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Harry_Potter's_Erect_Nipples
Harry_Potter's_Erect_Nipples
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
 
Limitless
Limitless
@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?
 
Harry_Potter's_Erect_Nipples
Harry_Potter's_Erect_Nipples
in the projective plane P^2(R)
 
Limitless
Limitless
@Harry_Potter's_Erect_Nipples, that makes much more sense. Hm. One second.
 
Harry_Potter's_Erect_Nipples
Harry_Potter's_Erect_Nipples
it vexes me
 
Limitless
Limitless
@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. $(x,y,z): (cx,cy,cz)=(x,y,z)$.
 
Harry_Potter's_Erect_Nipples
Harry_Potter's_Erect_Nipples
8:09 AM
I was thinking posting the question but I can't decide if it's sophomoric or not
 
Limitless
Limitless
I post quite a few sophomoric questions. I would not feel ashamed, @Harry_Potter's_Erect_Nipples.
 
Harry_Potter's_Erect_Nipples
Harry_Potter's_Erect_Nipples
ok then I'll post it
dans le zone euro il n'y a pas de convergence economique
 
Limitless
Limitless
Interesting. $\int_{-1}^{1}\sin^{-1}(x)dx=0$ by inspection.
 
Sanchez
Sanchez
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…
a Dangerous Idea
a Dangerous Idea
9:56 AM
@OldJohn wow! nice picture, how did you like the documentary?
 
Old John
Old John
@aDangerousIdea Thanks - I watched about half of the documentary, and hope to finish it today
It is good
 
Jayesh Badwaik
Jayesh Badwaik
10:20 AM
which documentary is being talked about?
@OldJohn Awesomeful photograph.
 
Old John
Old John
 
 
1 hour later…
Old John
Old John
11:27 AM
someone mention google images?
 
a Dangerous Idea
a Dangerous Idea
11:39 AM
@OldJohn I found that google images found your avatar
 
Old John
Old John
Interestingly, if you search my full real name, you find no images of me at all
 
a Dangerous Idea
a Dangerous Idea
I used google image search.
Hmm.. strange I can't reproduce the result...
 
Old John
Old John
me neither
@WillHunting I see you capped again today
 
user19161
@OldJohn I capped today, long ago!
 
Old John
Old John
@WillHunting I noticed a couple of hours ago - you have been busy!
 
user19161
11:45 AM
@OldJohn Rather, lucky!
 
Old John
Old John
I am only here briefly - taking a break from ceiling painting
 
user19161
Hey @anon!
 
Old John
Old John
back to the ceiling ...
 
a Dangerous Idea
a Dangerous Idea
12:02 PM
@OldJohn Here is what I found :)

https://www.google.ca/search?hl=en&tbo=d&bih=685&biw=1280&tbs=simg:CAESEgmipX-uqjrNJCFBaw5LmO07_1A&tbm=isch&dur=4075#hl=en&tbo=d&tbs=simg:CAQSEgmipX-uqjrNJCFBaw5LmO07_1A&tbm=isch&iact=hc&vpx=4&vpy=164&dur=2869&hovh=188&hovw=268&tx=70&ty=273&sig=111063255842295461445&ei=CS-7UMeAOoz0iwLWo4HACQ&page=1&tbnh=141&tbnw=200&ved=1t:2220,r:0,s:0&fp=1&bpcl=39314241&biw=1280&bih=685&cad=b&sei=7EC7UOOSEIfMigLEoYDYBQ&bav=on.2,or.r_gc.r_pw.r_cp.r_qf.&sei=_UK7UPKhO-itiAKS-YCgAw
 
 
2 hours later…
a Dangerous Idea
a Dangerous Idea
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!
 
amWhy
amWhy
@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!
 
amWhy
amWhy
@WillHunting Hmmm, may pose more of a distraction than a bonus!
 
user19161
@amWhy Did you see me in your dreams last night?
 
amWhy
amWhy
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. =)
 
amWhy
amWhy
@WillHunting Yummm, what flavor?
 
user19161
@amWhy Hmm, I like vanilla flavour, just like the great anon.
 
a Dangerous Idea
a Dangerous Idea
banana?
 
user19161
@aDangerousIdea Hahaha!
 
amWhy
amWhy
1:51 PM
@aDangerousIdea hmmm, I think not! How about butter pecan, or chocolate malt?
 
