Mathematics - 2019-11-20 (page 1 of 2)

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12:00 AM

LOL, that sounds like a language problem.

In some languages, "please" and "you're welcome" are the same, for example.

wait did i paste that?

that's so weird

i pasted the linke

link*

Apparently so.

user image

6

A: Does multicollinearity increase the variance of the beta for every covariate or just those that are collinear?

I'd vote for singular values/eigenvalues/eigenvectors over determinants and adjugates for the way to approach this. TLDR: standard errors increase as the eigenvalues of $X^TX$ get increasingly small and this corresponds to the formation of valleys in the loss surface representing our increasing ...

Here we go

I'm confused what it means to "approach" a non-trivial nullspace

I thought a matrix either has a nontrivial one

or it doesn't.

What does it mean to "approach"?

or should I just keep studying

and ask this question after doing more problems

best that comes to mind is if I had vectors e1,e2, and e1+t*e3 in RR^3

They're talking about varying $X$, aren't they?

12:02 AM

yes

Think about a continuous parametrized family $X_t$ of matrices.

So it would be like different matrices and the sequence of them approaches it

ok

So if $X_0$ has trivial nullspace, then for all small $t$ $X_t$ also does. But you can't take the converse.

that makes sense. great. that was what i was confused by :) super

So they're imagining some nonsingular matrices approaching a singular one, I guess.

12:04 AM

small meaning "within some neighborhood", to be clear?

Yes, obviously depending on the particular family, etc.

right

cool

Let $X,Y, Z \sim \mathbb{U}(0,1)$ what is $P(X + Y > Z)$

As an example I'm fond of, consider $$M(x,y,z)=\begin{pmatrix} 1 & x & y \\ x & 1 & z \\ y & z & 1\end{pmatrix}$$

The condition for it to be PSD is $(x,y,z)\in [-1,1]^3$ and $\det M=1-x^2-y^2-z^2+2xyz\geq 0$

12:09 AM

I am trying to think about the whole space as the unit cube and there must be some prism in which $X+ Y > Z$ and the probability will then be the volume of that prism over the volume of the cube

The set of (x,y,z) in [-1,1]^3 which fulfills that is shown here: images.app.goo.gl/efbcFwqfSVW6QfjL7

@Semiclassical interesting

I know how to do this problem with integration but a peer told me the prism will actually be a triangular pyramid. However, I do not see it

If you're in the cube but outside that figure, the matrix has at least one negative eigenvalue. If you're inside the figure, you have all positive eigenvalues. So you get a surface of singular matrices with nonnegative eigenvalues.

If you ignore the nonnegative part and ask for what (x,y,z) the matrix is singular, you get this instead:

(well, that figure rotated appropriately)

Lots of fun stuff there.

@genescuba I don't agree that P(X+Y>Z) will be the volume of a triangular pyramid. However, P(X+Y<Z) would be.

hmm, why is tthat

Simplest way to see that is to note that Z and 1-Z occur with equal frequency, so P(X+Y<Z) = P(X+Y<1-Z)=P(X+Y+Z<1).

12:19 AM

intuitively X+Y > Z seems like that 2 sides of a triangle must add up to be greater than 3rd

At which point one notes that the surface x+y+z=1 intersects with the vertices (1,0,0), (0,1,0), (0,0,1) of the cube.

so you get a tetrahedron with those as vertices along with (0,0,0)

and that's a triangular pyramid

volume works out to be 1/6 I should think?

Yes. P(X+ Y > Z) = 5/6 is the solution

you could also pull off this logic with the original version: the plane x+y=z intersects the cube at (1,0,1), (0,1,1), and (0,0,0)

ah I see.

with the point (1,1,0) being the other vertex

I like symmetry, so P(X+Y+Z<1)=1/6 is neat

12:24 AM

how is this point (1,1,0) on the x+y = z plane ?

it's not. But it is a vertex of the cube and it lies in the region x+y>z, so it's the other vertex of the tetrahedron in that case

ah makes sense, yup those will be the vertices of the region x + y > z

right.

clever solution that saves integration !

yeah. of course, you have to remember how to get the volume of a cube.

which is easy to do with integration but I can never remember how to figure that out without calculus

12:30 AM

Hey guys! Could somebody please give me a hand with the following question, please? I am not really sure how to start.

