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2:34 AM
@robjohn well maybe brute force is not the technical term, but there certainly isnt anything elegant about what I have in mind. In the past I wrote a (unoptimised) algorithm for visiting every face on a trinagle mesh, and was thinking of massaging the problem into that form...which I may do as a sanity check anyhow
there is a certain elegance to brute force
if brute force gets you there faster and without expending valuable brain cycles, how elegant can an elegant way really be
I think expending brain cycles is my goal
It is slowly turning to mush otherwise
I don't know how much of a hint would be too much here
Also I made an observation that I dont know how to formalise
If im on the right track maybe your silence is enough
If i am wildly off kilter perhaps silence is also good too
I want to explore my observation a little
I dont know how to say this without it sounding dumb, but I just noticed a relationship between the position of the verticies and wheather the line segments intersect inside the circle
2:46 AM
that is getting warm
Ok, i wont have time to solve it today (time is almost up before i turn into a grease monky again) but if i split the cicle into two sectors, how do I tell which sector a vertx is in?
My best guess is to use polar coordinates and define the sector by angle of the arc...
Ahhhh! A fly in my hotchoc...dammit maths
Got a way of expressing which sector the vertex is in, and now i must go. Thanks for the hint ;)
3:02 AM
another bad day.
i want a reset button.
3:15 AM
hi nick gurr
3:47 AM
How do you write $ds^2=\frac{dx^2}{x^2}+\frac{dy^2}{y^2}$ in matrix form? Is it $\frac{1}{x^2}$ and $\frac{1}{y^2}$ on the diagonal and 0 else-where? I need this to calculate the volume form.
$g=\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ is the matrix representation of $g=ds^2=dx^2+dy^2$
so for $ds^2=\frac{dx^2}{x^2}+\frac{dy^2}{y^2}$ my best guess at the matrix representation is $\begin{bmatrix} \frac{1}{x^2} & 0 \\ 0 & \frac{1}{y^2} \end{bmatrix}$
pretty good guess
4:04 AM
and then the volume form has a formula, $\omega=\sqrt{|g|}dx^1 \wedge dx^2\cdot\cdot\cdot dx^n.$
and I get $\omega=\frac{dx}{x}$ as an answer
so then my guess is that I should write integrals like this $\int f(x)\frac{dx}{x}$
i'm seeing an "n" where i think i should be seeing a 2 and the det does not appear to have been computed correctly
and some cdots that maybe aren't there
square root of that is 1/xy
ok but the notation is being mixed in weird ways. your det formula appears to be in notation that appears to be using x = (x^1, x^2) as a vector valued mapping, but up top with ds^2 you're using x as one coordinate
i'm playing "type mismatch" here instead of answering more directly because i think any confusion may be arising from this sort of thing and not deeper stuff
4:19 AM
yeah I don't know
Q: Volume form of manifold with metric $ds^2=\frac{dx^2}{x^2}+\frac{dy^2}{y^2}$

geocalc33How do you write $ds^2=\frac{dx^2}{x^2}+\frac{dy^2}{y^2}$ in matrix form? Is it $\frac{1}{x^2}$ and $\frac{1}{y^2}$ on the diagonal and $0$ else-where? I need this to calculate the volume form. $g=\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ is the matrix representation of $g=ds^2=dx^2+dy^2$ so f...

I typed up my thoughts and attempts in a question:)
if your surface is parametrized as the image of some two-variable function r(x,y), then just using x and y, your computations show that the surface element dS oughta be 1/|xy| dx dy and you'd integrate some function f defined on the surface with respect to dS by integrating f(r(x,y)) 1/|xy| dx dy. i think this is just the usual thing from multivariable calculus
but for single variable integration
people often say something like area element or something when studying surfaces in 3 space although volume form is the general term for "hypersurfaces" in n-space
if you want to do a 'line integral' of a curve on the surface you'd integrate with respect to ds, not dS.
i may be confusing this by introducing new notation dS when it seems like you may already be consulting sources using varied notation that might not be mutually compatible with one another
i recommend picking one book and just following through how they do it. doesn't matter which one, just something that has a consistent universe of notation and discourse. some guy around here wrote one.