a Dangerous Idea
a Dangerous Idea
Those are good too :)
 
user19161
@amWhy makes note
 
amWhy
amWhy
@WillHunting oop...I mistakingly clicked on the "moving target"!
 
user19161
@aDangerousIdea too
 
a Dangerous Idea
a Dangerous Idea
gn
 
user19161
1:52 PM
@amWhy OIC.
 
amWhy
amWhy
@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!
 
a Dangerous Idea
a Dangerous Idea
That's OK I don't mind :)
 
amWhy
amWhy
@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.
 
amWhy
amWhy
1:55 PM
@aDangerousIdea Thanks for being a sport. =)
 
a Dangerous Idea
a Dangerous Idea
np
 
user19161
@amWhy BBL=Be Back Later, BRB=Be Right Back, AFK=Away From Keyboard, CU=See You, NP=No Problem
 
amWhy
amWhy
@WillHunting Well, I just learned "OIC"....
:7096690 np: that one I do know!
 
a Dangerous Idea
a Dangerous Idea
(removed) = removed
 
Jayesh Badwaik
Jayesh Badwaik
$\color{grey} (removed)$
 
amWhy
amWhy
1:59 PM
And I know btw, BTW!
 
user19161
@amWhy You should go get some coffee now!
 
Jayesh Badwaik
Jayesh Badwaik
I give up
 
a Dangerous Idea
a Dangerous Idea
IGU
 
Jayesh Badwaik
Jayesh Badwaik
$ \color{grey} {\text{(removed)}} $
yeah, this is good now.
:P
 
a Dangerous Idea
a Dangerous Idea
IGU2
 
amWhy
amWhy
2:05 PM
@WillHunting Yes, coffee time!
 
a Dangerous Idea
a Dangerous Idea
ct!
 
amWhy
amWhy
Thanks, IGU, I should make a "cheat sheet" and study up!
 
a Dangerous Idea
a Dangerous Idea
plug your ears sonic boom is here :)
 
Jayesh Badwaik
Jayesh Badwaik
I'm blown away out of the chat. See you guys later.
 
a Dangerous Idea
a Dangerous Idea
later
 
sonicboom
sonicboom
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?
 
Ilya
Ilya
2:32 PM
evidence that did knows Pink Floyd
 
Khromonkey
Khromonkey
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?
 
Alexei Averchenko
Alexei Averchenko
@Khromonkey dd :)
 
Khromonkey
Khromonkey
what?
 
Alexei Averchenko
Alexei Averchenko
boot from a live cd, and use dd to copy the disk volume to a normal file
 
Khromonkey
Khromonkey
I dont have a live cd
 
Alexei Averchenko
Alexei Averchenko
2:43 PM
dunno then
 
Khromonkey
Khromonkey
my computer came with windows 7
 
Alexei Averchenko
Alexei Averchenko
less badass than dd, but will work nicely too :)
 
Khromonkey
Khromonkey
thanks
 
Alexei Averchenko
Alexei Averchenko
np
 
N3buchadnezzar
N3buchadnezzar
Hweh
WOT
NEw people in chat!
 
robjohn
robjohn
3:08 PM
@aDangerousIdea But it has a nose! ;-)
 
Alexei Averchenko
Alexei Averchenko
3:20 PM
math.stackexchange.com/questions/249163/… - i gotta talk to people more :(
 
KaliMa
KaliMa
3:55 PM
anyone know polar coords and ellipses
 
N3buchadnezzar
N3buchadnezzar
4:12 PM
@robjohn How is it going?
 
user1230219
user1230219
Question, does anyone know where I could find an english copy of the Rudolphine Tables or how to intepret the Latin version?
 
N3buchadnezzar
N3buchadnezzar
Huh, what is it?
 
user1230219
user1230219
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.
 
Alexei Averchenko
Alexei Averchenko
admirable :)
 
Old John
Old John
Hi all
 
Alexei Averchenko
Alexei Averchenko
4:21 PM
good luck with this!
hi
 
Argon
Argon
Hi
 
Old John
Old John
@KaliMa most people here, probably - feel free to ask
 
Alexei Averchenko
Alexei Averchenko
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?
 
KaliMa
KaliMa
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
 
Alexei Averchenko
Alexei Averchenko
@KaliMa are you sure that that last part is solvable in radicals?
 