Here's the question: If f is periodic with a period of $2a$ for some $a>0$, then $f(x)=f(x+2a)$ for all $x∈R$. Show that if f is continous, there exists some $c∈[0,a]$ such that $f(c)=f(c+a)$.

oh boy

with regards to the starred comment: I meant to say "volume of a cone". i certainly know how to justify the volume of a cube :S

I'm aware of the fact that the Intermediate Value Theorem should be used here, but I'm not sure how to use it in this instance.

12:50 AM

@Abwatts can you justify it using words?

@LeakyNun The question makes sense to me since I know that the y values in periodic functions repeat at some point, but I'm not sure what to do since the proof should work for all real numbers.. Am I supposed to use induction here?

If there is an exact sequence of $A$-modules: 0->M->N->N->0 with N finitely generated, do we have $M=0$?

I have the following calculation for an isometric embedding (into Minkowski space). Although I understand the calculation steps, I don't know why they are done. Could anybody clarify those steps?

In particular: understand the first two lines (how to equate the two metrics/ line elements), but then...
(i) why do we need those (two) requirements/ where are they coming from, and
(ii) why does it make sense to propose this solution ... and
(iii) why for $t>=-1$?

Here is the link to the calculation: https://www.dropbox.com/s/qmcefg3idmxyom6/Photo%20Nov%2019%2C%2016%2019%2017.jpg?dl=0

I am trying to show that for a surjective endomorphism M->M of a finitely generated A-module $M$(say by n elements), the kernel of the induced endomorphism A^n->A^n is the same as the kernel of M->M.

The above exact sequence is what I get after applying snake lemma.

@Abwatts: Here's a big hint. How can you rephrase the equality of two numbers?

1:03 AM

@Abwatts I think you misunderstood the question. They don't just need to "repeat at some point"

surely they repeat at some point, since e.g. f(0) = f(2a)

The question is why it repeats distance $a$ apart SOMEWHERE.

Still, you should answer my hint.

@TedShifrin When two numbers are pointing to the same value and thus, they explicitly said to be equal?

Pointing? I'm asking you to algebraically rearrange what it means to say $u=v$.

Or difference of two functions perhaps?

Aha.

$u=v$ if and only if $u-v=0$. So what function should you look at for your problem?

1:10 AM

$g(x)=f(x+a)-f(x)$?

I guess you looked at the answer someone posted to your question? But yes.

On what interval?

Re @Eric

Yes :P.. [0, a] where a > 0?

or simply all positive real numbers?

What interval do you want $c$ to be in?

$[0,a]$ for $a > 0$?

You don't need to repeat that about $a$. It is in the original statement.

OK, so that's a good interval to use. Now figure out how the intermediate value theorem gives you what you want.

1:15 AM

@TedShifrin Let $X$ be a smooth projective curve over $k=\Bbb F_q$ and $K$ be the function field and $\Omega_K$ be the field of meromorphic $1$-forms; why $\Omega_K \otimes_K K_v = k_v((u)) \ \mathrm du$ for a point $v$ on $X$?

Can I pick both $0$, and $a$ since they're in the interval c is in, and show where the function $g(x)$ is both above and below in that interval?

I thought about splitting the proof into 3 cases where the first 2 clearly prove the question right away, and the third one where we have to do some rearrangements to apply the IVT thm. Would that be a valid approach for this kind of question?

If you have that $X\sim\text{Gamma}(n,p)$ then how would you find the distribution of $Y=(X,\text{log}(X))^T$?

I wouldn't

(Note: The end goal here is to find the moment generating function of $Y$, but you have to calculate the Expectation of an exponential function in order to get that, which requires knowledge of the distribution, unless there is some other method I am unaware of)

you can use LOTUS

1:29 AM

I suppose. But that $\text{log}(X)$ in $E(e^{t_1X+t_2\text{log}(X)})$ is making me wary

LOTUS is LOTUS.

Also, doesn't, $\int_0^\infty e^{t_1x+t_2x}f_{(X,\text{log}(X))}(x,\text{log}(x))dx$ still require me to first find the distribution of $Y$?

$\displaystyle E\left(e^{t_1 X + t_2 \log(X)}\right) = \int_0^\infty e^{t_1 x + t_2 \log(x)} f_X(x) \ \mathrm dx$

Aaaaah, I'm dumb

That looks gross

1:35 AM

I failed to notice that $e^{t_1X+t_2\text{log}(X)}$ is just a function of $X$

In multivariable functions when we do a change of variables, the scaling factor is the absolute value of the determinant of the jacobian

why in single variable functions for change of variables (u-sub) do we not take the absolute value?

we do

though I guess the Gamma distribution has nice expectation values for both X and log(X), so I guess that integral could be tractable

@Semiclassical what

that's just another gamma integral

I guess. I haven't done a lot of gamma integrals.

