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Lucas Henrique
Lucas Henrique
11:00 PM
I have some solutions already
 
Ted Shifrin
Ted Shifrin
Then why are you asking, Lucas?!!
 
Lucas Henrique
Lucas Henrique
They are not mine, lol
And I didn't read them. I was asking if my method was right
 
Ted Shifrin
Ted Shifrin
Well, my solution isn't yours, either :P
Oh.
You had a method?
That limit with product thing? Nah, I don't see any way to fix that.
Take logs and use what I just said.
 
Lucas Henrique
Lucas Henrique
Do you guys want to see one of the solutions? I just read it
 
Ted Shifrin
Ted Shifrin
I'm fine with mine, thanks :)
 
GFauxPas
GFauxPas
11:04 PM
Ted isn't that only for $x$ near $-1$?
 
NaCl
NaCl
Why is $$\sum^n_{j=1}\cos\left(\left(j-\frac{1}{2}\right)x\right)=\frac{\sin\left(nx\ri‌​ght)}{2\sin\left(\frac{1}{2}x\right)}$$?
 
Lucas Henrique
Lucas Henrique
Lacks rigorosity, tho
 
Ted Shifrin
Ted Shifrin
No, @GFauxPas, it's for $x$ near $0$.
Oh, you can make mine totally rigorous.
That's what Taylor's Theorem is for (error estimates in Taylor polynomials).
 
Lucas Henrique
Lucas Henrique
@TedShifrin how?
 
GFauxPas
GFauxPas
in that case, we have $\displaystyle \sum_{i=1}^N \ln \left({1 + \frac i {N^2}}\right) \sim \sum_{i=1}^N \frac i {N^2}$?
 
Ted Shifrin
Ted Shifrin
11:06 PM
Right. And then you need to look at an error estimate.
 
GFauxPas
GFauxPas
Im not used to having both bounds of summation in the summand
 
Ted Shifrin
Ted Shifrin
Pull the $N^2$ in the denominator out, @GFauxPas.
 
GFauxPas
GFauxPas
oooh
triangular number
 
Ted Shifrin
Ted Shifrin
Right.
 
GFauxPas
GFauxPas
$\frac 1 2 N^{-2} \cdot N \cdot (N+1)$
$= \dfrac 1 2 \dfrac {N+1}{N}$
$\to \frac 1 2$
 
Lucas Henrique
Lucas Henrique
11:08 PM
yes
 
GFauxPas
GFauxPas
so $\ln y = 1/2$, that is, $y = e^{1/2}$
okay, so let's talk error
 
Ted Shifrin
Ted Shifrin
Right. Now show that the error is insignificant (i.e., limits to 0).
 
Lucas Henrique
Lucas Henrique
If you guys are interested, I do also have another question
But I think this one is... a lot harder
(Without gamma function)
Prove that $\sqrt{2} = \frac{2 \cdot 2}{1\cdot 3}\cdot\frac{6 \cdot 6}{5\cdot 7} \cdot \frac{10 \cdot 10}{9\cdot 11}\dotsb$
 
Ted Shifrin
Ted Shifrin
So when you approximate $\log(1+x)$ by $x$, what's the maximum error (when $|x|<1$)?
 
GFauxPas
GFauxPas
$E_n = \frac 1 {(n+1)!} (x-1)^{n+1}D^{n+1}\ln \eta$
 
Ted Shifrin
Ted Shifrin
11:11 PM
No, that's not right, @GFauxPas.
 
Lucas Henrique
Lucas Henrique
@LucasHenrique this one is hell...
 
GFauxPas
GFauxPas
for some $\eta$ in $[0..1]$
is that right now?
 
Lucas Henrique
Lucas Henrique
And sorry @AkivaWeinberger, but I really can't understand why I was wrong
 
Ted Shifrin
Ted Shifrin
Not $(x-1)$, since I'm doing the function $f(x)=\log(1+x)$.
And we're doing $n=1$.
I would rather write $f^{(n+1)}(\eta)$ as opposed to your way.
 
GFauxPas
GFauxPas
I thought we need to consider all $n > 1$
okay sure
 
Ted Shifrin
Ted Shifrin
11:13 PM
No, I'm just doing linear Taylor approximation here. Right?
 
GFauxPas
GFauxPas
oh right
 
Lucas Henrique
Lucas Henrique
That $a_1 \to a_2 \to \dots \to 1 \implies a_1 \cdot a_2 \cdots = 1^n = 1$
 
GFauxPas
GFauxPas
agreed
 
Ted Shifrin
Ted Shifrin
That's wrong, @Lucas. Look at $\lim (1+1/n)^n$.
 