KaliMa
KaliMa
4:25 PM
what?
 
Old John
Old John
@KaliMa there is a standard matrix for rotating the axes by an angle $\theta$
 
KaliMa
KaliMa
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
 
user1230219
user1230219
@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.
 
Alexei Averchenko
Alexei Averchenko
how do you express integer points on an ellipse?
 
Old John
Old John
@KaliMa pretty sure that is the way to do it
 
Alexei Averchenko
Alexei Averchenko
4:26 PM
@user1230219 sure, but what about integer points on their intersection?
 
user1230219
user1230219
@AlexeiAverchenko, I guess you look at the discriminant of the resulting quartic?
A not so elegant solution (and probably not effective) I guess.
 
Old John
Old John
I think you might have to bite the bullet and do the algebra here
 
Alexei Averchenko
Alexei Averchenko
can we enumerate pairs of ellipses such that their intersections are integer in a sane manner?
 
KaliMa
KaliMa
the points dont have to be integral
just the distances to those points
 
Alexei Averchenko
Alexei Averchenko
oh
looks like it's going to come down to pythagorean triples
 
KaliMa
KaliMa
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)
 
Alexei Averchenko
Alexei Averchenko
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?
 
Old John
Old John
@KaliMa shortest distance to intersection from where exactly? and how can two integers be separated by 90 degrees?
 
KaliMa
KaliMa
i'm not sure
@OldJohn The distance from origin to closest intersection = b, the other one = c
 
Old John
Old John
@KaliMa that's better
 
Alexei Averchenko
Alexei Averchenko
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
 
Old John
Old John
4:36 PM
how do we know that the two vectors representing $b$ and $c$ are definitely at right angles?
 
Alexei Averchenko
Alexei Averchenko
then you'll know the axes of symmetry are orthogonal (being the diagonals of that rhombus)
@OldJohn ^^^^^^^^^^^
 
Old John
Old John
right - so we don't know yet
 
robjohn
robjohn
@N3buchadnezzar Pretty well. Just got back from the park.
 
Alexei Averchenko
Alexei Averchenko
you'll only need symmetry to prove that
 
N3buchadnezzar
N3buchadnezzar
@robjohn Any tips for headache/ from overworking?
 
KaliMa
KaliMa
4:40 PM
I'm pretty sure angle BOC will always be 90 degrees
 
Alexei Averchenko
Alexei Averchenko
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?
 
KaliMa
KaliMa
i just tried a bunch of examples
 
Alexei Averchenko
Alexei Averchenko
come on, i just basically laid out the proof for you
do some work ffs
 
KaliMa
KaliMa
no need to be mean
i just told you i tried a bunch of examples
 
Alexei Averchenko
Alexei Averchenko
but what if you missed the one counterexample?
 
MSEoris
MSEoris
4:42 PM
hello there
 
KaliMa
KaliMa
because i am pretty sure it doesnt exist
 
Alexei Averchenko
Alexei Averchenko
how can you be certain that you didn't miss it unless you actually prove that it doesn't exist
?
 
KaliMa
KaliMa
because i'm willing to take the risk
 
Alexei Averchenko
Alexei Averchenko
@KaliMa there's no "pretty sure" in mathematics
 
KaliMa
KaliMa
i dont possibly see how a counterexample can exist
 
MSEoris
MSEoris
4:43 PM
that doesnt mean it doesnt :)
 
KaliMa
KaliMa
it would require really wonky workarounds
 
Alexei Averchenko
Alexei Averchenko
i don't care what you can or cannot see, prove it rigorously
 
KaliMa
KaliMa
i can't "prove it" rigorously because i am not familiar with formal proofing
i just know it by exhaustion
 
robjohn
robjohn
@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.
 
Alexei Averchenko
Alexei Averchenko
@KaliMa yes you do, if you ever studied geometry
 
user1230219
user1230219
4:44 PM
Exhaustion is a formal method - but have you "exhausted" al possibilities @KaliMa
 
KaliMa
KaliMa
trying a bunch of integral and irrational / rational rotations, 1000000 examples by simulation, they all had 90 degree separation
 
robjohn
robjohn
@N3buchadnezzar Otherwise, I just tough it out.
 
N3buchadnezzar
N3buchadnezzar
@robjohn Yeah, I do not have any of those
 
Alexei Averchenko
Alexei Averchenko
@KaliMa what if you missed the one that has a different angle?
 
