 PM
theres a planet, on which aliens with three arms and some antennas live. one day all of them decided to hold hands so that:
$1)$every arm was being held
$2)$every alien was holding hands with exactly three other aliens
$3)$if two aliens held hands, then one of them had six times less antennas than the other alien.
Can the total amount of antennas be $9001$?
 
 
2 hours later…
6:12 PM
@r9m: I don't see anything in the chat log that would scare Chris's sis away. I see that she has been on main within the hour.
but it has been 2 days since she was on chat
I'm away for a bit. BBL
 
later
 
r9m
6:36 PM
@robjohn :O .. okay now I'm a bit surprised too :{
 
:-)
 
r9m
@skullpatrol :D
 
but not always a "friendly" ghost :-)
 
r9m
Casper = Collaboration For Astronomy Signal Processing :)
ghost .. I don't think so :) .. very real :)
@robjohn there is a way I could draw Chris'ssis to chat :P
 
@blue
 
r9m
6:53 PM
does @Balarka have a blog ? :)
 
No.
I don't.
 
r9m
'kay
 
I think I need to learn quite a bit of stuff before blogging all over.
Anyway, so , where was I? Ah, yes. @blue.
Consider Spec Z. Consider Spec Z[i]. Latter is "sitting over" the other in the sense that multiplying out conjugate primes gives primes in Z. For example, ideals (2 + i) and (2 - i) are sitting above 5 = (2 + i)(2 - i). Interestingly, for 2, say, one has (1 + i) and (1 - i) but both are the same ideals (multiply out by an appropriate unit).
So this "natural" map Spec Z[i] --> Spec Z looks like a ramified 2-sheeted covering of topological spaces, doesn't it? What am I missing?
Oh hang on a second. Some primes don't cover at all! Yeah, how can I forget them : the ideal (3) just sits there. It's a prime ideal in both Z and Z[i]. Then the fibres over (3) is just itself and nothing else and... that means there is a single cover around that ideal after all... ? Am I making sense?
I guess I have to think about this.
 
7:41 PM
@BalarkaSen that's right
 
8:23 PM
@blue Ahem. I am not sure if it can be realized as coverings of Riemann surfaces after all.
'cause in such a covering everything away from the ramification points has multiple covers as their fibre spaces.
@685-252 You're the troll, no?
I mean, skullpatrol.
 
he is the jaden smith of the chatroom
 
jaden smith?
 
A smith made of jade.
Not that it makes sense.
 
I'll quote pedro : "PUPPNUNPUNPUNUPPUNU"
 
@BalarkaSen When did he say that?
 
8:36 PM
Loads of times.
 
Link or it didn't happen.
 
Jul 10 at 2:36, by Pedro Tamaroff
PUPNUPNUPUPNUPNUPNUPNUPNUPNUPUNUNPNUPNU
Just an example.
 
But that's not what you said. You misquoted him.
 
I somehow knew you'd say that.
 
why did you misquote him?
 
8:43 PM
Stop trolling, @685-252
 
stop misquoting, @BalarkaSen
 
@BalarkaSen Probably ELIZA, best to ignore.
 
@DanielFischer Haha
Right.
Heya @MickLH!
 
hey balarka :)
 
how's the project going? or have you given up on it?
 
8:48 PM
going well, it's going slow though as I've been busy trying to keep a roof over head
 
right. yeah.
if i am not wrong, you have implemented BBP-type algos to evaluate transcendental constants, right?
 
9:03 PM
Maybe a couple
Sorry I'm a bit busy even now
 
 
1 hour later…
10:13 PM
Who upvotes that??? It's not even complete yet.
 
@blue Do you recall a way to describe a nonstandard field embedding $\mathbb R \hookrightarrow \mathbb C$?
 
besides "pick transcendence bases," no
 
alas
 
10:39 PM
@DanielFischer There are lots of people who upvote poor questions just for the sake of trolling
 
If I could believe that it was trolling, that would be at least a somewhat rational reason. Alas, I suspect it's mere stupidity.
 
Ahhh I got to touch the Westminster cathedral organ today :)
@DanielFischer @Hakim They get badges for upvotes no?
 
@Alizter Yes, like the Electorate badge.
 
@Alizter You get the badge also for downvoting. And downvoting questions doesn't even cost you the measly $1$ rep.
 
Hello all.
 
10:50 PM
Hi @5space
 
yeah
This organ I have played on. It is very nice.
 
11:21 PM
@Alizter Is that you?
 
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