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21:07
Boom shaka, my bros
happy april fools' day pal
RH solved, have you heard?
It just happen to come out on 4/1, complete coincidence
21:23
@EnjoysMath By who?
lol
I sometimes have a feeling that I will be the one to solve RH, lol.
I got bookcovers for my softcover books, yay!
21:41
and you don't even like number theory, @Jasper :P
mr @Pedro
@JasperLoy Did you get the pdf?
I think I sent it.
@TedShifrin Hello there.
you full of knowledge today, mr @Pedro?
@TedShifrin I'm full of a headache.
Not totally grumpy though.
hands @Pedro a martini
I am having coffee and cookies. =D
Could've put some liquor in it.
21:43
oh well
So since you're having dessert, I won't offer you dinner.
@TedShifrin Any news?
These things you post always have 30 seconds or more of ads ... so I don't watch.
@TedShifrin You need to be stronger or the interwebs will get you.
21:46
Meh.
I'm totally depressed at the lack of 4/1 day posts on MSE. Get in there guys, be creative, already deleted mine
I had 2.5 hours of lecture, 1.5 hours of advising, and 2 hours of office hours, so I'm bitchy.
@TedShifrin What did you lecture about?
Hmm, geodesics and geodesics on surfaces of revolution in geometry; in the multi class, finished the proof of Stokes's Theorem and did the classical versions thereof. More physics tomorrow.
Assigned a problem that drove all my students crazy, so I probably will remove it from the assignment: Prove that a $k$-dimensional manifold is orientable if and only if there is a nowhere-zero $k$ form on it. (Yes, I told them about partitions of unity.)
@TedShifrin Let them suffer!
21:49
You never like it when I make you suffer, let it be noted.
Not entirely true. =D
But suffering is needed.
No pain no gain.
No sacrifice, no victory.
Yeah, well, they're still math babies.
Well they are not freshmen are they?
Yea... if someone could ping me when there's a 4/1 day post up, that would be great... thanks.
@EnjoysMath ???
21:51
About half are freshmen, actually.
4/1 day = april fools day, "Riemann Hypothesis Solved", "Twin Prime Theorem" etc. With a link to a funny video, and make it look real.
@Enjoys, are you still 12?
I never did 4/1 day when I was young, I guess we're backwards
well, I guess so
@TedShifrin you could probably come up with a good one. Go for it, bro!
21:53
@TedShifrin Like this?
Oh.
@Pedro, stop doing everyone's homework for them, damn it.
@TedShifrin But I...
I didn't.
Was sharing the proof I know/like.
Note I didn't provide a full solution, though.
Sit still, @Jasper.
@pedro I did not get it. You need to send to my latest email.
@JasperLoy The gmail one?
21:57
@PedroTamaroff Yes, the one I just used to send to you.
I am going to listen to this for the next 2 hours.
@Pedro: I am however pleased that you instructed the OP to draw a picture. I have influenced you. :D
@TedShifrin Certainly!
@TedShifrin I like books without pics because I can draw them on my own, lol.
well, @Jasper, I've learned that a lot of "readers" don't know how to draw them or don't bother to, so I believe in drawing them in my books and lectures.
@PedroTamaroff Are you sending now?
22:00
OK, going to a concert tonight, so I have to have dinner and skedaddle. G'night.
@JasperLoy I sent it.
@TedShifrin Cheers.
@PedroTamaroff Sorry, but still nothing here, lol.
@PedroTamaroff My address is [email protected].
WAT
I was sending it to [email protected]
@PedroTamaroff Wrong, I deleted that long ago.
OK sent.
22:06
@PedroTamaroff Received, thanks!
I missed Ted again.
@Mike Thank god you missed, I do not want to see you harm him again..
2
22:22
@Mike Jesus Mike, I still can't get it, it's driving me crazy
@N3buchadnezzar What?
@PedroTamaroff Like with a bow gun or car etc
........................................
@N3buchadnezzar What did Mike do?
@PedroTamaroff He missed ted
I don't know if Red can take many more hits from me.
22:42
Hey could like two of you give this answer an upvote, math.stackexchange.com/a/735243/66223 someone really helped me, I upvoted, then they made an edit that I want to say thanks for but I cannot twice!
@Mike I made comments colorful.
Oh is that that unicorn thing?
Change the colour of the top bar. Should be a feature
I asked Mike earlier, but I still can't get the second one, can someone help me?
I took two derivatives and ended up with the squared (k-nt) but I can't get the rest...
@Anthony can I have a copy to do? I'm always looking for practice!
Also LOL Adobe reader, I remember the days my computer would grind to a hault and suck when I opened a PDF.
Whatchu mean?
22:55
Well it's a set of analysis questions.
Oh sure
Give me a link!
uh
I actually don't know how
Okay can you email it to me? You know how to attach stuff?
That sucked
22:56
It did yes.
What's your email?
First letter of my first name, dot, my surname @warwick.ac.uk
But don't write it here! These logs are public!
Spambots are everywhere >.>
lol
I think I sent it
but damn I can't get this stupid part
it's not even hard
jesus
No seriously @Anthony it's really easy, just run /[a-z0-9-.]+@[a-z0-9-]+(\.[a-z]{2,3})+/i
Anyway did you email it to me?
I think I did.
22:59
Thanks
Nope failed
oh first letter
@PedroTamaroff you there?
Could you help me out?
I could.
Might you?
23:01
I might.
Will you? <3
Got it @Anthony thanks
Haha, OK.
Could I have the other 7 please?