1:37 AM

you can literally absorb t1 and t2 inside

if you remember the gamma pdf

oh, right. $e^{t_2 \log(X)}=X^{t_2}$

fair enough

Yeah, you distribute the $e^{\text{junk}}$. It actually is simplifying fast

@Rithaniel maybe looking for a kernel will help here after expanding

I think you might be able to manipulate the expression to get the gamma kernel

1:51 AM

End result looks like: $\frac{\lambda^n\Gamma(n+t_2+1)}{(\lambda-t_1)^{n+1}\Gamma(n)}$
Unless I made some mistake somewhere

What expectation value is that supposed to be?

though I guess I should also ask which parametrization you're using for Gamma

That's the moment generating function of $(X,\text{log}(X))$ where $X\sim\text{Gamma}(n,\lambda)$ and $f_X(x)=\frac{\lambda^n}{\Gamma(n)}x^{n-1}e^{-\lambda x}$

Good evening good friends. In Wikipedia is stated that in d demensional Vectorspace of
finite field with n elements there is exactly
$\prod\limits_{k=1}^{n-1} (q^n-q^k) $ so much basis.

Can you please redirect me to a topic/video/page where this is discussed and showed how one reaches such result?
My love and thanks.

Opsy dayzy that's not quite correct let me rephrase.

$\prod\limits_{k=1}^{d-1} (q^d-q^k) $ for d dimensions and q elements.

2:09 AM

Hi!
(sorry for the repost)
I have the following calculation (see image) for an isometric embedding (into Minkowski space). Although I understand the calculation steps, I don't know why they are done. Could anybody clarify those steps?

In particular: I understand the first two lines (i.e. how to equate the two metrics/ line elements), but then...
(i) why do we need those (two) requirements (line 3-4)/ where are they coming from, and
(ii) why does it make sense to propose this solution (line 5-6)... and

2:58 AM

@Rithaniel actually, that can't be quite right: If that's the moment generating function, then you should recover E[1]=1 when t1=t2=0

whereas it looks to me like you don't quite cancel properly to get that

Ah, I see my mistake. I assumed my integral was the expectation, but it was actually just an integral over the whole domain of the function

$\frac{\lambda^n\Gamma(n+t_2)}{(\lambda-t_1)^{n}\Gamma(n)}$

Let $M \in \operatorname{GL}_2(\Bbb Q)^+$ (positive determinant) and let $\Gamma \subset \operatorname{SL}_2(\Bbb Z)$ be a subgroup of finite index. Suppose $M\Gamma M^{-1} \subset \operatorname{SL}_2(\Bbb Z)$ also. Then $M\Gamma M^{-1}$ obviously has finite index in $\operatorname{SL}_2(\Bbb Z)$.

Consider the left coset spaces $\Gamma \setminus \operatorname{SL}_2(\Bbb Z)$ and $M\Gamma M^{-1}\setminus \operatorname{SL}_2(\Bbb Z)$. The map sending $\Gamma\cdot A \mapsto M\Gamma M^{-1}\cdot A$ for all $A \in \operatorname{SL}_2(\Bbb Z)$ is a bijection I think.. it's obviously surjective.. I just need to show that the kernel of that guy is empty and I think this follows from $\Gamma \cap M\Gamma M^{-1} = \varnothing$, does that sound reasonable? lol

Also hi :)

@Rithaniel nice

that's still only a necessary condition, of course: it's correct at t1=t2=0

Well, I suppose I should add some domain limitations. Like, $t_1<\lambda$

Hmm well actually $\Bbb I_2 \in \Gamma \cap M\Gamma M^{-1}$

3:12 AM

yeah. valid in some neighborhood of t1=t2=0

$\frac{\lambda^n\Gamma(n+t_2)}{(t_1-\lambda)^{n}\Gamma(n)}$ where $t_1>\lambda$

Or $\frac{\lambda^n\Gamma(n+t_2)}{(\vert\lambda-t_1\vert)^{n}\Gamma(n)}$ in general

Wait, no, that doesn't work

The $e^{(-\lambda+t_1)x}$ is fixed, and so $\lambda-t_1$ has to be positive

Yeah, the answer is $\frac{\lambda^n\Gamma(n+t_2)}{(\lambda-t_1)^{n}\Gamma(n)}$

@Rithaniel One more sanity check: If you set $t_1=0$ in that, then it's not a function of $\lambda$ anymore

Does it make sense that E[e^(t2 log X)]=E[X^t2] would be independent of $\lambda$?

