User wang tao - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T04:29:12Z http://mathoverflow.net/feeds/user/11131 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47557/regarding-kolmogorovs-superposition-theorem Regarding Kolmogorov's Superposition Theorem Wang Tao 2010-11-28T05:22:24Z 2013-03-22T23:08:59Z <p>Hi Experts,</p> <p>I have question regarding Kolmogorov's Superposition Theorem:</p> <p>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:</p> <p>${f(x_1,x_2,...,x_m)= \sum_{q=0}^{2m} \Phi_q (\sum_{p=1}^m \phi_{p,q}(x_p))}$</p> <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 ${f(x_1,x_2,...,x_m)}$.</p> <p>Question is: Is it possible to find inner functions ${\phi_p{(x_p)}}$ which is independent of $q$, that satisfies the superposition theorem:</p> <p>${f(x_1,x_2,...,x_m)= \sum_{q=0}^{2m} \Phi_q (\sum_{p=1}^m \phi_p (x_p))}$ </p> <p>Where ${\Phi_q, \phi_p, N}$ can be selected and defined where appropriate.</p> <p>It is critical to our works on nonlinear control, and we look forward to your advises on possible solutions, tips, related documents,etc.</p> <p>Thank You! Wang Tao </p> http://mathoverflow.net/questions/47778/solving-functional-equation Solving Functional Equation Wang Tao 2010-11-30T10:01:25Z 2012-07-12T10:22:01Z <p>Continue with my previous question <a href="http://mathoverflow.net/questions/47557/regarding-kolmogorovs-superposition-theorem" rel="nofollow">“Regarding Kolmogorov's Superposition Theorem”</a>, here are some further questions:</p> <p>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 $\frac{du(x)}{dx}\ge 0$; $\frac{\partial w(x, y)}{\partial y}\ge 0$</p> <p>Question-2 Can these functions $\Psi(x), w(x,y), u(x)$ be solved, analytically or approximately?</p> <p>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.</p> <p>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...</p> http://mathoverflow.net/questions/48123/regarding-a-feature-of-multivariate-real-function Regarding a Feature of Multivariate Real Function Wang Tao 2010-12-03T02:16:38Z 2010-12-03T11:51:44Z <p>Any real function can be expressed as a function of the sum of two monotonic real functions? </p> <p>More precisely, for real function p(x, y), there exist continuous real functions P(x), h(x,y), g(x) such that:</p> <p>$p(x,y)=P(h(x,y)+g(x))$</p> <p>Where $P(x), h(x,y), g(x)$ are arbitrary satisfying $\frac {d(g(x))}{dx}>0$, $\frac {\partial h(x,y)}{\partial y}>0$ </p> <p>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> <p>$p(x,y)=\frac {u(f(x,y))}{u(x)}=F[ln(w(x,y))-ln(u(x))]=F(ln \frac {w(x,y)}{u(x)})=\Psi (\frac {w(x,y)}{u(x)})$</p> <p>Thank you very much!</p> http://mathoverflow.net/questions/47778/solving-functional-equation/47843#47843 Answer by Wang Tao for Solving Functional Equation Wang Tao 2010-12-01T00:15:12Z 2010-12-01T03:12:21Z <p>So far what I can think of on this question is:</p> <p>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.</p> <p>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.</p> <p>Question: how to find the appropriate $w(x,y),u(x)$ to span into $\Phi(\frac {u(f(x,y))}{u(x)})$</p> http://mathoverflow.net/questions/47557/regarding-kolmogorovs-superposition-theorem/47674#47674 Answer by Wang Tao for Regarding Kolmogorov's Superposition Theorem Wang Tao 2010-11-29T13:26:51Z 2010-11-29T13:26:51Z <p>Hi AgCl</p> <p>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?</p> <p>Warm Regards Wang Tao</p> http://mathoverflow.net/questions/48123/regarding-a-feature-of-multivariate-real-function Comment by Wang Tao Wang Tao 2010-12-04T02:40:26Z 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: $\frac {\partial lnx}{\partial x}=\frac {1}{x}&gt;0$, $\frac {\partial lny}{\partial y}=\frac {1}{y}&gt;0$ for $x&gt;0, y&gt;0$. http://mathoverflow.net/questions/47778/solving-functional-equation Comment by Wang Tao Wang Tao 2010-12-01T05:44:30Z 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? http://mathoverflow.net/questions/47778/solving-functional-equation/47843#47843 Comment by Wang Tao Wang Tao 2010-12-01T05:39:16Z 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 &quot;Add Another Answer&quot;. But I will keep your advice in mind.