Mathematics - 2018-10-29 (page 3 of 3)

IP Address

21:08

How do I get $x$ the subject of $y=e^{x}-e^{-x}$? Any tips

user76284

$2 \sinh x = e^x - e^{-x}$

Also let $z = e^x$.

IP Address

Actually I was gonna use that to work out the inverse on $\sinh{x}$

but it turns out to be a quadratic

didnt spot it until just now

user76284

Yeah, $y = z - z^{-1}$.

IP Address

appreciate the reply tho

user193319

I'm trying to compute the limit $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \ln(k)$. My thought was to write this as an integral, but I can't quite find the right function. I tried partitioning $[0,1]$ into subintervals of length $1/n$ and integrating $\ln x$ over $[0,1]$. Unless I made a mistake, this gives $\int_{0}^{1} \ln x = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \ln(\frac{n}{k})$, which isn't quite what I want...

I could use some help on writing that limit as an integral.

user76284

21:19

$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \ln k = \lim_{n \rightarrow \infty} \frac{1}{n} \ln \prod_{k=1}^n k = \lim_{n \rightarrow \infty} \frac{1}{n} \ln n!$

user193319

Yeah...$\lim_{n \rightarrow \infty} \frac{1}{n} \ln n!$ is what I'm trying to compute, and I've reduced it to the limit I mentioned above (actually, I'm really trying to compute $\lim_{n \to \infty} (n!)^{1/n}$).

Semiclassical

I'm guessing "Stirling's formula" isn't going to suffice

user76284

By Stirling's approximation, $\ln n! = n \ln n - n + O(\ln n)$ so the limit must diverge.

user193319

If it did, I don't think I can use it.

Semiclassical

right

i mean, that's a decent enough way to check your answer

user193319

21:21

True.

Semiclassical

@user76284 actually, tho, this remark is pertinent. that limit diverges

which in fact shouldn't be surprising. stirling says, to leading order, that $n!\sim n^n$

taking $1/n$ gets you down to linear divergence, but that's still divergence

user193319

If it's not possible to write that limit as an integral, is it possible to bound that limit below by some by integral which does diverge to infinity?

user76284

I don't think you need an integral.

user193319

What should I do, then?

user76284

Lower bound the factorial.

$n! \geq (n/2)^n$ I believe.

I think I got it finally haha.

user193319

21:28

Oh, so $\frac{n}{2} \le (n!)^{1/n}$.

which obviously shows that $(n!)^{1/n} \to \infty$.

user76284

Right, $\frac{1}{n} \ln n! \geq \frac{1}{n} \ln \left(\frac{n}{2}\right)^n = \ln \frac{n}{2}$ which clearly diverges.

user193319

Ah, I see. Thanks!

user76284

No problem. Anyone here good at Riemannian geometry? :-)

Mike Miller

I'm not, but I know something. I didn't answer your previous question about numerical invariants measuring distortion of diffeomorphisms since I wasn't sure what to say.

Semiclassical

21:59

@user193319 you should, of course, establish that this bound is actually true.

Ted Shifrin

@user76284: I'm a differential geometer but not a Riemannian geometer. But ask.

user76284

The $n! \geq (n/2)^{n/2}$ bound might be easier (you're just using the middle element of the product $n/2$ times, which is clearly less than the second half of the product).

@TedShifrin I was wondering what might be the right way to quantify the curvature of a deformation of the plane (see my messages and picture above).

I'm thinking of taking the Frobenius norm of the logarithm of the Jacobian matrix and integrating it over the manifold (just the unit square in this case).

Ted Shifrin

So you're making a family of surfaces in $\Bbb R^3$?

What do you mean by Jacobian matrix? If you're mapping $\Bbb R^2\to\Bbb R^3$, that isn't a square matrix.

user76284

I have a map $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$.

The interpretation of this map is that it's deforming the plane.

Ted Shifrin

Your pictures make absolutely no sense to me.

They're pictures of surfaces in $\Bbb R^3$.

So you are doing the flow of a vector field, and you're asking something about local geometry of that flow, not about the global picture of the region.

user76284

22:11

I can see why you're confused. They look 3D.

One moment.

Ted Shifrin

So you're basically asking locally how your flow distorts area?

user76284

These are the originals.

I guess it's more of a soft question since I'm thinking about how to quantify the scrunchiness/complexity/Dirichlet energy (?) of such a map.

