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12:01 AM
@Obliv yeah, gradients are perpendicular to level curves
@usukidoll it should be in your favorites list no?
 
found it. The question I have is related to this I've found
http://math.stackexchange.com/questions/192452/whats-the-proof-that-the-euler-totient-function-is-multiplicative
 
That doesn't make sense..
 
a unit occurs when there's a multiplicative inverse.. so the Cartesian product I have is $R_{1} \times R_{2}$.. So if $R_{1} \times R_{2}$ then $R_{1}$ is a ring and $R_{2}$ is a ring
 
@usukidoll if $R_1\times R_2$ what?
 
isn't if $R_{1}$ is a ring and $R_{2}$ is a ring, then wouldn't $R_{1} \times R_{2}$ be a ring too?
 
12:07 AM
wait I'm really unsure @arctic could you do me a solid and straighten me out? Would $\nabla f \cdot g = 0$ give the equation for the tangent plane b/t them? I guess I should work an example
 
the plane tangent to the level curve $\nabla f=c$ at the point $u$ will be $\nabla f(u)\cdot (x-u)=0$
 
gradients?
 
math.stackexchange.com/questions/1972196/… I wish to make sure this question is labelled properly. Is this a differential geometry question?
 
I never took that course so I wouldn't know the category
 
12:10 AM
me neith
*neither
 
@arctic c is a vector, correct?
 
I just know that differential geometry has a lot to do with coordinate "frames"
 
@Obliv woops I mean the curve $f(x)=c$, sorry
 
oh damn so I did give the wrong definition lol..
So if $f(x,y)$ and $g(x,y)$ shared the same tangent plane i.e were both orthogonal to $\nabla f$ or $\nabla g$, then would the equation of the tangent plane be given by $\nabla f \cdot (x-u) = 0$ for any point u ?
 
what do you mean by share the same tangent plane
 
12:16 AM
lie on the same tangent plane **
x-u gives an equation of a line right?
 
say P is the plane tangent to z=f(x,y) at some point (a,b,f(a,b)) and Q is the plane tangent to z=g(x,y) at some point (c,d,f(c,d)). you're hypothesizing that P=Q are the same plane?
 
hmm, so wouldn't $\nabla f \cdot g = 0$ work? for whenever g is orthogonal to it?
 
the gradient orthogonality thing is for level curves, not 3D graphs z=f(x,y)
unless you want to clarify what you mean by a plane being tangent to a function
 
that's like the dot product of gradient and g
it's been a while since I've done Calculus IV material
hi @0celo7
 
usukidoll: I've never heard of calc 4.
do you mean differential equations?
 
12:21 AM
:S
no
I guess it depends on the country
 
I'm not following entirely @arctic $\nabla f$ is orthogonal to $f(x,y)$ only at level curves?
 
multivariate calculus?
 
^ yup
 
i.e when there is a maximum/minimum? I knew it..
 
12:22 AM
That was my confusion
That started this whole thing.
 
where I'm from that is calculus 3
 
@Obliv the vector $\nabla f(u)$ is orthogonal to the level curve $f(\vec{x})=c$ at the point $u$
 
I've heard in some countries all the material in calculus 4 is bundled into 3
 
notice how I keep mentioning level curves, I don't say "it's orthogonal to f(x)"
@usukidoll um, what does your country put into calc 4?
 
is calculus 4 vector calculus?
 
12:23 AM
calculus 4 where I'm from is only 2 chapters in the textbook :P
 
USA goes up to calc 3 (vector calculus). after that it just gets called differential geometry.
 
