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5
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edited Dec 1 2010 at 3:12 |
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 $(\Phi(\frac{f(x,y)}{u(x)})$ (\Phi(\frac{u(f(x,y))}{u(x)})$ is a solution.
For example, given $f(x,y)=ax+by$, select $u=x$, then:
$\Phi(\frac{f(x,y)}{u(x)})= \Phi(\frac{u(f(x,y))}{u(x)})= \Phi(a+b\frac{y}{x})$ 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 $\Phi(\frac {u(f(x,y))}{u(x)})$
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4
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edited Dec 1 2010 at 0:31 |
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 $(\frac{w(x,y)}{u(x)}$ (\frac{w(x,y)}{u(x)})$ span into $(\Phi(frac{f(x,y)}{u(x)})$ (\Phi(\frac{f(x,y)}{u(x)})$ is a solution.
For example, given $f(x,y)=ax+by$, select $u=x$, then:
$\Phi(frac{f(x,y)}{u(x)})= \Phi(\frac{f(x,y)}{u(x)})= \Phi(a+b\frac{y}{x})$ 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 $\Phi(\frac {u(f(x,y))}{u(x)})$
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3
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edited Dec 1 2010 at 0:26 |
So far what I can think of on this question is:
Rewrite $u(f(x,y))=u(x) \Psi(\frac{w(x,y)}{u(x)})$ 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 $(\Phi(frac{f(x,y)}{u(x)})$ is a solution.
For example, given $f(x,y)=ax+by$, select $u=x$, then:
$\Phi(frac{f(x,y)}{u(x)})= \Phi(a+b\frac{y}{x})$ 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 $\Phi(\frac {u(f(x,y))}{u(x)})$
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2
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edited Dec 1 2010 at 0:20 |
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:
$\Phi(\frac {u(f(x,y))}{u(x)})=\Gamma({\frac \Phi(\frac{u(f(x,y))}{u(x)})=\Gamma(\frac {w(x,y)}{u(x)})$
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 $(\Phi(frac{f(x,y)}{u(x)})$ is a solution.
For example, given $f(x,y)=ax+by$, select $u=x$, then:
$\Phi(frac{f(x,y)}{u(x)})= \Phi(a+b\frac{y}{x})$ 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 $\Phi(\frac {u(f(x,y))}{u(x)})$
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1
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answered Dec 1 2010 at 0:15 |
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 $(\Phi(frac{f(x,y)}{u(x)})$ is a solution.
For example, given $f(x,y)=ax+by$, select $u=x$, then:
$\Phi(frac{f(x,y)}{u(x)})= \Phi(a+b\frac{y}{x})$ 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 $\Phi(\frac {u(f(x,y))}{u(x)})$
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