6 hours later…
10:24 AM
Any help with notation : I have a vector field $X: \mathbb{R} \times \mathbb{R}^d \to \mathbb{R}^d$ and the book says $X\in C(\mathbb{R};L)^N$ but has no reference for notation...
Sorry $C(\mathbb{R};L)^d$
hi, why is it true that if $h \neq 0$ almost everywhere on some interval $(a,b)$, $-\infty \leq a < b \leq +\infty$, and $fh = 0$ almost everywhere on $(a,b)$, then $f = 0$ a.e on $(a,b)$?
couldn't we have say, $h = 1$ on $[0,1]$ and $0$ outside, and $f = 0$ on $[0,1]$ and $1$ outside, and then this would be false?
10:43 AM
@Monty what is $L$?
the underlying question i wanted to ask was, if I know that the fourier transform of a product of two functions is zero, and I know that at least one of the factors of the product is nonzero on a set of positive measure, can I conclude the other factor is zero a.e?
@porridgemathematics well, if $h=1$ on $[0,1]$ and $0$ outside, then it certainly isn't the case that $h\neq0$ almost everywhere for any interval properly containing $(0,1)$
and on the interval $(0,1)$, the conclusion holds, as it should
oh I mistyped, I meant to say $h$ is not 'zero almost everywhere', i.e. $h$ is not zero in $L^p$
then your counter-example works and the claim is false
$L^p$ is not an integral domain
right, my underlying question was assume the fourier transform of a product of two functions in $L^2(a,b)$ is zero, and suppose one of the factors is not equal to $0$ in $L^2(a,b)$, is the other factor necessarily $0$ in $L^2(a,b)$?
10:47 AM
that fails for the exact same reason
oh, right, we could just take the fourier transform of the product I proposed
is that what you meant by same reason?
your $f$ is not in $L^2$, but yeah, the point is the same
well okay, set $(a,b) = (-2,2)$
take disjoint sets $A,B$ with finite, positive measure, then $1_A,1_B$ are in $L^p$, neither is zero in $L^p$, but their product is zero
okay great
2 hours later…
12:44 PM
Say I have two objects in a 3d euclidean space and I want to know if they intersect. Would it be a good idea to quickly negate intersection by bounding the objects in a convex hull and check if the two convex hulls dont intersect?
Because if the boundry volume does not intersect, the objects themselves surely dont
2 hours later…
2:33 PM
Suppose that I have a metric space $X$ with metric $d:X\times X\to [0,\infty]$. Is it true that $d$ is continuous on $X$?
I suppose not because degenerate metric space, which separates distinct points by $1$ and same points by $0$ is a counterexample.
So it seems that $d$ is not continuous in general.
can someone help
i am a bot :(
And if that is the case then, if I have two continuous functions $f$ and $g$ from metric space $X$ to metric space $Y$ with $E\subset X$ dense in $X$ and that $f(x)=g(x)$ for all $x\in E$
I want to prove that $f(x)=g(x)$ on $X$
@Euler2 2nd row , first 2. Then , 3rd row - 3rd column.
Had $Y$ been $\mathbb R$, then things would’ve been much easier. As I could’ve defined $h(x)=f(x)-g(x)$ and proceeded to prove the result.
yeah it works thanks
2:43 PM
However, in this case I believe that I can’t let $h(x)=f(x)-g(x)$ as - may not be defined in $Y$.
So I define $h(x)=d_Y(f(x),g(x))$ for $x\in X$ to prove the result.
But I have difficulty proving it continuous. Any hints/suggestions? @leslietownes @TedShifrin @robjohn
I have already proven that $f(E)$ is dense in $f(X)$.
