So I have seen this prove by induction which confuses me to a certain extend. Especially the base case where they set n = 1. $$f^1(x) = f(x) = \frac{x}{\sqrt{1+x^2}} = \frac{x}{\sqrt{1+1x^2}$$. My problem is that it doesn't work n = 2. The beginning fuctions where f: IR -> IR: x-> $$\frac{x}{\sqrt{1+x^2}}$$ and $$f^n(x) = \frac{x}{\sqrt{1+nx^2}}$$. Would be really nice if someone could help me.

So if n = 2 doesn't work the induction should not work, right? But the solution says otherwise.

In a metric space with topology induced by the metric, is it true that every compact set is contained in some open balls with finite radius?

yes. let $K \subset X$ be compact and pick $p \in X$. There is a function $f: K \to [0,\infty)$ given by $f(x) = d(x,p)$, continuous in $x$. Because $K$ is compact $f$ has a maximum, say $M$. Then $K$ is contained the closed $M$-ball around $p$, or the open $M+\varepsilon$-ball around $p$.

Wow

This is a really elegant proof

I thought I need some covering argument.

Can I then define a continuous bump function f (i.e. 0 on outside of K) on $K$ such that f is integrable on the measure space $(V,B(V),d)$, where $V$ is the metric space I started with and $d$ its metric?

Hello. I need to prove that if A and B are matrices in row-reduced echelon form (finite matrices) and also A and B are row equivalent then A=B. I have a basic prrof in my head, but it is all in verbal english, i would like my proof to be a little "constructive", if you know what i mean ( somewhat compact and to the point). Any hints appreciated.

is there a way to ask wolframalpha for the singularity locus of an algebraic surface ?

@quallenjäger any continuous function with compact support is integrable, with integral of its absolute value bounded by its maximum * measure of the compact set.

@MikeMiller Sorry I mistyped, I mean a bump function on the ball around K.

And out of my own interest, do I need a specific measure to make it work? Or does it hold for a general measure.

That happens only if your measure plays well with the topology, they might be completely unrelated in general!

@quallenjäger what if the ball around K is the whole space?

@quallenjäger pick a point $x$ and consider the balls $B(x,n)$ for $n \in \Bbb N$

The set of points with $d(x,K) < c$ again form a compact set as long as $c$ is chosen small enough (this requires a proof; to see the potential difficulties, think of a compact ball inside of the unit disc in $\Bbb C$.

@quallenjäger As long as every point has a neighborhood basis of finite measure sets (as long as every point has a neighborhood of finite measure) this is true by a covering argument

$<c$ or $\leq c$?

$\leq $ my bad

how do measure and metric go together?

is it the Haar measure?

Idunno

no that doesn't even make sense

Hausdorff measure for example.

You usually ask for a radon or borel regular or some other weirder condition measure

this doesn't make sense, $d$ isn't a measure

So that it plays well with the topology and you have approximation theorems for measurable or Borel sets in terms of Borel or compact sets

Ok I meant the Hausdorff measure, which should be compatible with the metric.

also if you take $K = \{0\} \subseteq \Bbb R = V$ then I think you can see that you can't define such a bump function

He corrected himself to mean on a ball around $K$.

Of course the actual support of a function whose support is contained in a closed set must be contained in the interior of that closed set.

I don't think so

unless we have different definitions of "support"

hi @loch

Hi @LeakyNun

@loch so the dRc-2 of $\Bbb R^3 \setminus \{0\}$ is $\Bbb R$ right

@MikeMiller What if the support is the closed set its self?

What is dRc-2 ?

de Rham cohomology 2

Oh ya

People usually write $H^2_{dR}$

so there should be a closed 2-form that is not exact, right

i.e. a vector field whose div is 0 but is not the curl of any vector field

@quallenjäger Be careful about what you mean by 'support'. Either it means $S = f^{-1}(\Bbb R \setminus \{0\})$ or $\overline S$.

In the first case the support is open. In the second case it is the closure of its interior.