Hmmmmm, that's an interesting question. I'm not sure

3:31 AM

So you'd have, for instance, $\displaystyle E[X]=\int_0^\infty x\frac{\lambda^n}{\Gamma(n)}x^{n-1}e^{-\lambda x}\,dx$ be a function of $n$ alone

Easiest way to see if that makes sense is to consider the substitution $u=\lambda x$

Well, $E(X^t)=\int_0^\infty \frac{\lambda^n}{\Gamma(n)}x^{n+t-1}e^{-\lambda x}dx=\frac{\Gamma(n+t)}{\Gamma(n)}\int_0^\infty \frac{\lambda^n}{\Gamma(n+t)}x^{n+t-1}e^{-\lambda x}dx=\frac{\Gamma(n+t)}{\Gamma(n)}$

Wouldn't that last integral need to be $\int_0^\infty \frac{\lambda^{n+t}}{\Gamma(n+t)}x^{n+t-1}e^{-\lambda x}\,dx=1$?

Ah, shoot, yeah.

Right. So you endn up with $E(X^t) = \frac{\Gamma(n+t)}{\Gamma(n)}\lambda^{-t}$

which makes sense if only for dimensional reasons: If $x$ and $\lambda$ have units, then $\lambda x$ should be dimensionless (otherwise $e^{-\lambda x}$ is rather nonsense)

so $E((\lambda X)^t)$ should be dimensionless, but not $E(X^t)$

Newest form: $\frac{\lambda^n\Gamma(n+t_2)}{(\lambda-t_1)^{n+t_2}\Gamma(n)}$

3:38 AM

Okay. That one matches what I had in Mathematica :)

Excellent use of technology to head me off at the pass :P

yep. but you'll note that the sanity checks didn't really involve any tech

Indeed, but you gotta know to look for them

definitely

So that's the lesson: If you have something (and aren't pressed for time) examine it with a fine-toothed comb

3:42 AM

eh, the lesson for me is to consider simple limiting cases

also, when I see a distribution like $f_X(x)\,dx=\frac{\lambda^{n}}{\Gamma(n)} x^{n-1}\,dx$, my immediate impulse is to consider the substitution $u=\lambda x$

that gives $f_X(x)\,dx = \frac{1}{\Gamma(n)} u^{n-1}\,du$, which is $\lambda$-independent

So $$E[e^{t_1 X}X^{t_2}] = \int_0^\infty e^{t_1 x}x^{t_2}f_X(x)\,dx = \lambda^{-t_2}\int_0^\infty e^{(t_1/\lambda)u}u^{t_2} \frac{1}{\Gamma(n)}u^{n-1}\,du$$

oh, bleh. I missed something obvious in there

$f_X(x)\,dx = \frac{\lambda^n}{\Gamma(n)}x^{n-1}e^{-\lambda x}\,dx=\frac{1}{\Gamma(n)}u^{n-1}e^{-u}\,du$

Main thing is that the integral is only a function of $\lambda$ via $t_1/\lambda$

(which, comparing with your earlier form, corresponds to writing the result as $\lambda^{-t_2}(1-t_1/\lambda)^{-n-t_2}\frac{\Gamma(n+2)}{\Gamma(n)}$

4:06 AM

I assume that $\Gamma(n+t_2)$ in the numerator?

woops, yeah

Alright, then woo, it checks out

So, here's something I puzzled out earlier today: If $N,M$ are random variables with joint distribution $n^m$ where $M$ is discrete with support across the natural numbers (including 0) and $N$ is continuous with support on the interval $(x,1-x)$, then what is $x$?

well, one should presumably have $\sum_{m=0}^\infty \int_x^{1-x}n^m\,dn=1$

So $\sum_{m=0}^\infty n^m = \frac{1}{1-n}$

and thus $1=\int_x^{1-x}\frac{dn}{1-n}=\left[-\ln(1-n)\right]_x^{1-x}=\ln(1-x)-\ln x=\ln(1/x-1)$

so $1/x-1=e\implies x=1/(1+e)$

So that's neat

Yeah, that's what I got. Also, if it's support is $(0,x)$ then $x=1-\frac{1}{e}$. It's relatively easy for people at this level of math, but what do you think about it for an undergrad assignment?

eh. I think it's fine: The first thing you should know about distributions is that they're normalized.

