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Let $x,y\in R$ be two arbitrary scalar numbers. $f_1(x,y),...,f_n(x,y)$ are n scalar functions of $x,y$. Note $x,y$ only have two degree of freedom (DOF), so I think the n functions $f_1(x,y),...,f_n(x,y)$ at most have two DOF. That probably means the vector $[ f_1(x,y),...,f_n(x,y) ]^T$ can not be an arbitrary vector in $R^n$ space. My question is: how to determine whether two functions are independent? And how to determine which functions are redundant among the n functions?

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Do we have any constraints on these functions (i.e. are they required to be continuous, differentiable, etc.)? –  Alex Becker Mar 30 '11 at 14:00
    
What do you mean by "whether two functions are independent"? Do you mean "linearly independent"? If so, two functions are linearly independent if one is not a multiple of the other. –  joriki Mar 30 '11 at 14:21
    
Hi Alex: I think these are just general functions. Does the constraints really matter? –  Shiyu Mar 30 '11 at 14:59
    
Hi joriki: Maybe the word "independent" is not so appropriate. What I mean is like this. $x,y$ are totally free. So the vector $[x,y]^T$ can be an arbitrary vector in $R^2$. However, the vector $[f_1(x,y),f_2(x,y)]^T$ may not be an arbitrary vector in $R^2$. For example, $f_1=cos(x+y), f_2=sin(x+y)$ can be arbitrary vectors in $R^2$ (up to a scale). But $f_1=cos(x+y), f_2=cos^2(x+y)$ is not. I just don't know whether there are concepts such as DOF or independence for functions. –  Shiyu Mar 30 '11 at 15:06
    
Yes, the constraints matter. Look up "space-filling curves". You can have $n$ continuous functions $f_j(t)$ of one variable such that $[f_1(t), \ldots, f_n(t)]$ for $t \in \mathbb R$ fills all of $\mathbb R^3$. –  Robert Israel Mar 30 '11 at 15:07

2 Answers 2

if you want $f_1,f_2,f_3$ st $\{(f_1(x,y),f_2(x,y),f_2(x,y)) : x,y\in\mathbb{R}\}=\mathbb{R}^3$ then probably not. each equation $f_i(x,y)=a_i$ is basically a curve in the plane and the intersection of three curves may not exist.

you could (artificially) make 3 functions that do satisfy this however. establish a bijection $B=(B_1(x,y),B_2(x,y),B_3(x,y)):\mathbb{R}^2\to\mathbb{R}^3$, then let $f_i=B_i$ so that for any point in $\mathbb{R}^3$ you can look up the corresponding $x,y$. however i doubt these functions will be continuous.

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Thanks for your answer yoyo. But the problem I concern is not to construct some functions such that they are independent to each other. On the contrary, I am given many (say 100) functions. How can I determine which functions are redundant and can be removed? If the independent variable of these 100 functions are merely 2, then there probably exist only 2 independent functions and all the other 98 functions are redundant. –  Shiyu Mar 30 '11 at 15:25
up vote 0 down vote accepted

Thanks everyone. I come back again. I think I didn't express myself well in the question. Here is my own answer to this question. What I use is 'Diffeomorphism'.

Definition: A function $\phi: R^n \mapsto R^n$ is called diffeomorphism if it is smooth and if its inverse $\phi^{-1}$ exists and is smooth.

Given a function, we can use the singularity of its Jacobian matrix to determine whether the function is diffeomorphism. Here is an example.

$$\left[ \begin{array}{ccc} z_1 \\\ z_2\end{array} \right] =\phi (x) =\left[ \begin{array}{ccc} 2x_1+5x_1x_2^2 \\\ 3\sin x_2\end{array} \right]$$ Then the Jacobian of $\phi (x)$ is $$J=\left[ \begin{array}{ccc} 2+5x_2^2 & 10x_1x_2 \\\ 0 & 3\cos x_2 \end{array} \right]$$ It is apparent that the Jacobian is singular if and only if $\det (J)=3(2+5x_2^2)\cos x_2=0$. That is $x_2=k\pi+\pi/2$. Hence the inverse function $\phi ^{-1}$ exists only when $x_2\neq k\pi+\pi/2$.

This might be the right way to understand the question.

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