Ted Shifrin

So how it distorts area is the best I can think of.

user76284

Does the deformation thing make sense though? It's like a diffeomorphism on $\mathbb{R}^2$ but without the invertibility condition.

Ted Shifrin

Are you actually flowing by the vector field $f$ for different times, or are you just applying $f$?

user76284

22:14

Yeah I think that makes sense which is why I'm thinking of using the Jacobian matrix.

Just applying $f$.

Ted Shifrin

Flow of a vector field is going to be a diffeomorphism.

user76284

Not flowing by $f$.

Ted Shifrin

So you're not thinking of a vector field at all. You're just looking at a mapping.

OK, I get it now.

user76284

Yeah that's a better way to put it.

I actually have an open question regarding vector fields and diffeomorphisms: math.stackexchange.com/questions/2976528/….

Ted Shifrin

Don't distract me now with that.

user76284

22:15

But in this case I'm just using the "displacement field".

Ted Shifrin

You want a global measurement (like an integral) as opposed to an infinitesimal measurement?

user76284

Correct

A functional, essentially.

Ted Shifrin

OK, so I see. I'm not sure why you were taking logs. I would be inclined to do some integral of the norm (to an appropriate power) of the Jacobian.

logs get you in trouble if you're constant or near constant anywhere.

I don't see that this has anything to do with curvature of Riemannian manifolds, however.

So that was a red herring.

user76284

Isn't it a differentiable endofunction on $\mathbb{R}^2$? I was interpreting the latter as a Riemannian manifold in case thinking more generally helps.

Ted Shifrin

If you're doing this on a manifold, you're still asking about the distortion coming from $f$ and not the intrinsic geometry of the manifold.

user76284

22:20

Can you clarify what you mean by "logs get you in trouble if you're constant or near constant anywhere."?

Ted Shifrin

It seems like you want something like the harmonic map functional. If you have a more general Riemannian manifold, the metric tensor comes into "measuring" the derivative of $f$.

All I'm saying is that you couldn't use logs if the function is constant. Wouldn't you be trying to take log of the zero matrix? I don't know where logs came from in the first place.

user76284

Yeah, logarithm of a non-invertible matrix would be undefined. What's the geometric interpretation of a non-invertible Jacobian? Determinant = zero, which corresponds to squashing neighboring points really close together, right?

Ted Shifrin

Sure.

Sharath Zotis

Is "for every" equivalent to "for all"

user76284

The reason I was thinking of taking logs is because it would take the Jacobian of the identity transformation (namely the identity matrix) to the zero matrix. More generally it's mapping the linear transformation represented by the Jacobian "back" to the Lie algebra.

Ted Shifrin

22:23

@Sharath, yes, but I would rather see context to be sure.

user76284

Does this kind of make sense?

It's a bit fuzzy.

e.g. a rotation matrix is mapped to the skew-symmetric matrix that generates it.

Ted Shifrin

I'd rather do something like $\|df - I\|$, I think.

But if you're mentioning that kind of thing, you really want a $1$-parameter group of maps to differentiate.

But I see your point. You want distortion from the identity.

user76284

I guess that works too, it's the right idea. Basically quantifying deviation from the identity at a local/infinitesimal level and then integrating it over the whole map.

Yeah exactly.

Ted Shifrin

I think I would start with something like harmonic mappings and the energy functional, but do $\|df-I\|^2$ rather than $\|df\|^2$.

If you just stretched things, without twisting, my suspicion though it that you'd still want that to be basically low energy. You're looking for twisting more than anything else? I'm not sure.

user76284

Yeah, low energy seems to correspond intuitively to what I'm looking for.

Ted Shifrin

22:32

But it's like you want to separate out stretching and twisting? Of course, other than for conformal maps, we don't get a super-nice decomposition. I'm still not exactly getting what you're trying to measure other than what I've suggested.

The answer to your linked question is that the flow of a $C^r$ vector field is a $C^r$ diffeomorphism.

user76284

That sounds like the Riemann tensor = Weyl tensor + Ricci tensor decomposition?

Adam

@CaptainAmerica16 what did MIT do? I thought it was the brave members of the secret service that were responsible for making that whole Aaron Swartz problem "go away". And by go away, I mean arrange for his assassination and the necessary cover up there after. I guess I'm going to be banned from chat again oh well

Ted Shifrin

(It's got more differentiability in the $t$ variable.)

user76284

I'm new to this whole field so I'm still getting a hang of the terminology as you can probably tell :-)

So the exponential of a $C^r$ vector field is always a $C^r$ diffeomorphism then? Nice.

