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I have the following special function.

$$f(x) = \sum _{i=1}^n \left\{\frac{(x - z_i)_+^2}{1+ 2*z_i+(x - z_i)_+^2}\right\} - \left\{(\frac{x^3}{3} - \frac{(x - z_i)_+^3}{3})\right\} $$

which + means If $z_i$ is bigger than x its equal $z_i - x$ and else it's equal zero. Also $z_i$ is a vector of value. How can I find the root of this cubic function?

rose
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  • It's not a cubic function, not with that $x^2$-term in the denominator. – Gerry Myerson Oct 07 '13 at 05:24
  • How can I get the root(s) of this function? – rose Oct 07 '13 at 05:33
  • If $n$ is, say, 5, then I see this as a polynomial of degree 13 --- actually, a collection of polynomials of degree 13, with one polynomial giving way to the next at each $z_i$. So, how do you usually solve polynomials of degree 13? Unless they have some very special structure, you hand them to a computer, which applies some numerical method, such as Newton's Method. Have you had a look at what happens, even for $n=1$? – Gerry Myerson Oct 07 '13 at 05:39

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