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02:00 - 19:0019:00 - 00:00

19:00
@LeakyNun oh what
yeah
but yeah one needs more calculation for that I guess
Yeah it's not obvious to me at all
But I see what you mean
It's believable
did Naka draw the arrows to explain this lol
no he made the moves on the analysis board
so I think there are 4 moves to analyse
1... Raxc8 2. dxc8=Q Rxc8 3. Rxc8
1... Rdxc8 2. dxc8=Q Rxc8 3. Rxc8
1... Rdxe8 2. dxe8=Q+ Kg8 3. Qg8#
(I completely missed this line and discarded it as just being the second line lol)
1... Rxd7 Bxd7
19:06
Ah OK
and in all the lines (except the third obviously) you lose a rook
so it was in this position that Hikaru Nakamura resigned the game
lmao
nice reference
@LeakyNun about your example (x^2+y^2-1)^2 = 0. Because it is polynomial it is differentiable everywhere and in x,y=sqrt(1/2) the gradient is 0. Did I get it right?
the gradient is 0 everywhere on the curve
but yes you did get it right
but why is it a curve?
it is a circle
19:09
@BalarkaSen in the first line you have a nice x-ray
@FarhadRouhbakhsh exactly
@LeakyNun where did you get this example from?
my head
thank you math nerd!
@BalarkaSen thank you
When you're solving cos(z) = 2i, how come for some solutions it has ln and others it has log
I already got the answer but I'm comparing it to online solutions I found and they all look different
some people write ln as log
19:13
because some people treat log and ln and the same
how are they the same though? cause isnt ln base e and log is base 10
Usually, mathematicians write log unless it's a calculus class. Engineers and physicists continue to write ln.
computer scientists also use lg
for log base 2
@HokieFan7 symbols are arbitrary
No, log is the way we write ln. We write $\log_{10}$ if we ever want that (and we don't).
19:14
and different conventions exist
oh ok, thanks all
@LeakyNun Let $M$ be a scheme and $\theta$ be the structure sheaf
ok, go on
What happens if $M$ happens to be an abelian variety? Then what, mister?
Hahaha
Great minds, @Ted
19:15
is that a reference
M could also be a moduli space
Abelian varieties have $\Theta$ divisors and $\theta$ has a meaning.
We should link Leaky to that $[\gamma]$ guy
Let $M$ be a scheme and $\theta$ be the structure sheaf
let n_1,..,n_k be the prime factors of an integer p
19:17
$\gamma$ be a path, $[\gamma]$ be the derivative, $\gamma'$ another curve and let $\gamma^t = \gamma'|_{(-\epsilon, \epsilon)}$
2
$(-\gamma_\gamma, \gamma_\gamma)$
@BalarkaSen Sadly, that was the "script" and not the poor OP's fault.
@LeakyNun That reminded me of something, 1 sec
Feb 28 '18 at 20:43, by Balarka Sen
$\gamma^X_p[0, t] \cup \gamma^Y_{\gamma^X_p(t)}[0, t] \cup \gamma^X_{\gamma^Y_{\gamma^X_p(t)}(t)}[0, -t] \cup \gamma^Y_{\gamma^X_{\gamma^Y_{\gamma^X_p(t)}(t)}(-t)}[0, -t]$
oh my
'oly 'ell
19:21
i demand reparations for having this intrude my field of view
hahahah
Let $G$ be a grothendieck prime
a @Balarka!!! What the.
show that the set $G$ has one element
intrude upon, @Thorgott :D
19:23
Feb 28 '18 at 20:36, by Balarka Sen
@AkivaWeinberger Let $\gamma^X_p$ denote the integral curve of $X$ starting at a point $p$. Draw the square $\gamma^X_p[0, t] \cup \gamma^Y_{\gamma^X_p(t)}[0, t] \cup \gamma^X_{\gamma^Y_{\gamma^X_p(t)}(t)}[0, t] \cup \gamma^Y_{\gamma^X_{\gamma^Y_{\gamma^X_p(t)}(t)}(t)}[0, t]$, is what I meant, where $X$ and $Y$ are scaled to be unit vector fields, say.
Feb 28 '18 at 20:36, by Balarka Sen
I'm become the god of horrendous notation
@TedShifrin dont ask
Somehow I've managed to teach and write homework questions about what you were discussing without ever vomiting on the board/paper.
hey that actually makes sense
@loch why genus 2 => hyperelliptic
@TedShifrin when you lecture, do you pronounce the German names "correctly"?
"used to" ?
Yes, within reason.
The big problem is Dirichlet. It's a French name that belonged to a German. How does one pronounce it?
hmm I thought about that before
19:29
I always pronounce it as Deereeshlay
I forgot my conclusion, if I had one to begin with
That's French, Balarka. I usually do, too. The German pronunciation is a harder "ch" but, correctly, it's not a "k" sound.
@TedShifrin But then in German you would also pronounce the t at the end
oh wiki says Duden says [ləˈʒœn diʀiˈkleː] is the "German" pronunciation
@TobiasKildetoft well it's a borrowed word so the t gets to be silent
I suppose so
But then the ch sound also gets to stay
19:31
it's like the h sound at the start of a British here
LOL ... hopeless.
well... language is not mathematics
I've never heard anyone pronounce the t at the end
I had a long discussion at one point about how to pronounce Sylow
Zooloff
19:31
Sülof
But no one in American would get that.
Thing is, he was Norwegian, but the name is Polish
Sülo
(also, it is not a regular l there, but I could not bother to find out how to put a line through it)
Well, I have a Russian name even though I'm 2nd generation American.
So you transliterate that as a "w," Tobias?
19:33
Right, no idea if he would have pronounced in as a Norwegian or a Polish name
@TobiasKildetoft wait, it's a Polish name?
@TedShifrin No, the l in the middle
Wait how do you pronounce Sylow
welcome to the discussion
@Tobias: Yes, I know. The Polish crossed-ell is pronounced like a w.
19:34
The standard confusion was Ceelow or Saaeelow I thought
Personally, I pronounce it as in Danish, which is a bit like Norwegian, but less nice
[ˈpeːtɛr ˈsyːlɔv]
@BalarkaSen well the "y" is definitely an ü...
@TedShifrin I see. I might misremember then. At least that was now what the people there who had some idea of Polish concluded
19:35
Can anyone help me with this problem, I need to find all the values of (1 - i*sqrt(3))^i in terms of x + yi
Write $1-i\sqrt 3$ as $e^\text{blah}$, @Hokie.
how do you pronounce
where $b=\ln 2$, $l=i$, $a=\pi$, and $h=-1/3$? @TedShifrin