Ted Shifrin

But I still say this has (virtually) nothing to do with the curvature tensor of the underlying manifold, @user76284. In particular, $\Bbb R^2$ is flat.

Zee

22:35

It’s a local diffeo

CaptainAmerica16

@Adam I genuinely don't know what you're talking about.

Lucas Henrique

Hey @Ted!

user76284

@TedShifrin I guess the "deformation" is a new manifold embedded in $\mathbb{R}^2$?

And I want to use the metric from the original ambient $\mathbb{R}^2$?

Ted Shifrin

Hi @Lucas.

Lucas Henrique

Could you please take a look at my answers. Tagged you Yesterday.

user76284

22:37

Like if I mapped $\mathbb{R}^2$ to the unit circle but kept the ambient metric. It seems to me that this new manifold would be "intrinsically" distorted, if that makes sense?

Ted Shifrin

Yeah, that doesn't really make sense, @user76284. The plane is the plane. You're trying to look basically at the pullback of the metric tensor by the mapping $f$. (You can't go the other way unless $f$ is a diffeomorphism.)

Lucas Henrique

TBH I couldn’t advance because of my potato abilities regarding complex analysis.

Ted Shifrin

@Lucas: What are we talking about? I've had too many people throwing things at me.

user76284

"Pullback of the metric tensor by $f$" sounds vaguely plausible :-) I'll have to look into it.

Ted Shifrin

BTW, it's more standard to say "the time-$1$ map associated to the vector field" than to use exp ...

user76284

22:40

Ah, ok. I took the idea of using exp from to the exponential map (since the length of the resulting geodesic is the magnitude of the vector).

Ted Shifrin

exp is only for geodesics ...

user76284

You can think of diffeomorphisms as a group and the vector fields that generate them as their Lie algebra generators though, right?

Adam

@CaptainAmerica16 lol well you need some latex gloves for starters, be competent in a form of martial arts that won't leave a forensic trace, and either some rope, or a bottle containing a cocktail of various sleeping pills

user76284

Pullback of the metric sounds about right, by the way: math.stackexchange.com/a/2849081/76284

Ted Shifrin

@Adam, maybe you should cease and desist.

Lucas Henrique

22:42

@TedShifrin I'm sorry... well, if you're interested, here.

Ted Shifrin

Oh, the FTA proof.

It's a reasonably standard proof based on the maximum value theorem for continuous functions on compact sets.

CaptainAmerica16

@Adam ??? I think it may be time for me to stop replying...

Ted Shifrin

@user76284 Yeah, you can also use the exponential to go from $\mathfrak g$ to $G$, but I'm telling you the way to say it so that a mathematician understands you.

@Lucas: I had to dig out the proof. By considering $g(z)=f(z+z_0)$, you still have a polynomial and now the point where $|g(z)|$ assumes its minimum value is at $z=0$.

user76284

Fair enough. By the way, what would be the right matrix norm for $\|J - I\|$? Does a Frobenius norm or operator norm make more sense?

Ted Shifrin

It depends what you're measuring. Operator norm gives the maximum stretch. Frobenius is not particularly allied to the underlying geometry. It's just treating the matrix as a giant vector, so you lose the meaning of the linear map, I suppose.

user193319

23:00

I'm trying to show that $n!$ diverges by bounding it below by something whose divergence is well-known. Someone in chat suggested $(n/2)^{n} < n!$, but this inequality fails for $n=10$ (perhaps even at smaller $n$)...What else could serve as a lower bound?

Ted Shifrin

No, the person suggested $(n/2)^{(n/2)}$.

You should understand why that is a lower bound!

user193319

Hmm...I tried that one, too; I tried proving it by induction, but I couldn't get it to work.

Ted Shifrin

You don't need induction.

user193319

$(n+1)! = (n+1)n! \ge (n+1) (n/2)^{n/2}$...but it wasn't clear how to proceed from this.

Why don't I need induction?

Ted Shifrin

Think about multiplying out what $n!$ is ... $n(n-1)(n-2)\dots (n/2)(n/2-1) \dots 2\cdot 1$. (Take the case $n$ even for convenience.)

user193319

23:04

Yes, by definition $n! = n(n-1)(n-2)\dots (n/2)(n/2-1) \dots 2\cdot 1$.

user76284

Oops, typo. You're taking the middle element and multiplying it by itself $n/2$ times. But this is clearly less than or equal to the second half of the whole product.

