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12:28 AM
What are you up to @usukidoll? Still need some help?
Though looknig over it it seems you reached the step before the final one.. Just need to use the definition of $+$ (symmetric difference I assume) again over that
 
qwr
I don't follow the forefront of mathematics, do you think any big math problems will be solved in the next 50 years?
 
Hi there
I need some help with Time Complexity stuff
what kind of complexity is O(N Log(4)) ?
 
qwr
4?
just 4?
 
well, every time I double N it takes four times more time
 
qwr
that's n^2?
 
12:41 AM
yes,
is like n^2
so it is quadratic time?
and if it is O( N Log(3) ) or N^1.585?
I'm not very good on all this but I'd like to know what is the complexity time for an algorithm I developed
 
hmm, I am a bit uncertain @JesusSalas
 
ok, I'll try to explain myself
 
But I -think- they usually take the n itself as it is, and disregard the rest, for guesstimating
Or well
Generally.
For example, if T(n)=log2*n^2+321321n
 
@qwe sure
 
They would say O(N)=n^2
We are sepaking about the same thing here, right?
 
12:46 AM
I have an algorithm that when I fee a set of 100 numbers to process takes 1 second when I double the numbers (200) takes 4 seconds, if I double again (400) it takes 16 seconds, for 800 takes 32 seconds
 
qwr
@Mike sure to my question?
 
and I'm wondering how to represent this time complexity using O(..)
@studen
 
qwr
@Mike like the riemann hypothesis or navier stokes equations. I would love to see it solved in my lifetime
 
@Studentmath I really don't know :) this is why I'm asking
 
I see, hmm
I would use recursive function to describe it, I think.
O(100)=1
Assuming of course the set of numbers is always at the doubles of 100
And then T(n)=4*T(n-1) for the rest
That would be
T(100)=O(1)
where n={100, 200, 300, 400....}
I am hoping I am not only confusing you, if it is not the case (doubles of 100) then it's probably not the right way to do it
 
12:51 AM
we can use 1 as a start point :)
1,2,4,8,16,32
everyt time I double N it takes 4 times the previous time,
 
Then yes, I will use recursive function for that.
 
T(2N) = 4T(N)
 
Hahhh... I see.
Disregard all I said before.
 
this sound more right to you?
 
It sounds right but it doesn't define it well, I think.
I would say:
 
12:54 AM
probably not :D
 
T(n)=c, n=1
T(n)=2T(n/2), n>1
where c is whatever T(1) is, of course
Looks very alike to the general merge-sort reccurence equation
 
Hey guys, quick error check if someone doesn't mind. I've got an array $L_x$ which represents the values of a function $W(x)$ at integer values
 
Does this make sense to you?
 
I've developed a routine to approximate $\int_a^b W(x) dx$
this is it:

$\sum_{x=\lfloor a\rfloor}^{\lfloor b\rfloor-1}(\frac{L_x + L_{x+1}}{2})
+\frac{(\lceil a\rceil-a)(L_{\lceil a\rceil} + L_a)}{2}
+\frac{(b-\lfloor b\rfloor)(L_{\lfloor b\rfloor} + L_b)}{2}$
Am I being stupid?
 
@Studentmath well, math never fails...
 
12:59 AM
Not if it is done right :P
 
My idea is that I am approximating the integral by summing the trapezoid formed by each two adjacent samples
 
so it seems complexity if O(N^LOG(4)) that result in O(N^2), what means quadratic time
and O(N^Log(3)) results in O(N^1.585) that means linearithmic time
 
As well as creating "edge" trapezoids to account for non-integer values of $a$ and $b$
 
Makes sense, assuming you use log with base 2
 
yes I do
 
1:02 AM
Alright, go on
 
Thank tou so much!
*Thank you
 
Ah, erm.. You're welcome I guess! Not that I did much but just follow your line of thought :P
 
1:33 AM
Hi guys, a quick question on terminology. What is that fundamental approximation called where you sandwich a value between a lower point and an upper point till you get a descent approximation.
 