GFauxPas
GFauxPas
so $f^{(2)}(\eta) = \dfrac {-1}{\eta^2}$?
 
Lucas Henrique
Lucas Henrique
11:14 PM
@TedShifrin I can't see why they relate
 
Ted Shifrin
Ted Shifrin
You can only do stuff like that for a fixed finite product.
 
GFauxPas
GFauxPas
oh, at $x = 1$
 
Ted Shifrin
Ted Shifrin
@GFauxPas, maybe $-1/(1+\eta)^2$?
 
GFauxPas
GFauxPas
right
 
Ted Shifrin
Ted Shifrin
But take absolute value and the biggest that can be is $1$.
 
GFauxPas
GFauxPas
11:15 PM
because $0 \le \eta \le 1$
 
Ted Shifrin
Ted Shifrin
@Lucas: Each $(1+1/n)$ goes to $1$, but what does the product $(1+1/n)^n$ go to?
Strict inequality, actually, @GFauxPas.
 
Lucas Henrique
Lucas Henrique
Damn, I'm dumb
 
GFauxPas
GFauxPas
because strictly monotonic?
 
Lucas Henrique
Lucas Henrique
thank you @TedShifrin, @AkivaWeinberger and @GFauxPas
 
Ted Shifrin
Ted Shifrin
No, because of what Taylor's theorem says. Plus things don't converge at $x=1$.
Well, they do, but ... never mind.
 
GFauxPas
GFauxPas
11:17 PM
oh, was misreading the theorem
 
Ted Shifrin
Ted Shifrin
@Lucas: You have to be very careful with things like that. You can only do what you wanted if the number of terms in the product is fixed.
 
Lucas Henrique
Lucas Henrique
Yes, I understood y'all point now
 
GFauxPas
GFauxPas
so we need $\dfrac{(x-1)^2}{2!}$ to go to $0$
as $x \to 1$
 
Ted Shifrin
Ted Shifrin
No, no: you have to use the error for each term of the sum, @GFauxPas.
 
GFauxPas
GFauxPas
which it does
 
Ted Shifrin
Ted Shifrin
11:18 PM
No, no, no.
There's no $(x-1)^2$. It's $x^2$.
 
GFauxPas
GFauxPas
:( it's been so long since I did this lol
 
Ted Shifrin
Ted Shifrin
We're approximating $\log(1+k/n^2)$.
Then adding 'em up.
 
GFauxPas
GFauxPas
why $x^2$ and not $(x+1)^2$?
 
Ted Shifrin
Ted Shifrin
Because my $f(x)=\log(1+x)$.
 
GFauxPas
GFauxPas
which is centered at $x = 0$
 
Ted Shifrin
Ted Shifrin
11:20 PM
Right.
You have to be consistent throughout.
 
GFauxPas
GFauxPas
$\dfrac {x^2}{2!}$
 
Ted Shifrin
Ted Shifrin
Right: That's an upper bound on our error. So what do we get for the sum?
 
GFauxPas
GFauxPas
geometric series
 
Ted Shifrin
Ted Shifrin
Huh?
 
GFauxPas
GFauxPas
oh, it's a function of $x$
 
Ted Shifrin
Ted Shifrin
11:22 PM
We're looking at $\sum_{k=1}^n \log(1+k/n^2)$.
@Alessandro!!!
 
Alessandro Codenotti
Alessandro Codenotti
Hi @Ted
 
Ted Shifrin
Ted Shifrin
Parked on top of any shops yet? :)
 
Alessandro Codenotti
Alessandro Codenotti
No, but I have another lesson tomorrow so there's a nonzero chance of that happening soon :P
 
Ted Shifrin
Ted Shifrin
Glad to hear it.
 
GFauxPas
GFauxPas
$\dfrac 1 2 \dfrac {k^2}{n^4}$?
 
Ted Shifrin
Ted Shifrin
11:23 PM
@GFauxPas, summed from $1$ to $n$, yes.
Guess you're way more interested than @Lucas is ...
 
GFauxPas
GFauxPas
I guess so
 
Lucas Henrique
Lucas Henrique
Hahahahah
 
GFauxPas
GFauxPas
$n^{-4} \to 0$ as $n \to +\infty$ then
 
Lucas Henrique
Lucas Henrique
I know nothing about error
 
Ted Shifrin
Ted Shifrin
No, @GFauxPas. Pay attention.
 
