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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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marked as duplicate by Did, Willie Wong Apr 30 '12 at 10:55

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How about $f: [10, 15]\to \mathbb R, x \mapsto 25$ –  martini Apr 30 '12 at 9:00
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Of course there is. For example the constant function $f(x) = 25$, for all $x \in \mathbb{R}$. –  Martin Wanvik Apr 30 '12 at 9:03
    
Exuse me i just asked a more specific question thank you –  Boubouh Apr 30 '12 at 9:28
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@Boubouh: As I said, edit your original question. There is no need to ask another one. –  Martin Wanvik Apr 30 '12 at 9:30
    
@Boubouh: You might also use $f(x)=25\cos (2\pi x)$ for $x\in\mathbb R$. –  Dejan Govc Apr 30 '12 at 9:48

1 Answer 1

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!

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+1. Nice observation! –  Dejan Govc Apr 30 '12 at 10:17

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