Return to Question

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.

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 https://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.

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.

EDIT: As a consequence of the development and the answers, we get the following interesting identity (by differentiating $min(x,x)$ which gives $1$):

$$ \sum_{k\in\mathbb N,k\;odd} \frac {\sin k\pi x} {\pi k} = \frac 1 4$$

for all $x\in\(0,1)$, and hence:

$$ \sum_{k\in\mathbb N,k\;odd} \cos k\pi x = 0$$

for all $x\in\(0,1)$. so:

$$ \sum_{k\in\mathbb N,k\;odd} \cos k = 0$$

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.

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.

EDIT: As a consequence of the development and the answers, we get the following interesting identity (by differentiating $min(x,x)$ which gives $1$):

$$ \sum_{k\in\mathbb N,k\;odd} \frac {\sin k\pi x} {\pi k} = \frac 1 4$$

for all $x\in\(0,1)$, and hence:

$$ \sum_{k\in\mathbb N,k\;odd} \cos k\pi x = 0$$

for all $x\in\(0,1)$.

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.

EDIT: As a consequence of the development and the answers, we get the following interesting identity (by differentiating $min(x,x)$ which gives $1$):

$$ \sum_{k\in\mathbb N,k\;odd} \frac {\sin k\pi x} {\pi k} = \frac 1 4$$

for all $x\in\(0,1)$, and hence:

$$ \sum_{k\in\mathbb N,k\;odd} \cos k\pi x = 0$$

for all $x\in\(0,1)$. so:

$$ \sum_{k\in\mathbb N,k\;odd} \cos k = 0$$