Alessandro Codenotti
Alessandro Codenotti
11:25 PM
I was just checking the chat before going to sleep, have a nice day everyone, I'm already going
 
GFauxPas
GFauxPas
sorry, as I said, havent done this in years
 
Ted Shifrin
Ted Shifrin
Night, @Alessandro.
You have to sum $k^2$, don't you? @GFauxPas
 
Lucas Henrique
Lucas Henrique
Either way, how do you show this?
 
GFauxPas
GFauxPas
isn't it constant?
we're working on it Lucas
 
Lucas Henrique
Lucas Henrique
19 mins ago, by GFauxPas
in that case, we have $\displaystyle \sum_{i=1}^N \ln \left({1 + \frac i {N^2}}\right) \sim \sum_{i=1}^N \frac i {N^2}$?
 
Ted Shifrin
Ted Shifrin
11:25 PM
Think back to how you did the first part, @GFauxPas.
@Lucas: Do you know any calculus?
 
Lucas Henrique
Lucas Henrique
Yes
 
Ted Shifrin
Ted Shifrin
What's the slope of the tangent line of $y=\ln(1+x)$ at $x=0$?
 
GFauxPas
GFauxPas
oh dur
it's another triangular number
?
 
Lucas Henrique
Lucas Henrique
well, y' = 1/(1+x) so y'(0) = 1?
 
Ted Shifrin
Ted Shifrin
No. You're looking at $\sum\limits_{k=1}^n k^2$ now.
 
GFauxPas
GFauxPas
11:27 PM
oh, theres a formula for that
 
Ted Shifrin
Ted Shifrin
Right, @Lucas. So $\ln(1+x)\approx x$ for $x$ near $0$.
 
GFauxPas
GFauxPas
let me look it up
 
Lucas Henrique
Lucas Henrique
@TedShifrin that's right.
 
GFauxPas
GFauxPas
$\dfrac {n(n+1)(2n+1)}{6}$
 
Lucas Henrique
Lucas Henrique
isn't this much more of a "numerical" solution
rather than an "analytical" solution?
 
Ted Shifrin
Ted Shifrin
11:28 PM
All you need, @GFauxPas, is that it is roughly a constant times $n^3$ ... And you have $n^4$ in the denominator. So, as $n\to\infty$ ...
 
GFauxPas
GFauxPas
Lucas that's why we're analyzing the error
 
Ted Shifrin
Ted Shifrin
No, @Lucas. It's a totally analytic solution if you do it correctly.
 
GFauxPas
GFauxPas
nice Ted, thanks for being patient with me
 
Lucas Henrique
Lucas Henrique
Oh, I got it
 
GFauxPas
GFauxPas
Lucas, when you have a Taylor polynomial, Taylor's theorem also gives you a bound on the error between the series and the function
 
Ted Shifrin
Ted Shifrin
11:30 PM
no, polynomial, not infinite series.
 
GFauxPas
GFauxPas
oops, right
we're showing that the error goes to $0$ as the index of summation goes to infinity
 
Ted Shifrin
Ted Shifrin
And indeed we're done.
Now write it all up :P
 
GFauxPas
GFauxPas
that's Lucas's job
:P
 
Ted Shifrin
Ted Shifrin
well, you have to help him.
 
Lucas Henrique
Lucas Henrique
Ok, good night guys
xD
 
Ted Shifrin
Ted Shifrin
11:31 PM
LOL
 
GFauxPas
GFauxPas
sure
 
Lucas Henrique
Lucas Henrique
just kidding.
 
Ted Shifrin
Ted Shifrin
Where are you, anyhow, Lucas?
 
Lucas Henrique
Lucas Henrique
Brazil.
 
Ted Shifrin
Ted Shifrin
Ah, cool.
 
Lucas Henrique
Lucas Henrique
11:32 PM
8:31pm here
 
GFauxPas
GFauxPas
proofwiki.org/wiki/Taylor%27s_Theorem Here's the formation I'm using, Lucas
 
Ted Shifrin
Ted Shifrin
BTW, my linear algebra book was published in Portuguese :P
 
Lucas Henrique
Lucas Henrique
@GFauxPas I'm familiar with Taylor's Theorem
 
GFauxPas
GFauxPas
and the error term?
 
Lucas Henrique
Lucas Henrique
@TedShifrin Ooooh!
@GFauxPas You mean, "complexity"?
 
GFauxPas
GFauxPas
11:32 PM
What is Taylor's theorem?
 
