Find the derivative of the root of cubic equation

Question

Consider the following cubic equation:

\begin{align} \alpha ^3-2 \alpha ^2- \alpha(6+4x^2) +8 x^3-4x&=0 \end{align} Where $x\in \mathbb{R}$.

The solutions $\alpha_1,\alpha_2,\alpha_3$ will depend on $x$. Solving this equation on Mathematica or Wolfram Alpha gives me complicated third roots.

I am not interested in the actual value of the roots, I am only looking for the value of $\frac{\mathrm{d}\alpha}{\mathrm{d}x}$. Does a method exist to obtain a "nice" expression for $\frac{\mathrm{d}\alpha}{\mathrm{d}x}$ without having to actually solve the cubic equation?

Edit: Applying the advice from the answer, I obtain: $$\frac{\mathrm{d}\alpha}{\mathrm{d}x}=\frac{(4 + 8 \alpha x - 24 x^2)}{(-6 - 4 \alpha + 3 \alpha^2 - 4 x^2)}$$ Is there a way to eliminate the $\alpha$ terms and have an expression $\frac{\mathrm{d}\alpha}{\mathrm{d}x}$ uniquely depending on $x$?

Answer 1

Implicit function theorem to the resccue!

We have :

$$ F(x, \alpha) = 0$$

Say that $\alpha(x)$ that is alpha is some function of x around some nice local interval which keeps us on the zero set of F.. and then take derivative

$$ \frac{dF}{dx} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial \alpha} \frac{d \alpha}{dx}$$

Hence,

$$ \frac{d \alpha}{dx} = \frac{-\frac{\partial F}{\partial x} }{ \frac{\partial F}{\partial \alpha}}$$

With,

$$ F(x, \alpha) = \alpha ^3-2 \alpha ^2- \alpha(6+4x^2) +8 x^3-4x$$

I'll leave the computation up to you :-)

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Going further, suppose we have a point on the curve $(x_o,\alpha_o)$ , then we can write a taylor expansion of $\alpha$ locally as:

$$ \alpha(x) = \alpha_o + \frac{d \alpha}{dx}|_{x_o,\alpha_o} (x-x_o) + \frac{1}{2!} \frac{ d^2 \alpha}{dx^2}|_{x_o,\alpha_o} (x-x_o)^2 + O \left( (x-x_o)^3 \right)$$

The catch is that this polynomial only behaves like a curve for some small segment of it (geometrically thinking). Also note that taking derivative of above polynomial yields $\frac{d \alpha}{dx}$ as a polnyomial in terms $(x-x_o)$

Answer 2

If you look at the determinant $$\Delta=-1472 x^6+1152 x^5+2944 x^4+1408 x^3+1488 x^2-992 x+1008$$ it is negative except for $-1.3< x < 2.1$.

So, if you exclude this region, using the hyperbolic method for one real root, you have $$\alpha=\frac{2}{3} \left(1+\sqrt{12 x^2+22} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{-54 x^3+18 x^2+27 x+31}{\sqrt{2} \left(6 x^2+11\right)^{3/2}}\right)\right)\right)$$

Write it as $$\alpha=\frac{2}{3} \left(1+f(x) \cosh \left(\frac{1}{3} \cosh ^{-1}(g(x))\right)\right)$$ from which $$\frac 32\frac{d\alpha}{dx}=f'(x) \cosh \left(\frac{1}{3} \cosh ^{-1}(g(x))\right)+$$ $$\frac {f(x)\,g'(x)}{3\sqrt{g^2(x)-1}} \sinh \left(\frac{1}{3} \cosh ^{-1}(g(x))\right)$$ Just a bit (!!) of patience.

Have fun. But, back to serious, enjoy the implicit function theorem.

Answer 3

This answer will overlap alot with @Claude_Leibovici 's answer, but also covers the case when the cubic has 3 real roots.

Starting with Nickalls' parameters related to the geometry of the cubic, applied to your cubic:

$$\begin{align*} \alpha_N &= \dfrac{-b}{3a} = \dfrac{2}{3} \quad \text{(abscissa of inflection point)}\\ \\ \left[\delta(x)\right]^2 &= \dfrac{b^2-3ac}{9a^2} = \alpha_N^2 - \dfrac{c}{3a} = \dfrac{4}{3}x^2+\dfrac{22}{9} \\ &\quad (\alpha\text{ distance squared from inflection point to turning point})\\ \\ y_N(x) &= f(\alpha_N) = \dfrac{2b^3}{27a^2} - \dfrac{bc}{3a} + d = 8x^3-\dfrac{8}{3}x^2-4x-\dfrac{124}{27}\\ &\quad \text{(ordinate of inflection point)}\\ \\ h(x) &= 2a\left[\delta(x)\right]^3 =2\left[\delta(x)\right]^3 = 2\left(\sqrt{\dfrac{4}{3}x^2+\dfrac{22}{9}}\right)^3 \\ &\quad \text{(y distance from inflection point to turning point)} \\ \end{align*}$$

