Problem.
By solving the cubic equation [imath]x^3+x+1=3+\epsilon[/imath], find the largest possible value of [imath]\delta[/imath] that works for any given [imath]\epsilon >0[/imath]. (35 (b) from Stewart Calculus.)
My attempt:
Any third degree polynomial [imath]x^3+px+q=0[/imath] with [imath]\Delta =\left(\frac{p}{3}\right)^3+\left(\frac{q}{2}\right)^2>0[/imath] has one real root
[math]x_1=\sqrt[3]{\frac{-q}{2}+\sqrt{\Delta}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\Delta}}[/math]and two complex conjugate roots
[math]x_2=\omega^2\sqrt[3]{\frac{-q}{2}+\sqrt{\Delta}}+\omega\sqrt[3]{\frac{-q}{2}-\sqrt{\Delta}}[/math] [math]x_3=\omega\sqrt[3]{\frac{-q}{2}+\sqrt{\Delta}}+\omega^2\sqrt[3]{\frac{-q}{2}-\sqrt{\Delta}}[/math]
In this case, the equation above can be written as
[math]x^3+x-(2+\epsilon)=0[/math]
where [imath]p=1[/imath] and [imath]q=-(2+\epsilon)[/imath]. Hence [math]\Delta=\left(\frac{1}{3}\right)^3+\left( \frac{2+\epsilon}{2}\right)^2>0[/math]
Substituting into Cardano's formula we have a real solution
[math]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}}[/math]
From a graphical perspective, we are finding the intersection of the polynomial [imath]y=x^3+x+1[/imath] and the horizontal line [imath]y=3+ϵ[/imath]. Finding the intersection point of these two graphs, aligns with finding the [imath]x[/imath]-coordinate of the intersection point as [imath]\epsilon[/imath] varies. For small [imath]\epsilon[/imath] we can approximate a solution as the equation becomes [imath]x^3+x-2=0[/imath] which has a real solution [imath]x=1[/imath]. This makes sense since [imath]\lim_{x \to 1} \left( x^3+x+1\right)=3[/imath]. This solution can be written as [imath]x=1+\delta[/imath]. If we substitute [imath]x=1+\delta[/imath] into the original equation [imath]x^3+x-(2+\epsilon)=0[/imath] we obtain
[math](1+\delta)^3+(1+\delta)-(2+\epsilon)=0[/math]Expanding
[math]1+3\delta+3\delta^2+\delta^3+1+\delta-2-\epsilon=0[/math]
For small [imath]\delta[/imath] we can ignore higher order terms [imath]\delta^2[/imath] and [imath]\delta^3[/imath].
[math]\delta\approx\frac{\epsilon}{4}[/math]Thus the approximate solution for the x-coordinate when [imath]y=x^3+x+1[/imath] intersects the horizontal line [imath]y=3+ϵ[/imath] is
[math]x\approx 1+\frac{\epsilon}{4}[/math]
By solving the cubic equation [imath]x^3+x+1=3+\epsilon[/imath], find the largest possible value of [imath]\delta[/imath] that works for any given [imath]\epsilon >0[/imath]. (35 (b) from Stewart Calculus.)
My attempt:
Any third degree polynomial [imath]x^3+px+q=0[/imath] with [imath]\Delta =\left(\frac{p}{3}\right)^3+\left(\frac{q}{2}\right)^2>0[/imath] has one real root
[math]x_1=\sqrt[3]{\frac{-q}{2}+\sqrt{\Delta}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\Delta}}[/math]and two complex conjugate roots
[math]x_2=\omega^2\sqrt[3]{\frac{-q}{2}+\sqrt{\Delta}}+\omega\sqrt[3]{\frac{-q}{2}-\sqrt{\Delta}}[/math] [math]x_3=\omega\sqrt[3]{\frac{-q}{2}+\sqrt{\Delta}}+\omega^2\sqrt[3]{\frac{-q}{2}-\sqrt{\Delta}}[/math]
In this case, the equation above can be written as
[math]x^3+x-(2+\epsilon)=0[/math]
where [imath]p=1[/imath] and [imath]q=-(2+\epsilon)[/imath]. Hence [math]\Delta=\left(\frac{1}{3}\right)^3+\left( \frac{2+\epsilon}{2}\right)^2>0[/math]
Substituting into Cardano's formula we have a real solution
[math]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}}[/math]
From a graphical perspective, we are finding the intersection of the polynomial [imath]y=x^3+x+1[/imath] and the horizontal line [imath]y=3+ϵ[/imath]. Finding the intersection point of these two graphs, aligns with finding the [imath]x[/imath]-coordinate of the intersection point as [imath]\epsilon[/imath] varies. For small [imath]\epsilon[/imath] we can approximate a solution as the equation becomes [imath]x^3+x-2=0[/imath] which has a real solution [imath]x=1[/imath]. This makes sense since [imath]\lim_{x \to 1} \left( x^3+x+1\right)=3[/imath]. This solution can be written as [imath]x=1+\delta[/imath]. If we substitute [imath]x=1+\delta[/imath] into the original equation [imath]x^3+x-(2+\epsilon)=0[/imath] we obtain
[math](1+\delta)^3+(1+\delta)-(2+\epsilon)=0[/math]Expanding
[math]1+3\delta+3\delta^2+\delta^3+1+\delta-2-\epsilon=0[/math]
For small [imath]\delta[/imath] we can ignore higher order terms [imath]\delta^2[/imath] and [imath]\delta^3[/imath].
[math]\delta\approx\frac{\epsilon}{4}[/math]Thus the approximate solution for the x-coordinate when [imath]y=x^3+x+1[/imath] intersects the horizontal line [imath]y=3+ϵ[/imath] is
[math]x\approx 1+\frac{\epsilon}{4}[/math]