(I can't do trigonometry)
Aren't you missing a $+$, @Leaky?
i remember liking trig identitites
19:38
1)h^0(canonical divisor) = 2 [because its genus = 2]
2) this means that two linearly independent sections s_0, s_1 of a line bundle
3) you define C -> \P^1 mapping p to [s_0(p): s_1(p)].
4) but wait - s_0 and s_1 might vanish simultaneously, in which case this map won't be defined (otherwise you should check that this is well-defined / i.e. choosing different trivializations of your line bundle won't change your map).
5) so what you have now is a rational map C -> \P^1, possibly not defined at points where s_0(P) = s_1(P) = 0
The discussion I had started because the chair at a conference was curious if he had pronounced my name correctly (he almost had). And I happened to have actually emailed two people whose work I was mentioning to figure out how to pronounce their names
@BalarkaSen That looks like standard differential geometry notation, what am I missing?
@TobiasKildetoft lemme ask my Polish friend
me too
@AlessandroCodenotti Lol
19:40
@Alessandro: smack
4
can anyone develop a crypto-system based on 1/x
You know I'm right
or do you need like pell's equation or some stuff
pells equation is x^2 -Dy^2=1
where D is a non square int.
@loch I've actually been tortured by 18.116 enough for you to skip from 1 all the way to 7
@LeakyNun good!
19:41
but I guess it's an analytic argument?
Ted Shifrin, how do you account for the i power though? thats throwing me off
@LeakyNun then not as good!
(or better depending on your religion i guess)
can you mod a hyperreal
Then raise to the $i$th power. Remember you're going to get countably many values for the answer, @Hokie.
19:42
@loch I mean, a meromorphic (read rational) function is just a holomorphic map (read algebraic map) to P^1 right
more GAGA
@loch @Leaky: I was going to comment that for curves there is no base locus. You just pull out $(z-z(P))^\mu$ from your homogeneous coordinates.
is there some coordinate chart that locally looks like a hilbert space
oh
@loch I implicitly interpreted 1 as "pick a non-constant function, look at the corresponding function with the prescribed poles, etc etc"
what does enrich mean in mathematics
Thanks Ted I think I'm on the right track
19:49
I never internalized Riemann Roch
I have come back to it a couple times but it just escapes my mind every time
I told you about Griffiths' interpretation in terms of the geometry of the canonical curve? Very beautiful.
Surely, it's classical Italian algebraic geometers, not Griffiths, but the modern sheaf era has eviscerated it.
Yeah I think you did mention it to me
@TedShifrin ah that makes it obvious
@TedShifrin what is it?
19:51
It's in at least one set of my lecture notes, @Balarka. Not sure if I sent you my complex geometry course from MIT.
I don't think you did haha
@Leaky: The interpretation of $h^0(K-D)$ is in terms of hyperplanes containing the images of the points of the divisor under the canonical mapping.
oh we're beyond curves now?
This is curves
No, we're on curves. Canonical mapping $\iota_K(C)$.
Non-hyperelliptic curves, though.
19:53
is it what 18.116 calls the Kodaira map
yeah I did a PSet on that
the restriction of the hyperplane bundle is the canonical bundle
I love it that a course number is calling things things.
@Zacky welcome
I'm taking Angles 230
@geocalc33 greetings!
19:55
and when I go to study to be a priest I will take Angels 500
Nosedive; you tried too hard
I'm not sorry
but yeah I did
okay next topic
@Balarka @Leaky @loch
@Zacky how did you become so adept with integration?
I have been summoned
19:57
user image
4
user image
3
Hmm, those graphics resized without my permission. Oh well.
Thank!!
Dank!
@LeakyNun I'm not listening to Salut d'Amour
why not
it's good
I'm judging a classical piece by it's name
20:01
also, today I realized why effects is called FX
balarka sen listen to Die Walkure
I love Wagner!
thanks for illuminating me as well leaky
I need to listen to more Wagner
Parsifal man
Banger
okay I'll look it up
20:04
has this chat heard of the French mathematician Pierre Cousin (1867 - 1933)? he doesn't have a wiki page so...
yeah i know Cousin
famous for the Cousin problems
oh he has a wiki page in fr.wiki
Pierre Cousin est un mathématicien français né le 18 mars 1867 à Paris et mort le 18 janvier 1933 à Arcachon. Il est notamment à l'origine du théorème et du lemme qui portent son nom. == Publications == « Sur les fonctions de n variables complexes », Acta Math., vol. 19,‎ 1895, p. 1-62 (DOI 10.1007/BF02402869) Voir en ligne « Sur les fonctions triplement périodiques de deux variables », Acta Math., vol. 33,‎ 1910, p. 105-232 (DOI 10.1007/BF02393214) Liste d'autres articles == Référence == Portail des mathématiques Portail de la France…
standard motivation for sheaf cohomology
I've only heard of Jaques Tits to be honest
Lol
it's Jacques fwiw
20:06
oh true
what is a sheaf in 5 words
no symbols
portal between local and global
there's a quora post by Tits' grandniece on how to pronounce his last name
is it just "tee"
@geocalc33 I don't really know the answer.. I always tried to do math for fun after high-school ended, and integrals happened to be cool.
20:08
No idea. I pronounce it as theeth with less emphasis on h for the first t
@geocalc33 presheaf with glue & stuff
portal between local and globa...presheaf with glue and other stuff...
what is the mechanism that ports the info back and forth?
small patch patch glue big
lol
@Zacky nice I respect that
Hi. I have to calculate the derivative of $f(x)=x^n$ via the abstract definition (without the differencial-quotient), i.e. I want for a point $x_0$ to reach $f(x) = f(x_0) + L(x-x_0) + R(x, x_0)$ where $L$ is a linear function and $R$ some residuum term that, taken the absolute value and divided by $|x-x_0|$ approaches $0$ for $x \rightarrow x_0$.

