last day (81 days later) » 

2:59 PM
So, how's things?
 
hows it going
 
Doing quite okay.
 
its cool on my end. the selection exams ended a few days ago
 
Studied a little for JEE, am confident I will be able to do, er, well this time. Not too badly, in any case.
 
u still preparing for cmi
ah
 
3:00 PM
I am going to go back to the CMI prep now.
And I'd like advice, which is why I'm here.
What are you doing?
 
Ah very nice, we might be able to work on schtick togather
I am working on this book called uh
"An Excursion In Mathematics"
 
(The JEE thing is a back-up. You can get an IISER with any not-disgusting Advanced rank. :) That's it.)
 
Yep, indeed
 
I have Challenge and Thrill.
 
Oh nice
 
3:01 PM
I guess you're not taking any coaching?
 
Nope
 
That eases my fears a little.
@BalarkaSen you mean preparing, or gulp at CMI?
 
haha preparing
I have no idea if I'll get in CMI
 
it's not easy, but it's no JEE
 
I really want to but I'm terribad
I didn't think JEE math was hard
the time constraint's a thing but it's really not hard
 
3:03 PM
Oh, it isn't, not morally.
 
right
on the other hand I do think CMI entrance is quite hard. much more fun, but hard
 
But if you want to perform at a similar level to what CMI needs, we're talking ranks of like 200, which is not easy.
(I mean, seat counts and so on.)
 
Oh but how are you comparing the two exams?
 
Nonscientically
:P
 
i suspected
 
3:05 PM
you know me
Anyway, last month, I felt really good after I finished most of the CTPC inequalities chapter. Towards the end I felt like I was learning to think a little like an analysis person.
So what are you doing now? Studying really hard?
 
I have been doing a chunk of number theory and combinatorics
so we have our roles reversed
@SohamChowdhury nah
i can't study hard
 
indeed, indeed
 
im biologically retarded
 
I am going to do a bit of "number theory" now
 
What's in CTPC inequalities?
 
3:08 PM
oh, not much. C-S, weighted AM-GM, a few inequalities that you can derive from them, etc.
mostly it's just clever algebra
 
ah-hah
 
but CMI is less inequality-heavy than ISI, say
 
Right
 
it's really all the "logic problems" that get you
prove this algorithm terminates, yada yada
 
I actually learnt to inequality-manipulate in my statistics courses in school
I'll have a look at CTPC's ineq chapter rn
well flick
libgen doesnt have it
 
3:10 PM
room topic changed to War Room: (no tags)
 
room mode changed to Gallery: anyone may enter, but only approved users can talk
 
danks
 
why did you think libgen would
 
iunno
 
3:12 PM
anyway, I will be ordering a copy of Engels again
I lost mine a few years back, it is Good
 
I have le Engels
it's quite nice
if only i read anything out of it
 
especially the combi chapter taught me how to think, but i've forgotten lmao
 
i'm shocked!!!1!
 
ah, it's good to be back
 
Truly
How is college?
 
3:14 PM
Bad
The analysis prof is good, but the others are stupid
 
Analysis guy knows his stuff, but, e.g. algebra teacher calls sets with just an operation on them "groupoids"
 
arent those magmas
 
linear algebra teacher says $(A,B) \mapsto \frac12 (AB-BA)$ is nonassociative because "matrix multiplication is noncommutative"
what if I change the $-$ to an $+$? "stop arguing"
@BalarkaSen yip
 
lol
welp
I learnt a lot of linear algebra
feelsgoodman
 
3:16 PM
and I had a really bad midterm because linalg teacher said "wtf is a two-sided identity", etc. and everyone marked me badly
 
not sure if it's of any use to CMI schtick though
 
what kind of thing?
 
Down to earth linear algebra which I knew only in half-assed format for all these years
 
half the tests here are Gaussian elimination
 
Structure theorem of matrices, etc etc
 
3:17 PM
oh, yes, that is a good feeling
remember when I learned multicalc
 
@SohamChowdhury Wait, what?
 
basically, the teachers here mark papers by comparing them to "known answers" in their heads
you'd think a linear algebra teacher would know that one-sided identities exist
"don't act smart"
I can't not write like that anymore :(
anyway
let's get this ball rolling
I have to leave this shithole
 
Yes. Yes you have to
and i have to not fall into it
this place sucks
Man I'd be glad if I at least get ISI
 
Apparently you can, like, go to IMSc all the time
someone (on Quora, where else) mentioned a first-year getting credit for a diffgeo course at IMSc
you should be that guy wink wink
 
I don't really care about going anywhere, I just want to get into a decent enough university where I can continue studying math
ISI is decent enough that I would not have to unlearn the math I know already to survive college
can u screenshot me some exercises from CTPC ineqs
if possible
 
3:48 PM
sorry, was away
 
I can take pictures, yes
 
cool
 
will send you some on gchat later.
what are you doing now?
 
nice
@SohamChowdhury vaguely fiddling with some problems about polynomials in Excursion
 
3:50 PM
gimme one or two pls, I'm bored but also don't want to get up
 
I also want to do some actual math
@SohamChowdhury Ah OK let's see
 
I have done zero actual math for a very long time
just looking at le algebraic geometry book once a month like it's some ritual
:(
 
I'll just send you whatever. Here's one which says "Find all poly's satisfying $p(x+1) = p(x) + 2x + 1$"
I have an idea and it's not hard
yeah ok i got it i think
Ugh the book does it by normie techniques
 
now you will realise how badly I suck
maybe differentiating will work?
 
