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I have a series of inequalities:

$$y_{1min} \leq y_{1}(x) \leq y_{1max}$$ $$y_{2min} \leq y_{2}(x) \leq y_{2max}$$ $$..$$ $$y_{nmin} \leq y_{n}(x) \leq y_{nmax}$$

Note that $x\in\mathbb{R}$

The question is, is there a general method that allows me to find the range of $x$ that satisfies the above inequalities?

Things are easy if there is just one inequality, and $y_1(x)$ is at most a quadratic function. But I am looking for a general solution here.

Edit: My $y$ function can roughly be written as


where $c$ and $e$ are small comparatively.$(dx+e)^g$ must be positive.

In one of the $y_i(x)$, $f=5/3, g=2/3$, but in another, $f=1, g=1$. It is safe to assume that both $f,g\in\mathbb{Q}$, that is they are rational numbers.

$b,c,d,e\in\mathbb{R}$ , and they vary from one $y_i(x)$ to another.

Hope this helps.

Edit:Let's make this question a little general; instead of letting $y(x)$ depends on one variable, what if $y$ depending on multiple variables? That is, how to find the range for $(x_1,x_2,x_3,..., x_i)$, given that

$$y_{1min} \leq y_{1}(x_1,x_2,x_3,..., x_i) \leq y_{1max}$$ $$y_{2min} \leq y_{2}(x_1,x_2,x_3,..., x_i) \leq y_{2max}$$ $$..$$ $$y_{nmin} \leq y_{n}(x_1,x_2,x_3,..., x_i) \leq y_{nmax}$$

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@Ngu Soon Hui: at the generality in which you put the question I doubt there is anything that can be said. Given a closed subset $F\subseteq\mathbb R$ is is easy to give a set of inequalities (in fact just one) involving continuous functions which has solution set precisely $F$... To turn this into something which can reasonably be expected to have a general solution you will need to be more specific. – Mariano Suárez-Alvarez Sep 21 '10 at 14:10
As mentioned already, solution strategies for such things are often tailored to the inherent structure of the problem. Not exploiting it can only be a profligate waste of computer time and effort. Please give more specifics about your function(s) of interest. Turning to the multivariate case, exploitation of structure becomes even more important, since you have much degrees of freedom. – J. M. Sep 21 '10 at 14:13
@J.M., question updated. – Graviton Sep 21 '10 at 14:23
How about you post the complete problem up!? Surely that would be easier rather than playing a cluedo game! – alext87 Sep 21 '10 at 14:55
Please state the problem in all its gory details, so that people do not need to guess at the statement. Otherwise, this is really not a real question... – Mariano Suárez-Alvarez Sep 21 '10 at 16:13
up vote 2 down vote accepted

If $f$ and $g$ are rational the problem is very easy. Multiply out by the denominator, take the $d^{th}$ power of both sides where $d$ is the least common denominator of $f$ and $g$, and you've reduced to the polynomial case, which is easy (check the extrema and so forth). This is why it is important to provide details.

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Yup, they are rational numbers. – Graviton Sep 22 '10 at 0:49

For a start, suppose we can find the interval $I_i\subseteq\mathbb{R}$ such that $y_{imin}\leq y_i(x)\leq y_{imax}$. Then the interval $I$ which satisfies all the constraints is

$I=\bigcap_{i=1}^n I_i$

Thus we can concentrate on finding $I_i$ separately for each $1\leq i\leq n$. Here I'm sure we will need some conditions on $y_i(x)$. Do your functions have compact support? Are they continuous? Do they oscillate wildly in any region?

In the general case - finding a feasible point You mentioned that $y_i(x)$ are continuous. Lets suppose that in addition they are continuously differentiable. I'll do the problem with $1$ constraint for easy of notion but the general case is exactly the same though more difficult to write down.

Rewrite the constraints as

$y_1(x)\geq y_{1min}$

$-y_1(x)\geq -y_{1max}$

Now let $A(x) = \left(\nabla y_1(x),\nabla(-y_1(x))\right)^T$ (the jacobian matrix of the constraints).

Also let $b=\left(y_{1min},-y_{1max}\right)^T$.

Then we need to find a $x_0$ such that $Ax_0\geq b$. We do this by solving the following (phase-1) problem.

let $x_{guess}\in\mathbb{R}^n$. Let $r=\min(b-Ax_{guess},0)$ and solve the problem

$\min_{x\in\mathbb{R}^n,\mu\in\mathbb{R}}$ $\mu$ subject to $Ax + \mu r\geq b$ and $\mu\geq 0$

Now, you may not think that we have come very far but because $(x,\mu)=(x_{guess},1)$ is an initial feasible point for this minimisation problem it can be solved by standard software. If the optimal point gives an optimal value $>0$ then the orignal problem is not feasible. If the optimal value is $0$ then the orignal problem is feasible and the optimal point is a feasible point for the orignal problem.

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I would like to clarify that the functions are well behaved, no singularity, continuous, even though they are not polynomials. – Graviton Sep 21 '10 at 14:04
not sure how would you handle multiple $x_i$ variables case? See the updated question – Graviton Sep 21 '10 at 14:08
not sure whether is there a matlab function that does this? – Graviton Sep 21 '10 at 14:33
Look up at the optimization toolbox in Matlab. – alext87 Sep 21 '10 at 14:46

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