2:58 PM
@Koro Suppose $E=\mathbb{Q}$ and $X=\mathbb{R}$. Can you do it in this case?
i know it's gauche, but you could also use a sequence-based approach
3:16 PM
@Koro: you know that $f(e)=g(e)$ for all $e\in E$. So pick some $x\in X$ and $\epsilon\gt0$. Can you show that $d_Y(f(x),g(x))\le\epsilon$? (I had $E$ and $X$ reversed; now it's fixed)
@robjohn: Yes!! I did.
Can you explain it here?
I just opened this chat and saw your hint and before that I’d finished. I’ll explain my working now.
I assumed for contradiction that there is an $x\in X\setminus$ such that , $d_Y(f(x), g(x))\gt 0$. It follows by continuity of $f$ and $g$ at $x$ that given $\epsilon\gt 0$, there exists $\delta\gt 0$ such that $d_X(x,t)\lt \delta\implies d_Y(f(t),f(x))\lt \epsilon/2$ and $d_Y(g(x),g(t))\lt \epsilon/2$. Since $E$ is dense in $X$, we know that there is a $p\in E$ such that $d_X(x,p)\lt \delta$ whence it follows that $d_Y(f(x),f(p))\lt \epsilon/2$ and similarly for $g$.
Then by triangular inequality $d_Y(f(x),g(x))\le d_Y(f(x),f(p))+d_Y(f(p), g(x))$ and noticing that $f(p)=g(p)$ , we get a contradiction.
immediately proving our theorem.
Sir @robjohn, I suppose that you were hinting me to do this only. Is my proof correct?
3:33 PM
It looks okay, but you need to define the $\epsilon$ you want to use.
One minor correction (not minor while writing an exam though): “ I assumed for contradiction that there is an $x\in X\setminus E$”
the vibe is right but i have not doublechecked the epsilonology
Sir @robjohn: I have taken that to be arbitrary.
For quadratic equations why do we divide the equation by a and why on earth do we need to have a perfect square?
@leslietownes Hi
3:37 PM
rajorishi, you don't have to do any of those things but it sounds like you are talking about a common derivation of the quadratic formula via "completing the square"
if so, at a high level, the idea is to reduce solving an arbitrary quadratic to taking square roots
i.e., turning an arbitrary quadratic into $x^2 = \text{something}$, which is a solved problem
So if my epsilon is arbitrary and I have some $|x|\lt \epsilon $ where $\epsilon \gt 0$ is arbitrary then $|x|=0$. I have used this result. Is my proof correct now sirs @leslietownes @robjohn?
Which is a solved problem @leslietownes .I could not really get the last line.
if you ask me to solve $x^2 = p$ the answers are $x = \pm \sqrt{p}$. i just take square roots
koro why do you get a contradiction? you are implicitly using the hypothesis that $d_Y(f(x),g(x)) > 0$ but it isn't explicit where. an evil instructor would say that you do not use that hypothesis at all.
why can't $d_Y(f(x),g(x)) < \epsilon$?
Because $\epsilon$ is arbitrary
@leslietownes How do they compare x and y in the cases? Please assist me with this.
3:45 PM
That means in particular $\epsilon = \frac{d_Y(f(x),g(x))}{2}$
think about the real case, suppose $f$ and $g$ are known to satisfy the hypothesis and also to be bounded by ten million. if epsilon is arbitrary, and i take epsilon = 20 billion, there's no contradiction whatsoever
ok, right. notice you didn't have that above. epsilon is arbitrary in applying the definition of the existence of the limit, but it isn't arbitrary in this proof, you're choosing that value of it
and the assumption that $d_y(f(x),g(x)) > 0$ is exactly what you use when you do that
And my above choice of epsilon gives reductio ad absurdum
if it wasn't assumed positive, it couldn't be plugged into the proof as an 'epsilon'
yeah, you've got it
The sequence approach I hinted at requires an understanding of the sequential consequences of density. Fix arbitrary $c$ in $X$. [Because $E$ is dense in $X$] there is a sequence $(e_n)$ in $E$ for which $e_n \to c$. Then $f(c) = \lim f(e_n) = \lim g(e_n) = g(c)$. This chain of equalities is justified by the continuity of $f$, $g$ (outer equalities) and because $e_n$ is in $E$ for all $n$ (inner equality; it means $f(e_n)=g(e_n)=0$ for all $n$).