Now, i "know", that the Linear function I want to get is $nx^{n-1}$, and I managed to get some hacky transformations to reach that. But how would I get there without "knowing my target" ?
While it is kind of obvious because the top of the fraction kind of gets zero in the limit fast than the bottom because of the exponent - I have no clue how to show this without L'Hospital or anything else (which I have to)
20:15
@T_01 $x^n - x_0^n = (x-x_0) (x^{n-1} + x^{n-2} x_0 + \cdots + x x_0^{n-2} + x_0^{n-1}) \approx (x-x_0) (nx_0^{n-1})$
well
yeah, turns out this is equivalent to the limit definition / approach
smells like binomial stuff but it isnt, is it?
it's the difference of squares on steroid
$a^2-b^2 = (a-b)(a+b)$
okay i see
20:22
also geometric series
true story. thank you - this seems clearer now. But any chance to get anywhe with my limit, mentioned above?
@TedShifrin nice!
once you write down the expression for $R$ then it shouldn't be so difficult
@T_01 I would write $x-x_0=h$.
does this help?
20:27
You want the $h$ stuff in the expansion $f(x_0+h)-f(x_0)$.
$\Psi_{\gamma,\kappa,\rho}(t)=e^{{\gamma}it}e^{(\kappa A+\rho A^*)}$
I think i am confused, sorry. Also I think I am kind of done if i can show $\lim_{x \rightarrow x_0} \frac{|x^n + (n-1)x_0^n - nxx_0^{n-1}|}{|x-x_0|} = 0$, or, am I?
That doesn't look right.
@BalarkaSen just as I said there are 7k viewers ytd on Hikaru's stream
But the limit is zero and in fact this seems to lead what I want
20:31
now he has 10k viewers
$(x_0+h)^n-x_0^n = nx_0^{n-1}h + \text{terms with } h \text{ with higher powers}$.
my calculation looks as follows
If you're writing it your way, you need to show $x^n-x_0^n - nx_0^{n-1}(x-x_0)$ goes to $0$ faster than $x-x_0$. I tell you it's a lot easier to write $h$ instead of $x-x_0$ for all the algebra.
Okay, I believe you
Oh, I see. What you wrote is the same as what I wrote, but multiplied out.
20:33
ah yeah
But it's not obvious how to get that limit is $0$ unless you do factoring and estimates. It's far easier to use binomial expansion as I did with the $h$ in there.
okay, I try it. Thank you
I need help here. So $A=\cos(\theta x-\psi)$ is an equation of a wave with phase $\psi$
and $B=\sin(\theta x -\phi)$ is another equation of a wave with phase $\phi$
is that good notation so far?
20:54
@TedShifrin look at this link. This is compatible with the STATEMENT of the implicit function theorem which I found in my Adams' calculus book, and also gives my answer.
1
Q: Explaining Tangent Planes With Implicit Differentiation for Multivariable Calculus