Akiva Weinberger
Akiva Weinberger
Of course, he's completely unaware of the numerous mistranslations
 
GFauxPas
GFauxPas
what does it say
 
Ted Shifrin
Ted Shifrin
sends DogAteMy to his Spanish lesson for 5 weeks
 
Lucas Henrique
Lucas Henrique
@GFauxPas A way to write infinitely differentiable functions as an polynomial series
 
Daminark
Daminark
@Ted Oh that's interesting
 
Ted Shifrin
Ted Shifrin
11:35 PM
Huh? Demonark
 
GFauxPas
GFauxPas
a way to approximate infinitely diff functions
 
Daminark
Daminark
Book published in Portuguese
 
Lucas Henrique
Lucas Henrique
@GFauxPas, well, at infinity it's true, isn't? :P
 
Ted Shifrin
Ted Shifrin
Oh.
 
GFauxPas
GFauxPas
not always
 
Ted Shifrin
Ted Shifrin
11:36 PM
Yeah, I guess W.H. Freeman had enough interest that they had it translated.
So I assume some schools in Portugal/Brazil are using it.
BTW, Demonark, you got my email?
 
GFauxPas
GFauxPas
and we were only using a linear approximation
so we need to know how good that approximation is, and Taylor's theorem tells us about the error
 
Lucas Henrique
Lucas Henrique
OMG
 
Daminark
Daminark
Yeah @Ted, I messaged you on there for reference
 
Ted Shifrin
Ted Shifrin
oh.
 
Lucas Henrique
Lucas Henrique
@TedShifrin you're a famous mathematician and I didn't even know it
Oh my God...
UGA math department?
 
Ted Shifrin
Ted Shifrin
11:38 PM
Formerly.
 
Lucas Henrique
Lucas Henrique
Oh my God. You're incredible, man
 
Ted Shifrin
Ted Shifrin
rolls several eyes
ok, Demonark, we should be all set ... except you still have to prepare your lecture.
 
GFauxPas
GFauxPas
loans Ted his eyes so he has more to roll
 
Ted Shifrin
Ted Shifrin
I actually rolled 9 the other day.
 
Daminark
Daminark
This is true
I'm TeXing up some lecture notes right now
 
Ted Shifrin
Ted Shifrin
11:42 PM
BTW, @Lucas, @EricSilva is also Brazilian (/American).
 
Lucas Henrique
Lucas Henrique
Well, @EricSilva, hi! Or should I say "olá"? :P
Boy, I'm so anxious
 
Ted Shifrin
Ted Shifrin
anxious?
 
Lucas Henrique
Lucas Henrique
My school managed to send me to IMPA: "Institute of Pure and Applied Mathematics"
 
Ted Shifrin
Ted Shifrin
oh cool ... I've known lots of people from there over the years.
 
Lucas Henrique
Lucas Henrique
Just in time when we'll have IMO here in Brazil
 
Ted Shifrin
Ted Shifrin
11:47 PM
my mathematical brother (and very well known geometer) Manfredo do Carmo was at IMPA. When I was a grad student, there were lots of dynamical systems people from IMPA who visited (and vice versa).
 
Lucas Henrique
Lucas Henrique
@TedShifrin nice. AFAIK IMPA is "relevant" in the international mathematics
 
Eric Silva
Eric Silva
ayyy um outro brasileiro
 
Ted Shifrin
Ted Shifrin
Very, yes.
 
Eric Silva
Eric Silva
Neves told me to do everything i can to retire to impa
 
Ted Shifrin
Ted Shifrin
I think it's a little early to be thinking about retiring, ERic.
 
Eric Silva
Eric Silva
11:48 PM
lol im just repeating his advice
 
Ted Shifrin
Ted Shifrin
But at the rate the US is going, maybe you want to get to Brazil in the next few months.
 
Lucas Henrique
Lucas Henrique
@EricSilva My academic plan is UFRJ -> IMPA -> Berkeley
 
Ted Shifrin
Ted Shifrin
Ah, @Lucas ... I was at Berkeley.
 
Eric Silva
Eric Silva
cool cool
 
Lucas Henrique
Lucas Henrique
But I have ENEM first so it's a pain in the ass.
@TedShifrin What did you think of it?
 
Ted Shifrin
Ted Shifrin
11:49 PM
You'd better understand the proof I just did if you have such aspirations.
 
Lucas Henrique
Lucas Henrique
:(
 
Ted Shifrin
Ted Shifrin
I liked it, but it's not a good department for people who are shy.
 
Lucas Henrique
Lucas Henrique
@TedShifrin I'm trying Princeton, then. :P
 
Ted Shifrin
Ted Shifrin
Um, it's difficult in different ways.
 