The roots of your cubic, $\alpha_k(x)$, $k \in\{0,1,2\}$, can be expressed as a hyperbolic function of a complex hyperbolic angle, $\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)$, described in my short write-up here, based off of Nickalls' and Holmes' work:

$$\begin{align*} \alpha_k(x) &= \begin{cases} \alpha_N + 2\delta(x)\cosh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right] \quad h(x) \ne 0\\ \\ \lim_{h(x) \to 0} \alpha_N + 2\delta(x)\cosh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right] \\ \\ \quad \quad \quad= \alpha_N + \sqrt[3]{\dfrac{-y_N(x)}{a}}\exp\left[\dfrac{1}{3}\left(i2k\pi\right)\right] \quad h(x) = 0 \\ \end{cases} \\ \end{align*}$$

with

$$\begin{align*} \psi_k(u) &= i\theta_k(u)+ \phi(u)= \begin{cases} i(2k+1)\pi + \cosh^{-1}(-u) & u \in \mathbb{R}, u < -1\\ \\ i2k\pi + i\cos^{-1}(u) & u \in \mathbb{R}, -1 \le u \le 1\\ \\ i2k\pi + \cosh^{-1}(u) & u \in \mathbb{R}, u > 1\\ \\ i\left(2k+\dfrac{1}{2}\right)\pi + \sinh^{-1}(-iu) & u \in i\mathbb{R}\\ \end{cases} \\ \end{align*}$$

It's worth noting here that the input argument of $\psi_k(u)$, $u = \dfrac{-y_N(x)}{h(x)}$, is actually a discriminant between the case of 3 real roots and the other cases of 1 real root.

So to find your desired derivative

$$\begin{align*}\dfrac{d}{dx}\alpha_k(x)&= \dfrac{d}{dx}\left(\alpha_N+2\delta(x)\cosh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right]\right)\\ \\ &=2\delta'(x)\cosh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right]\\ \\ &\quad + 2\delta(x)\sinh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right]\dfrac{1}{3}\dfrac{d}{d(-y_N/h)}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\dfrac{d}{dx} \dfrac{-y_N(x)}{h(x)}\\ \\ &=2\delta'(x)\cosh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right]\\ \\ &\quad + 2\delta(x)\sinh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right]\dfrac{1}{3}\psi_k'\left(\dfrac{-y_N(x)}{h(x)}\right)\\ \\ &\quad \quad \cdot \dfrac{y_N(x)h'(x)-h(x)y'_N(x)}{\left[h(x)\right]^2}\\ \\ \end{align*}$$

At this point, the computation of $\delta'(x)$, $y'_N(x)$, and $h'(x)$ should be fairly straightforward.

The derivative of the complex hyperbolic angle $\psi_k(u)$ is

$$\begin{align*} \psi_k'(u) &= \begin{cases} \dfrac{-1}{\sqrt{u^2-1}} & u \in \mathbb{R}, u < -1\\ \\ \dfrac{-i}{\sqrt{1-u^2}} & u \in \mathbb{R}, -1 \le u \le 1\\ \\ \dfrac{1}{\sqrt{u^2-1}} & u \in \mathbb{R}, u > 1\\ \\ \dfrac{-i}{\sqrt{1-u^2}} & u \in i\mathbb{R}\\ \end{cases} \\ \end{align*}$$

Then all that's needed is substitution of all of those functions of $x$ into the expression for $\alpha_k'(x), k \in \{0,1,2\}$.

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If you don't want to use the hyperbolic function expression for the cubic roots, then from Nickalls' 1993 paper, you can go with

$$\begin{align*}\alpha_0(x) &= \alpha_N + \sqrt[3]{\dfrac{1}{2}\left(-y_N(x)+\sqrt{\left[y_N(x)\right]^2-\left[h(x)\right]^2}\right)} + \sqrt[3]{\dfrac{1}{2}\left(-y_N(x)-\sqrt{\left[y_N(x)\right]^2-\left[h(x)\right]^2}\right)}\\ \\ \alpha_{1,2}(x) &= \alpha_N - \dfrac{1}{2}\alpha_0(x) \pm i\dfrac{\sqrt{3}}{2}\sqrt{\left[\alpha_0(x)\right]^2-4\left[\delta(x)\right]^2}\\ \end{align*}$$

So computing your desired derivatives:

$$\begin{align*}\alpha'_0(x) &= \dfrac{1}{3}\dfrac{\dfrac{1}{2} \left(-y'_N(x) + \dfrac{y_N(x)y'_N(x)-h(x)h'(x)}{\sqrt{\left[y_N(x)\right]^2-\left[h(x)\right]^2}}\right)}{\left[\sqrt[3]{\dfrac{1}{2}\left(-y_N(x)+\sqrt{\left[y_N(x)\right]^2-\left[h(x)\right]^2}\right)}\right]^2} \\ \\ &\quad + \dfrac{1}{3}\dfrac{\dfrac{1}{2} \left(-y'_N(x) - \dfrac{y_N(x)y'_N(x)-h(x)h'(x)}{\sqrt{\left[y_N(x)\right]^2-\left[h(x)\right]^2}}\right)}{\left[\sqrt[3]{\dfrac{1}{2}\left(-y_N(x)-\sqrt{\left[y_N(x)\right]^2-\left[h(x)\right]^2}\right)}\right]^2} \\ \\\end{align*}$$

$$\begin{align*}\alpha'_{1,2}(x) &= -\dfrac{1}{2} \alpha'_0(x)\pm i\dfrac{\sqrt{3}}{2}\dfrac{\alpha_0(x)\alpha'_0(x) -4\delta(x)\delta'(x)}{\sqrt{\left[\alpha_0(x)\right]^2-4\left[\delta(x)\right]^2}}\\ \end{align*}$$

Again, all that's needed is substitution of all of those functions of $x$ into the expression for $\alpha_k'(x), k \in \{0,1,2\}$.