I will shamefully admit that i forgot how div and curl works so im slightly unsure about your second line - but yes to your first line

then follow-up to your yes to my first line is whether you could give an explicit example

$*d(1/r)$. here $r = \sqrt{x^2 + y^2 + z^2}$, so $d(1/r) = r_x dx + r_y dy + r_z dz$. the rule for the Hodge star is that it is linear and $*dx = dy \wedge dz, *dy = dz \wedge dx, *dz = dx \wedge dy$. This definition works uniformly to give a closed but not exact $(n-1)$-form in each $\Bbb R^n \setminus \{0\}$.

To see that it is not exact, integrate it against the unit sphere.

thanks

This generalizes uniformly to all dimensions, and is a rephrasing of the way the example is usually presented.

would 1/r^2 work?

Nope. The rule I am using here is that 1) $*$ is an isomorphism, 2) upto sign, $*d*df = \Delta f$, where $\Delta$ is the Laplacian on functions. So $*df$ is a closed form if and only if $f$ is a harmonic function.

If $f$ is a function that depends only on the radius $r$, then the same is true of its Laplacian, and that is (as a function of $r$), $$\frac{1}{r^{n-1}}\left(r^{n-1}f'\right)'.$$

here $n = \dim \Bbb R^n$.

thanks

So what you want is for $r^{n-1} f'$ to be constant in $r$; and you want your function to be $f(r) = \frac{c}{r^{n-2}}$ unless $n = 2$, in which case you want $f(r) = |\log r|$.

I see

So I guess there will be fewer square roots flying around when $n$ is even.

Hello

I asked a question here yesterday, but there was no response, thought I would try it again today..

Is it possible to find an equation of a rotated ellipse knowing three points that lie on its curve and the center point?

@MikeMiller It can be open in the first case but it must not be isn't it?

Sorry I mean it has not to be.

But I still can't follow you, if the support is contained in a close set, why should it be contained in its interior.

Am I better off submitting a proper mathematics.stackexchange question?

@Cosinux yes, because you have three equations and three unknowns

@loch everything is algebraic so you might have some overkill

@quallenjäger the inverse image of an open set under a continuous function is open

The interior is the largest open set contained in a set. If the support is an open set contained in another set, it must be contained in that set's interior.

If you use the second definition of support (closure of the set of points where $f$ is nonzero), then the point is that the support-set must have nonempty interior. Certainly $\{0\} \in \Bbb R$ does not.

I was just trying to explain why Leaky's example worked.

@LeakyNun How do I actually do it? I know I can find an equation for an unrotated ellipse by solving the system of equations, but I'm not sure how I would do it for a rotated ellipse, since I don't know what the actual angle is..

let's just solve $ax^2+bxy+cy^2+dx+ey+f=0$

I should probably projectivize this

we actually have $6$ unknowns here but only $4$ equations, and this is a linear situation

wait no

ok not all of those are ellipses

I should find a pair of orthogonal vectors

say $(a,b)$ and $(-b,a)$

I mean the inverse image of an open set under a continuous function is open only if the function is defined on the open set isn't it.

then our ellipse is $A(ax+by+C)^2 + B(-bx+ay+D)^2 + E = 0$ right

Anyone here think they can derive the formula for arclength of a curve on the surface of the unit sphere?

great, now I have $7$ unknowns

@loch help

@Cosinux let your centre be $(p,q)$

translate it to the origin

Ok, it will always be in the origin

i.e. let $X=x-p$ and $Y=y-q$

so we want to solve $A(ax+by)^2 + B(-bx+ay)^2 + C = 0$

@MikeMiller For example, $f:[1,2]\rightarrow [1,2], f(x)=x$. Under the first definition it would be $[1,2]$ which is closed.

I really should projectivize this

@Cosinux and if we plug in the three points, we would have 3 equations

what are the a and the b?

(a,b) is the direction of the major or minor axis

@quallenjäger That's open in the topology of $[1,2]$.

Ok I see,

You only ever talk about the topology on the space your function is defined on

Thanks @MikeMiller

yup

@LeakyNun how would I compute the (a, b) given the 3 points?