4:22 AM

Also, in the $(0,x)$ case, we have that $E(N)=\frac{1}{e}$ and $E(M)=e-2$
in the $(x,1-x)$ case, $E(N)=\frac{2}{1+e}$ and $E(M)=e-\frac{1}{e}-1$

nice

1 hour later…

5:43 AM

Hello

1 hour later…

6:55 AM

@JoeShmo Ah yes, I forgot lol

1 hour later…

8:25 AM

Can every compact set in a manifold be written as a finite union of compact sets each of which lie in a chart?

Here is my approach: let $\{U_\alpha\}$ cover $K$, where $U_\alpha$'s are charts. Then some $U_1, \dots, U_n$ cover $K$. Let $K_i = \overline{K\cap U_i}$. Then clearly $K = K_1 \cup \dots \cup K_n$.

But how do I show that $K_i$ is in some chart?

Alessandro Codenotti

@feynhat the whole manifold if it's compact

Or a non simply connected one, like a circle on the torus

?

Are you giving counter-examples?

@AlessandroCodenotti

Alessandro Codenotti

8:59 AM

Yes

9:11 AM

@AlessandroCodenotti I don't see how circle is a counter example.

Alessandro Codenotti

9:43 AM

Not any circle, but one of the nontrivial ones, like turning around the torus

So the red and magenta circles in the image at the top

It's not contained in a single chart because they are homeomorphisms with $\Bbb R^2$, which is contractible

1 hour ago, by feynhat

Can every compact set in a manifold be written as a finite union of compact sets each of which lie in a chart?

union

Alessandro Codenotti

Wait, I thought you wanted the union to be disjoint

No. I just want each $K_i$ to be inside a chart.

Alessandro Codenotti

Oh ok then it looks doable

10:45 AM

@AlessandroCodenotti This answer seems to work. But I am not so sure about what he means by balls.

Are those balls taken in some local coordinate about x?

...or is he thinking of $M$ as being embedded in some $\mathbb{R}^l$ and taking balls there?

Alessandro Codenotti

Every topological manifold is metrizable (if you ask topological manifolds to be second countable)

Answering your doubts from the comments @feynhat

11:11 AM

@AlessandroCodenotti I don't think I have sufficient knowledge of metrization theorems. Can you point me to the theorem which says that second countable => metrizability.

Alessandro Codenotti

11:22 AM

Urysohn metrization theorem

Hausdorff+regular+second countable implies metrizable

To show that a manifold is regular you can use that locally euclidean implies locally compact and that Hausdorff+locally compact implies completely regular

@BaiduryaMathaddict But if that is the case, i.e., if lim_{x->c} f(x)^{g(x)} = lim_{x->c} f(x)^{lim_{x->c} g(x)} when why do I often see the 'exponential-log technique' used to evaluate these limits? That is, we evaluate lim_{x->c} f(x)^{g(x)} using lim_{x->c} Exp[ g(x) Log(f(x)) ]. The first technique is simpler so why does this 'exponential-log technique' get used?

@AlessandroCodenotti Thanks a lot.

But, I think for each $K_i$ to lie in a chart, we should take the balls in local coordinates.

What I mean we cover $K$ with $B(x, d/3)$ where each ball is taken in local coordinates about $x$. Then we can cover $K$ with finitely many of these say $B_i$. And then, set $K_i = \overline{K \cap B_i}$. Clearly, $K = \cup K_i$, and $K_i \subset B(x, 2d/3)$ which is in a chart.

12:26 PM

user image

Sets from top to bottom:
1. Countably infinite/Countable (included here for completeness)
2. omega finite
3. Stackel finite
4. Tarski finite
5. Bounded, Streamless or Noetherian (From intuitionist set theories)
6. Amorphous
7. Δ2
8. Δ3, which includes Motowoski finite
9. Δ4
10. Δ5
11. Kuratowski finite
12. Hyperset (from non well founded set theories)
13. Fuzzy set
(NB I knew they exists, but I am too lazy to find the axioms that knocks out the Deltas one by one)
I wonder if there is a way to prove, that omega finite is the smallest possible finite in the hierarchy of finite definitions...