Don't worry about it. I used an idea that is a personal favorite of mine; we can work together so we both improve at the stuff we succ
@Soham Close
 
3:57 PM
"succ"

I see you are a man of culture etc
 
You can read that off from my profile description in chat
I have convinced myself that memes are true form of art
thou art meme
 
sometimes substituting roots of unity works well
 
this doesn't look like that though
 
yeah
i'll give you a hint. i think about forward differencing a lot
 
4:02 PM
I'm trying something like $\sum_n a_n ((x+1)^n-x^n) = 2x + 1$ rn
anyway, you don't have to get me to solve anything now. i'll remember, and maybe tell you I've got it after two days :)
 
That's cool by me
 
throw a few more at me if you like, that way I have something to do the rest of the evening
(btw, I really need to learn geometry, I don't know shit about that. somehow I think that not all Euclidean geometry can be $\mathbf C$-ed away, lol.)
 
Ok, let me find some more
I haven't done anything beyond this problem so far so I'll be random
@SohamChowdhury Oh me too man
I don't know heck about Euclidean geometry. I suck at it
I can do coordinate reasonably well though
 
@BalarkaSen at this college I'm having to do a lot of "classical" coordinate geometry
if I stay here for three years I'll turn into a fucking 19th century Italian geometer
@BalarkaSen shoot
 
Doing multivariable calculus taught me a lot of vector geometry, and I'm glad for it
fduck the internet
Hmmm this looks like a weird ass problem
For any positive integer $n$, prove that there exists a polynomial $P(x)$ of degree at least $8n$ such that $$\sum_{k = 1}^{(2n+1)^2} |P(k)| < |P(0)|$$
o lawd how do i
 
4:17 PM
this looks like a complex analysis problem, but then that's how I react when I see inequalities of this nature
is this Excursion?
 
hm
yep
The first thing to notice is that I just want to bound $| \sum_k P(k)|$ anyway
 
... doesn't that weaken the inequality? after all $$\left|\sum P(k)\right| \le \sum |P(k)|$$
 
I mean if $\sum |P(k)|$ is bounded by $|P(0)|$, so is $|\sum P(k)|$, by what you wrote
and it's easier to work with $|\sum P(k)|$
 
yes, but bounding the easier expression doesn't imply bounding the given one ... am I being dense?
 
I don't mean that if I bound the mod-sum by |P(0)| then I bound sum-mod too
no, no, you're right to be confused. I didn't mean that
 
4:22 PM
you meant testing hypotheses?
 
ah, got it
 
I write down trivialities a lot. Sometimes they lead to a good guess
In this case I want to construct a $P$ so that seems like a good approach
Do you think it's a good idea to do this for a specific $n$?
as a test case
This is fucking strange so I have no idea
 
$8$ is big
it says Z^+ but you can try 0
 
how about $n = 1$... lol
 
4:26 PM
actually, wait, that won't work
(the lhs is then empty, lol)
 
mhm
 
"degree at least 8n" wtf
 
so I want a $P$ of degree at least $8$ such that $\sum_{k = 1}^9 |P(k)| < |P(0)|$
@Soham the book makes you be like that sometimes
 
I think maybe most of the coefficients will be 0, idk
 
prolly, prolly
wait, $P(0)$ is the constant coefficient of $P$
 
4:28 PM
wait
 
do you not think it's kind of weird to be bounded by the constant coeff?
 
you can probably triangle-inequality things
$|\sum P(k)| \le \sum |a_i k^i| = \sum |a_i| k^i$
 
What if I let all the coefficients to be the same
 
collect all the terms with the same $a_i$, you get a sum of the $k^i$ which you can bound somehow
 
like, all of them are the constant coeff, $P(0)$
right
 
4:30 PM
that is my best guess for a solution unless some trick like this ^^ holds
 
So, $P(x) = P(0)(1 + x + \cdots + x^N)$?
i.e., $= P(0) (x^{N+1} - 1)/(x - 1)$
for some $N$
Well $\sum_k (k^{N+1} - 1)/(k-1)$ is going to be fucking huge m8
Maybe I have to divide $P(x)$ by something big ass
 
I have to eat
cya
 
bubye
 
(oh, btw, the $p(x+1)$ problem gives a kind of telescoping sum that you can use for integers)
 
Yep
If you do it for all integers you also do it for all real numbers
'cuz polynomial is determined by it's values on a finite number of integers
 
4:37 PM
yep
 
half the problems in one chapter in Niven-Zuckermann-Montgomery were doable by forward differencing and telescoping
good times
 
that inequalities chapter too
 
ahh
 
4:50 PM
Hmm, what if I use $p(x) = (x - 1)(x - 2) \cdots (x - 8)$
$|p(0)|$ is just $8!$
And $p(k)$ for $k = 1, \cdots, 8$ is just $0$
but eh
Oh no that almost works, $p(9) = 8!$
I am guessing something of this sort is going to do ze trickery
Well, yeah, that immediately gives you a solution
'cuz you are allowed to have degree greater than $8n$
 

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