@Nick Squeeze theorem or related?
 
$9 < (3.5)^2 < 16$
and you work your way to finding the middle value
 
@Nick I'm still guessing here but I'd call that "bounding the result". You know it's between two values so you keep going until you find something in the bounds.
 
Yeah, good enough
@BrandonEnright: Newton's approximation for a square root is kinda like that, right?
... oh nope, he used calculus more than I thought.
 
@Nick I dunno. I'd call that iterative convergence.
 
1:40 AM
@BrandonEnright: Good name. :D
@Brandon: How would you go about finding the value of $10^{0.3010}$
 
@Nick Put it into a calculator
 
@Nick I'd plug it into a calculator and let somebody else worry about how the algorithm to produce arbitrary precision.
Too 100 decimal places: 1.999861869632744111148164903112008935482332675039764696793459968140147522407917‌​045132454483145909836
 
@Mike: Yeah, but I want to break free
 
why?
 
1:50 AM
when would you possibly need to know the value of 10^.301
 
@Nick "This video is not available in your country."
 
@BrandonEnright: eh, what is that, China?
 
@Nick nope, US. YouTube doesn't like my US IP for some reason.
 
eh, here it is again
@Mike: ... When I want to approximate large numbers quickly and I only have the log table part of my log table.
 
lol
 
1:54 AM
@Mike: Not a joke. I was in that situation once.
 
If $A$ is an infinite subset of the natural numbers can I say that $A$ is countably infinite or do I need to display a bijective map?
 
ugh I missed it !!!!!!!!
$(A \cap C) \cup (B \cap C) \backslash (A \cap C) \cap (B \cap C)$
$(A \cap C) + (B \cap C)$ @Studentmath
 
@Alex do you have any theorems about cardinality?
 
oohhh I think I see it now ^^
 
The definition I have is that a set is countable if it is finite or it is the same size as the natural numbers
If $A$ is a subset of the natural numbers, its cardinality cant be larger than the natural numbers is how I was thinking about it
 
2:14 AM
@Mike: I've just asked my question on main
 
@Nick HAI
DAFAQ
Where is Ted
 
@Pedro: A happy Dafaq to you too.
Where is Ted?
 
Where is Ted?
 
@Pedro: Wait, is this the Ted that you're looking for?
 
Where is Ted?
I was talking about Ted Shifrin,.
@Nick I do watch TED.
 
2:19 AM
@Pedro: Well, you're not watching him right now, otherwise you'd know where he is.
He's probably at work
 
lol @ nick
 
2:54 AM
Hey all.
 
3:17 AM
Hey.
 
How's it going?
Could you humor me on this question? math.stackexchange.com/questions/698414/…
 
@Mike
 
3:37 AM
Nooo Mike will destroy me
 
Ehehehehe
 
@seaturtles Just wrote down all subgroups of $S_4$. 'd love to draw a lattice but I fear it is too hard.
 
@Anthony what do you mean by "does not recover the original equation"?
 
Er, if you expand a square wave, don't you get bumps by the edges?
 
are you talking about gibbs phenomenon?
 
3:45 AM
Yeah
 
why didn't you just say that to begin with?
 
I don't really know... I mean it's just a name right?
I assume it happens with other functions too.
@PedroTamaroff Real analysis pregunta
 
@Anthony hehe
 
and yes, the approximations will get bumps at the edges, and these bumps will be squished into nothingness in the limit. so what is your question?
 
What is it?
 
3:47 AM
@seaturtles I thought that they stayed forever
 
@seaturtles le drámá
 
right
 
@PedroTamaroff is S is in T is in X, how can I show that the closure of S is in the Closure of T?
 
fight*
 
@Anthony they are always there in the approximations, but they are getting squished into arbitrarily tight neighborhoods of the discontinuity, so that in the limit every point that is not the discontinuity itself gets to where it needs to be
 
3:48 AM
@Anthony They do for poor physicists, not for mathematicians.
 
