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i want to know if there is a function for example that gives the same result for a set of noncontiguous reals. i want to have these results

f(10)=25 f(25)=25 f(34)=25 f(85)=25 f(14)=25 f(13)=25 for example

in fact i have: - A number between 1 and 536870912 - Several sets of noncontiguous numbers
- A black box which must do a transformation to the sets - Each transformation must be unique and gives a number between 1 and 255 I want to know which thing (function or anything else ) that i can have in the black box that does this transformation?

{4, 212, 10, 35000} ==> f(x) ==> 250
{584, 12, 140, 5} ==> f(x) ==> 15 I mean passing a given set by f(x) gives always the same result for the passed set or i can imagine that for each element of a set i can have f(4)=250, f(212)=250, f(10)=250, f(35000)=250

Thank you

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For your example, that would be the constant function $f(x)=25$... If you want to get different results, see my answer. – Dennis Gulko Apr 30 '12 at 9:38
This is extremely unclear - while Dennis' function satisfies the first requirements you have give, I can't figure out what would satisfy the rest of them. For example, how is this number between $1$ and $536870912$ supposed to fit into it? What do you mean by doing a "transformation to the sets", and so on. – Martin Wanvik Apr 30 '12 at 9:41

Note that a function $f: A \to B$, where $A$ and $B$ are sets, is a rule that assigns to each element $a$ in $A$, a single element $f(a)$ in $B$. So your new question is no less trivial than the first one you asked, in which one solution was the constant function $f(x) = 25$, defined, say on $\mathbb{R}$. It is just a matter of writing down the appropriate rule (though it is a wholly different matter if you want it to be, say, continuous, differentiable or polynomial - in that case, something like Dennis Gulko's answer is more appropriate).

If I'm interpreting your question correctly, you have a collection $A_1, A_2, \ldots, A_n$ of finite sets of real numbers, and you want the function to assign to each member of the set $A_i$ the number $a_i$. So, we can define a function $f: A \to \mathbb{R}$, whose domain $A$ is $A_1 \cup A_2 \cup \ldots \cup A_n$, where $$ f(x) := a_i, \text{ if } x \in A_i $$ Note that in order for this to be unambiguous, no two sets can have an element in common (in other words, $A_i \cap A_j = \emptyset$ whenever $i \neq j$).

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Yes you are interpreting my question well Thank you, so what should be f(x) in this case – Boubouh Apr 30 '12 at 11:14
@Boubouh: I said what $f(x)$ is, didn't I? – Martin Wanvik Apr 30 '12 at 11:19

You should use polynomial interpolation.
If you want a polynomial $f(x)$ that at the points $x_1,...,x_n$ has value $y_1,...,y_n$ respectively, define the following: $$p_i(x)=\frac{(x-x_1)\cdot...\cdot\widehat{(x-x_i)}\cdot...\cdot(x-x_n)}{(x_i-x_1)\cdot...\cdot\widehat{(x_i-x_i)}\cdot...\cdot(x_i-x_n)}$$ (where $\widehat{(x-x_i)}$ denotes a term that is not included) Now $p_i(x_j)=0$ for all $j\neq i$ and $p_i(x_i)=1$.
Define $f(x)=\sum_{i=1}^n y_ip_i(x)$.

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