Michael SemkoLet $f : D \subseteq\Bbb R^3 \to \Bbb R$ be a $C^1$ function. A point $(a, b, c) \in \Bbb R^3$ is a critical point of $f$ if $$ \nabla f(a, b, c) = (0, 0, 0). $$ The corresponding number $k = f(a, b, c)$ is called a critical value: any real number $c$ that is not a critical value is called a re...

@TedShifrin Never keep arguing with anybody.
Please follow the Be Nice policy of the chat. If some answer suited better for you than another answer there's no reason to attack the other person.
3
@Balarka Sen And also tell everybody not to label other people notes by saying "Your notes are no good"
There's no labeling involved. Everyone's entitled to their opinions as long as it doesn't directly attack anyone. Be Nice.
@BalarkaSen Not understanding the content of my notes and saying "your notes are no good" when they are actually helpful is nothing but labeling.
Truth is not a matter of opinion, judgement and labeling is.
Thank you for your attention
Everything is opinion. It's perfectly ok to say a certain book or notes are not good if the person believes so. If you don't believe it you simply ignore the opinion.
I think Apostol's Calculus is a terrible book. Many will disagree. So what?
just Be Nice when stating opinions
21:05
@BalarkaSen finally got my bullet over 1600 again after two weeks of limited playing
@BalarkaSen "Everything is opinion". I love this sentence
@BalarkaSen All Im saying is to say people should not judge before they understand the subject they are talking about
If Ted didn't understand multivariable calculus we'd be in trouble
That does seem to describe you pretty well
in this case my problem could be solved with the help of the notes I had, and as you see in the link with the "implicit function theorem" that I had in mind. If he cannot help me with the material I have in mind, then I see no reason for him to say "Your notes are no good".
Its better for him to say "I dont understand how your problem can be solved with the concept of "implicit function theorem" you have in mind"
or silence also is a good answer.
Well I'm glad you were able to help yourself! Your demeanor suggests you will have to get used to that
21:13
@MikeMiller Let me tell you again that I can never cope up with judgements which are far from reality.
If anybody has got used to these kinds of things, I wish good luck for them in their life.
of course Ted is a professor and he knows a lot more than me- a simple calculus student
but that should never lead us to say that his helps are ALWAYS working or his comments are ALWAYS suitable
I have a simple question regarding Mathematica, how do I plot a parametrized elipse as such puu.sh/FDimw/c8a14aaeee.png but not having to copy paste the code 4 times for all sign combinations (in this case the +- doesnt work)