I didn't knooooow.
Whoops.
 
@Anthony Well.
 
Maybe I should delete that.
 
What is your definition of closure?
@Anthony No, you shouldn't.
 
Aight
closure is set of limit points
What is this song?
 
3:50 AM
@Anthony It is awesome.
 
I see.
 
look what you people have done to my starboard
 
@Anthony Well, can you show that if something is a limit point of $S$ and $S\subseteq T$, then that thing is also a limit point of $T$?
@seaturtles Pats in the back.
@seaturtles CCCCCCCCCCCCOMBO BREAKER.
 
Oh I guess that must be true...
 
@seaturtles Do you know the Heisenberg groups?
@Mike $\uparrow$
 
3:58 AM
@PedroTamaroff upper triangular matrix groups?
 
@seaturtles upper triangular with ones in the diagonal. So actually a subgroup of ${\rm SL}(n,k)$.
 
yes
 
Rotman calls it the "unitriangular matrices"
 
that's the word I was looking for
 
3:59 AM
In simplifying $1-2csc^2(x)+csc^4(x)$, is the first step to just convert it into sines?
 
trying to remember something
errrr I wouldn't convert that would be $\frac{1}{sinx}$
 
Keith Conrad has something on the Heisenberg group, because for $p$ odd prime it is one of the two nonabelian groups of order $p^3$.
 
I know ! Let $ csc x = r$ and then factor it out
 
@Tanner sometimes in math there isn't "the" first step; there are many possible things. have you tried anything?
 
so it would be like $1-2r^2+r^4$
 
4:00 AM
@usukidoll indeed :) (although you would say factor it, not factor it out)
 
ooo a turtle eats it
 
this is why I have the ability to retract into my shell
 
$(-2)^2-4(1)(1) = 4-4 =0$
hmm we can factor it
 
@PedroTamaroff what about showing the interior is contained in the interior?
 
4:02 AM
@Anthony Pick an interior point $x$ of $S$.
 
$r^4-2r^2+1$
$(r^2-1)(r^2-1)$
 
Then you can find a nbhd $V_x\subseteq S$.
 
I've tried simplifying it, but I got stuck at $\dfrac{sin^2(x)(2sin(x)+1)}{\sin^6(x)}.
 
oy no!
 
Since $V_x\subseteq T$ too, for $S\subseteq T$, $x$ is also an interior point of $T$.
@Anthony.
 
4:03 AM
I think I see identities popping up ...
$(csc^2x-1)(csc^2x-1)$
hmm wait one of the identities is $1+cot^2x=csc^2x$
 
@PedroTamaroff That was a dumb question.
I'm sorry.
 
yeah $csc^2x-1=cot^2x$
 
yes
 
eats turtle
 
@Anthony There's a special ring of hell for people like you Anthony.
LOL JK. You're good.
 
4:05 AM
@usukidoll impossibru
 
so $(cot^2x)(cot^2x)$
 
@PedroTamaroff :(
 
@Tanner what was your original problem again?
 
lol
$1-2csc^2(x)+csc^4(x)$
 
I'm asking because I'm wondering if it had a $=0$ meaning to solve or just that which is ok we can factor
yay @Ted is back$
 
4:07 AM
Only briefly :)
 
so we can just factor then @ Tanner
@Ted I think I got the symmetric difference problem... should i latex again to see if its right?$
 
Factor from what? I'm still at the original problem :P
 
omg Tanner can you plz look up at what I wrote ^^^^^^^^^^^
sighhhh eats turtle
 
Well you confused me with the r's
 
4:08 AM
@Tanner can you factor 1-2r+r^2 ? perhaps it would look easier writing r^2-2r+1
 
Should be very brief after what we discussed, @usukidoll.
 