Thanks @arctic I understand now.
 
really? my uni still has calculus 1-4 x.x
I think we're behind .... which is typical
 
(though the professors were a bit slow last semester so we ended at integrals)
i never learned vector calculus
 
12:24 AM
my linear algebra professor didn't teach me eigenvalues and eigenvectors. I had to learn it by myself
 
oh wow
 
seems like a pretty big omission
 
which was kaka because everyone else taught it
 
oh wow
nah
vector calculus is kind of a last minute thing
 
12:25 AM
I don't mind learning something by myself as long as I know and master what's going on.
 
the way i understand
 
vector calculus is mvp... ezpz
 
the calculus textbook only has a brief one chapter overview of vector calculus
and since it isn't used immediately (or in some cases never), most classes using it as a per-requisite just relearn it.
@usukidoll I cannot wrap my head around the details on my own, but yeah it does seem in theory simple.
 
there's one part of the question I have that's throwing me off... ugh these iff statements if that wasn't around I would be done already
 
what question?
 
12:27 AM
@arctictern thanks for the help
 
your politeness is odd. I've rarely been able to help you. :\
 
@arctictern I am really having trouble trying to do something. Are you familiar at all with rotations about axes?
 
@arctictern is a lifesaver :3
 
@TheGreatDuck sure. might not be fun though.
 
math.stackexchange.com/questions/1972196/… I cannot figure this out and it is a major roadblock in what I'm doing.
 
12:30 AM
at least it's not Euclidean Geometry... it's like a dead math topic :S
 
I heard that the set of three rotations is unique
but i don't know how to find them
 
@arctic Would it be fair to say that f(x,y) is maximal by a constraint function g(x,y) = k when they intersect at a level curve (at which the gradient will be orthogonal to both)?
 
I honestly have no idea how to even go about attempting this.
I could make a whole bunch of statements about it I suppose
but I suspect it is a massive dead end
 
@TheGreatDuck so you want to send the three coordinate basis vectors to three other arbitrary vectors?
just use the three other vectors as the columns as a 3x3 matrix
 
no
@arctictern any coordinate basis is the result of rotating the three x y z axes by some number
I want to find those three angles
the graphics system I have only accepts angular rotations about axes
 
12:32 AM
oh, you want to find the Euler angles?
 
is that what they're called?
I have no experience in this subject.
I just tend to think to high for my level of thinking sometimes. :p
 
not a fan of the question unfortunately :P
 
i.e at the same point $u = (x_1,x_2)$, $\nabla f \cdot (x - u) = \lambda \nabla g \cdot (x - u)$ ?
 
@arctictern damn. Thanks for the term though. Google has been uncooperative. There might be a formula somewhere.
 
@Obliv are you trying to figure out lagrange multipliers?
 
12:35 AM
yes. @arctic
 
@arctictern oh my lord. I finally found something relating to quaternions.
The Euler angles are three angles introduced by Leonhard Euler to describe the orientation of a rigid body. To describe such an orientation in 3-dimensional Euclidean space three parameters are required. They can be given in several ways, Euler angles being one of them; see charts on SO(3) for others. Euler angles are also used to describe the orientation of a frame of reference (typically, a coordinate system or basis) relative to another. They are typically denoted as α, β, γ, or φ, θ, ψ. Euler angles represent a sequence of three elemental rotations, i.e. rotations about the axes of a coordinate...
 
yes, quaternions do 3D and 4D rotations
I love explaining that
 
@arctictern usually you've given me a hint at what I should be looking at; here product integrals w values in a lie algebra
as long as the subdivision is sufficiently small I can log back to a Lie algebra
 
I cannot describe the feeling I just got from skimming that article in the last 10-20 seconds. I would be punished for foul language.
XD
 
ah good ol' wikipedia
 
12:39 AM
@usukidoll welp, it looks like what I wish to do is far too complex atm.
so I'll do the lazy option.
I'll just keep accumulating rotations and hope it works. XD
 
I think i'm overthinking
because the proof doesn't look that hard but I'm interpreting one part of the question as something else :P
 
@usukidoll That article isn't for your problem. It is summarizing my problem: I wish to find the euler angles of a basis.
and it is apparently NOT trivial.
:p
ummm... lol?
 
that iff part threw me off a bit... or maybe I slept too late as usual
 
check that image.
i think you messed up the link.
 
