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21:01
Can someone help with rotating some vectors in space?
@PaulManta Do you know what the rotation matrix is?
@PeterTamaroff In principle.
It is $[\sin \phi, \cos \phi][-\cos \phi, \sin \phi]$ if I'm not wrong, for two dimesions.
A matrix that, when multiplied by a vector, gives a transformed vector.
@Eugene have you read the synopsis on IMDB?
21:03
@PaulManta Well, yes. But it only rotates it.
You can probably work it yourself.
@robjohn yes. i have read a little bit of the plot as well on wikipedia but that didn't seem very compelling. yet this movie is raved about by most people who have seen it
@PeterTamaroff $\begin{bmatrix}\sin(\phi)&\cos(\phi)\\-\cos(\phi)&\sin(\phi)\end{bmatrix}$
@robjohn Thanks!
I love the Feynman problem solving algorithm.
@N3buchadnezzar your question has capped me for today :-)
21:06
@robjohn =)
@JonasTeuwen what is that?
@robjohn hahah. that's the how many-th day in a row now?
@robjohn:
*The Feynman problem solving algorithm:*
1. Write down problem;
2. Think very hard;
3. Write down solution.
@PeterTamaroff but I usually swap the sin and cos so that a rotation of 0 gives the identity
@JonasTeuwen Ooh, that is good :-)
@robjohn If you are Gauss you can skip step 2. 8-).
21:07
@JonasTeuwen there really is no better way
@robjohn Oh, that was my slip, it is cos sin -sin cos
@PeterTamaroff $\begin{bmatrix}\cos(\phi)&\sin(\phi)\\-\sin(\phi)&\cos(\phi)\end{bmatrix}$
@robjohn I really need to learn that code.
jordan ellenberg's answer is still the best
21:09
That's the one step in problem-solving/checking that I really wish could be drilled home better somehow - 'test on specific data'. Does this equation make sense when $x=0$, or $x=1$, or as $x\rightarrow\infty$, etc etc...
"+50 Slowly and with difficulty, just like amateur mathematicians."
@Eugene mathematicians learn old things and discover new things :-)
@Eugene Who is he?
@robjohn also true.
@PeterTamaroff a professor at wisconsin
@Eugene Where Rudin is, if I remember correctly
21:11
Super good number theorist.
Rudin is dead :(
@DylanMoreland Ack!
@DylanMoreland What?
JSE is a cool dude. Whenever I see him we talk about Pavement.
@robjohn rudin is dead. the math club went to visit him in the hospital before he died
@robjohn A couple of years ago, I think.
21:11
@Eugene I missed that. It is very sad.
@robjohn yes it was. EVERYONE wanted to visit him but it was forbidden since we didn't want to disturb him with too many visitors
he still has an office on the 6th floor (if i'm not mistaken) with his wife.
@Eugene You knew the author of Principles of math. Analysis, or was he another guy?
@PeterTamaroff well i didn't know him personally if that's what you meant
@Eugene Oh, OK.
@robjohn i just noticed that they brought a pirated copy of baby rudin!!!
@DylanMoreland yeah. JSE is a very nice guy. i don't know who pavement is though. recently they brought melanie wood into the department. she's a superstar.
21:21
@Eugene You guys know a lot of famous math people!
@PeterTamaroff well i know them. whether THEY know me is an entirely different matter.
Melanie graduated from Duke University where she won a Gates Cambridge Scholarship, Fulbright fellowship, and a National Science Foundation graduate fellowship in 2003, in addition to becoming the first American woman and second woman overall to be named a Putnam Fellow in 2002.[5][6][7] During the 2003–2004 year she studied at Cambridge University.

In 2004, she won the Morgan Prize for work in two topics, Belyi-extending maps and P-orderings, making her the first woman to win this award.[3][7] Her paper on the second topic was published in the Journal of Number Theory.
@Eugene That is so cool! Are you in the picture?
@robjohn no. i didn't make the cut... =(
@Eugene what, you had to prove yourself to be able to go?
21:27
@robjohn you had to be in the upper echelons of the math club ie committee member
@Eugene I assume that is his wife behind him.
