8:00 AM
can anyone explain the geometric intuition behind the fact that y-x^3=0 has a singularity at its point at infinity?
dfdafdsafas

@Harry_Potter's_Erect_Nipples, I'm not following what you mean.
You have a cubic curve $y=x^3$. What does 'point at infinity' mean?

in the projective plane P^2(R)

@Harry_Potter's_Erect_Nipples, that makes much more sense. Hm. One second.

it vexes me

@Harry_Potter's_Erect_Nipples, it is vexing me too. I only know a light bit of projective geometry, and this came from the study of homogeneous coordinates. i.e. $(x,y,z): (cx,cy,cz)=(x,y,z)$.

8:09 AM
I was thinking posting the question but I can't decide if it's sophomoric or not

I post quite a few sophomoric questions. I would not feel ashamed, @Harry_Potter's_Erect_Nipples.

ok then I'll post it
dans le zone euro il n'y a pas de convergence economique

Interesting. $\int_{-1}^{1}\sin^{-1}(x)dx=0$ by inspection.

so if you write down the projective equation
it's yz^2 - x^3 = 0
When z = 0 (i.e. at infinity), x = 0
so the point at infinity is [0,1,0]
Now let's localize to the x-z plane: which means we set y = 1
and we look at the point (0,0) in x-z plane, the equation is now z^2 - x^3 = 0. This should look singular using what definition you have in mind.
@Ha
@Harry_Potter's_Erect_Nipples
5

2 hours later…
9:56 AM
@OldJohn wow! nice picture, how did you like the documentary?

@aDangerousIdea Thanks - I watched about half of the documentary, and hope to finish it today
It is good

10:20 AM
which documentary is being talked about?
@OldJohn Awesomeful photograph.

1 hour later…
11:27 AM

11:39 AM

Interestingly, if you search my full real name, you find no images of me at all

Hmm.. strange I can't reproduce the result...

me neither
@WillHunting I see you capped again today

user19161
@OldJohn I capped today, long ago!

@WillHunting I noticed a couple of hours ago - you have been busy!

user19161
11:45 AM
@OldJohn Rather, lucky!

I am only here briefly - taking a break from ceiling painting

user19161
Hey @anon!

back to the ceiling ...

12:02 PM
@OldJohn Here is what I found :)

2 hours later…
1:39 PM
@robjohn Here is a visually similar one of your gravatar :)

https://encrypted-tbn2.gstatic.com/images?q=tbn:ANd9GcSxkBkHKqfgZ7gHw-1IqlllGbg6wlUb5PXXoIx8jvVoofw7KNX5MA

user19161
@amwhy Congrats on getting 10k! Yay!

@WillHunting I was surprised this morning, first logging in a little bit ago! Thanks!
@WillHunting Congrats on exceeding 200!

user19161
@amWhy Now you can see deleted posts with your X-ray eyes!

@WillHunting Hmmm, may pose more of a distraction than a bonus!

user19161
@amWhy Did you see me in your dreams last night?

1:49 PM
@WillHunting Yup...I waved, and you waved back! ;-)

user19161
@amWhy Hahaha, OK. Let's have some ice cream in the next dream. =)

@WillHunting Yummm, what flavor?

user19161
@amWhy Hmm, I like vanilla flavour, just like the great anon.

banana?

user19161

1:51 PM
@aDangerousIdea hmmm, I think not! How about butter pecan, or chocolate malt?

Those are good too :)

user19161
@amWhy makes note

@WillHunting oop...I mistakingly clicked on the "moving target"!

user19161

user19161
1:52 PM
@amWhy OIC.

@aDangerousIdea I like that: "OIC"...I had to sound out the letters, and "aha!"

user19161
@amWhy I think you clicked on the wrong line again!

That's OK I don't mind :)

@WillHunting Whoa...I think I need to fully waken up. Perhaps you can enlighten me: "lol" = ?: lot's of laughs? laughing out loud?

user19161
@amWhy I think you are better than me at these things. It should be the latter.

1:55 PM
@aDangerousIdea Thanks for being a sport. =)

user19161
@amWhy BBL=Be Back Later, BRB=Be Right Back, AFK=Away From Keyboard, CU=See You, NP=No Problem

@WillHunting Well, I just learned "OIC"....
:7096690 np: that one I do know!

