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I think I may be able to construct a solution for this, using the Implicit Function Theorem (IFT), if I can prove that a certain complex derivative is nonzero everywhere.micromass said:2. Show that the n roots of a real/complex polynomial of degree n depend continuously on the coefficients.
I will use commas to denote concatenation of compatible vectors.
We represent a degree-n complex polynomial with vector of complex coefficients (c_0,...,c_n)=\mathbf c\in \mathbb C^{n+1} by p[\mathbf c]:\mathbb C\to\mathbb C.
Then create a function f:\mathbb R^{2n+4}\to \mathbb R^2 such that for any \mathbf a,\mathbf b\in\mathbb R^{n+1} and x_1,x_2\in\mathbb R:
$$f(\mathbf a,\mathbf b, x_1,x_2)=(Re\ (p[\mathbf a+i\mathbf b](x_1+i\ x_2)),\ Im\ (p[\mathbf a+i\mathbf b](x_1+i\ x_2)))$$
Given any coefficient vector \mathbf c\in\mathbb C^{n+1} set \mathbf a=Re\ \mathbf c,\ \mathbf b=Im\ \mathbf c and let \mathbf r=(r_1,r_2,...,r_n)\in\mathbb C^n be the vector of roots of p[\mathbf a+i\mathbf b], ordered by the dictionary order on the real then imaginary components.
Let k be an arbitrary positive integer not exceeding n. Then we have
$$f(\mathbf a,\mathbf b,Re\ r_k,\ Im\ r_k)=0$$
To use the IFT, we need to show that the 2\times 2 matrix with i,j element D_{2n+2+j}f_i(\mathbf a,\mathbf b,Re\ r_k,\ Im\ r_k) is invertible, where the operator D_i indicates differentiation wrt the ith real, scalar component of the argument to the function to which the operator is applied.
This matrix is the Jacobian of the function
$$(Re\ z,Im\ z)\mapsto (Re\ (p[\mathbf c](z)),\ Im\ (p[\mathbf c](z)))$$
which, by the Cauchy-Riemann equations, has form \begin{pmatrix} u&v\\-v&u\end{pmatrix} iff the function
$$z\mapsto p[\mathbf c](z)$$
is holomorphic. It is known that all complex polynomials are holomorphic, so the Jacobian has the required form, which means its determinant is equal to u^2+v^2, which is 0 iff D_jf_i(\mathbf a,\mathbf b,Re\ r_k,\ Im\ r_k)=0 for all i,j\in\{1.2\}. So the matrix is invertible everywhere in coefficient space \mathbb C^{n+1}, except where the complex derivative is zero.
Then the IFT tells us that, where the Jacobian is invertible, there is a neighbourhood U of (\mathbf a,\mathbf b)\subset \mathbb R^{2n+2}, a neighbourhood V of (Re\ r_k, Im\ r_k)\subset \mathbb R^{2} and a unique, continuously differentiable function g:U\to V such that
$$\{(\mathbf x,g(\mathbf x))\ |\ \mathbf x\in U\}=
\{(\mathbf x,\mathbf y)\in U\times V\ |\ f(\mathbf x,\mathbf y)=\mathbf 0\}$$
That is, for any coefficient vector \mathbf c(\mathbf x) derived from \mathbf x by the formulas
$$Re\ c_{j-1}=x_{j},\ Im\ c_{n+j}=x_{n+1+j}$$
(note that in accordance with convention, the indices of components of \mathbf c start at 0 whereas those of \mathbf x start at 1),
the kth largest root is a continuously differentiable function of \mathbf x. Courtesy of the diffeomorphism from \mathbb C^{n+1} to \mathbb R^{2n+2} we can conclude that it is also a continuously differentiable function of \mathbf c.
Since both \mathbf c and k were chosen arbitrarily, we can conclude that there is everywhere a continuous dependence for all roots.
That leaves the matter of points where the Jacobian is zero, to be considered. In fact, what concerns us is points where both the polynomial and its derivative are zero.
I think these will be points where the coefficients make the kth largest root have a multiplicity at least two, and I expect we might be able to use that fact. More later.
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