KaliMa
KaliMa
because i didnt
 
N3buchadnezzar
N3buchadnezzar
4:45 PM
I sort of really need to work too, exam on the 5th..
 
Alexei Averchenko
Alexei Averchenko
@KaliMa you can't know that unless you prove it
 
robjohn
robjohn
@N3buchadnezzar Then get hydrated.
 
KaliMa
KaliMa
then please show me the one i missed, because i think it's pretty clear it doesn't exist
 
user1230219
user1230219
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.
 
N3buchadnezzar
N3buchadnezzar
Just really worn out from the exam yesterday, it was really tough. And made me somewhat depressed-
 
Alexei Averchenko
Alexei Averchenko
4:45 PM
@KaliMa there's no "pretty clear" in mathematics
mathematics is a precise science
 
user1230219
user1230219
@KaliMa, have you studied geometry?
 
MSEoris
MSEoris
what course @N3buchadnezzar
 
Alexei Averchenko
Alexei Averchenko
it either is or it isn't
 
KaliMa
KaliMa
@user123021 In high school, but we never did anything as complex as this
@AlexeiAverchenko So please show me the counterexample
 
N3buchadnezzar
N3buchadnezzar
@MSEoris Complex anlysis, laplace, pde and fourier transforms.
 
Alexei Averchenko
Alexei Averchenko
4:46 PM
@KaliMa i can prove it doesn't exist
but you can't
 
KaliMa
KaliMa
okay how about this
 
Alexei Averchenko
Alexei Averchenko
and that's the problem
 
KaliMa
KaliMa
consider two ellipses on top of each other
 
MSEoris
MSEoris
thats a very wide range of topics :)
 
KaliMa
KaliMa
let's focus on the top/down/left/right points of intersection
i rotate one by theta
 
N3buchadnezzar
N3buchadnezzar
4:47 PM
@robjohn We got this integral on our exam $$ \int_0^{2\pi} \frac{\cos^2\theta}{2 + \sin \theta}\,\mathrm{d}\theta $$
 
KaliMa
KaliMa
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
 
Alexei Averchenko
Alexei Averchenko
@KaliMa wtf is that even supposed to mean?
 
robjohn
robjohn
@N3buchadnezzar Did you get it?
 
KaliMa
KaliMa
@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
 
Old John
Old John
@KaliMa sorry - I don't see that as a proof at all
 
KaliMa
KaliMa
4:49 PM
@OldJohn Why not?
 
N3buchadnezzar
N3buchadnezzar
@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...
 
user1230219
user1230219
@KaliMa, try using congruent triangles.
 
Alexei Averchenko
Alexei Averchenko
@KaliMa ok, that's a nice one, but it requires some work
 
KaliMa
KaliMa
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
 
Old John
Old John
@KaliMa you have given no mathematical reason why the movements of the intersections should be "exactly offset" - whatever that means
 
Alexei Averchenko
Alexei Averchenko
4:51 PM
yep, this argument is correct
 
robjohn
robjohn
There is always the Weierstrass substitution as well, but your class is about contour integration?
 
KaliMa
KaliMa
@OldJohn Because of symmetry, how on earth can one side move more than the other?
 
N3buchadnezzar
N3buchadnezzar
Indeed
 
Alexei Averchenko
Alexei Averchenko
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
 
Old John
Old John
@KaliMa are we talking about 2 points which (at least appear to be) rotated by 90 degrees?
 
KaliMa
KaliMa
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 $a$
 
Old John
Old John
@KaliMa It is quite possible no such integral $b$ and $c$ exist
 
KaliMa
KaliMa
right
 
robjohn
robjohn
Did you get $$
\frac{2i}{4}\oint\frac{(u+1/u)^2}{4i+(u-1/u)}\frac{\mathrm{d}u}{iu}
$$
 
KaliMa
KaliMa
i am trying to generate unique values of valid $a,b,c$
 
Alexei Averchenko
Alexei Averchenko
@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?
 
robjohn
robjohn
4:58 PM
@N3buchadnezzar That's using $u=e^{i\theta}$
 
Ilya
Ilya
good evening
 
Old John
Old John
@AlexeiAverchenko definitely computable - do you mean - can I remember the formula?
 
robjohn
robjohn
@Ilya Hey there!
 