yeah sorry
 
12:42 AM
"This image was removed"
oh ok.
:)
 
uuuh...
 
usually I don't test these because I assume that they work like they supposed to
 
i think you are doing something beyond my level
that ain't multivariate calculus
 
yeah... so anyway I think I finally saw through this
 
12:44 AM
"R1 and R2 be rings"
 
a unit occurs when there is a multiplicative inverse
like ax = 1 where x =$a^{-1}$
 
"rings"
are you in grad school?
im pretty sure ring theory is far above the level im at
 
so I'm thinking if $R_{1}$ is a ring and $R_{2}$ is a ring then the cartesian product $R_{1} \times R_{2}$ is a ring.
no I'm not in grad school ;p
 
Rings are not grad school.
 
@0celo7 well I've never seen them. I just know the term exists and it means something weird.
but i dont even remember
 
12:46 AM
when you study higher level math you're gonna cross into these
 
Weird?
They're just about the most natural things
 
$\mathbb{Z,Q,R,C},M_n(F)$ are all rings
 
@0celo7 weird in that the name ring makes little sense and I cannot remember even what they are.
some kind of function thing, right?
 
@TheGreatDuck No.
 
$a_{1}$ is a unit so there exist s an inverse.
$a_{2}$ is a unit so there exists an inverse
so $a_{i}$ in general is a unit so there exists an inverse
not a function thing at all
ring R has like 12 axioms...
then you get something like R[x], Q[x]
 
12:48 AM
oh
nevermind I just looked up my old question
a ring is a set upon which addition and multiplication have the commutative property
 
yeah
 
@TheGreatDuck multiplication doesn't need to be commutative in a ring. e.g. matrices.
 
how do you have addition and multiplication of a cross product
 
huh?
 
@arctictern I mean multiplication distributes over addition
 
12:51 AM
well, the cross product isn't associative, so that wouldn't make R^3 a ring
it makes it a lie algebra
 
@arctictern no you misunderstand. @usukidoll wants to know if R^2 for instance, is a ring
my statement is
 
WHAT! WHOA!
 
"how do you multiply elements of r^2"
@usukidoll what's the problem?
 
No one saw that.
 
did usukidoll ask if R^2 was a ring?
 
12:52 AM
@0celo7 R^n is automatically a vector space?
 
whoa whoa whoa whoa whoa... there's a difference between $R_{2}$ and $R^2$ I just have 2 rings and the Cartesian product of 2 rings and there's units
I never asked R^2 at all
 
@TheGreatDuck You did not see that!
 
@arctictern no I didn't :S!!!
 
@TheGreatDuck Yes.
 
@usukidoll that's what I thought
 
12:53 AM
@usukidoll I am paraphrasing. Isn't R a ring?
and isn't R^2 the cartesian product of R and R?
 
R is a ring in general yeah .. there's 12 axioms
omg no x(!!!
 
most people would not use R^2 for a direct product R x R of rings, they'd use R^2 maybe as a free R-module
 
^ this this is gold
 
oh
i don't have any experience of ring theory
other than the obvious knowledge of the term
XD
 
I got $R_{1}$ is a ring... $R_{2}$ as a ring so if both of them are rings then $ R_{1} \times R_{2}$ is a ring... so if there are units that means multiplicative inverses exist
 
12:54 AM
that's why I said R^2 couldn't be a ring, cause addition in R^2 makes sense, but multiplication makes no sense.
after all, C is sometimes used in analogue of R^2
and everyone agrees that between C and vectors, multiplication is... complex and sometimes varies
(after all 2d vectors and complex numbers both serve as ways to view R^2)
(at least... that's how my non-euclidean geometry class looked at it. Take it with a grain of salt and assume we aren't doing complicated vector analysis)
 
Could it be that the unit for $R_{1} \times R_{2}$ is the ordered pair ($a_{1},a_{2}$) and $a_{1}$ is the unit for $R_{1}$ and $a_{2}$ is the unit for $R_{2}$?
 
wait
what is R_1 and R_2?
or are they just dummy variables referred to ring 1 and ring 2?
 
