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Is it assured that if we are making only small changes of coefficients in a polynomial equation

$x^n+a_{n-1}x^{n-1}+... + a_1x +a_0=0$

roots for these equations are changing also by small values? and .. for this case do we have also assured continuity of changes in roots, also in the case when a root is changing from a real value to imaginary one? (i.e. function $\mathbf{x}=f(\mathbf{a})$ - where $\mathbf{x}$ is vector of roots $\mathbf{x}= [x_1,x_2,..x_n]^T$ and $\mathbf{a}$ is vector of coefficients $\mathbf{a}= [a_0,a_1,..a_{n-1}]^T$ - is continuous, smooth, makes no great differential changes in values)

  • What theorem does describe this situation ?

Edit (after 9 hours)..

So far there are no answers...
I wonder whether use of Jacobi matrix and Vieta's formulas for the function $\mathbf{a}=f^{-1}(\mathbf{x})$ would not be a good approach.


Below I'm developing this approach:

The equation $x^n+a_{n-1}x^{n-1}+... + a_1x +a_0=0$
can be written as

$\mathbf{x}=f(\mathbf{a})$

where $\mathbf{x}$ is vector of roots and $\mathbf{a}$ is vector of coefficients.

In general case it's not easy to obtain function $\mathbf{f}$, but it is possible from Viete'a formulas

$\begin{cases} x_1 + x_2 + \dots + x_{n-1} + x_n = - {a_{n-1}} \\ (x_1 x_2 + x_1 x_3+\cdots + x_1x_n) + (x_2x_3+x_2x_4+\cdots + x_2x_n)+\cdots + x_{n-1}x_n = {a_{n-2}} \\ {} \quad \vdots \\ x_1 x_2 \dots x_n = (-1)^n {a_0} \end{cases}$

to obtain function $g(\mathbf{x}):\ \mathbf{a}=f^{-1}(\mathbf{x}) = g(\mathbf{x}) $ and from this: $ d\mathbf{a}=J(\mathbf{x})d\mathbf{x}$

For this function we can calculate Jacobi matrix

$\mathbf J = \begin{bmatrix} \dfrac{\partial \mathbf{g}}{\partial x_1} & \cdots & \dfrac{\partial \mathbf{g}}{\partial x_n} \end{bmatrix} = \begin{bmatrix} \dfrac{\partial g_1}{\partial x_1} & \cdots & \dfrac{\partial g_1}{\partial x_n}\\ \vdots & \ddots & \vdots\\ \dfrac{\partial g_n}{\partial x_1} & \cdots & \dfrac{\partial g_n}{\partial x_n} \end{bmatrix}$

$ =\begin{bmatrix} \\ (-1)^nx_2 x_3 \dots x_n & \dots & (-1)^nx_1 x_2 \dots x_{n-1} \\ \vdots & \ddots & \vdots\\ \ x_2 + x_3 + \dots + x_n & \dots & x_1 + x_2 + \dots + x_{n-1} \\ -1 & \cdots & -1 \end{bmatrix}$

Now the problem is to have finite (and possibly small) $\mathbf {J}^{-1}$ $ \dots$ (we have $ d\mathbf{x}=\mathbf{J}^{-1}(\mathbf{x})d\mathbf{a}$)
what is possible only if $det{\mathbf {(J)}}\neq{0}$.

  • How to analyze this determinant to get the best results for the starting question?
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    $\begingroup$ Mildly relevant: Wilkinson's polynomial. $\endgroup$ – Karolis Juodelė Nov 24 '16 at 20:22
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    $\begingroup$ The roots depend continuously on coefficients, but changes can be quite abrupt, while still mathematically smooth. See for example Continuity of the roots of a polynomial in terms of its coefficients. $\endgroup$ – dxiv Nov 24 '16 at 20:42
  • $\begingroup$ @dxiv very interesting ... thank you... $\endgroup$ – Widawensen Nov 24 '16 at 21:06
  • $\begingroup$ @KarolisJuodelė It is very relevant, thank you.. $\endgroup$ – Widawensen Nov 25 '16 at 7:01
  • $\begingroup$ $x^2 + a_1x + a_0$, small $\pm$ changes arount $a_0 = 0$ where $a_1 = 0$?! $\endgroup$ – Pieter21 Nov 28 '16 at 10:40

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