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Possible Duplicate:
Is there a function or anything else that gives the same result on a set?

I want to know if there is a function, for example, that gives the same result on the interval $[10,15]$. I want to have these function values

$$f(10)=25,\, f(11)=25,\, f(12)=25,\, f(13)=25,\, f(14)=25,\, f(15)=25$$

Thanks.

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    How about $f: [10, 15]\to \mathbb R, x \mapsto 25$2012-04-30
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    Of course there is. For example the constant function $f(x) = 25$, for all $x \in \mathbb{R}$.2012-04-30
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    Exuse me i just asked a more specific question thank you2012-04-30
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    @Boubouh: As I said, *edit* your original question. There is no need to ask another one.2012-04-30
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    @Boubouh: You might also use $f(x)=25\cos (2\pi x)$ for $x\in\mathbb R$.2012-04-30
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    From the OP's question it appears that the question http://math.stackexchange.com/questions/138801/is-there-a-function-or-anything-else-that-gives-the-same-result-on-a-set is meant to be a refinement/clarification of this one. Closing as duplicate.2012-04-30
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    what are you hoping to achieve with this? what's the larger picture? are you familiar with the concept of [level sets](http://en.wikipedia.org/wiki/Level_set)?2012-04-30

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Actually, following Dejan Govc's lead, you could take any Fourier Series $$ f(x) = a_0 + \sum_{n=1}^\infty a_n \cos\left(2\pi nx\right) + b_n \sin\left(2\pi nx\right) $$ whose cosine coefficients $a_n$ sum to $25$ (the $b_n$ are completely free). The constant function corresponds to $(a_0,a_1,\dots)=(25,0,0,\dots)$ and Dejan's to $(0,25,0,\dots)$, both with all $b_n=0$. The complete function space would look like the set of periodic functions on the unit interval with the boundary condition $f(0)=f(1)=f(k)=25$ for all $k\in\mathbb{Z}$. This is a pretty large space of functions!