(removed) = removed

$\color{grey} (removed)$

1:59 PM
And I know btw, BTW!

user19161
@amWhy You should go get some coffee now!

I give up

$\color{grey} {\text{(removed)}}$
yeah, this is good now.
:P

IGU2

2:05 PM
@WillHunting Yes, coffee time!

Thanks, IGU, I should make a "cheat sheet" and study up!

plug your ears sonic boom is here :)

I'm blown away out of the chat. See you guys later.

later

2:09 PM
If we have the the integral of the function f(x) = 1/x over the open interval (0, 1) can we say that as f(x) -> infinity as x -> 0, the integral will be undefined?

2:32 PM
evidence that did knows Pink Floyd

Hi guys
Im going to install mint on my computer and give it my whole drive
But how can I make a backup of windows?
In case I want to install it again?

@Khromonkey dd :)

what?

boot from a live cd, and use dd to copy the disk volume to a normal file

I dont have a live cd

2:43 PM
dunno then

my computer came with windows 7

less badass than dd, but will work nicely too :)

thanks

Hweh
WOT
NEw people in chat!

3:08 PM
@aDangerousIdea But it has a nose! ;-)

3:20 PM
math.stackexchange.com/questions/249163/… - i gotta talk to people more :(

3:55 PM
anyone know polar coords and ellipses

4:12 PM
@robjohn How is it going?

Question, does anyone know where I could find an english copy of the Rudolphine Tables or how to intepret the Latin version?

Huh, what is it?

Uh, tables made by Kepler using Brahe's data - they note planatery and stellar positions and allow one to predict (fairly accurately) the position of the bodies in the future.
In essence, I'm trying to use newton's classical laws to show that his and kepler's laws are equivalent, but I'm having difficulty finding a starting point.
the table is in the following link, http://docs.lib.noaa.gov/rescue/Rarebook_treasures/QB41K431627.PDF
if you scroll down to around 2/5 from the bottom you'll see tables.

Hi all

4:21 PM
good luck with this!
hi

Hi

@KaliMa most people here, probably - feel free to ask

I find myself often writing Bourbaki-esque things like this:

Below, "mapping" will mean "continuous mapping" unless explicitly stated otherwise. "Embedding" will mean "regular monomorphism", and "quotient mapping" will mean "regular epimorphism"; in other words, these terms will be used a bit more generally than they are usually used, in order to do as little evil as possible.
is this a common practice?

i have an ellipse of form x^2+4y^2=4a^2 and I am trying to find an equation for a rotated ellipse of angle theta, and the solutions for the intersection between the rotation and the original such that the distances from the origin to intersections are integral

@KaliMa are you sure that that last part is solvable in radicals?

4:25 PM
what?

@KaliMa there is a standard matrix for rotating the axes by an angle $\theta$

i've been googling like mad to find some framework or appoach that'll let me do this
i tried using newX = x cos(theta) - y sin(theta), newY = y cos(theta) + x sin(theta) but it got messy

@AlexeiAverchenko, I've read that intersecting conics (particularly ellipses) result in quartic equations, which are solvable in radicals, if that's what you are both talking about.

how do you express integer points on an ellipse?

@KaliMa pretty sure that is the way to do it

4:26 PM
@user1230219 sure, but what about integer points on their intersection?

@AlexeiAverchenko, I guess you look at the discriminant of the resulting quartic?
A not so elegant solution (and probably not effective) I guess.

I think you might have to bite the bullet and do the algebra here

can we enumerate pairs of ellipses such that their intersections are integer in a sane manner?

the points dont have to be integral
just the distances to those points

oh
looks like it's going to come down to pythagorean triples

4:29 PM
this isn't a solution but i am saying let $b$ be the shortest distance to intersection and $c$ be the distance to the other
$b$ and $c$ must be integers
(b and c will also be separated by 90 degrees)

you got two axes of symmetry here
so it's going to boil down to two cases
can you say something about the sum of these distances, or their squares?

