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I have recently asked a question related to an inverse function which was not so obvious to calculate:

Inverse function of $y=W(e^{ax+b})-W(e^{cx+d})+zx$

Now I would like to learn;

Given $$f(x)=f_1(x)+f_2(x)$$

What are the conditions imposed on $f_1$ and $f_2$ such that

$${f}^{-1}(x)={f_1}^{-1}(x)+{f_2}^{-1}(x)$$

and when is such an approximation is not good at all?

Thanks alot in advance-

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Well that would mean $x = f_1^{-1}\left(f_1(x)+f_2(x)\right) + f_2^{-1}\left(f_1(x)+f_2(x)\right)$ so it'd really be strange.

Assuming there are solution, the set of all solutions $S$: $$S = \left\{ ( f_1, f_2 ) \in \left( \mathbb{R}^{\mathbb{R}} \right)^2, f_1 : \mathbb{R} \leftrightarrow \mathbb{R}, f_2 : \mathbb{R} \leftrightarrow \mathbb{R}, {f}^{-1}(x)={f_1}^{-1}(x)+{f_2}^{-1}(x) \right\}$$

The neutral element for $+$ would be $( x \mapsto 0, x \mapsto 0 )$ but $x \mapsto 0$ isn't bijective so it can't be in a pair of your set.

The neutral element for $\times$ would be $( x \mapsto 1, x \mapsto 1 )$ but $x \mapsto 1$ isn't bijective so it can't be in a pair of your set.

The neutral element for $\circ$ would be $( x \mapsto x, x \mapsto x )$ but $\frac{1}{2}x \not= x + x$ so it's not in the set either.

So the set would be unlikely to have any interesting properties and you probably wouldn't be able to determine if a function is in the set before calculating its inverse, making it completely useless...

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how did you get your first equation? –  Seyhmus Güngören Sep 26 '12 at 20:28
    
I replaced $x$ by $f(x)$ in your second equation so I got $f^{-1}(f(x))=x$ on the left side and on the right side, I did the same thing with the right members of your equations (which means I did the same thing and then used your first equation to replace $f(x)$). –  xavierm02 Sep 26 '12 at 21:45
    
I see and what are the next steps? what are $+,\times ,o$? $(x\rightarrow 0,x\rightarrow 0)$? Is it possible to give more details as I dont understand what you mean. –  Seyhmus Güngören Sep 26 '12 at 22:03
    
Well assuming you did have solutions, you would be able to put all those solutions in a set. Then, you would want to know properties of that set. Such as $\left(x,y\right) \in S^2 \implies x + y \in S$. I checked if $(S, +)$, $(S, \times)$ or $(S, \circ)$ was a group. And none of these 3 is since it would have to contain the neutral element for the operation. –  xavierm02 Sep 27 '12 at 17:59

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