# The simple roots of a polynomial are smooth functions with respect to the coefficients of the polynomial? [closed]

Why the following is true?

«The simple roots of a polynomial are smooth functions with respect to the coefficients of the polynomial?»

## closed as off-topic by Bobson Dugnutt, Silvia Ghinassi, Daniel W. Farlow, Stefan Mesken, Michael AlbaneseMar 12 '16 at 2:59

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Let $F:\mathbb{C}^{n+1} \rightarrow \mathbb{C}$ be the map that takes the coefficient vector $\mathbf{a}$ of a polynomial $p_\mathbf{a}(z)$, and a point of evaluation $z$, and evaluates the polynomial at that point, $$F(\mathbf{a},z) := p_\mathbf{a}(z) = a_0 + a_1 z + a_2 z^2 + \dots + a_n z^n.$$ We locally parameterize the zero set of this map, $$\{(\mathbf{a}, z) : F(\mathbf{a},z)=0\}.$$ In particular, let $(\mathbf{b}, r_b)$ be a point in this zero-set (that is, $r_b$ is a root of the polynomial with coefficient vector $\mathbf{b}$). The implicit function theorem tells us that if the partial derivative $\frac{\partial F}{\partial z}(\mathbf{b},r_b)$ is invertible, then there exists a function $r:\mathbb{C}^n \rightarrow \mathbb{C}$ such that the zero set is the graph of this function, $$0 = F\left(\mathbf{a}, r(\mathbf{a})\right), \quad (\text{locally})$$ for all $\mathbf{a}$ in some neighborhood of $\mathbf{b}$. Moreover $r$ is differentiable in this neighborhood, with gradient, $$\nabla r(\mathbf{a}) = - \left(\frac{\partial F}{\partial z}(\mathbf{a},r(\mathbf{a}))\right)^{-1} \nabla_a F(\mathbf{a},r(\mathbf{a})) , \quad (\text{locally}).$$ All that remains to be checked is that $\frac{\partial F}{\partial z}(\mathbf{b},r_b)=p_\mathbf{b}'(r_b)$ is nonzero. To see this, note that if $r_b$ is a simple root of $p_\mathbf{b}(z)$, then $p_\mathbf{b}$ can be factored as follows, $$p_\mathbf{b}(z) = (z-r_b)q(z)$$ for some polynomial $q(z)$ that does not have $r_b$ as a root. Directly computing the derivative with the product rule, we get, $$p'_\mathbf{b}(z) = q(z) + (z-r_b)q'(z),$$ and then substitute in $r_b$ to get, $$p'_\mathbf{b}(r_b) = q(r_b) + (r_b - r_b)q'(z) = q(r_b) \neq 0.$$
• Thanks. Let inverse function theorem to a function $f$ at point $x$, is holds. Why the questions is done? – H.S Mar 20 '16 at 18:53
• -If $\frac{\partial F}{\partial z}(\mathbf{b},r_b)=p_\mathbf{b}'(r_b)$ be nonzero, why does guaranteed to be the case whenever $r_b$ is a simple root.? – H.S Mar 20 '16 at 22:13
• Edited to discuss this. If $r_b$ is a simple root, you can factor the polynomial, then take the derivative with the product rule. Intuitively, near a simple root a polynomial looks like a line with nonzero slope. – Nick Alger Mar 20 '16 at 22:44