@KaliMa shortest distance to intersection from where exactly? and how can two integers be separated by 90 degrees?

i'm not sure
@OldJohn The distance from origin to closest intersection = b, the other one = c

@KaliMa that's better

consider the rhombus defined by your four points
i mean, consider a convex polygon with these points as vertices
and prove that it is indeed a rhombus

4:36 PM
how do we know that the two vectors representing $b$ and $c$ are definitely at right angles?

then you'll know the axes of symmetry are orthogonal (being the diagonals of that rhombus)
@OldJohn ^^^^^^^^^^^

right - so we don't know yet

@N3buchadnezzar Pretty well. Just got back from the park.

you'll only need symmetry to prove that

@robjohn Any tips for headache/ from overworking?

4:40 PM
I'm pretty sure angle BOC will always be 90 degrees

prove that the symmetry group acts transitively on the sides of the polygon, and that the opposite angles are equal
@KaliMa but did you check?

i just tried a bunch of examples

come on, i just basically laid out the proof for you
do some work ffs

no need to be mean
i just told you i tried a bunch of examples

but what if you missed the one counterexample?

4:42 PM
hello there

because i am pretty sure it doesnt exist

how can you be certain that you didn't miss it unless you actually prove that it doesn't exist
?

because i'm willing to take the risk

@KaliMa there's no "pretty sure" in mathematics

i dont possibly see how a counterexample can exist

4:43 PM
that doesnt mean it doesnt :)

it would require really wonky workarounds

i don't care what you can or cannot see, prove it rigorously

i can't "prove it" rigorously because i am not familiar with formal proofing
i just know it by exhaustion

@N3buchadnezzar When I get a bad headache, I take 2 Advil (ibuprofen) 2 Tylenol (acetaminophen), and 2 Sudafed (pseudoephedrine) (decongestant). Within an hour or two things are much better.

@KaliMa yes you do, if you ever studied geometry

4:44 PM
Exhaustion is a formal method - but have you "exhausted" al possibilities @KaliMa

trying a bunch of integral and irrational / rational rotations, 1000000 examples by simulation, they all had 90 degree separation

@N3buchadnezzar Otherwise, I just tough it out.

@robjohn Yeah, I do not have any of those

@KaliMa what if you missed the one that has a different angle?

because i didnt

4:45 PM
I sort of really need to work too, exam on the 5th..

@KaliMa you can't know that unless you prove it

then please show me the one i missed, because i think it's pretty clear it doesn't exist

It's not called exhaustion just because you've done 100000 examples.It's exhaustion when you've shown that all cases don't have that certain thing.

Just really worn out from the exam yesterday, it was really tough. And made me somewhat depressed-

4:45 PM
@KaliMa there's no "pretty clear" in mathematics
mathematics is a precise science

@KaliMa, have you studied geometry?

it either is or it isn't

@user123021 In high school, but we never did anything as complex as this
@AlexeiAverchenko So please show me the counterexample

@MSEoris Complex anlysis, laplace, pde and fourier transforms.

4:46 PM
@KaliMa i can prove it doesn't exist
but you can't

and that's the problem

consider two ellipses on top of each other

thats a very wide range of topics :)

let's focus on the top/down/left/right points of intersection
i rotate one by theta

4:47 PM
@robjohn We got this integral on our exam $$\int_0^{2\pi} \frac{\cos^2\theta}{2 + \sin \theta}\,\mathrm{d}\theta$$

the distance of intersection a point travels is oppositely offset by the point's movement on the other side of the ellipse
left vs right, top vs down
therefore COB will always be 90 degrees
however you can translate that "formally" is what i mean when i say "i am pretty sure." just because i can't write it down doesn't mean i can't understand what's going on

@KaliMa wtf is that even supposed to mean?

@AlexeiAverchenko It means in a default ellipse, the extremities are orthogonal by definition. If we rotate one, we know that their intersections with the original will be exactly offset by the other sides of the ellipse, and therefore retain their orthogonal nature
so we know it's 90 degrees

@KaliMa sorry - I don't see that as a proof at all

4:49 PM
@OldJohn Why not?

@robjohn Nooo... I turned it into a complex integral by using $z = e^{i\theta}$. Then I found the poles. But was not able to find the residues, as the whole thing got really messy. Somewhat evil to give on an introductory exam...

@KaliMa, try using congruent triangles.