Eric Silva
Eric Silva
scary
 
Lucas Henrique
Lucas Henrique
11:50 PM
@TedShifrin b-but I'm just a highschooler
 
Eric Silva
Eric Silva
princetons colors give me a headache
 
Lucas Henrique
Lucas Henrique
And we do not have calculus here in Brazil
 
Ted Shifrin
Ted Shifrin
There goes Eric with his school colors obsession.
 
Lucas Henrique
Lucas Henrique
@EricSilva how old are you?
 
Ted Shifrin
Ted Shifrin
What year of high school, Lucas?
 
Eric Silva
Eric Silva
11:51 PM
@Ted I have synesthesia and certain colors combined with certain letters give me a headache! so i have a reason!
@Lucas 20
 
Ted Shifrin
Ted Shifrin
rolls a few eyes at Eric
 
Eric Silva
Eric Silva
it's true! it's neurological man
 
Ted Shifrin
Ted Shifrin
You overdo it.
 
Lucas Henrique
Lucas Henrique
@TedShifrin US-Brazil school systems are different. I'd say I have this and next year to pass all my exams to got to uni
 
Akiva Weinberger
Akiva Weinberger
What color is a P
 
Eric Silva
Eric Silva
11:52 PM
maybe a bit
@Akiva pinkish
 
Akiva Weinberger
Akiva Weinberger
What about R
 
Eric Silva
Eric Silva
like between pink and purple
 
Ted Shifrin
Ted Shifrin
LOL
 
Eric Silva
Eric Silva
red
 
Ted Shifrin
Ted Shifrin
Enjoy your intensive Spanish, DogAteMy!
 
Akiva Weinberger
Akiva Weinberger
11:53 PM
So drawing that little extra line changes its color, cool :P
 
Ted Shifrin
Ted Shifrin
OK, I'm outta here. Have to have dinner and go play bridge for the evening.
Happy learning, @Lucas.
 
Lucas Henrique
Lucas Henrique
@EricSilva are you at university?
 
Typhon
Typhon
You ever have one of those moments where you cannot remember something important?
 
Akiva Weinberger
Akiva Weinberger
(Not sarcastic, I bet it actually is cool, though the headaches are less so)
 
Ted Shifrin
Ted Shifrin
It's called old age, Typhon.
 
Typhon
Typhon
11:53 PM
and it drives you nuts?
 
Eric Silva
Eric Silva
@Lucas I go to university of chicago
 
Lucas Henrique
Lucas Henrique
@TedShifrin does it mean I should (or I must) learn it "quickly"?
 
Typhon
Typhon
@TedShifrin 21 is not old age.
 
Ted Shifrin
Ted Shifrin
No, just learn, Lucas.
 
Eric Silva
Eric Silva
@Akiva it's cool, i think it helps me remember things, the headaches suck but i get a LOT of those anyway
 
Akiva Weinberger
Akiva Weinberger
11:54 PM
remains confused about Typhon's age
 
Typhon
Typhon
@AkivaWeinberger just ignore it. uses Jedi mind trick
 
Eric Silva
Eric Silva
didn't Typhon say he was in high school
 
Lucas Henrique
Lucas Henrique
@EricSilva how was the SAT and the application process for you?
 
Ted Shifrin
Ted Shifrin
Bye, all.
 
Typhon
Typhon
im a bit older
 
Eric Silva
Eric Silva
11:55 PM
Bye @Ted
 
Typhon
Typhon
dont overthink it
i said i didnt do well
 
Lucas Henrique
Lucas Henrique
@TedShifrin I will, professor, I will. :)
Bye @TedShifrin!
 
Eric Silva
Eric Silva
@Lucas Well for the SAT i didnt study, but that was a bad thing. The application process was annoying because I waited until the last minute bc I didn't know if i could go to college until after the application process was open
 
Semiclassical
Semiclassical
hi chat
 
Eric Silva
Eric Silva
all in all i was lucky and am not someone to look to as an example or anything
 
GFauxPas
GFauxPas
11:56 PM
hi semi
 
Lucas Henrique
Lucas Henrique
hello!
 
Eric Silva
Eric Silva
any advice i would give is to start earlier than me and actually study for things
 
Lucas Henrique
Lucas Henrique
@Semiclassical I remember that you helped me before
 
Semiclassical
Semiclassical
heh, glad i was memorable.
 
GFauxPas
GFauxPas
Semi were you there for the $e^{1/2}$ product-series shenanigans?
Ted finally figured it out lol
 
Lucas Henrique
Lucas Henrique
11:57 PM
actually, I don't frequent this chat because the stuff here is hard and smart people are scary. heheh
 
GFauxPas
GFauxPas
you should hang out here anyway
:)
 
Semiclassical
Semiclassical
nah, I wasn't.
 
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