$R_{1}$ is a ring and $R_{2}$ is ring like my picture stated
both of them are rings
 
ok
i have no idea how to help you, tbh.
 
I may have an idea.. thanks for trying though
 
1:02 AM
im going to go see if I can get some other stuff done 'round the site
 
it's ok.
 
thanks for giving me that "euler angle" stuff though
 
@arctictern This can't possibly work. Suppose g is in the group. Then int_0^t g dx should be a path going from e to g, presumably the exponential of some linear subspace of mathfrak g. But the exponential map isn't injective so that's not well-defined.
 
right
 
1:20 AM
how do you combine the traditional infinite series to do something like get a nice series representation of e*pi?
 
Can someone explain the Fréchet derivative to me?
It seems rather odd the way its formulated.
@PVAL-inactive For someone who's inactive, you seem to have logged on.
 
I use this chat still.
I am refraining from answering or commenting on main though.
 
Good to not entirely lose you, then.
 
i'm just going to foil normally and switch to ceasero sums
 
1:36 AM
@shaihorowitz You're going to foil two infinite series together for that purpose? Sounds like the bruteforce way to do it.
Look into the Cauchy product: en.wikipedia.org/wiki/Cauchy_product
 
Any one here know section formula?
Why AP:PB = x_1 - x : x - x_2 ?
(pinging @DHMO)
 
@Ramanujan $AP$ is the subsegment from $A$ to the segment's division point $P$. $PB$ is the subsegment after the division point between $P$ and $B$. The lengths of their horizontal components are $x_1 - x$ and $x - x_2$ respectively.
One thing to note is that the horizontal components can be seen as the bases of right triangles for which $AP$ and $PB$ are the respective hypotenuses. These triangles are similar, so their bases are the same ratio as their hypotenuses are with each other.
Found a perfect picture to describe it:
 
2:19 AM
@Axoren so we can use that ratio of sides of triangle instead of general section formula ? Of m:n?
And will that give same ratio as m:n?
 
@Ramanujan Yep. That's a property of similarity.
 
@Axoren thanks,time for college
 
 
4 hours later…
6:29 AM
i found that 10*n>nth prime for all n less then 10^6
much nicer bound then 2^n
 
 
3 hours later…
9:46 AM
if $p_n$ is the nth prime, $p_n\geq n\ln(n)$ for all n>1
so that boud is going to eventually fail
 
10:07 AM
Hello all! I've got a really quick, dumb question that I guess is fairly perennial about standard terminology -- if, in some well behaved vector space, a function maps numbers to numbers, operators map functions to functions, and functionals map functions to operators, what is the name given to an object that maps operators to operators?
*Sorry, I of course meant to type that functionals map to numbers
 
user228700
Hi everyone :-)
 
user228700
I've a quick question about circles.
 
user228700
 
user228700
It's given that $AP^2=AD.AE$. How come?
 
In elementary plane geometry, the power of a point is a real number h that reflects the relative distance of a given point from a given circle. Specifically, the power of a point P with respect to a circle O of radius r is defined by (Figure 1) h = s 2 − r 2 , {\displaystyle h=s^{2}-r^{2},} where s is the distance between P and the center O of the circle. By this definition, points inside the circle have negative...
 
user228700
10:39 AM
Thank you! :-) [Could've done this myself, sheesh. Sorry :/]
 
user228700
Actually, no, I still have a small question...
 
user228700
In that diagram, it makes sense for us to equate $(s-r)(s+r)$ and $PA.PB$. How is this equivalent to $PM.PN$?
 
user228700
Isn't $s$ a specific constant? ie. Length of $OP$
 
user116211
10:56 AM
@Kaumudi Got that in geometry while studying circles back in 10.
 