Akiva Weinberger

12:44 PM

@feynhat Yes

3 hours later…

Akiva Weinberger

3:24 PM

Where's the center of mass of a hemisphere

(just the 2D surface, not the 3D solid)

Is it just halfway up?

hi

Akiva Weinberger

hı

user image

I made a graph

Akiva Weinberger

Nic Cage voice ♫ Look at this

Wait

Was that not Nick Cage

Apparently it was not

look at this graph

3:32 PM

It definitely was Nick Cage

It reminds me of G-bach's C

Akiva Weinberger

Apparently it's Chad Kroeger, whoever that is

hes the singer for Nickleback

Chad Kroeger is Nick Cage's evil twin

Akiva Weinberger

but it's from Nickelback which makes me feel like, at least in some weird sense, I was close

Not OK:

If you plug "If you plug $x$ into itself, the resulting sentence is false" into itself, the resulting sentence is false

OK:

If you substitite "If you substitute $x$ into itself, the resulting string of symbols cannot be obtained from the axioms" into itself, the resulting string of symbols cannot be obtained from the axioms

3:40 PM

I maintain that that's gross

@AkivaWeinberger Yes, because surface area of a zone of a sphere is dependent only on the "height" of the zone.

Akiva Weinberger

The latter is how Gödel's First Incompleteness Theorem is proven

@TedShifrin Ah, yeah, 'cause of the area-preserving map to the cylinder

Here's something I'm seeing in a question and I feel like I should know of

DogAteMy, ayup.

First, a familiar observation: Consider the space curve $x\mapsto (x,x^1,x^2,\cdots, x^{d})$ in RR^{d}$. Then any $(d-1)$-plane intersects this moment curve in at most $d$ points.

Does that still work if we replace $x^k$ with $f_k(x)$ where $\{f_k(x)\}$ is a sequence of orthogonal polynomials?

(The specific question is for the case of Chebyshev polynomials of the first kind.)

3:48 PM

The original is based on the Vandermonde determinant, but that can be rephrased as a linear independence statement in function space, so probably yes.

Right.

Oh, I'm being dumb. It's just the fundamental theorem of algebra when you pull back the equation of the plane.

Ah. That seems to match the answer so far: math.stackexchange.com/a/3443587/137524

Though I agree with Omnomnomnom that the answer hasn't yet verified that the determinant is nonzero anywhere. (I'm sure it is, but their last comment seems to miss the point.)

Akiva Weinberger

> First, a familiar observation: Consider the space curve $x\mapsto (x,x^1,x^2,\cdots, x^{d})$ in $\mathbb{R}^{d}$. Then any $(d-1)$-plane intersects this moment curve in at most $d$ points.

Missing $

sigh

this is the danger in being too lazy to put mathjax on

The answer's last comment shows that the product $\prod_{j<k}(y_j-y_k)$ is nonzero. But the question is whether the determinant is nonzero, and I don't see how their comment addresses that.

3:53 PM

Because that's the Vandermonde determinant.

But their matrix isn't the Vandermonde matrix.

Oh, they're just arguing by the root-factor theorem for polynomials.

If $f$ is a polynomial and $f(x)=f(y)$, then $x-y$ is a factor of $f$.

Right. But that only shows $p(x)=A q(x)$.

could still have $p(x)=0$.

You just have to show it's nonzero somewhere.

Yes, I agree. I just don't see where they've done that in their answer.

3:56 PM

You're right, there's a gap in the argument.

Okay.

Now I'm gonna eat breakfast :P

lol

I mean, I would very much doubt that the determinant vanishes identically. But just because I find it implausible for X to be false, doesn't mean I know how to prove it's true :P

Oh, but in the last comment, the answerer addressed it. Is he wrong?

It was shortly after your objection.

What I see is him addressing that $\prod_{j<k}(y_j-y_k)$ is nonzero.

But the point to fill in is whether the determinant itself is nonzero.

3:58 PM

No, look at the final comment.

Oh, actually, I don't know where that comes from.

Yeah.

Yeah, I think he's addressing the original determinant.

Hmm.

He's telling you which $x_i$ give a specific nonzero result.

Well, he's telling me what $x_i$ -should- give a nonzero result.

4:00 PM

Well, if you verify his product formula for the $\sin x_i$ ...

Anyhow, bye for now.

later.