Oh dear gimme some time @AlecTeal I need to get my HW done
23:02
Sure
And @PedroTamaroff I posted up a ways, it's an easy thing with the binomial theorem
But I don't know how to get it!
@Anthony which Q?
@Anthony ?
The last sum
Q5.a?
23:03
jah
Dude-.
It's not that hard.
I know!
I believe you!
But I can't get it :/
It's driving me crazy.
$(1)$ you see right?
yeah
23:04
Thank god.
I will use a different notation.
Just to confuse you.
$$B_k^n(t)=\binom nk x^k(1-x)^{n-k}$$
JK. It's because they are called Bernstein polynomials.
@Anthony what polynomial did the sum come from, it's the expansion of (t+(1-t))^n that's just 1^n
So it must be 1
So expanding $(k-nt)^2=k^2-2tnk+n^2t^2$.
23:06
Indeed.
So let's work with each sum separately.
Okay.
@AlecTeal Yeah I meant the second part of 5a.
Consider $$P(x,y)=(x+y)^n=\sum_{k=0}^n\binom nk x^ky^{n-k}$$
Then $$(xD_x)^2 P(x,y)=\sum_{k=0}^n \binom nk k^2x^ky^{n-k}$$
But at the same time, $xD_x=nx(x+y)^{n-1}$, and so $$(xD_x)^2 (x+y)^n=nx(x+y)^{n-1}+n(n-1)x^2(x+y)^{n-2}$$
Correct?
I'm sorry, what is xDx?
oh
Differentiate w.r.t to x, multiply by x.
23:09
Yes.
Well, now plug in $y=1-x$ in both sides.
What do you get?
nx + n(n-1)x^2
Right, $$\sum\limits_{k = 0}^n \binom nk {k^2}{x^k}{y^{n - k}} = nx + n(n - 1){x^2}$$
Oh... Alright.
Should I do the same thing for the other terms?
Right, the other terms are easier.
23:12
Thanks so much Pedro.
Note I just followed the hint in your book. =)
I just see it though.
Yeah I was taking the derivative of the sum
:/
hey how do you guys pronounce LaTeX
$\LaTeX$
can you write out how you would say it phonetically
23:19
I have never checked how to pronounce it as I've never referred to it that way
It's impossible in english I think.
With that said, I read it as "Lay Tech's" haha
me 2, though when I say it around people not familiar with that sort of thing they think I'm talking about the synthetic rubber
and I sound like a weirdo
It's mathjax here anyway
would you say Lay Tech?
23:20
@Ethan "Lay Tek"
yea thats what I was guessing
Btw, it's 4:20 in the part of the USA that has legal weed
blaze it
So is it "Math Jak"?
23:32
If I have $|\Sigma(a(b-c))|$ how can I pull the absolute value into the sum?
@Anthony, triangle inequality is one way.
Let $(X,||⋅||)$ be a norm space over $ℝ$, and $S = \{x_i | i ∈ ℕ\}$ be a countable set, where $span_ℝ(S)$ is dense in $X$. I need to prove that $X$ is separable. The only way I can come up with is to show that $span_ℚ(S)$ is countable and that it is dense in $span_ℝ(S)$ using Cauchy sequence convergence. Does anyone have a better way?
There is no code for span.
Do \operatorname{span}.
That's the standard argument, @GregRos.
23:38
Hmm. Alright. I was kind of scared I was missing something because I had no idea we're supposed to know about that sort of construction in this course. Thanks.
Sigh, answer a question, and nobody even looks at it :(
@DanielFischer What happened?
@PedroTamaroff Well, nothing. Nobody viewed it after I answered.
@PedroTamaroff Did you just serially upvote all my recent answers?
@DanielFischer Just the ones I deemed worthy of an upvote.
It wasn't serial upvoting!
Let's see what the script says.
Ram
Ram
23:45
Hi All. I have question.
Let A(X) affine coordinate ring. And m_p is a maximal ideal at p \in X i.e., { f/g st. g(p) is non zero; f,g are \in A(x)}. As m_p is maximal. We have A(X)/m_p is field. (say k the affine base field here). And we have evaluation maps which map A(X)/m_p to k. How to prove this is isomorphism ?
@DanielFischer LOL. I was looking at localization of modules, and I said... hmm this looks like tensor products. Next page I see $S^{-1}M\simeq S^{-1}R\otimes_R M$. Happy Peter is happy.
You're getting prophetic, @Pedro.
@DanielFischer Well, I guess I am developing so new "inuition", which is cool.
And useful.
I think a good deal of algebra is understanding beyond the make believe constructions.
That is, what things behave like, not just what they are concretely.
Am I making sense?
23:50
Yes, you are.
What Pedro said is too deep for me, lol.
@DanielFischer I hope you didn't think I was mindlessly voting you, Daniel. I upvote a lot.
Of course, when I deem things worth upvoting.
@PedroTamaroff Hmm, you should upvote me instead, lol.
Everytime someone asks for a book recommendation, the old ones are listed. Not many are aware of the new ones.
@PedroTamaroff No, I assume you don't vote without reading. It was just a rather fast succession.
@DanielFischer I am a fast reader =D
23:58
I like to upvote my friends now and then, sorry.
@JasperLoy Perhaps the old ones are simply better than the new ones? (Anyway, people know them better.)
I better not discuss votes too much, in case people complain again that I am cheating, lol.
There is a user whom I shall not name who posts on meta that I and another user have discussions about mutual upvoting.
And then he says that we conclude that that max number of votes is what we will take advantage of so that system does not revoke them.

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