user116211
There were more theorems like that.
 
user228700
Hm, maybe I did too, but since I don't like geometry all that much, I've forgotten all the properties/theorems... (excluding the imp. ones)
 
those are all the same property but in school they treat you like you are too dumb to notice
 
user116211
Secant-Tangent theorem; tangent-tangent theorem.
 
user116211
And all that.
 
user228700
11:01 AM
@Mussulini Which properties are u talking about?
 
power of a point
 
user116211
Also remembered square of sum less square.
 
user228700
Can u tell me how that's true?
 
user228700
15 mins ago, by Kaumudi
In that diagram, it makes sense for us to equate $(s-r)(s+r)$ and $PA.PB$. How is this equivalent to $PM.PN$?
 
user116211
@Kaumudi Which diagram?
 
11:07 AM
@Kaumudi By similarity. Join AN and MB.
 
user116211
So, you were talking about the wiki picture.
 
user116211
Well, they are equivalent since the external point is common.
 
user116211
The theorem can be proved for both cases when the secant passes through the center as well as for the case when it doesn't.
 
user228700
@BalarkaSen Ah, OK, I will try that. Thanks :-)
 
user228700
@MAFIA36790 Yep.
 
user116211
11:23 AM
Oh, BTW, @Kaumudi, this is called intersecting chord theorem.
 
I never remember any of the names.
 
$(x+1)(x+2)\dotsb(x+n)=x^n+a_1 x^{n-1}+a_2 x^{n-2}\dotsb$
I'm trying to find the a_k sequence
 
it is actually a doubly sequence in both n and k
@BalarkaSen yes I did use that but it's not the same thing
 
The elementary symmetric polynomials are expressible in terms of the sum-of-powers things. That's called the Newton's formulas.
 
11:27 AM
$a_1=\frac{n(n+1)}{2}$, $a_2=\frac{(n-1)n(n+1)(3n+2)}{24}$, $a_3=\frac{5n^6+19n^5+15n^4-15n^3-20n^2-4n-7440}{240}$
yes they are sums of sums of powers, that's how I obtained these. I'm trying to find out if there's a closed form for the a_k
I proved by induction that a_k is a polynomial in n of degree 2k
 
user228700
@MAFIA36790 Ah, OK...
 
user228700
@BalarkaSen I don't remember the theorems either, lol. However, if I do remember 'em, I don't tend to remember their names.
 
11:44 AM
I just try to prove them myself, since I can't remember them.
 
user116211
I'll not say I remember every name, but these are some of the few in my head.
 
12:38 PM
Many thanks for your feedback @user1952009
 
The function does not need to be continuous to admit parcial differentiation?
 
nope
 
@HiHello Not in general. Take eg, $f(x, y) = 1$ if $y = x^2 \neq 0$ and $f(x, y) = 0$ otherwise.
 
@HiHello there are pretty pathological functions that admit all directional derivatives, and are linear in the vector, and are not even continuous
 
good morning
 
12:49 PM
morning, @SamuelY
 
hey @Balarka
 
'sup
 
how's math going?
also, not much is up
 
pretty good
 
learn any cool theorems recently?
 
12:50 PM
The derivates will not exists <=> lim does not exists?
The definition limit
 
Yes
 
Thanks
 
Hi! $(2x^2-1)/((x^2-1)(2 x^2+3))$ find partial fraction? Any hint, please?
 
@SamuelYusim a few, yeah
 
anything I might understand?
@Mithlesh first factor x^2-1
 
12:56 PM
I think you'll understand most of what I have learnt in differential geometry. I like Hilbert's theorem, that there are no compact everywhere negatively curved surfaces in $\Bbb R^3$.
 
@SamuelYusim , can I write this as (x+1)(x-1)?
 
11 hours ago, by Ramanujan
user image
 
11 hours ago, by Axoren
user image
 
in topology, I learnt about spin structures, which might be a bit technical but still quite cool.
 
