User ohad asor - MathOverflow

RFC for definite integral connection to second derivative

2013-01-12T00:31:08Z

Hi,

During my research I found an interesting fact, and I'd like to know if it's interesting for others as well. Find a function $g(x,t):[0,T]\times[0,T]\rightarrow[0,T]$ such that for any twice differentiable $f(x):[0,T]\rightarrow[0,T]$ such that $f(0)=f'(0)=0$, the equality $$ f(x)=\intop_0^Tf''(t)g(x,t)dt$$ holds. Note that $g$ is independent of $f$.

I found such a $g$, and I'll post it as an answer soon. I'd like to know if this is simple/known/interesting.

Elaborating Mercer's theorem (RKHS) on Cameron-Martin space $k(x,y)=\min(x,y)$

2013-01-08T21:04:35Z

Hi,

I'm trying to employ Mercer's theorem on the kernel $k(x,y)=\min(x,y)$. It is known (and easy to verify) that this is a nonnegative-definite kernel over $[0,T]$ for any $T>0$.

Fix $T>0$. Let's calculate the eigenfunctions of the transformation $ \mathscr T_kf=\intop_{0}^T k(x,y)f(y)dy$:

$$ \lambda\psi(x)=\intop_{0}^T \min(x,y)\psi(y)dy=$$ $$ \intop_{0}^x \min(x,y)\psi(y)dy - \intop_{T}^x \min(x,y)\psi(y)dy=$$ $$ \intop_{0}^x y\psi(y)dy - x\intop_{T}^x \psi(y)dy\implies$$ $$ \lambda\psi'(x)= x\psi(x)-x\psi(x)-\intop_{T}^x \psi(y)dy\implies$$ $$ -\lambda\psi''(x)= \psi(x)\implies$$ $$\psi(x)=C_1\sin\frac x {\sqrt\lambda} + C_2\cos\frac x {\sqrt\lambda}$$

it seems like we're allowed to pick $C_1=1$ and $C_2=0$. So we pick $\psi_n(x)=\sin nx $ and $\lambda_n = n^{-2}$. Then Mercer's theorem actually says that we should get: $$ \min(x,y)=\sum_{n=1}^\infty n^{-2}\sin nx\sin ny$$

this all seem very nice, but when evaluating this numerically, it doesn't work. I tried also to normalize $\psi$ by dividing by its norm which is $\sqrt {\frac 1 {4n} (2nT-\sin2nT)}$, and it didn't help.

I also tried to substitute the original solution with $C_1,C_2$ in the original eigenvalue problem equation, and then to calculate $C_1,C_2$, but they turned out to depend on $x$, which is of cource unacceptable.

So what's wrong here?

I also asked the question on http://math.stackexchange.com/questions/272857/elaborating-mercers-theorem-rkhs-on-cameron-martin-space-kx-y-minx-y, but no answers there. I'll post in either forum immediately if an answer will come.

Injective with Finite Discontinuities Mapping from $\mathbb R^n$ to $[0,1]$

2013-01-02T06:26:14Z

Hi,

as a continuation to the fully answered question:

http://mathoverflow.net/questions/117609/injectiveintregrable-mapping-from-mathbb-r3-to-mathbb-r

Can one think of an injective $f:\mathbb R^n\rightarrow[0,1]$ that has only a finite number of discontinuities? Or maybe one can come with some topological claim showing it is impossible?

Injective&Intregrable Mapping from $\mathbb R^3$ to $\mathbb R$

2012-12-30T10:15:32Z

Hi,

Can one think of an injective and Riemann integrable $f:\mathbb R^3\rightarrow\mathbb R$? (of course it cannot be continuous)

Cameron-Martin like RKHS

2012-12-23T11:57:02Z

Hello,

I know that $k(x,y)=min(x,y)$ is the reproducing kernel of the Cameron Martin space of all i.i.d. RVs of Brownian motion at different times, with the $cov$ inner product.

What is the RKHS generated by $k(x,y)=min(f(x),f(y))$ for some $f$?

