User wang tao - MathOverflow

Regarding Kolmogorov's Superposition Theorem

2010-11-28T05:22:24Z

Hi Experts,

I have question regarding Kolmogorov's Superposition Theorem:

It is known that: Let ${f(x_1,x_2,...,x_m): \Re^m :=[0,1]^m \to \Re}$ be an arbitrary multivariate continuous function. From Kolmogorov’s Superposition Theorem we have the following representation:

${f(x_1,x_2,...,x_m)= \sum_{q=0}^{2m} \Phi_q (\sum_{p=1}^m \phi_{p,q}(x_p))}$

with continuous one-dimensional outer functions ${\Phi_q}$ and inner functions ${\phi_{p,q}}$. All these functions are defined on real line. The inner functions ${\phi}$ are independent of function ${f(x_1,x_2,...,x_m)}$.

Question is: Is it possible to find inner functions ${\phi_p{(x_p)}}$ which is independent of $q$, that satisfies the superposition theorem:

${f(x_1,x_2,...,x_m)= \sum_{q=0}^{2m} \Phi_q (\sum_{p=1}^m \phi_p (x_p))}$

Where ${\Phi_q, \phi_p, N}$ can be selected and defined where appropriate.

It is critical to our works on nonlinear control, and we look forward to your advises on possible solutions, tips, related documents,etc.

Thank You! Wang Tao

Solving Functional Equation

2010-11-30T10:01:25Z

Continue with my previous question “Regarding Kolmogorov's Superposition Theorem”, here are some further questions:

Question-1 Is it true: for any given $C^1$ continuous real function $f(x, y): \Re^2 \to \Re^1$, there exist $C^1$ continuous real functions $\Psi(x), w(x,y), u(x)$ such that: $u(f(x,y))=u(x)\Psi(\frac{w(x, y)}{u(x)})$ where $\frac{du(x)}{dx}\ge 0$; $\frac{\partial w(x, y)}{\partial y}\ge 0$

Question-2 Can these functions $\Psi(x), w(x,y), u(x)$ be solved, analytically or approximately?

It is found that constraint $\frac{\partial w(x,y)}{\partial y}\ge 0$ is hard to deal with. Without this condition, there is one obvious solution $u(x)=x, w(x, y)=f(x,y), \Psi(x)=x$, where $\frac{du(x)}{dx}=1>0$, but $\frac{\partial w(x, y)}{\partial y}\ge 0$ may not satisfy.

One may find by letting $w(x,y)=w(y)$, this becomes the question we asked in "Regarding Kolmogorov's Superposition Theorem" but since the answer is NO found by AgCl, we want to remove the constraint $w(x,y)=w(y)$ hoping solution exist...

Regarding a Feature of Multivariate Real Function

2010-12-03T02:16:38Z

Any real function can be expressed as a function of the sum of two monotonic real functions?

More precisely, for real function p(x, y), there exist continuous real functions P(x), h(x,y), g(x) such that:

$p(x,y)=P(h(x,y)+g(x))$

Where $P(x), h(x,y), g(x)$ are arbitrary satisfying $\frac {d(g(x))}{dx}>0$, $\frac {\partial h(x,y)}{\partial y}>0$

This is equivalent to mine another question “Solving Functional Equation”. By letting $h(x,y)=ln(w(x,y)), g(x)=-ln(u(x)), p(x,y)=\frac {u(f(x,y))}{u(x)}$, we have:

$p(x,y)=\frac {u(f(x,y))}{u(x)}=F[ln(w(x,y))-ln(u(x))]=F(ln \frac {w(x,y)}{u(x)})=\Psi (\frac {w(x,y)}{u(x)})$

Thank you very much!

Answer by Wang Tao for Solving Functional Equation

2010-12-01T00:15:12Z

So far what I can think of on this question is:

Rewrite $u(f(x,y))=u(x)\Psi(\frac{w(x,y)}{u(x)})$ as: $\frac {u(f(x,y))}{u(x)}=\Psi(\frac {w(x,y)}{u(x)})$ Since $\Psi$ can be arbitrary, we can further rewrite it as: $\Phi(\frac{u(f(x,y))}{u(x)})=\Gamma(\frac {w(x,y)}{u(x)})$ where $\Psi=\Phi^{-1}(\Gamma)$ Above equation shows that any $w(x,y),u(x)$ which makes $(\frac{w(x,y)}{u(x)})$ span into $(\Phi(\frac{u(f(x,y))}{u(x)})$ is a solution.

For example, given $f(x,y)=ax+by$, select $u=x$, then: $\Phi(\frac{u(f(x,y))}{u(x)})= \Phi(a+b\frac{y}{x})$ which can be spanned by $\frac {y}{x}$, so $w=y$ is a solution.

Question: how to find the appropriate $w(x,y),u(x)$ to span into $\Phi(\frac {u(f(x,y))}{u(x)})$

Answer by Wang Tao for Regarding Kolmogorov's Superposition Theorem

2010-11-29T13:26:51Z

Hi AgCl

Thank you very much for the reply. It helps to save my time and effort. As this topic is important to me, I want to think about my answer to you carefully. Allow me some days to draft a proper reply to you?

Warm Regards Wang Tao

Comment by Wang Tao

2010-12-04T02:40:26Z

By setting $P(x)=e^x, g(x)=ln(x), h(x,y)=ln(y)$, we have: $p(x,y)=xy=e^{ln(xy)}=e^{ln(x)+ln(y)}=P(g(x)+h(x,y))$ Where: $\frac {\partial lnx}{\partial x}=\frac {1}{x}>0$, $\frac {\partial lny}{\partial y}=\frac {1}{y}>0$ for $x>0, y>0$.

Comment by Wang Tao

2010-12-01T05:44:30Z

Yes. This yields one solution. However any more functions other this? I gave one example in belowing post, hope to have your advices?

Comment by Wang Tao

2010-12-01T05:39:16Z

Hi Willie, Thanks for the advise. I intended to do so but found the words exceeded the limit, so I changed to use the "Add Another Answer". But I will keep your advice in mind.