Iterated functions

The use of zeroes to obtain a factored form for a polynomial is well known, as is the expansion of the coefficients as symmetric functions of the roots. If

$$(z-z_1)(z-z_2)\ldots(z-z_n)$$ P(z) = (92)

$$z^n + a_1z^{n-1}+\cdots+a_n$$ = (93)

then

$$z_1+z_2+\cdots+z_n$$ a₁ = (94)

$$z_1z_2 + z_1z_3 + \cdots + z_{n-1}z_n$$ a₂ = (95)

$$z_1z_2z_3+z_1z_2z_4+\cdots$$ a₃ = (96)

$$\ldots$$

$$z_1z_2 z_3 \cdots z_n$$ a_n = (97)

In particular, if any z_i = 0, then a_n = 0 and so on for additional vanishing roots.

Of course, the reverse process, determining the zeroes given the coefficients, is more difficult and is the subject of much numerical analysis.