12:59 PM
So what will be the ratio when line is (triangle area = 0) vertical? @DHMO @BalarkaSen @SamuelYusim @HiHello
 
(a spin structure on an $n$-manifold $M$ is essentially a choice, upto homotopy, of $n$ independent nowhere zero vector fields along the 1-skeleton of $M$ which extends to $n$ independent nowhere zero vector fields along the 2-skeleton of $M$)
$n \geq 3$
@Ramanujan Please don't ping random people to answer your questions.
11
 
Can you help me
 
I don't want to, sorry
 
@BalarkaSen I can understand there are some people who don't want to share knowledge,sorry
 
I guess they call them spin structures because you end up filling disks with swirly things?
 
1:10 PM
@SamuelYusim hah. not quite (good picture though!); the thing is that the tangent bundle $TM$ is, in some natural way, a principal $\text{Spin}(n)$-bundle.
Spin(n) is the simply connected double cover of the group SO(n), $n \geq 3$.
 
hm. that's probably a better reason to call them that
 
That is to say the fibers aka tangent spaces admit a smoothly varying Spin(n)-action. (they naturally admit an SO(n)-action, using action by orthonormal orientation-preserving matrices on the tangent spaces - modulo an inner product, so the point is this action lifts to an Spin(n)-action)
 
makes sense
 
sup everyone
 
hey
 
1:19 PM
spin strutures? sounds interesting
If we have an $A$-module $M$, and $S$ a multiplicatively closed subset of $A$
nah sry
 
@Ramanujan undefined
 
@SamuelY so what's new with you?
 
just doing assignments and stuff
combinatorial optimization is surprisingly cool
 
1:34 PM
what's combinatorial optimization about?
 
essentially it's about a really weird correspondence between problems in graph theory (and other stuff) and problems about polyhedra in $\mathbb{R}^n$, so that either one will solve the other
 
sounds fun!
 
yeah, it's cool
 
can you give an example?
 
sure.
 
1:38 PM
@Ramanujan I recommend you do your computations in the scheme Spec R[T]
 
given a graph $G$ with each edge $e$ given a weight $c_e \in \mathbb{R}^+$, you want to find a spanning tree of $G$, i.e., one that hits every vertex, with the smallest possible sum of edge weights. We want to encode our answer as $x^T \in \mathbb{R}^{|E(G)|}$ where $x_e = 1$ if we pick $e$. We can write down a polyhedron which has at least one minimum weight spanning tree on its boundary using some clever inequalities, which I can show you if you want, but will take some explaining. (not done)
Now we usually want to find algorithms for solving problems like this. The common one for minimum-cost spanning trees is Kruskal's algorithm, which goes like this: order the edges from lowest cost to highest, say $e_1, \dots, e_m$, and set $F = \emptyset$ and $H = (V(G), F)$. For $i = 1, \dots, m$, if the ends of $e_i$ are in different components of $F$, put $e_i$ in $F$. Stop when $H$ is connected.
I forgot to mention that you also force the polyhedron to not have any non-minimal spanning trees on its boundary. You can now prove validity of this algorithm by showing that any possible output it gives will be on the boundary of our cleverly chosen polyhedron, using some geometry/optimization techniques
 
interesting
 
from what I can tell, every combinatorial optimization problem (even NP-hard ones) gets something out of trying this on it
and even if you can't prove an algorithm is completely correct, you can often prove that algorithms will give upper or lower bounds on your numbers
so you can get nontrivial results about, say, the travelling salesman problem out of this
 
1:56 PM
nice!
 
@JuanFran I don't know what scheme Spec R[T]
 
do you, like, prove that such a polyhedron exists or can actually construct it? that is, does that polyhedron technique give you an explicit spanning tree?
 
you actually construct it, and yes it does
 
curious. can you explain in a few words why this technique is efficient?
intuitively I want to understand how the polyhedron gets in the story
 
efficient in what sense?
 

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