@robjohn yup. that's mary ellen rudin. she is a topologist
What's the deal with de Brange's proof of the GRH?
@PeterTamaroff Does he have one? If it isn't reccent, then there must be something wrong with it.
@robjohn 2004 it says. Was it withdrawn?
See here the second parahraph under "Work".
21:31
@PeterTamaroff An Apology
@PeterTamaroff brian conrey proved that his approach cannot work
@robjohn Isn't that a defense in favour of his work? (As opposed to apologizing)
@PeterTamaroff Louis de Branges (1992) showed that the Riemann hypothesis would follow from a positivity condition on a certain Hilbert space of entire functions. However Conrey & Li (2000) showed that the necessary positivity conditions are not satisfied.
@Eugene I see.
Li is the same one that attempted a proof but had a mistake, right?
In arXiv
i don't know
yes it's the same one apparently
mistake on page 29
@robjohn lol
@PeterTamaroff well, that's what it says.
@robjohn yes, i know
21:42
@PeterTamaroff it was only meant humorously.
@robjohn That Katznelson book is pretty kickass!
somehow i am not surprise that wikipedia has this
@Eugene It cites Carleson's theorem so it must be good!
@JonasTeuwen sorry i've never heard of it
@robjohn Okay. That theorem is cool, but I would really like it only to diverge on $E$.
So, this requires us to understand how badly Carleson's theorem can fail 8-).
So, that probably means I have to dive into Lacey-Thiele-(Hytönen) 8-).
22:01
parametric functions confuse me, I want to find a graph where f and g are both in terms of x?
this is like inception
I can't figure out how to manipulate $\frac{dy}{dt} = \frac{dy}{dx}*\frac{dx}{dt}$ to solve for $\frac{dy}{dx}$
It is so incredibly hard to work with an x and a t variable, they all look the same
@JonasTeuwen Go to it! :-)
anyone know how to manipulate that? My book is doing some magic with no explanation
@robjohn But... If somebody already solved it that would suck :-(.
@Jordan you mean like dividing both sides of the equation by $\frac{\mathrm{d}x}{\mathrm{d}t}$?
They all look the same? Mathematically speaking they are the same, except that usually you restrict $t$ to the positive real line.
@robjohn I have too many ideas of problems to solve!
22:11
@JonasTeuwen dude. that's life. i've been scooped twice already
@JonasTeuwen That's good! I never had enough.
My advisor had to calm me down a bit today. 8-). He softly suggested I should look at special cases first.
Not to general $C_0$-semigroups with kernels.
@robjohn I am not sure but the book gets $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
Hmm, I have some ideas how to solve the problem about UMD spaces being interpolation spaces of some UMD space and some Hilbert space. It will probably fail, but, still...
2 mins ago, by robjohn
@Jordan you mean like dividing both sides of the equation by $\frac{\mathrm{d}x}{\mathrm{d}t}$?
@JonasTeuwen It is good to devise problems. Really.
22:13
@robjohn I am not sure but the book gets $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
@BenjaminLim looks like the salad wore off.
how many times can we repeat this?
probably a lot of times
@robjohn :-). Yes. But at least, I think my attempt will learn me more about tensor products of such weird things...
@JonasTeuwen and that is a wonderful thing!
22:15
Yes.
I spent all my money putting out contracts on tensor indices.
I think the book got the formula wrong, it doesn't match
@Jordan Yes it does!
no it doesn't, I just tried it out and it doesn't
@Jordan eh?
22:19
It should be dy/dx * dx/dt
they have it wrong
I just wrote a proof without words, and of course it seems to be taking off. Unfortunately, I capped a while ago :-(
@Jordan that is $\frac{\mathrm{d}y}{\mathrm{d}t}$
not according to the book
@robjohn anon has a trick of deleting it and undeleting it after the cap is over
the book gets dy/dt / dx/dt
which is dy* dt / dx dt
@Jordan that's $\frac{\mathrm{d}y}{\mathrm{d}x}$
22:21
I don't see how it is
@Eugene that works?! I'll have to try it :-)
@Eugene I need a second look at something.
@robjohn it doesn't give you back the points you got but it prevents you from losing more.