Ahh I got it I think @leslietownes
it's sometimes fun if you prove something about metric spaces using open balls, to try to find a sequence proof, and vice versa. different people come at it different ways. sometimes it's easier to find one type of proof than the other, although which type may vary by person and problem
3:49 PM
So I needed to give that particular value of epsilon also which brings contradiction. Right?
yes. if epsilon is arbitrary there is no way to decide whether there's anything wrong with $d_Y(f(x),g(x)) < \epsilon$. if $\epsilon = d_Y(f(x),g(x)) + 10$ for example, no contradiction
@leslietownes meanwhile I’ll take a look at your sequential proof also. Thank you!
Noted @leslietownes. Many thanks :)
Thanks sirs @leslietownes and @robjohn
Why did my instrucotr multiply by 2 to relate x and y?
some people hate sequence proofs because they don't generalize to every conceivable space, and (particularly if the sequence has to be constructed in some complicated way) require reference to infinite sequences of things when (as in the open ball based proof) only a few parameters are often needed to get the job done.
Please refer to the last picture posted.
3:53 PM
rajorshi, this picture is unclear to me
I agree that sometimes sequences proofs work like miracle for example while proving discontinuity of $\sin (1/x)$ at $ 0$.
I am saying why the multiplied by a factor in the numerator and denominator?
while proving discontinuity of Dirichlet function (or is it also called Lebesgue function?). I mean the function on set of reals which assumes value 1 at rationals and 0 otherwise.
yeah, in some instances it's just easier to identify relevant sequences. the form of the function will suggest something specific that you don't have with just some arbitrary function on a metric space
as an exercise, come up with an open ball based proof for the dirichlet function :)
I’ll try that :)
3:58 PM
anyone know straight edge and compass geometry books
the books on geometry I see keep slipping in algebra somewhere, the greek in me is saddened by this
hartshorne's "geometry: euclid and beyond" has a very good discussion of them that would not offend you in its first chapters. it doesn't have a huge amount of exercises, but what's there is very good
Coxeter has good ones. Introduction to Geometry is very nice. David Hilbert has a good geometry book. Robin Hartshorne has a nice geometry book
lol! yes, its a very nice book!
thank you all, perhaps there shall be some light on the plane anew
I tried kiselev, but there's too much algebra
4:02 PM
I mean, Euclid is still quite nice also. Once you get into used to the writing style, it is very lovely.
kiselev introduces the geometric mean theorem with a square root
γιατί το κάνει αυτό?????
@Quin yeah I've spent most of my time on the elements
hartshorne goes through the elements in some detail, pointing out where euclid cheats a little by using stuff that doesn't follow from his axioms or not handling all cases of a proof.
it's funny how every commentary or undergrad-level approach to the elements i have ever seen stops before the 3D stuff. i think by that point every author stares at that stuff and just thinks, if you want to do this, good luck.
yeah and hartshorne also recommends sections of euclid that are more worth reading. some sections introduce very fundamental ideas, others are just a grind. it seems hartshorne helps point this out
i mean they might refer to specific results but they don't go through each proposition.
all the ink has been spilled by then
i'll check out hartshorne!
4:07 PM
see if you can find my copy of the pre-springer edition that the postal service lost when i moved. it's probably in some used bookstore somewhere after some crook postal worker made a fortune off of selling it.