My own inclination would be to observe that the leading term of T_n(y) is (2y)^n/2 (for n>0) and argue that the leading term of the determinat is therefore blah blah blah

4:17 PM

Hello, I got a question

How can we convert a proper fraction transfer function to state space

say $G(s) = \frac{s^3+s^2 + 3}{2s^3 + 4s + 7}$

This will cause $\frac{Y(s)}{U(s)} = G(s) = \frac{a_1 s^2 + a_2 s + a_3}{2s^3 + 4 s + 7} + D$ but now how to proceed

I should ask on site

wikipedia seems to have an entry on this subject: en.wikipedia.org/wiki/Realization_(systems)

@Semiclassical It has for case of strictly proper

ah. I'm guessing this isn't because there's no quartic term in the denominator?

yes, i mean I can do it when power of s on numerator is less than that on denominator, but if it is equal then its a problem

for me

One does have $$G(s)=\frac{s^3+s^2+3}{2s^3+4s+7}=\frac{1}{2}+\frac{2s^2-4s-1}{4s^3+8s+14}$$

So if one can get rid of that constant $1/2$, then one is back to a proper transfer function?

4:25 PM

Yes! but how is it to be done, i mean getting rid of that 1/2

yeah

I want to say it's just "take D=1/2"

so that $y(t) = Cx(t)+\frac12 u(t)$

I'm not conversant with the conventions, though. So it's not clear when D is a matrix vs. a scalar when I read that WP entry

looking at the article: A is n-by-n, B is 4-by-1, C is 1-by-4 and D is 1-by-1

so taking D=1/2 seems reasonable enough

I mean if I only had $\frac{Y(s)}{U(s)} = \frac{2s^2-4s-1}{4s^3+8s+14}$ then I get $4y''' + 8y'+14y = 2u''-4u'-u$ right but here due to extra constant 0.5 how can we directly conclude your statement, I am not able to understand that

4:46 PM

@Semiclassical according to some people, the volume of $[0,1]^n$ is $1/n!$

:/

volume a cube is a tricky thing

lol

i meant to say volume of a cone :P

(and how to get it without calculus)

@jeea looking at wikipedia, I'm tempted to suggest changing variables from $y$ to $z=y-u/2$

there's a convention in which the volume element is $n!\,dx^1\wedge\cdots\wedge dx^n$

awful!

simply because, if $y(t)=Cx(t)+Du(t)$ with $D=1/2$, then $z(t) = Cx(t)$ i.e. $D=0$

I think it really does just come down to $D\neq 0$.

4:57 PM

@semiclassical thanks a lot I got it!

neat

5:35 PM

@RyanUnger Blah.

amusingly, 1/n! is the volume of the cone I was having in mind :P

I did manage to convince myself how you'd dissect a cube to get the usual result, though.

@Semiclassic: You can do a synthetic geometric argument that three pyramids fit in a cube. And then it's standard to approximate cones by pyramids.

Hmm. I managed to get 6 pyramids into a cube, though I could just as easily have done 3 pyramids into a triangular prism.

also, I'm being sloppy and equating cone with pyramid.

Oh, oops, if I'm doing triangular pyramids, I guess I'm wrong.

I guess we could look at Euclid and see what he did for this.

hmm

Book XII, Proposition 7: Any prism with a triangular base is divided into three pyramids equal to one another with triangular bases.

5:40 PM

Oh, you beat me.

I was looking at an old-fashioned paper book.

huzzah for google-fu

he actually has it in more generality than I did: I only tried to do the case of a right triangular prism whose cross-section was a 45-45-90 triangle

you have officially snatched the pebble from the master's hand grasshopper

you may leave the temple :P

Oh, look, it's skull.

eh. except that Euclid beat us all to it :P

hi, all

5:43 PM

I wonder if that dissection works in higher dimensions.

I know that there's Dehn stuff going on there

that is: Can I always dissect the n-cube into n! congruent polyhedra?

In the 3-D case, you have to change one of the pyramids. Are they all actually congruent?

I just think they have equal volume.

Hmm.

@Semiclassical wouldn't that make Euclid the grand master :P

I think they're all the same in 3D up to reflection

can't say I'm totally sure about that tho

I thought about this years ago, but don't berember.

5:52 PM

@TedShifrin did you keep any of the Elements after retirement?

:^)

Yes, I kept the book that was given to me as a gift years ago by a grateful student.

We are certainly grateful that you and Semiclassical come to the chat room regularly :-)

As one of my good friends likes to say, blah, blah, blah :)

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