@Tanner that's subsitution to make your life easier silly ^^
$ (A+B) \cap C = (A \cap C) + (B \cap C)$
so the symmetric difference of $A+B$ is $A \cup B \backslash A \cap B$
 
@sea Yes, although is r supposed to be an abbreviation for something?
 
and then we have
$[(A \cup B) \backslash (A \cap B)] \cap C$
 
r9m
Anyone interested in a bit of euclidean geometry ? :)
 
4:11 AM
By distributive law we have
$(A \cap C) \cup (B \cap C) \backslash (A \cap C) \cap (B \cap C)$
 
@Tanner yes. you have $1-2\csc^2(x)+\csc^4(x)$ and we wrote $1-2r+r^2$. what do your detective skills tell you we are using "r" to represent?
 
applying the symmetric difference would be...
$(A \cap C) + (B \cap C)$
@r9m aw hex no.. I heard it's tough
$(A \cap C) + (B \cap C) = (A \cap C) + (B \cap C)$
 
Once you prove the lemma we were discussing earlier, yes @usukidoll
 
so I shouldn't have taking the complement def of $A+B$ in the first place... I should've left it alone.. expand the symmetric difference and applied the distributive law for $ \cap C $ and then symmetric difference again
 
Why do you write the same thing 3 times? @usukidoll
 
4:14 AM
...ummm because if I have to write a proof I need to put words?
 
Right, so we would then have 1-2r(1+r) if we factored it
 
@Tanner no.. $(r^2-1)(r^2-1)$
which is really $(r^2-1)^2$
 
You still need the lemma. Either way you have $(X\backslash Y)\cap Z$.
 
r9m
anyone seen this problem before or encountered something close to this just let me know :)
http://math.stackexchange.com/questions/698478/a-problem-with-concyclic-points-on-mathbbr2
I have been banging my head with this one since 10 grade :P
 
@seaturtles
 
4:16 AM
hmm where is the lemma?!
 
@Pedro
 
@TedShifrin Yes?
 
Howdy
 
was that $(X \cap Z \backslash Y \cap Z)$?
 
@TedShifrin Hello!
 
4:16 AM
Yes @usukidoll
 
I was a bit distracted with group theory.
 
oh that lemma of $(X \backslash Y ) \cap Z = X \cap Z \backslash Y \cap Z$?
 
mmm dips turtle in soy sauce
 
@usukidoll You're nasty.
 
4:18 AM
XD
bunny rabbit..... oh new av...
 
@usukidoll @seaturtles Got it, cot^4(x). Thanks for pointing me in the right direction!
 
@PedroTamaroff ?
 
@Pedro thinks a new avatar a day keeps the doctor away ;)
 
eyup
oh hi again * grabs turtle with chopsticks*
 
@seaturtles I think I forgot my question. You took too long! =D
Oh!
I remember now.
 
4:20 AM
now I'm trying to remember 2 questions oh man!!! I think the last one was about proving or disproving that two injective composite functions are injective
 
Suppose someone gives you $\langle a,b\mid a^{p^2}=b^p=1\;[b,a]=a\rangle$, @seaturtles
 
wasn't there a collary or something? about injective functions?
all I know is that it's one to one
 
And without showing a "concretization" of that, you'd like to show that group "exists".
 
of course it exists
if you mean finite, that's quite intuitive
using the "commutators are fudge factors" notion
 
@seaturtles Well, in this case you have a small amount of relations to check.
 
4:23 AM
one can "slip a's past b's" as long as one tallies up extra a's. in the end everything is a power of a times a power of b
 
But if you had a larger group or whatnot, you'd have a bit more work yes?
 
almost surely
 
And sometimes the relations might lead to absurdities.
Or the trivial group.
 
?
yes, the trivial group
you won't get contradictions
 
I meant $\langle a,b\mid a^{p^2}=b^p=1\;[b,a]=a^p\rangle$.
Heh, I wonder what the other one is.
 