If I define $M$ as a mapping from $\mathbb{R}^d \to \mathbb{R}^d$, which is just matrix multiplication i.e $x\mapsto Ax$ for some invertible matrix $A$, then can If I define the pushforward of a (absolutely cont) probability measure $\mu$ on $\mathbb{R}^d$ as $M_{\#} \mu $ then what can we say about this guy ? is it a probability measure? hows it density relate to $\mu$
4:19 PM
@leslietownes: I thought a lot about arbitrary epsilon. In fact, there is even a theorem for set of real numbers that if we have for some $x\in \mathbb R$, $0\lt x\lt \epsilon $ for any arbitrary $\epsilon \gt 0$, then $x=0$
@Monty Think about determinants.
So my question is back to square 1. Why do I have to specifically mention for what $\epsilon \gt 0$, I’ll have contradiction? $\epsilon \gt 0$ being arbitrary is itself an overkill.
@TedShifrin Yeah im just doing a change of variable
Hi @TedShifrin
Hi Koro
4:22 PM
unless determinant is 1, no longer a prob
say $\mu$ has a density $f$ will the density of $M_{\#}\mu$ just be $ f \circ A^{-1} $
sorry not a density
but radon nikodym derivative
koro your proof derives a contradiction from the fact that (after some inequalities) a deduction that $d_Y(f(x),g(x)) < \epsilon$. This will be a contradiction for some values of epsilon, but not others. You choose a value of epsilon that gives you a contradiction.
Assuming $\det A=1$?
4:24 PM
@leslietownes: note that $x$ is fixed here.
@koro if you ask me to derive a contradiction from the fact that some nonnegative function of a fixed $x$ is less than $\epsilon$, i can't do that without some idea of what $\epsilon$ is. if it's "arbitrary" it might well be a contradiction at all.
Oh, no more probability then. Then you need a factor if $|\det A|$ or its reciprocal, no?
I understand for proving Cauchy sequence for example, if we have $ \sum _{n\gt N} x_n \lt \epsilon $ then here $N$ depends upon epsilon. We can’t use the aforementioned theorem. But in this case, x is fixed.
when people say epsilon is arbitrary in this context, they mean, because i have some positive number in mind, and the limit definition provides me with something for every positive epsilon, i can take epsilon to be that positive number. i don't have to worry that the limit definition won't provide me a delta in that context, because all i need to worry about is epsilon being positive. the limit definition will give me a delta as long as that happens.
@leslietownes: Please note that $\epsilon $ is not arbitrary. I emphazise that $\epsilon \gt 0$ is arbitrary.
4:28 PM
ok, put that in everywhere i said it, same message.
if i make some assumption for the purpose of contradiction and deduce that 2 < epsilon, i don't know if i'm done unless i know why there's some reason why 2 can't be less than epsilon.
But since $\epsilon$ is any arbitrary chosen positive number. It’s clear that it can be $1$ also.
PS: I’m still going thru your last message.
if i'm allowed to put epislon = 1/2 into the beginning (which i might be), then i get my contradiction. but if i'm doing that i should put that at the beginning of my proof. or say something like "since epsilon > 0 is arbitrary, and 1/2 is a positive number, we can put epsilon = 1/2 into the above analysis and deduce a contradiction" at the end of the proof.
You guys are making me hate $\epsilon$.
i'll clear the air. $g_{ijk} = \sum_{a,b,c} \Gamma_{jk}^{a,p,q} \nabla_X(\langle g_{ab}, g^{ab}\rangle) [X^{pq}, Y_{pq}]$.
Garbage, but nice try.
4:33 PM
@leslietownes: I would like to emphasize that limit definition is one thing as there we have delta depending upon epsilon . In my case above my $x$ is fixed and a fixed non negative quantity less than arbitrary positive values of epsilon implies that number has to be zero . I’m really interested in knowing how you can contradict this.
If $x$ is fixed, give it a fixed sort of name.
x is fixed non negative number @TedShifrin
I thought $x$ was in a metric space?