Convolutional Equation

2012-11-13T03:36:50Z

Hi,

Given C>0. Let $f,g,h$ be $L^2$ functions such that $f,g,h$ have a compact and finite-measure support, and $f*f(x)=g*g(x)=h*h(x)=f*g(x)=g*h(x)=f*h(x)=0$ (where $*$ is the convolution) for all $|x| > C$.

Does that imply $f(x)=g(x)=h(x)=0$ for all $x$? What is I won't require $f*f=g*g=h*h=0$? What if we had two functions $f,g$ instead of three $f,g,h$, and $f*f(x)=g*g(x)=f*g(x)=0$ for all $|x| > C$? Will it imply they both identically zero everywhere?

If not, please provide (or hint to) a counterexample.

Space of Bandlimited Functions

2012-11-12T07:28:08Z

Hi,

I was asking myself about some necessary and/or sufficient conditions for a function to be bandlimited (i.e. its Fourier transform is zero t residing out of [-B,B] for some B>0). Of course, if a function is bounded (timelimited), it cannot be bandlimited. But for a non-bounded function, how can we tell if it's bandlimited or not? Or, how can we know if a function is both time- and band- unlimited? Any sufficient/necessary conditions. I'll be glad if you can share with me some known results.

Thanks!

Comment by Ohad Asor

2013-01-12T17:22:17Z

Thanks Gerald I appreciate your help

Comment by Ohad Asor

2013-01-12T15:30:21Z

I got $-min(x,t)$

Comment by Ohad Asor

2013-01-12T15:06:15Z

can you express it in terms of max/min?

Comment by Ohad Asor

2013-01-12T13:12:55Z

I dont understand where $f$ comes into the picture

Comment by Ohad Asor

2013-01-12T13:11:22Z

so maybe I didn't get it.. doesn't Dirac and Heavyside discontinuous?

Comment by Ohad Asor

2013-01-12T11:31:16Z

Hint: $g$ is continuous.

Comment by Ohad Asor

2013-01-09T10:22:43Z

now it really works :)

Comment by Ohad Asor

2013-01-09T08:56:47Z

Thanks for your help! Unfortunatly, this doesn't work as well. Here's my code: #include <iostream> #include <cmath> using namespace std; long double phi(long double x, long double n, long double t = 2) { static long double pi = acos((long double)-1); return sin((n+.5)*pi*x/t)*sqrt(t*2.)/(pi*(n+.5)); } int main(int, char**) { long double sum = 0, x=.8,y=.3; for (long double n = 1;n< 100000;n++) { sum += phi(x,n)*phi(y,n); if (((int)n)%1000) cout<<n<<' '<<sum<<endl; } return 0; }

Comment by Ohad Asor

2013-01-08T22:01:05Z

here's my code: ## #include <iostream> #include <cmath> using namespace std; long double phi(double x, double n) { return sin(n*x)/n; } int main(int, char**) { long double sum = 0, x=8,y=1; for (long double n = 1;n< 100000;n++) { sum += phi(x,n)*phi(y,n); if (((int)n)%1000) cout<<n<<' '<<sum<<endl; } return 0; } ##

Comment by Ohad Asor

2013-01-08T21:50:05Z

wow that's amazing. you actually got $ \min(x,y)=\sum_{n=1}^\infty n^{-2}\sin nx\sin ny$?

Comment by Ohad Asor

2013-01-04T20:09:26Z

Of course, but once it's answered in either locations, I'll immediately post the answer on the second one. I have no intention to abuse (and will not, of course).

Comment by Ohad Asor

2013-01-04T13:48:19Z

Most Significant Bit (the leftest one)

Comment by Ohad Asor

2013-01-02T14:59:29Z

yes, I do understand now, many thanks!! btw when I practically want to use what we learnt here (especially in the previous question), I encounter a problem that $f(x)$ for some $x\in\mathbb R^n$ requires $nd$ digits to be represented numerically, when $d$ is the maximum number of digits in the elements of $x$. so this can be numerically unstable. I'm now trying to find such injective&integrable function that would be more numerically stable for computers.

Comment by Ohad Asor

2013-01-02T07:25:59Z

i agree i dont have a full understanding, but this will come with time... i'm doing my best

Comment by Ohad Asor

2013-01-02T07:12:34Z

Thanks! Please note that in the mentioned question (in that link), there is an example with only countably many discontinuities.