Alexei Averchenko
Alexei Averchenko
@OldJohn well, i remember that the arc length is not computable, so I wondered about the horde
 
robjohn
robjohn
@Ilya It is morning here, and a rainy one, at that.
 
Old John
Old John
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
 
Ilya
Ilya
@robjohn pity
 
Alexei Averchenko
Alexei Averchenko
@OldJohn well, no turing machine can compute it in the finite time
that's what I meant
 
Old John
Old John
can a turing machine compute $\sqrt{2}$ in finite time?
 
N3buchadnezzar
N3buchadnezzar
@robjohn Yes
 
robjohn
robjohn
@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.
 
Alexei Averchenko
Alexei Averchenko
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
 
Ilya
Ilya
@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.
 
KaliMa
KaliMa
it also seems to be the case that 5b^2*c^2 = 4a^2(b^2 + c^2)
 
Alexei Averchenko
Alexei Averchenko
@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
 
KaliMa
KaliMa
yes, a, b, and c are integers
 
Alexei Averchenko
Alexei Averchenko
$\frac{4}{5} a^2$ has to be integer
wait...
right
and $a$ also has to be integer
i'm not sure if i'm right here
 
KaliMa
KaliMa
5:06 PM
i am wondering if this is something i can rewrite and solve as a pell equation
 
Alexei Averchenko
Alexei Averchenko
doesn't make a lot of sense to mee
 
KaliMa
KaliMa
perhaps reparameter with two variables
 
Alexei Averchenko
Alexei Averchenko
oh, so $a$ has to be divisible by 5
that's interesting
 
robjohn
robjohn
@Ilya yes, you can smell rain coming here. It is very noticeable.
 
KaliMa
KaliMa
a doesn't have to be divisible by 5
 
robjohn
robjohn
5:07 PM
$$
\frac12\oint\frac{(u^2+1)^2}{u^2+4iu-1}\frac{\mathrm{d}u}{u^2}
$$
 
KaliMa
KaliMa
valid a,b,c=209,247,286
 
Alexei Averchenko
Alexei Averchenko
if it's not divisible by 5, then $\frac{4}{5} a^2$ and $a^2$ cannot be simultaneously integer
 
KaliMa
KaliMa
or 341,374,527
 
Alexei Averchenko
Alexei Averchenko
@KaliMa how do you know?
 
KaliMa
KaliMa
using a program to test brute force
 
robjohn
robjohn
5:08 PM
@N3buchadnezzar The roots of the denominator are $(-2\pm\sqrt{3})i$
 
Ilya
Ilya
@robjohn or especially the first rain in the end of August, after 2 months of drought
 
KaliMa
KaliMa
but trying to find a way to generate solutions
 
Alexei Averchenko
Alexei Averchenko
was this program using floating point arithmetic?
cause if so, then it's not exactly reliable :)
 
KaliMa
KaliMa
i dont need to
i just calculate each side, check if equivalent
 
Old John
Old John
@KaliMa here is a method which might help:
 
Alexei Averchenko
Alexei Averchenko
5:09 PM
ok, then i'm confused
what are $b$ and $c$?
 
KaliMa
KaliMa
b is the shortest integral distance from origin to the intersection of an ellipse and its rotation
 
Old John
Old John
if $x^2+4y^2=4a^2$ with $a$ an integer, then x must be even
 
KaliMa
KaliMa
c is the distance to the other intersection
 
robjohn
robjohn
@Ilya Only two months? We don't have rain from the middle of spring to the end of autumn
 
N3buchadnezzar
N3buchadnezzar
@robjohn and 0
 
KaliMa
KaliMa
5:10 PM
(there are four intersections but they have symmetry)
 
Alexei Averchenko
Alexei Averchenko
@KaliMa and what did you calculate to check things, exactly?
 
Ilya
Ilya
@robjohn at all?
 
robjohn
robjohn
@N3buchadnezzar I don't see any $0$...
@Ilya not usually.
 
N3buchadnezzar
N3buchadnezzar
@robjohn Rewrite the expression...
 