Let me give the reference of the theorem that I’m referring and sources also.
koro, in reference to your first proof, you just said "we get a contradiction" after deducing $d_Y(f(x),g(x) < \epsilon$. the arbitrary thing came later.
4:35 PM
No @TedShifrin. Suppose x is real number.
had you included a line, in that same set of lines, at the end saying "because epsilon > 0 is arbitrary, we get a contradiction" or something like that, i would have felt better about the proof.
I guess I will stay out of it.
Ahh @leslietownes now it sounds correct :)
@Koro since you have assumed that $d_Y(f(x),g(x))\gt0$, why not take $\epsilon=d_Y(f(x),g(x))/2$?
Oh, is leslie helping? i just got back from snapping owls, so I don't want to interrupt
it's also just clearer if you provide an example. unstated in that line of reasoning is "if $a \geq 0$ and $a < \epsilon$ for all $\epsilon$ then $a = 0$, and we assumed up above that $a$ was not zero." these are the jumps that many textbooks make in the later chapters but i would not make these jumps if the exercise is in chapter 1.
4:37 PM
@robjohn: Sir, there’s absolutely no harm in taking that also but I just thought about using any arbitrary epsilon.
@leslietownes: I see what caused confusion. I had that arbitrary epsilon thing in my mind but forgot to write that when I posted my working above
@Koro but that makes it hard to coordinate that with a contradiction based on $d_Y(f(x),g(x))\gt0$.
@robjohn snapping owls?
I would use an arbitrary $\epsilon$ when doing a direct proof. With an indirect proof, I would set $\epsilon$ to a useful value.
koro, that's the idea. there are two games, one to understand how to solve the problem, and two to include enough appropriately arranged detail to convince a reader, perhaps a very skeptical reader, that you haven't missed something. i think you won the first game and a lot of the above discussion was about the second.
and one or two lines, a change in order of wording, and things like that, can really affect game 2.
I understand @leslietownes @robjohn
4:45 PM
@TedShifrin owl
One final question for closure: Had I added this line "Since $\epsilon \gt 0$ is any arbitrarily chosen positive number, it follows that $d_Y(f(x),g(x))\epsilon \implies d_Y(f(x),g(x))\epsilon=0$ as metric $d_Y$ by definition can't be negative.", would my proof be correct/complete?
@robjohn @leslietownes
Oh, snapping = photoing.
I was thinking a turtle had morphed into an owl.
That is an odd condition @Koro
@copper.hat why??
I want to understand that
What does $d_Y(f(x),g(x))\epsilon \implies d_Y(f(x),g(x))\epsilon=0$ mean?
4:47 PM
OMG, I forgot to put signs this time.. Arghh
Please treat $d_Y(f(x),g(x))\epsilon$ as $d_Y(f(x),g(x)\lt \epsilon $.
@copper.hat . @robjohn and @leslietownes are aware of the background
ok, i'll step aside
@copper.hat: You may also provide your inputs please.
@Koro does that mean you want to say $d_Y(f(x),g(x))\lt\epsilon=0$
because $\epsilon\gt0$
4:50 PM
i think you have a pretty top notch team there already :-)
Copper, I stepped aside ahead of you, so don't feel lonely.
some owls do snap if they feel menaced by you. my daughter is obsessed with them, we watch youtube clips
@leslietownes we have a family of 4 owls in the park. two fledglings and their parents
No no. I don't know why I'm typing so wrong today. $d_Y ((f(x),g(x))\lt \epsilon \implies d_Y (f(x),g(x))=0$ because $\epsilon \gt 0$ is any arbitrary positive number.
4:52 PM
so you've gone from an indirect proof to a direct proof?
robjohn that's great. sometimes great horned owls perch in our neighborhood and fill the night with hoots but that seems to be a winter thing. no nesting.
I missed that
while walking the dog in the park late one night, i thought i saw a child standing in the middle of the field. as i got closer, it took off. it was a massive great horned owl. it was so cool
@leslietownes they hoot when deciding territory, that is in the winter
One owl used to sit at my window every morning and if I would look at it, it would look back at me and keep staring :)
4:54 PM
they usually show up here in November and stay until July.