@Cosinux do you have those actual points lol

I do, but they can be anywhere on the curve of the ellipse

can you give me some points

And I don't know by what amount the ellipse is rotated

Oh, you mean that way.. no, I don't have them. They are a part of a program I'm writing. They change

do you have some examples

I can try computing some

how about this

let the angle be $\theta$

actually no i won't depart from the algebra land

$A(ax+by)^2 + B(-bx+ay)^2 + C = 0$ as before

and assuming non-degeneracy I can set $C = -1$

I can also set $a^2+b^2=1$

so that's 4 equations and 4 unknowns

@Cosinux ok?

([0.4733333333333335, 0.8993731914031947], [0.7685005686342001, 0.7664996029980944], [0.6292915940880721, 0.8380878981613663])

Here is an example

what is the format of the output?

([x1, y1], [x2, y2], [x3, y3])

output

What do you mean by that?

Oh.. well, I would like to draw that ellipse

how should I give you the ellipse

Hmm.. what's the best format to representa rotated ellipse?

maybe $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ lol

Something that I can arrange in y = +/-function(x)

oh that might be more difficult

I don't think that's possible also

I think a better idea would be to parameterize it via the parameterization of the circle and apply an affine transformation

You are unlikely to do much better than that as Leaky is about to point out

See, I'll need to iterate through the x coordinates and calculate the two y's to draw each pixel..

if you use parametrization then everything is good

The problem is eg the ellipse you get by shearing the unit circle. It won't have reflectional symmetry anymore, just a 180-degree rotational symmetry

$(x,y)^T = A(\sin t,\cos t)^T$

where $A$ is a constant matrix

I'm slightly lost

is this loss

It should've been gain

Well.. I'll give you a different question: is it possible to find the angle of an ellipse by knowing three points?

yes

That way I could perform an inverse rotation on the three points and just compute it normally

really, everything boils down to solving a bunch of quadratic equations

which neither of us three are willing to do

Could you at least point me in the right direction?

$A(ax+by)^2 + B(-bx+ay)^2 = 1$ and $a^2+b^2=1$

plugging the three points separately into the first equation gives you three equations

so in total you have 4 equations

I'm not sure I believe it's possible to know something about an elliipse by knowing 3 points.

we know that its centre is the origin

Oh

That way I would find the a and b right?

yes

and the angle is either atan2(a,b) or atan2(b,a)

and I cannot be bothered to figure out which one it is

ok.. I'll give it a try

I think that given any ellipse through the origin there is a unique upper triangular matrix $A$ with $A\cdot C = E$, where $C$ is the unit circle.

I can believe that

but the choice of $A$ is not continuous

with respect to $E$

hi @Daminark

@LeakyNun I don't think that's true.

let's say we have the ellipse $\frac14x^2 + y^2 = 1$

hmm ok

idk

I should have said "...with positive determinant" up above. To do it, first scale to an ellipse with volume 1. Then the matrix is the upper triangular matrix with $\det = 1/\text{vol}(E)$ and second column equal to the uppermost point of the ellipse.

So precisely you have a continuous (smooth, if you like) map $\text{Ellipses} \to \Bbb H$, the upper half-plane, sending an ellipse to its uppermost point; write it as $(x(E), y(E))$, $y(E) > 0$, as well as $\text{Ellipses} \to \Bbb R^+$, the volume. Then the matrix is $$\frac{1}{\text{vol}(E)}\begin{pmatrix}1/y(E) & x(E) \\ 0 & y(E)\end{pmatrix}$$

I see

o..o

what

nothing

of course this is ellipses centered at the origin.

a closed subvariety :P

the point here is that we have decomposed $GL(\Bbb R) = \Bbb R^+ \cdot T(2) \cdot O(2)$, where $H(2)$ is upper-triangle matrices with det = 1 and positive first column, and let it act on the circle. we know that every ellipse centered at the origin is the image of the circle under a linear transformation, so this map $GL(\Bbb R) \to \text{Ellipses}$ is surjective. $O(2)$ acts trivially, $\Bbb R^+$ acts by scaling (and scales the volume the same amount), and what is left is the set $T(2)$

consider the function $x(y+z)$ defined on the unit sphere $S^2$

hmm, it's a compact manifold

can we somehow define the gradient and the hessian to find the stationary points and classify them?