Ilya
Ilya
@robjohn that's the feature of zeros :)
2
 
Old John
Old John
5:11 PM
so put $x=2z$, to get $z^2+y^2=a^2$
 
Alexei Averchenko
Alexei Averchenko
can anyone please just cruch groebner basis so that we could be done with it? :D
 
N3buchadnezzar
N3buchadnezzar
@robjohn You forget that the nominator is $(1+1/u)^2$
 
Old John
Old John
sorry - I am finding point with integer coords - not integer distances :(
 
KaliMa
KaliMa
yeah x and y can be anything
a,b,c are integers
 
Alexei Averchenko
Alexei Averchenko
@KaliMa well, if the equation you gave me is correct, $a$ has to be divisible by 5
that's just it
 
N3buchadnezzar
N3buchadnezzar
5:13 PM
@robjohn Also $\mathrm{d\theta} = \mathrm{d}u/iu$
 
Alexei Averchenko
Alexei Averchenko
either your computations are mistaken, or the equation you gave me
 
KaliMa
KaliMa
i just gave you clear examples that show it doesnt have to be divisible by 5 though
 
Alexei Averchenko
Alexei Averchenko
@KaliMa how did you compute them?
give me the formulas
 
robjohn
robjohn
7 mins ago, by robjohn
$$
\frac12\oint\frac{(u^2+1)^2}{u^2+4iu-1}\frac{\mathrm{d}u}{u^2}
$$
 
KaliMa
KaliMa
via the COB=90 degree rule
 
robjohn
robjohn
5:14 PM
@N3buchadnezzar :-)
 
Alexei Averchenko
Alexei Averchenko
concrete formulas please
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
 
N3buchadnezzar
N3buchadnezzar
@robjohn It boils down to $$ \frac{1}{2}\cdot \oint \frac{(u^2+1)^2}{u^2(u^2+4iu-1)} \mathrm{d}u$$
 
Alexei Averchenko
Alexei Averchenko
ok, let me guess
we have two symmetries
and four points
 
N3buchadnezzar
N3buchadnezzar
@robjohn Indeed, that is where the 0 comes from.
 
Alexei Averchenko
Alexei Averchenko
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?
 
KaliMa
KaliMa
5:17 PM
h/o trying to verify i didnt make a dumb mistake
 
N3buchadnezzar
N3buchadnezzar
@robjohn So it has two singularities inside the unit circle..
 
robjohn
robjohn
@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
 
N3buchadnezzar
N3buchadnezzar
@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
 
Alexei Averchenko
Alexei Averchenko
@KaliMa actually i was wrong
 
robjohn
robjohn
@N3buchadnezzar expanded the numerator and divided by the part of the denominator other than $u^2$
 
Alexei Averchenko
Alexei Averchenko
5:27 PM
either $a^2$ or $b^2 + c^2$ must be divisible by 5
 
Argon
Argon
@N3buchadnezzar 4 hours? That's brutal
 
Alexei Averchenko
Alexei Averchenko
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 :)
 
N3buchadnezzar
N3buchadnezzar
@Argon Indeed it was le horrible
 
KaliMa
KaliMa
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
 
robjohn
robjohn
@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.
 
N3buchadnezzar
N3buchadnezzar
5:34 PM
@robjohn Laurent, derivation is covered.
 
KaliMa
KaliMa
anyone know to to generate solutions to 4a^2(b^2 + c^2) = 5b^2c^2 ?
 
robjohn
robjohn
@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 $2i\sqrt{3}$
 
N3buchadnezzar
N3buchadnezzar
@robjohn Yeah, but you have to admit it is a bit of an evil integral to give at an exam yes?
 
robjohn
robjohn
@N3buchadnezzar It depends on what was gone over in class and what else was on the exam, but it is not a simple integral.
 
KaliMa
KaliMa
auh
 
robjohn
robjohn
5:44 PM
@N3buchadnezzar It was 4 hours?
 
N3buchadnezzar
N3buchadnezzar
@robjohn Yes
 
robjohn
robjohn
@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.
 
Argon
Argon
Good job, guys
 
robjohn
robjohn
@N3buchadnezzar what is the title of the course?
 
N3buchadnezzar
N3buchadnezzar
 
Argon
Argon
5:56 PM
Hellø
 
N3buchadnezzar
N3buchadnezzar
@Argon I fixed the link =)
 
Argon
Argon
Hahaha
 
N3buchadnezzar
N3buchadnezzar
@Argon You heard about J, K integrals?
 
Argon
Argon
@N3buchadnezzar J, K integrals? What are they?
 
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