# Inverse function theorem regarding solutions to a cubic equation

Let $c_1, c_2, c_3 \in \mathbb{R}$ be such that the equation $X^3-c_1X^2+c_2X-c_3=0$ has got three distinct solutions $s_1, s_2, s_3 \in \mathbb{R}$.

Show that there exists a neighborhood $U \subset \mathbb{R^3}$ of $(c_1, c_2, c_3)$ such that the solutions $x_1, x_2, x_3 \in \mathbb{R}$ of the equation $X^3-y_1X^2+y_2X-y_3=0$ depend continuously differentiably and bijectively on $(y_1, y_2, y_3) \in U$.

I think that I can use the inverse function theorem here, however my problem is that I am asked to show the existence of an inverse function $g: \mathbb{R^3} \rightarrow \mathbb{R^3}$, while the given cubic function maps to $\mathbb{R}$. How do I construct the function I want? I am really confused by the notion that my function has to depend on solutions to an equation.

Do I need to set up an equation system?

As I am stuck on the approach, any clarifications and hints are welcome.

My attempt:

I have set up the function

$$F(b,a)=F((b_1,b_2,b_3), (a_1,a_2,a_3))= \begin{bmatrix} a_1^3-b_1a_1^2+b_2a_1-b_3 \\ a_2^3-b_1a_2^2+b_2a_2-b_3 \\ a_3^3-b_1a_3^2+b_2a_3-b_3 \end{bmatrix}$$

Now I have

$$\frac{\partial F}{\partial a}= \begin{bmatrix} \frac{\partial F_1}{\partial a_1}& \frac{\partial F_1}{\partial a_2} & \frac{\partial F_1}{\partial a_3} \\ \frac{\partial F_2}{\partial a_1}& \frac{\partial F_2}{\partial a_2} & \frac{\partial F_2}{\partial a_3} \\ \frac{\partial F_3}{\partial a_1}& \frac{\partial F_3}{\partial a_2} & \frac{\partial F_3}{\partial a_3} \end{bmatrix}= \begin{bmatrix} 3a_1^2-b_1a_1+b_2 & 0 & 0 \\ 0 &3a_2^2-b_1a_2+b_2 & 0 \\ 0 & 0 &3a_3^2-b_1a_3+b_2 \\ \end{bmatrix}$$

Now the determinant of the Jacobian should not be $0$ for $b=(c_1, c_2, c_3), a=(s_1, s_2, s_3)$. However I do not see how that is given.

• Are $(y_1, y_2, y_3)$ supposed to be in a neighborhood of $(s_1, s_2, s_3)$, or should that be a neighborhood of $(c_1, c_2, c_3)$? – Michael Seifert Jul 19 '18 at 15:42
• @MichaelSeifert My bad, correcting it! – idle mathematician Jul 19 '18 at 15:43
• Might it work better to treat $c_1, c_2, c_3$ as fundamental symmetric polynomials of the roots, and send $(b_1, b_2, b_3), (a_1, a_2, a_3)$ to $(a_1 + a_2 + a_3 - b_1, a_1 a_2 + a_1 a_3 + a_2 a_3 - b_2, a_1 a_2 a_3 - b_3)$? – Daniel Schepler Jul 19 '18 at 23:19
• I think that if you set up the function $(s_1,s_2,s_3)\mapsto(c_1,c_2,c_3)$, the Jacobian determinant should be something like the discriminant of the polynomial. If this is right, you pretty much have the problem solved. – Lubin Jul 19 '18 at 23:27
• @DanielSchepler After you've set up the function I can see that it will work, but what is the idea behind this approach? I have never heard of fundamental symmetric polynomials before and this seems like a neat trick for this problem. – idle mathematician Jul 19 '18 at 23:37

We know that if $x_1, x_2, x_3 \in \mathbb{R}$ are the three roots of $X^3 - y_1 X^2 + y_2 X - y_3 = 0$, then $y_1 = x_1 + x_2 + x_3$, $y_2 = x_1 x_2 + x_1 x_3 + x_2 x_3$, and $y_3 = x_1 x_2 x_3$. This suggests to study the function $F : \mathbb{R}^3 \to \mathbb{R}^3, (x_1, x_2, x_3) \mapsto (x_1 + x_2 + x_3, x_1 x_2 + x_1 x_3 + x_2 x_3, x_1 x_2 x_3)$. Now, the Jacobian matrix of $F$ is:
$$JF(x_1, x_2, x_3) = \begin{bmatrix} 1 & x_2 + x_3 & x_2 x_3 \\ 1 & x_1 + x_3 & x_1 x_3 \\ 1 & x_1 + x_2 & x_1 x_2 \end{bmatrix}.$$
It is now straightforward to check that the determinant of this matrix is equal to $(x_1 - x_2) (x_1 - x_3) (x_2 - x_3)$. (One possible approach, which is somewhat roundabout but avoids tedious computations: note that the determinant is a polynomial in $x_1, x_2, x_3$ which is homogeneous of degree 3. On the other hand, if $x_1 = x_2$ then the first two rows are equal so the determinant is zero, implying that $x_1 - x_2$ divides this polynomial. Similarly, $x_1 - x_3$ and $x_2 - x_3$ also divide the polynomial; and these are pairwise relatively prime. Thus, $(x_1 - x_2) (x_1 - x_3) (x_2 - x_3)$ divides the determinant polynomial; and by comparing degrees and then coefficients of $x_1^2 x_2$ you see the quotient must be 1.)
Therefore, if $s_1, s_2, s_3$ are distinct then $JF(s_1, s_2, s_3) \ne 0$, implying that $F$ has an inverse function in some open neighborhood of $(c_1, c_2, c_3)$ such that $F^{-1}(c_1, c_2, c_3) = (s_1, s_2, s_3)$. Then in reverse, this implies that if $F^{-1}(y_1, y_2, y_3) = (x_1, x_2, x_3)$, then $(X - x_1) (X - x_2) (X - x_3) = X^3 - y_1 X^2 + y_2 X - y_3$ as polynomials in $\mathbb{R}[x]$, so $x_1, x_2, x_3$ are the roots of $X^3 - y_1 X^2 + y_2 X - y_3$ as desired.