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- 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.

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