Of course $y(E) > 0$ is a consequence of this, so the next point was to figure out the geometric meaning of that second column, since it's our parameter

Hey!

hi

In general I guess this says you need the center, topmost point, and volume of an ellipse to find a unique affine transformation from the circle to it

not so many parameters

hi everyone

hi @MatheinBoulomenos

Hello

@MatheinBoulomenos you have a rotated ellipse centred at the origin

you are given three points on the ellipse

can you deal with situation from an algebraic geometric point of view?

I don't know

ok

@MikeMiller can we find stationary points on a manifold by differentiating it once and classify them by differentiating it twice?

that's way too classically-geometric for what I know about algebraic geometry. I usually just think in terms of generalizations of stuff in commutative algebra or algebraic number theory (hopefully Ted doesn't read this, lol)

@MatheinBoulomenos was ist am importantesten wann man einen Sprach lernen?

I'm not really good at learning languages

I like dead languages

ik kan et sehen..

Stationary points of gradient flow?

just stationary points

Of what

of e.g. the function $x(y+z)$ defined on the manifold $S^2$

What is a stationary point of a function

I guess use the charts to bring the situation back to $\Bbb R^2$

What is a stationary point of a function on R^2

a place where the gradient is zero, I guess

hi @AlessandroCodenotti

Hi @Mathei

@LeakyNun 'There is a fairly standard determinant trick, where if the points are in "general position" (if not, you can find more points in straightforward ways), then reflect the points about the origin and take determinants to find the coefficients of the quadratic, analogous to what is done here for the circle: ambrsoft.com/trigocalc/circle3d.htm (random site I found on google).

@LeakyNun So what you really want is to discover the notion of gradient vector field on manifolds (with Riemannian metric) which still makes sense. :)

You use the inner product to turn the 1-form df into a vector field grad f. When you have a vector field you can flow.

A zero of df = a zero of grad f is a critical point. The flow doesn't move those.

Note that critical points are independent of metric. The flow depends on the metric, though.

I really need to learn differential forms

@KarlKronenfeld thanks

If the critical points are isolated and non-degenerate, then D(grad f) at these points is a symmetric bilinear form. These are classified by their number of positive and negative eigenvalues.

And in this case you can find a chart on which your function is $x_1^2 + \cdots + x_k^2 - \cdots -x_n^2$

you mean my function $x(y+z)$?

Any function on any manifold

k is the number of positive eigenvalues

I see

Your function has 6 critical points, corresponding to when grad f in R^3 is in the same direction as the normal field to the surface

So look for when grad f = c(x,y,z) on S^2

but can we do it without going to R^3?

otherwise it would be no different than just doing Lagrange multiplier

Well you've defined the function in R^3 so it seems easiest that way, but of course one can always use charts

You're computing zeroes of df, which is defined locally

is it df that is hard to compute?

It's not hard to compute, you just have to do it

well what is our basis?

Basis for what?

well normally in R^3 I would express df in terms of dx and dy and dz right

So on S^2 you would do that with dx and dy where x and y are the coordinates of a chart.

I see

in R^3 is "dx" and "dy" and "dz" just illusions?

in reality they are not constant right

I mean, "dx" is an element of $T_p^\ast\Bbb R^3$

"dx" can be interpreted as "the differential applied to the coordinate function x". So the notation is not just a name given to a basis element, it has an actual meaning.

You are right that you cannot say it is the same at every point without being able to compare tangent spaces at those points.

Or rather cotangent spaces.

I see

With a Riemannian metric you naturally get an object called a connection, which is a rule that takes as input a piecewise smooth path in M and spits out an isomorphism $T_{\gamma(0)} \to T_\gamma(1)$. If the connection came from the metric this isomorphism will be orthogonal.

hmm

This rule satisfies three properties: concatenating paths composes the corresponding isomorphisms; it doesn't depend on the parameterization of the path (well, you can't swap endpoints, but oriented parameterization); and it is continuous in paths.