4:24 AM
pedro do you know functions?
 
yes he does
(soaked in hot sauce)
 
@usukidoll Well, some of them I do.
 
haha
 
well I'm trying to remember a question oh man... it had something to do with proving or disproving that two injective composite functions are injective
what the hell... we have like a one to one situation twice
 
so what issue are you having?
 
4:25 AM
$f,g$ injective implies $f\circ g$ injective?
 
I think so
too bad I don't have my damn paper back... I'm trying to remember the last two questions
nughhh the struggle is real
see everyone pretty much f bombed on the midterm so the prof is allowing corrections...only problem is that I don't have my paper because I lost my voice over the weekend. the only questions I could remember on the top of my head are the set theory questions...I only remember a part of the function questions
 
On $sec^4(x)+sec^2(x)-2$, I'm trying to use the Pythagorean identity, and I got to sec^2(x)(sec^2(x)+2)
 
so now I'm kicking myself for losing my voice over the weekend
 
@Tanner given our other problem involved factoring, have you tried factoring?
 
either I have to wait until Wednesday to grab my stuff back and do the last two remaining questions or some kind soul from class would email the last two questions.
I can't remember every detail for those function questions.. except fragments.. the second to the last question was $X \rightarrow Y$ ... power set X -> Power set y onto... which means that there is a surjection
so I can't fully ask that question because what if I mess it up?
 
4:30 AM
@seaturtles I tried factoring first, and that didn't yield anything, so I tried the Pythagorean identity.
 
$(A+B) \cap C = (A \cap C ) + (B \cap C)$
$(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$ only things I remembered
I messed up on the second one.. I've used set union and set intersection and got the symbols or something lost...
I only received like half the points for that one.. ... but wow I can't believe I just... : / so you can prove the first one after all...because complement defintion and symmetric difference are two different things...
 
Wait, I did that wrong... that wasn't an identity
 
what the heck bro?
original question plz
 
$sec^4(x)+sec^2(x)-2$
 
let $sec^2x = r$
$r^4 +r^2 -2$
by discriminant we have
$1^2-4(1)(-2) = 1+8=9$
so we can factor
$(r^2-1)(r^2+2)$
wait backwards
factoring again we have $(r-1)(r+1)(r^2+2)$
 
4:36 AM
hmm we didn't learn discriminant yet
 
then what are you doing with these problems?
 
Simplifying
 
oops we let $sec x = r$
oh no no no no nononononoononoo
$(secx -1)(secx+1)(sec^2x+2)$
discriminant $b^2-4ac$ in the form of $ax^2+bx+c$
 
 
2 hours later…
6:14 AM
 
6:25 AM
lol
 
6:40 AM
@Nick Am I an alligator?
@Nick Besides, Rahane and Pujrara (and Kohli) are much better than Sachin.
 
r9m
@Sawarnik interested in euclidean geometry ?
@usukidoll what was that ?? I did'nt see it
 
now can u see?
 
r9m
@usukidoll you've got a bunch of those .. don't you ?? :P
 
heeheh
 
7:28 AM
@r9m What kind of questions? When the word Euclidean comes, I m ... well, scared.
 
my classmate said that it's hard as hell
 
@usukidoll what was that ?? I did'nt see it .. tooo
 
rarity
 
@Mike!
 
oh no
 
7:42 AM
@Mike nah
 
my feet hurt
 
@r9m Sorry, but it is beyond me.
 
r9m
@Sawarnik have you seen Sylvester Gallai Problem in A.E. ? Its a variant of that .. I am looking for a proof with the extremal principle :)
 
@Sawarnik Hullo
 
7:57 AM
Why @Mike?
 
who knows dude
these are hard questions
 
:(
>:(
 
@Sawarnik?
Why are you not talking much to me these days?
(Don't get me wrong, I would be glad to keep it this way =D)
 

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