Ted Shifrin

Now what can you say about the first $n/2$ terms in that product?

user76284

Since the middle element is less than or equal to any element of the second half of the product.

Ted Shifrin

We're working on the understanding, @user76284.

user193319

First $n/2$ terms? Reading from the left to right?

Ted Shifrin

23:05

Left to right as I wrote it, yes.

user330477

For this question, math.stackexchange.com/questions/505209/…, i don't understand why $h_i \circ f$ is a polynomial. It is clear to me that it is a function in $n$ variables. It gives the reason that composition of two polynomials is a polynomials. which is true, but here we have a polynomial map not a polynomial function. Can anyone care explaining me?

Ted Shifrin

@user330477: Every component function of a polynomial map is a polynomial function.

user193319

Well, for $k = 0,1,...,n/2$, we have $n-k \ge n/2$, because that holds iff $n/2 \ge k$. Right?

Ted Shifrin

Damn, that sure makes it difficult.

But yes.

user330477

@TedShifrin Yes, but how am not sure how does this help.

user193319

23:07

How else would I prove the first $n/2$ terms are $\ge n/2$?

Ted Shifrin

Because if $k\ge n/2$, then $k\ge n/2$?

@user330477: You were the one who said you had a polynomial map instead of a polynomial function. What is the problem you're having, then?

Hi, Eric and Mike.

user330477

@TedShifrin You certainly cannot write $h_i(f_1(p),\cdots,f_m(p))=((h_i \circ f_1)(p),\cdots, (h_i \circ f_m(p))$, right?

Eric Silva

h

Ted Shifrin

No, that makes no sense.

user330477

@TedShifrin I only know that composition of two polynomials is a polynomial. But what to do in case when we have a polynomial map?

Ted Shifrin

23:11

But if $h(x,y)$ is a polynomial, what is $h(f(u),g(u))$, if $f$ and $g$ are polynomials?

MatheinBoulomenos

@user193319 Another lower bound: $e^n=\displaystyle \sum_{k=0}^\infty \frac{n^k}{k!}>\frac{n^n}{n!} \Rightarrow n!>(\frac{n}{e})^n$

user330477

@TedShifrin It is that $f(u)$ , $g(u)$ are points in $K$ and so $h(f(u),g(u))$ is a polynomial in $K$. Is it that easy or am I missing something?

$K[x,y]$ instead of $K$.

Ted Shifrin

If you put in polynomials for the variables in a polynomial expression and expand everything out, you get a giant polynomial. Write out the formula explicitly.

user330477

@TedShifrin I understand that but I don't want to do explicit calculation. Is the argument given above correct?

Ted Shifrin

I don't know, actually. You'd have to make everything very explicit.

user330477

23:19

@TedShifrin So, how would I explain it? The explicit way or is there an alternative?

Ted Shifrin

If you want something other than the sentence you in the post, you need to tell me that $h_i$ are polynomials in $m$ variables, and then you're substituting $f_j$ for the $j$th variable, where $f_j$ is a polynomial in how many variables?

MatheinBoulomenos

@user330477 sums, products and scalar multiples of polynomials are again polynomials, plugging in some polynomials into a (multivariable) polynomials is the same as doing a combination of those

user193319

@TedShifrin I'm having trouble with the odd case: $n! = n(n-1)! \ge n (\frac{n-1}{2})^{\frac{n-1}{2}}$

Ted Shifrin

You don't need to worry about the odd case, @user193319.

user193319

Why? I don't understand.

Ted Shifrin

23:22

Because if $n$ is even, then $(n+1)!>n!$ and you can use the same lower bound for $(n+1)!$, can't you?

user330477

@TedShifrin It is given that $h_i$ are polynomials. Sorry, if I didn' mention that.

user193319

...D'oh!

user330477

@MatheinBoulomenos I see this via a few examples. But how do I go about proving this?

Ted Shifrin

@user330477 I assumed that. But you didn't tell me how many variables were in the $f_j$ polynomials.

user330477

$f_i$ are polynomials in $n$ variables and there are $m$ of them?

. not ?

Ted Shifrin

23:28

OK.

MatheinBoulomenos

@user330477 which part? there are formulas for sums and products of polynomials. The part that plugging in polynomials into another polynomial is the same as multiplying, adding and scalar multiplying the inputs together in some way is just true "upon inspection". If $f(x_1, \dots, x_n)=\sum \lambda_{i_1, \dots, i_n} x_1^{e_1} \dots x_k^{e_n}$, then you get $f(h_1(x), \dots, h_n(x))= \sum \lambda_{i_1, \dots, i_n} h_1(x)^{e_1} \dots h_n(x)^{e_n}$,
this is a sum of products of polynomials, thus a polynomial

user330477

@MatheinBoulomenos Thank you so much. Now everthing is clear.

