The solution to a cubic equation

[Charles Link]

TL;DR Summary

The solution of $x^3-x+.1=0$ is of interest. A quick approximation to the 3 real roots is found. This is compared with the general solution that is found to the equation in the form $x^3+px=q$, using the substitution $x=w-\frac{p}{3w}$, (the standard way of solving this type of equation), and obtaining a quadratic in $w^3$.

The equation $x^3-x+.1=0$, has 3 real roots that can be quickly approximated as follows: Writing the equation $x=.1+x^3$, iterative methods quickly indicate that there is a root $r_1$ near $x=+.1$, and more accurately $r_1 \approx + .101$. $\\$ Doing an approximate calculation of factoring out $x-.1$ from $x^3-x+.1$ yields $(x-.1)(x^2+.1x-1 )=0$ (approximately). $\\$ The quadratic formula can be quickly applied to the quadratic factor to give $r_2 \approx +.95$ and $r_3 \approx -1.05$. $\\$ These results were found to be in agreement with the standard solution to this cubic equation using Vieta's substitution $x=w-\frac{p}{3w}$. $\\$ Here $p=-1$, and $q=-.1$, for the equation in the form $x^3+px=q$. $\\$ The substitution gives a quadratic in $w^3$ that is $w^6-qw^3-\frac{p^3}{27}=0$. $\\$ The number $w^3 \approx .192 e^{i (\theta +n 360^{\circ}) }$ where $\theta \approx 105.1^{\circ}$ and $\theta \approx 254.9^{\circ}$, and $n=0,1,2$. $\\$ The complex number $w$ is then found as $w= (.192)^{1/3} e^{i (\theta/3+n 120^{\circ})}=.577 e^{i (\theta/3+n 120^{\circ})}$. $\\$ There are only 3 distinct $w's$. $\\$ When the computation is then made for $x$ from $w$, the result is $x=w+\frac{1}{3w}\approx 2(.577) \cos{\phi}$ , and $x$ is real as it must be. The three roots $x \approx .101, .95, -1.05$ are then obtained as above. $\\$ Note: I could have kept more decimals in these calculations, but I did them mostly by hand, using Taylor series expansions, etc. to get them reasonably precise. The goal here was simply to show agreement with the very quick solution above.

 

[Charles Link]

One comment on the above: I did this problem mostly for my own interest. I wanted to come up with a cubic equation where iterative methods would supply the first root. The other two roots can then be quickly found (approximately) by factoring out $x-r_1$, and solving the quadratic expression. $\\$ I had to google the method for the general solution to the cubic equation. I had seen it in high school, (at least one method of solution), but that was about 45 years ago for me. $\\$ One additional comment I have that perhaps someone can give some feedback=I made the statement that there are only 3 distinct $w's$: I think there might be 6 distinct $w's$. The pair of $w's$ with $\phi=$35 degrees $=105/3$ and $325$ degrees $=255/3+240$ give the same root $x=+.95$, but they are distinct. The quadratic in $w ^3$ is a 6th power equation, so it comes as no surprise that there could be 6 different $w's$. $\\$ Likewise, $\phi= 105/3+120=155^{\circ}$ and $255/3+120=205^{\circ}$ both form the x=-1.05 root, while 105/3+240=275 and 255/3=85 give the x=+.101 root. $\\$ I think we can conclude there are 6 distinct $w's$.$\\$ [Edit: The quadratic equation for $w^3$ gets two separate solutions for $w^3=re^{i \theta}$, (two different $\theta's$), and in subsequently solving for $w$, the integer $n$ above gives 3 distinct values for each $\theta$, for $n=0, 1, 2$. (Of course $n$ can be any integer, but only $n=0,1,2$ give distinct values for $w$.)]

 

[A.T.]

I had to google the method for the general solution to the cubic equation. I had seen it in high school, (at least one method of solution), but that was about 45 years ago for

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