Solving a cubic equation

[Aion]

Problem.

By solving the cubic equation $x^3+x+1=3+\epsilon$, find the largest possible value of $\delta$ that works for any given $\epsilon >0$. (35 (b) from Stewart Calculus.)


My attempt:

Any third degree polynomial $x^3+px+q=0$ with $\Delta =\left(\frac{p}{3}\right)^3+\left(\frac{q}{2}\right)^2>0$ has one real root

$$x_1=\sqrt[3]{\frac{-q}{2}+\sqrt{\Delta}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\Delta}}$$and two complex conjugate roots
$$x_2=\omega^2\sqrt[3]{\frac{-q}{2}+\sqrt{\Delta}}+\omega\sqrt[3]{\frac{-q}{2}-\sqrt{\Delta}}$$ $$x_3=\omega\sqrt[3]{\frac{-q}{2}+\sqrt{\Delta}}+\omega^2\sqrt[3]{\frac{-q}{2}-\sqrt{\Delta}}$$
In this case, the equation above can be written as
$$x^3+x-(2+\epsilon)=0$$
where $p=1$ and $q=-(2+\epsilon)$. Hence $$\Delta=\left(\frac{1}{3}\right)^3+\left( \frac{2+\epsilon}{2}\right)^2>0$$
Substituting into Cardano's formula we have a real solution
$$x_1=\sqrt[3]{\frac{2+\epsilon}{2}+\sqrt{\left(\frac{1}{3}\right)^3+\left( \frac{2+\epsilon}{2}\right)^2}}+ \sqrt[3]{\frac{2+\epsilon}{2}-\sqrt{\left(\frac{1}{3}\right)^3+\left(\frac{2+\epsilon}{2}\right)^2}}$$
From a graphical perspective, we are finding the intersection of the polynomial $y=x^3+x+1$ and the horizontal line $y=3+ϵ$. Finding the intersection point of these two graphs, aligns with finding the $x$-coordinate of the intersection point as $\epsilon$ varies. For small $\epsilon$ we can approximate a solution as the equation becomes $x^3+x-2=0$ which has a real solution $x=1$. This makes sense since $\lim_{x \to 1} \left( x^3+x+1\right)=3$. This solution can be written as $x=1+\delta$. If we substitute $x=1+\delta$ into the original equation $x^3+x-(2+\epsilon)=0$ we obtain

$$(1+\delta)^3+(1+\delta)-(2+\epsilon)=0$$Expanding
$$1+3\delta+3\delta^2+\delta^3+1+\delta-2-\epsilon=0$$
For small $\delta$ we can ignore higher order terms $\delta^2$ and $\delta^3$.

$$\delta\approx\frac{\epsilon}{4}$$Thus the approximate solution for the x-coordinate when $y=x^3+x+1$ intersects the horizontal line $y=3+ϵ$ is

$$x\approx 1+\frac{\epsilon}{4}$$

 

[Aion]

Problem.

By solving the cubic equation $x^3+x+1=3+\epsilon$, find the largest possible value of $\delta$ that works for any given $\epsilon >0$. (35 (b) from Stewart Calculus.)


My attempt:

Any third degree polynomial $x^3+px+q=0$ with $\Delta =\left(\frac{p}{3}\right)^3+\left(\frac{q}{2}\right)^2>0$ has one real root

$$x_1=\sqrt[3]{\frac{-q}{2}+\sqrt{\Delta}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\Delta}}$$and two complex conjugate roots
$$x_2=\omega^2\sqrt[3]{\frac{-q}{2}+\sqrt{\Delta}}+\omega\sqrt[3]{\frac{-q}{2}-\sqrt{\Delta}}$$ $$x_3=\omega\sqrt[3]{\frac{-q}{2}+\sqrt{\Delta}}+\omega^2\sqrt[3]{\frac{-q}{2}-\sqrt{\Delta}}$$
In this case, the equation above can be written as
$$x^3+x-(2+\epsilon)=0$$
where $p=1$ and $q=-(2+\epsilon)$. Hence $$\Delta=\left(\frac{1}{3}\right)^3+\left( \frac{2+\epsilon}{2}\right)^2>0$$
Substituting into Cardano's formula we have a real solution
$$x_1=\sqrt[3]{\frac{2+\epsilon}{2}+\sqrt{\left(\frac{1}{3}\right)^3+\left( \frac{2+\epsilon}{2}\right)^2}}+ \sqrt[3]{\frac{2+\epsilon}{2}-\sqrt{\left(\frac{1}{3}\right)^3+\left(\frac{2+\epsilon}{2}\right)^2}}$$
From a graphical perspective, we are finding the intersection of the polynomial $y=x^3+x+1$ and the horizontal line $y=3+ϵ$. Finding the intersection point of these two graphs, aligns with finding the $x$-coordinate of the intersection point as $\epsilon$ varies. For small $\epsilon$ we can approximate a solution as the equation becomes $x^3+x-2=0$ which has a real solution $x=1$. This makes sense since $\lim_{x \to 1} \left( x^3+x+1\right)=3$. This solution can be written as $x=1+\delta$. If we substitute $x=1+\delta$ into the original equation $x^3+x-(2+\epsilon)=0$ we obtain

$$(1+\delta)^3+(1+\delta)-(2+\epsilon)=0$$Expanding
$$1+3\delta+3\delta^2+\delta^3+1+\delta-2-\epsilon=0$$
For small $\delta$ we can ignore higher order terms $\delta^2$ and $\delta^3$.

$$\delta\approx\frac{\epsilon}{4}$$Thus the approximate solution for the x-coordinate when $y=x^3+x+1$ intersects the horizontal line $y=3+ϵ$ is

$$x\approx 1+\frac{\epsilon}{4}$$

I also noticed that as $\epsilon \to 0$ we have
$$x=\sqrt[3]{1 + \sqrt{\frac{1}{27} + 1}} - \sqrt[3]{-1 + \sqrt{\frac{1}{27} + 1}}=1$$which also makes sense in this context.