# Easy to find roots

Is there a smooth function $f:\mathbb{R} \to \mathbb{R}_{\geq 0}$ such that:

1) $\lim_{x \to \infty} = \lim_{x \to -\infty} = 0$

2) $\forall x > 0$, $f'(x) < 0$

3) $\forall x < 0$, $f'(x) > 0$

4) $\forall a_1, \ldots, a_n \in \mathbb{R}, K \in \mathbb{R}_{\geq 0}$ the roots of $g(x) = (\sum_{i=1}^n f(x - a_i)) - K$ are "easy to find" (i.e. have an explicit formula in terms of $a_i$ and $K$ for each of them).

My initial guesses were $f(x) = \frac{1}{x^2+1}$ and $f(x) = \exp(-x^2)$ but both fail on part 4.

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Are you sure you want to do a 'straight addition' of your bump functions? If some kind of 'superposition' is sufficient, the problem might be easier. I also suggest that you work backwards: start with a formula for the roots (as generic as possible) and then apply all the constraints above. –  Jacques Carette Jun 20 '10 at 13:22
Regarding the tag: Could you elaborate on the connection with Galois theory? –  S. Carnahan Jun 20 '10 at 13:27
Sorry, I thought the tag was relevant (although it's probably not), please feel free to re-tag as appropriate. –  Mark Bell Jun 20 '10 at 13:33
Perhaps you meant to exclude this by specifying smoothness, but a $C^0$ solution is $f(x) := e^{-\lvert x \rvert}$. Let $\sum_{i=1}^n f(x_*-a_i) = K$. To find the formula for $x_*$, suppose w/l/o/g that $a_i >x_*$ for $i=1\dots,m$ and $a_i < x_*$ otherwise. Now $\sum_{i=1}^n f(x_*-a_i) = \sum_{i=1}^m e^{a_i - x_*} + \sum_{i=1}^m e^{x_*-a_i} = (\sum_{i=1}^m e^{a_i})e^{-x_*} + (\sum_{i=1}^m e^{-a_i})e^{x_*}$ which is a quadratic in $e^{x_*}$ and easy to solve. –  Steve Huntsman Jun 20 '10 at 15:31
I meant to have a different range of summation ($m+1$ to $n$) in a couple of those sums, but this doesn't affect anything else. –  Steve Huntsman Jun 20 '10 at 18:14