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Is it possible to represent the following function with a single formula, without using conditions? If not, how to prove it?

$F(x) = \begin{cases}u(x), & x \le 0, \ v(x) & x > 0 \end{cases}$

So that it will become something like that: $F(x) = G(x)$ With no conditions?

I need it for further operations like derivative etc.

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    The only thing you can do is writing something like $F=u\cdot 1_{(-\infty,0]}+v\cdot 1_{(0,\infty)}$.2011-10-12

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What operations are allowed in the formula?

$G(x) = \frac{x + |x|}{2x} v(x) + \frac{x - |x|}{2x} u(x)$ will work (away from 0), but any "trick" along these lines is not going to help make taking derivatives any easier.

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    I read the OP's question and thought, "he wants a 'single formula' (non-piecewise) reformulation of his $f$." It didn't even cross my mind at the time that there might be other interpretations; I acknowledge that the question I answered is not the same as finding a $G$ that is *only* a function of $u$ and $v$. I never would have imagined this question would be so controversial, and apologize if my posts here have been bizarre or aggressive. I hope there are no hard feelings moving forward.2011-10-13
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Note: This answers the original question, asking whether a formula like $F(x)=G(u(x),v(x))$ might represent the function $F$ defined as $F(x) = u(x)$ if $x \leqslant 0$ and $F(x)=v(x)$ if $x > 0$.
The OP finally reacted to remarks made by several readers that another answer did not address this, by modifying the question, which made the other answer fit (post hoc) the question.


Just to make sure @Rasmus's message got through: for any set $E$ with at least two elements, there can exist no function $G:\mathbb R^2\to E$ such that for every functions $u:\mathbb R\to E$ and $v:\mathbb R\to E$ and every $x$ in $\mathbb R$, one has $G(u(x),v(x))=u(x)$ if $x\leqslant0$ and $G(u(x),v(x))=v(x)$ if $x>0$.

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    @maximus, This is shocking, and akin to *rewriting history*. You should have posted the second question as another post. (And we still do not know whether you realized the two versions of the question are completely different, or not.)2011-10-13