@KaliMa ok, that's a nice one, but it requires some work

anyways i am not working on this to prove that COB is 90 degres, i'm trying to find integral distances for C and B

@KaliMa you have given no mathematical reason why the movements of the intersections should be "exactly offset" - whatever that means

4:51 PM
yep, this argument is correct

There is always the Weierstrass substitution as well, but your class is about contour integration?

@OldJohn Because of symmetry, how on earth can one side move more than the other?

Indeed

if you can prove that the lines passing through your points and the origin are in fact symmetry axes of the figure
which is really trivial to prove

@KaliMa are we talking about 2 points which (at least appear to be) rotated by 90 degrees?

4:55 PM
@OldJohn tinyurl.com/ccksfoe I am saying given an ellipse x^2+4y^2=4a^2 for a given $a$, a copy of the ellipse can be rotated by 0<theta<90. now let $b$ be the dist to shortest intersection, $c$ be the other. I am looking for any integral $b$ and $c$ given $a$

@KaliMa It is quite possible no such integral $b$ and $c$ exist

right

Did you get $$\frac{2i}{4}\oint\frac{(u+1/u)^2}{4i+(u-1/u)}\frac{\mathrm{d}u}{iu}$$

i am trying to generate unique values of valid $a,b,c$

@OldJohn $b^2 + c^2$ is the square of the length of horde connecting the two points
do you remember a length of such a horde computable?

4:58 PM
@N3buchadnezzar That's using $u=e^{i\theta}$

good evening

@AlexeiAverchenko definitely computable - do you mean - can I remember the formula?

@Ilya Hey there!

@OldJohn well, i remember that the arc length is not computable, so I wondered about the horde

@Ilya It is morning here, and a rainy one, at that.

5:00 PM
@AlexeiAverchenko the arc length is computable - just that the integral cannot be expressed in terms of elementary functions - it is computable in terms of elliptic functions

@robjohn pity

@OldJohn well, no turing machine can compute it in the finite time
that's what I meant

can a turing machine compute $\sqrt{2}$ in finite time?

@robjohn Yes

@Ilya Oh, but I like rain. We get way to little here, and I like the chance to build a fire and sip tea, chai, or hot chocolate in front of the fire during the rain.

5:02 PM
What I'm really trying to say is that heuristically it shouldn't be likely that we can enumerate all angles that correspond to integer distatnces from these points to the origin, because it's related to computing these arcs

@robjohn in that case rain is also appreciated. I loved it much more back in Volgograd - there you can also smell that it rains, the air smells different - very freshly. Here though because of the constant humidity rain does not smell - and that I miss a lot.

it also seems to be the case that 5b^2*c^2 = 4a^2(b^2 + c^2)

@OldJohn no, irrationals cannot be represented in general in a turing machine
although $\mathbb{Q}(\sqrt{2})$ certainly can be
@KaliMa well, this makes it easier
if $b$ and $c$ are both integer, then it is clear that $a^2$ has to be integer, so $a$ too has to be integer
oops

yes, a, b, and c are integers

$\frac{4}{5} a^2$ has to be integer
wait...
right
and $a$ also has to be integer
i'm not sure if i'm right here

5:06 PM
i am wondering if this is something i can rewrite and solve as a pell equation

doesn't make a lot of sense to mee

perhaps reparameter with two variables

oh, so $a$ has to be divisible by 5
that's interesting

@Ilya yes, you can smell rain coming here. It is very noticeable.

a doesn't have to be divisible by 5

5:07 PM
$$\frac12\oint\frac{(u^2+1)^2}{u^2+4iu-1}\frac{\mathrm{d}u}{u^2}$$

valid a,b,c=209,247,286

if it's not divisible by 5, then $\frac{4}{5} a^2$ and $a^2$ cannot be simultaneously integer

or 341,374,527

@KaliMa how do you know?

using a program to test brute force

5:08 PM
@N3buchadnezzar The roots of the denominator are $(-2\pm\sqrt{3})i$

@robjohn or especially the first rain in the end of August, after 2 months of drought

but trying to find a way to generate solutions

cause if so, then it's not exactly reliable :)

i dont need to
i just calculate each side, check if equivalent

@KaliMa here is a method which might help:

5:09 PM
ok, then i'm confused
what are $b$ and $c$?

b is the shortest integral distance from origin to the intersection of an ellipse and its rotation

if $x^2+4y^2=4a^2$ with $a$ an integer, then x must be even

c is the distance to the other intersection

@Ilya Only two months? We don't have rain from the middle of spring to the end of autumn

@robjohn and 0

5:10 PM
(there are four intersections but they have symmetry)

@KaliMa and what did you calculate to check things, exactly?