@Eugene Oh, you mean deleting some of the answers that have been upvoted (with points) and then undeleting them later.
@robjohn Yeah!
22:23
@robjohn yes
I am applying the Feynman problem solving algorithm to the problem I posed.
I'm already at the second step! 8-).
@robjohn so right now you've capped but in 2 hours you'll have uncapped. undelete it later and you'll get all the upvote points that people give you from that point on.
I am still getting dy/dt * dt/dx pretty vertain the book has to be wrong
actually I think I might have got it, it is a really strange way they chose to do is because there are so many other ways to represent this
not sure why they chose this one way
@Eugene I should delete this answer that would free up 80 points :-)
@robjohn oh now. i don't think it reduces your rep points. or does it? i'm not sure.
22:31
Of course then I would end the day under 200.
actually it just might
@Eugene If the OP accepts it, I will at least get 15 points :-)
I am not a fan of introducing integration through positive and negative parts... I do not really follow why they should be treated separately other than technical difficulties.
@robjohn yes. actually that might work. i've deleted answers before and lost the rep. i wonder if you get it back when you undelete it
@Eugene You do.
22:34
So why not start by defining linear functionals on the the space of compactly supported continuous functions on some locally compact space? 8-).
I don't understand why a square root that is squared can be + or -, it has to be positive
@robjohn we found a way to game the rep system then. =)
@Eugene but it feels like cheating... I may not do it.
@robjohn what i've tried is deleting an answer, then undeleting it when the rep cap ends so i can just get the rep points from the subsequent upvotes.
@robjohn it's probably not cheating because if you think about it, it affects your next day earnings.
This is wrong
22:36
@Eugene yeah, but I would get 50 more points tomorrow that I wouldn't have gotten. and I could keep paying forward until a dry day comes.
@Jordan It is the square root of a square that can be positive or negative. Say $\sqrt{x^2} = x$.
If I replace the $x$ on the left-hand side with $-x$ and the same with the right hand side...
@JonasTeuwen $\sqrt{x^2}=|x|$
@robjohn that is true. but you never really have dry days though do you? =)
@robjohn Yes, that is what I am trying to explain!
@Eugene Oh, most of them are. Just look at my rep record.
@JonasTeuwen sorry, I'll be quiet :-x
22:38
@robjohn You do know that I know that, right? 8-).
@robjohn i see some in may. but since june i'd hardly call 100+ a dry day.
@JonasTeuwen Of course.
I don't understand how you can evaluate a negative square root
@JonasTeuwen but we all make typos
22:39
@robjohn Phew 8-).
I have $\sqrt{3}^2$
he earned double the rep cap in one day? how?
@Eugene Lately, he has rarely gotten below 300 a day
@robjohn today might be that day though.
I have negative days.
22:40
I have minus days.
ah that fateful day when the user was deleted
@skullpatrol That is pretty scary dude.
@robjohn how does he have 450 a day? i thought it caps out at 200
@Eugene I lost 55 points that day. Ooh, maybe that will happen again today. I have a 50 point buffer :-)
22:42
@robjohn lol!
math be wily
@Eugene I would have 320 today, if it weren't for the cap.
@JonasTeuwen I'm a pretty scary dude.
@robjohn that's the thing. how does brian still get his?
@skullpatrol Cool. Do you have a picture?
My voice has not fully recovered from the harmonic analysis conference yet!
22:43
@Eugene he gets a lot of acceptances and bounties, I guess.
I don't understand parametric equations and the formula for finding tangents and such, the variables dy and dx are made up and can be anything, so how do I know which fucntion goes on top and which on the bottom?
@robjohn brian scott the bounty hunter?
@Jordan Uh, just think about what happens if you do one of the two? Then try the other if it does not make sense.
I did, seems the same to me
@Eugene well, he has a lot of acceptances.
22:45
@robjohn that is also true. he's a night owl though it seems. i don't see much activity from him in the afternoon
I mean if I swap the x and y then the horizontal tangent will be different
@Jordan Yes.
so which goes on top?
there is no way to determine that
@robjohn you were right. he has to sleep sometime.