They do like to stare
One day, I opened window it looked at me, I looked back at it. It flew away and never came back :'(
especially the young ones
Yes it was very little
Now a crow has taken its place.
we had crows outside my home office last year. love em. my wife hates their begging calls.
They seem to be very smart, you know.
The crow I mentioned seems to know where we'll put food for it outside window.
4:56 PM
because he trained you to put it there.
It is just there at the right time (morning around 9:00 AM) and if nobody keeps food for it outside, it tries to knock the window. That's so cute :)
Hahah. Or seeing another way, we trained it to be there :)
@robjohn: Sir, I didn't quite understand: switching from indirect to direct proof?
There's no switching as I have already taken for contradiction some $x$
and not epsilon.
@Quin This picture is an North American Great Horned Owl, the European versions are supposedly a bit bigger.
Im in AZ so I would guess that what I saw was the north american great horned. it was probably because it was night, but when just standing out on the ground it looked so tall!
@robjohn that is a really great picture! i love the facial expression
I used this result: Suppose that $s$ is a non-negative real number such that for any arbitrary positive $\epsilon\gt 0$, we have $s\lt \epsilon$, then we have $s=0$. Proof: Suppose that $s\gt 0$ then since $\epsilon \gt 0$ is arbitray, let $\epsilon =\frac s2$ and by given hypothesis, we must have $s\lt \frac s2$ which means $2s-s\lt 0$ that is $s\lt 0$ which is a contradiction and therefore $s=0$.
@robjohn @leslietownes
I'm confused. I request your help in this @robjohn @leslietownes @copper.hat @TedShifrin
@Koro That is correct
5:08 PM
koro one reason there was a lot of static above is that it was not clear at the outset that you were using this result. from a classroom standpoint we also don't know whether you are 'allowed' to use this result (i.e., whether it was previously proved and citable, or whether you would be expected to include the proof you just gave as a sub-part of your argument). it's helpful to explicitly write out whatever path you are taking.
@Koro I am not really sure what it is that you are asking about? It is not easy to follow the thread of conversation to untangle exactly what you are trying to accomplish.
Yes. I accepted that earlier. I forgot to mention that in my first message explaining my working. And that's exactly what caused confusion. Now, my final question: with this idea included in my proof, my proof is complete. Right?
@copper.hat: I'm trying to prove that if $f,g$ are two continuous functions on metric space $X$ and let $E\subset X$ is dense subset of $X$ and given that $f(x)=g(x)$ for all $x\in E$ then prove that $f(x)=g(x)$ for all $x\in X$.
i think you've got it.
Thanks a lot @leslietownes. :)
I'm sorry for not making my arguments/ideas clear enough at the very beginning and causing lot of confusion to you also
Do I have to show that the set is non-empty while proving Euclid's division lemma this way?
5:14 PM
i'm not sorry. we got to spam the chat with about 50 epsilons. that's a good day for me.
Thanks a lot sirs @robjohn @leslietownes. My query is now answered.
Haha @leslietownes. Spamming with $\epsilon $. One day $\delta$'s day will also come I guess :D
i would also accept $N$s.
Does mathjax have $\varepsilon$? or just $\epsilon$
that answers that!
varepsilon is just epsilon in a fedora
5:23 PM
@Quin: It's "varepsilon" and is there in mathjax I think
wow laughing at that one!
5:48 PM
@Quin you could play with that... does mathjax have $\aleph$?
:D $\mathfrak{g} \xrightarrow{\text{ad}} \mathfrak{g}$
wow it can do a lot! can it do tikz?
@Quin cool that it stretched to cover the error
yeah super cool! i never really know what is "built-in" and what is some random package im importing, i may play with this!