@robjohn at all?

@N3buchadnezzar I don't see any $0$...
@Ilya not usually.

@robjohn Rewrite the expression...

@robjohn that's the feature of zeros :)
2

5:11 PM
so put $x=2z$, to get $z^2+y^2=a^2$

can anyone please just cruch groebner basis so that we could be done with it? :D

@robjohn You forget that the nominator is $(1+1/u)^2$

sorry - I am finding point with integer coords - not integer distances :(

yeah x and y can be anything
a,b,c are integers

@KaliMa well, if the equation you gave me is correct, $a$ has to be divisible by 5
that's just it

5:13 PM
@robjohn Also $\mathrm{d\theta} = \mathrm{d}u/iu$

either your computations are mistaken, or the equation you gave me

i just gave you clear examples that show it doesnt have to be divisible by 5 though

@KaliMa how did you compute them?
give me the formulas

7 mins ago, by robjohn
$$\frac12\oint\frac{(u^2+1)^2}{u^2+4iu-1}\frac{\mathrm{d}u}{u^2}$$

via the COB=90 degree rule

5:14 PM

how did you obtain the equations involving a, b, and c?
it's really hard for me to believe there is any equation that you can express in these terms
although i don't know that

@robjohn It boils down to $$\frac{1}{2}\cdot \oint \frac{(u^2+1)^2}{u^2(u^2+4iu-1)} \mathrm{d}u$$

ok, let me guess
we have two symmetries
and four points

@robjohn Indeed, that is where the 0 comes from.

so my guess is that you need two algebraic curves symmetric under both reflections to describe your points
that's two algebraic equations
what are those?

5:17 PM
h/o trying to verify i didnt make a dumb mistake

@robjohn So it has two singularities inside the unit circle..

@N3buchadnezzar Look at $$-\frac{1+2u^2+u^4}{1-4iu-u^2}=-(1+4iu+\dots)$$
@N3buchadnezzar that divided by $u^2$ has a residue at $0$ of $-4i$
The only other root in the unit circle is $(-2+\sqrt{3})i$ and it is a single root

@robjohn how did you get this?
@robjohn But finding it is still ugly =) At least for a 4 hour exam. This was one of 7 questions

@KaliMa actually i was wrong

@N3buchadnezzar expanded the numerator and divided by the part of the denominator other than $u^2$

5:27 PM
either $a^2$ or $b^2 + c^2$ must be divisible by 5

also either $b$ or $c$ has to be divisible by 2
@KaliMa both your examples fit these conditions, so you have good chances of being correct :)

@Argon Indeed it was le horrible

I am trying to determine if there's a way to generate valid solutions to 4a^2(b^2 + c^2) = 5b^2c^2
it isnt a normal pell i think

@N3buchadnezzar It is not a pretty answer, but there are short-cuts to finding the residues. I don't know if they were covered in your class.

5:34 PM
@robjohn Laurent, derivation is covered.

anyone know to to generate solutions to 4a^2(b^2 + c^2) = 5b^2c^2 ?

@N3buchadnezzar For the other root, I used $$\frac{(u^2+1)^2}{u^2(u+(2+\sqrt{3})i)}$$ evaluated at $u=(-2+\sqrt{3})i$
which gives $2i\sqrt{3}$

@robjohn Yeah, but you have to admit it is a bit of an evil integral to give at an exam yes?

@N3buchadnezzar It depends on what was gone over in class and what else was on the exam, but it is not a simple integral.

auh

5:44 PM

@robjohn Yes

@N3buchadnezzar Multiplying by $2\pi i$ and dividing by $2$, I get $\pi(4-2\sqrt{3})$
Mathematica gets $-2 (-2 + \sqrt{3}) \pi$
@N3buchadnezzar I would hate to have to do partial fractions on that.

Good job, guys

@N3buchadnezzar what is the title of the course?

5:56 PM
Hellø

@Argon I fixed the link =)

Hahaha

@Argon You heard about J, K integrals?

@N3buchadnezzar J, K integrals? What are they?