@TedShifrin Thank you so much for your help.

MatheinBoulomenos

glad to help

Ted Shifrin

@Mathein is always very helpful. I didn't help too much. But you're welcome.

MatheinBoulomenos

@Ted thanks :)

Leyla Alkan

23:33

Hi. Can anyone check if I've done the proof correctly.

this is an example of how subformulas are computed

user76284

The metric tensor is just the product of the Jacobian with its transpose, right?

Ted Shifrin

That's the pullback of the metric tensor, thought of as a matrix, yeah.

Mike Miller

@TedShifrin Ah! Thanks for clarifying the $C^r$ stuff. I was confused about that.

Ted Shifrin

That's the deep part of the theorem ... is the dependence on initial conditions being $C^r$.

Mike Miller

I was thinking about how the diff eq gained a derivative, so was confused about why it wasn't $C^{r+1}$. The $t$-direction comment was crucial.

Eric Silva

23:42

wut u talkin bout

Ted Shifrin

ohhh ...

right.

heya Eric.

Eric Silva

hlo

i see DE i perk

Ted Shifrin

what if I mumble "stochastic"? :D

Mike Miller

flowing a $C^r$ vector field

user76284

@MikeMiller Is this referring to my question?

Eric Silva

23:43

@TedShifrin depends on if the stochasticisity is additive or not

Mike Miller

yep

Eric Silva

if it's additive ill perk

otherwise ill run away

Ted Shifrin

LOL, Eric. You're so funny.

Eric Silva

i live for the laffs

Ted Shifrin

I understand. :)

Eric Silva

23:46

i like food and laffs in that order

user76284

The flow of a vector field $f$ on a manifold $\mathcal{M}$ can be defined by $\phi(0) = \mathrm{id}_\mathcal{M}, \dot{\phi}(t) = f \circ \phi(t)$. The geodesic flow on $\mathcal{M}$ can be defined by $\phi(0) = \mathrm{id}_\mathcal{M}, \dot{\phi}(0) = f, \dots$?

I know for an individual geodesic the second-order condition is $\nabla_\dot{\phi} \dot{\phi} = 0$.

Ted Shifrin

I'm not going to go into that now. It's too involved.

I need to leave in a few minutes, anyhow.

Riccardo

hi, can someone give me some feedback on the following statement? given a Fredholm operator $L : E \to F$, we can perturb it with a compact operator $k$ in such a way that $L+k$ is Fredholm and surjective. it's the analogue statement for families of Fredholm operators (indexing space is compact) true?

i.e. can find a continuous family of compact operators on the same indexing space ($X$) that would make my Fredholm operators $L_x$ surjective?

user76284

@TedShifrin Got time for one quick question? :-)

Mike Miller

I'm confident that's true, but I don't know a reference. @Riccardo

Maybe in Bleecker Booss-Bavnek's big book on index theory

Riccardo

23:52

oh, I'll check that book!

Mike Miller

Probably also in some old Atiyah paper.

Ted Shifrin

What's the question, @user76284?

Mike Miller

(As are most things known to mathematicians.)

user76284

Is it geometrically more "natural" to measure the deviation of the Jacobian $J$ or the metric tensor $G = J^T J$ to the identity matrix $I$?

Ted Shifrin

Well, I was suggesting norm squared earlier, so it amounts to the same thing.

Eric Silva

23:54

@MikeMiller if u say that name to a rhythm u got urself a sick beat buddy

user76284

Wouldn't the norm of $J = -I$ be nonzero in the first case and zero in the second, though?

Daminark

Public service announcement: Arrondo's projective varieties book sucks

Eric Silva

but what if it is you who does the sucking, eh?

bet you didnt think of that

Ted Shifrin

Yeah, if you're subtracting off $I$ as I suggested at one point, it's gonna be different.

Daminark

:0

Ted Shifrin

23:56

puts the room in time-out

Eric Silva

smart move

Mike Miller

pot calling the kettle nerd was the best burn of my life

Eric Silva

i try

user76284

@MikeMiller I guess the other thing about measuring the deviation of the metric tensor rather than the Jacobian is that all orthogonal matrices would end up with a norm of zero (since $X^T X = I$).

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