so as such puu.sh/FDir8/762e4a2005.png but for all 4 quadrants
@MikeMiller @BalarkaSen @LeakyNun Sorry for being sarcastic guys. But you shall know the truth, and the truth shall set you free.
Have a nice time
 
1 hour later…
22:33
17
A: Some way to integrate $\sin(x^2)$?

robjohnCheck out this answer for a real method to evaluate this integral. First note that $$ \int_0^\infty e^{-x^2}\,\mathrm{d}x=\frac{\sqrt\pi}2\tag{1} $$ The integral $$ \int_\gamma e^{-z^2}\,\mathrm{d}z=0\tag{2} $$ over the curve $\gamma$ consisting of the line from $0$ to $R$ then counterclockwise...

I don't understand why we can change the path to $0$ to $\pi/4$
22:46
Quick question if anyone is here, if I have that g(z) = 1/f(z) and I want to find g'(z), does it just become 1/f'(z)?
Does that work for real calculus?
Since $1/x$ is $1$ over $x$, is the derivative $1$ over $dx/dx = 1$?
It doesn't I think you need to use the chain rule
Plus remembering that the derivative of $1/x$ is $-1/x^2$.
oh wait, so it becomes (-1/f(z)^2)*f'(z)
0
Q: Connected sequences and pariwise intersection

topologicalorientablesurfaceProblem: Suppose $X$ is a topological space. Let $(A_n)_n$ be a sequence of connected subsets of $X$, for which, $A_n\cap A_{n+1}\neq \varnothing$. Then, $A=\bigcup_n A_n$ is connected. My attempt: Suppose $\bigcup_n A_n$ is disconnected, so there exists a non-constant continuous function $f:\b...

22:56
@Hokie: Which should remind you a lot of the quotient rule.
ohhh wait i can see it now, thanks!
r9m
r9m
23:17
@Simple it's explained in the answer I think .. are you comfortable with cauchy residue theorems and such?
I haven't heard about it
r9m
r9m
Then it might be tough to explain without some knowledge of complex analysis .
I know the definition of holomorphic function
r9m
r9m
okay .. do you know the integral of holomorphic functions over simple closed curves is $0$?
yes
the function is path independent
then $\int_{\gamma}f=F(z_1)-F(z_0)=0$ since end pint and start point are equal
r9m
r9m
23:24
Yes .. That's what is being used .. he denotes the closed curve described in the answer as $\gamma$ and wrote $\int_\gamma e^{-z^2}\,dz = 0$ using the Cauchy's theorem
I see Dough's answer already has the explanation of the missing bit .. why the integral over the circular arc goes to $0$ as we let $R \to \infty$
@TedShifrin Idk why .. but suddenly my youtube suggestion is flooded with your lecture videos! :-)
3 mins ago, by r9m
I see Dough's answer already has the explanation of the missing bit .. why the integral over the circular arc goes to $0$ as we let $R \to \infty$
oh, didn't scroll down
23:58
0
Q: Proper subsets of connected spaces and proper product

topologicalorientablesurfaceLet $A$ be a proper subset of $X$ and $B$ a proper subset of $Y$. If $X,Y$ are connected. Show that $X\times Y\backslash (A\times B)$ is connected. Lemma: Let $X$ be a space and $A_1,A_2...,A_n$ a finite sequence of connected subsets in $X$. If $A_j\cap A_{j+1}\neq \varnothing$ for each $j=1,2....

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