MathJax does load some packages, so it is not vanilla
but it does not support some more common things like \vspace though it does support \hspace
that makes sense considering \vspace is usually for the formatting of a whole document. hopefully people dont post things of that length as questions here!
im not sure i would have the patience to read it all
5:54 PM
@Quin there are some pretty long questions and answers posted on main
Though, I do often want to add a bit of vertical space to a post and I have to use non-standard means to get it.
@robjohn yeah it makes more sense on main with the longer format. Are the same packages loaded on main as here? I really don't know very much about how mathjax works. I am procrastinating by reading mathjax documentation right now ha.
chat does not support $\LaTeX$, but the ChatJax bookmark uses the same header as main. Although, ChatJax still uses 2.7.1 and main uses 2.7.5. I should update the bookmark, but then people may see different things until they update.
ahhh yes, that may be hard to get people to switch
6:26 PM
I should check the change log for MathJax from version 2.7.1 to 2.7.5 to see if it really makes a difference.
Geez! it's up to version 3.0 and it looks like it works differently.
This sounds like the way I lost Illustrator and Acrobat.
6:46 PM
@TedShifrin I won't go to version 3 unless SE does. I think there is so much content that might fail, I don't think they will.
Interesting. What do you suspect?
what do I suspect from what?
7:02 PM
Will be the changes of impact.
I don't know. They just said that things might break, but they were not specific.
Ah, lovely.
Hopefully that will change, or they will update 2 if problems arise. Or there will be better documentation about what needs to be changed in terms of content.
I will create a bookmark using version 3 at some point and see what happens.
Oh, that’s scary :)
8:00 PM
Hi everybody! Equations look nice when you're on the phone with the full version of the site. But it looks awful on the phone version. Why is that? I don't know what to do.
A: Minimizing $a_1x_1^2 + a_2x_2^2$ for positive $a_i$, where $a_1x_1+a_2x_2=B$

lone studentHere is the solution in a standard way: We have, $$\begin{align}a_1x_1^2+a_2x_2^2&=a_1\left(\frac{B-a_2x_2}{a_1}\right)^2+a_2x_2^2\\ &=\frac{B^2-2Ba_2x_2+a_2^2x_2^2}{a_1}+a_2x_2^2\\ &=\left(a_2+\frac{a_2^2}{a_1}\right)x_2^2-\left(\frac{2Ba_2}{a_1}\right)x_2+\frac{B^2}{a_1}\end{align}$$ This is th...

8:16 PM
Hello I have a question. In this paper, page 978 eq. 32. For N\to\infty, \epsilon\to0. I know that the sum of the integral converges for N\to\infty. But the right hand side does not converge since \sum_{n\ge1}\mu(n) does not exist. This is all happening in one line and I cannot locate my mistake.
@vitamind Hello! Can I ask a question? The comments made here cannot be deleted after a short time, is it correct?
@lonestudent Yes, correct. Approximately a 1-3 min time span.
@vitamind there are multiple summands on the RHS, just because one of those summands doesn't converge on its own doesn't mean the sum doesn't converge
@Thorgott The other summands do not depend on N.
$\epsilon(x,N)$ doesn't depend on $N$?
the formula in the next line suggests otherwise
8:24 PM
@vitamind Thank you very much!!. It is also valid for Both delete and edit?
@Thorgott epsilon goes to zero as N goes to infinity. Right after line 33, quote :"Thus epsilon\to 0, as N\to\infty "
@lonestudent Yes.
@vitamind Thank you for information.
then it seems as if you have to be wrong about the LHS converging
@Thorgott It's well known that it converges.
8:41 PM
it's so well known that there might not be one, thorgott.
@Thorgott It represents the sum over the trivial zeros in Riemanns explicit formula for the prime counting function. See here.
8:56 PM
that doesn't look like the same term to me
@Thorgott A proof can probaby be found in Edwards. I'm sure that the term on the LHS is 1. convergent 2. the same as the sum which runs over the trivial zeros.
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