@Jordan Yes there is. The derivative is $\frac{dy}{dx}$
Not $\frac{dx}{dy}$
22:50
for parametric equations
I guess I see it now, it depends on dy/dx
@PeterTamaroff what's a derivative?
@PeterTamaroff what's a derivative?
@Eugene Why are you asking that?
@PeterTamaroff because i don't know
22:53
@Eugene The derivative of $f$ ,is a function $f'$, such that $$f(x+h)=f(x)+hf'(x)+o(h)$$ for $h \to 0$
@PeterTamaroff ?
$f(a) = o(g(a))$ when $a \to 0 $ means that $f/g \to 0$
In this case you have
@PeterTamaroff this is too hard for me
@Eugene This is the bibliography of my Univ's Number Theory course
Kenneth Ireland, Michael Rosen, A Classical Introduction to Modern Number Theory, GTM Springer, 1982.
Anthony W. Knapp, Elliptic Curves, Princeton University Press, 1992.
Daniel A. Marcus, Number Fields, Springer, 1977.
Joseph H. Silverman, John Tate, Rational Points on Elliptic Curves, UTM Springer, 1992.
Ian Stewart, David Tall, Algebraic Number Theory, Chapman & Hall, 1987.
@PeterTamaroff what is all this?
22:58
@Eugene The bibliography of my Univ's Number Theory course.
@PeterTamaroff what is number theory?
@Eugene It's not funny anymore, dude!
@PeterTamaroff =D
@Eugene Do you know any of those books?
with the exception of knapp those are all good books
i've never read marcus
23:00
@Eugene It seems it doesn't focus on Analytic number theory, but rather Algebraic NT
which is cool
@PeterTamaroff of course i'm still more of the opinion for you to take your time in your learning
@Eugene Oh. Well, yes you're right. I think I'll ease off on NT and focus on Linear Algebra now, since it is required to have cleared Algebra I and Algebra II for NT
@PeterTamaroff linear algebra is good. this was what i used
I got it in a DJVU
I'll hit it on the vacations.
@PeterTamaroff linear algebra then rings fields modules and groups and you should be good to go
oh
and complex analysis
23:06
@Eugene OK.
Though it is not needed for Algebraic NT right? (It does not appear n the course info)
@PeterTamaroff complex analysis? not really. but to understand cyclotomic fields you do need roots of unity.
@Eugene But that isn't that hard!
@PeterTamaroff no it's not.
@PeterTamaroff but CA is very useful to have in number theory.
@Eugene Right.
@PeterTamaroff and now i must away. bye
23:10
@Eugene Bye.
what is a parameter?
like a variable?
I have no idea what this question is asking really
@Jordan A "parameter" is what they call a "dummy" variable.
so t?
is $d^2 y $ just a fancy way to say the second derivative?
23:25
sort of...
there are so many things not once explained in this book, incredibly frustrating
I am suppose to find $\frac{d^2 y}{dx^2}$
Partly as integration drill and partly to make a point about the use of "dummy variables", I'd call on several students, one after another, and demand that they tell me what is ∫dx/x, ∫du/u, ∫dz/z, ∫da/a, and then, as the clincher, I'd ask about ∫d(cabin)/cabin. Some of them would grin amiably and shout out "log cabin", and they were surprised when I told them that I didn't agree. The right answer (as I learned when I was learning calculus) is "house-boat", "log cabin plus sea".
strange
@ZhenLin hey
@skullpatrol hey
23:32
@ZhenLin been looking at the quotient topology
Can't really visualise what's going on
@BenjaminLim Yo
for example with $X$ as the unit circle
you say every point in the interior is equivalent to itself
and the whole boundary is equivalent to itself
what is the equivalence relation that we are quotiening out by?
@skullpatrol watch this: youtube.com/watch?v=OyQVjGdJ60g
I get some pretty strange answers
trying to decipher ESL answers is hard enough with mathese thrown in
Sometimes it just feels pointless asking questions here, I ask something simple and I get an answer that requires 45 minutes of learning to comprehend when all I wanted was to do what the book wanted me to do
23:56
Then stay focused on "what the book wants you to do."

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