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I am trying to solve this fourth-degree equation: $\omega^4+iB_3\omega^3+B_2\omega^2+iB_1\omega+B_0=0$, where coefficients $B_{0,1,2,3}$ are real, and $i$ is the imaginary number. The numerical values of these coefficients are: $B_0\approx10^6$, $B_1\approx10^7$, $B_2\approx-10^6$ and $B_3\approx-2$. Since it is a fourth-degree equation, so it must have only four solutions which are listed below. \begin{multline} \omega_{1}=-i \frac{B_3}{4}+\frac{1}{2}\ell -\frac{1}{2}\biggl(-\frac{4B_2}{3}-\frac{B_3^2}{2}-\frac{(12B_0+B_2^2+3B_1B_3)}{3\digamma}\\ -\frac{\digamma}{3} +\frac{i(-8B_1+4B_2B_3+B_3^3)}{4\ell}\biggr)^{1/2},\label{sol-1} \end{multline} \begin{multline} \omega_{2}=-i \frac{B_3}{4}+\frac{1}{2}\ell +\frac{1}{2}\biggl(-\frac{4B_2}{3}-\frac{B_3^2}{2}-\frac{(12B_0+B_2^2+3B_1B_3)}{3\digamma}\\ -\frac{\digamma}{3} +\frac{i(-8B_1+4B_2B_3+B_3^3)}{4\ell}\biggr)^{1/2},\label{sol-2} \end{multline} \begin{multline} \omega_{3}=-i \frac{B_3}{4}-\frac{1}{2}\ell -\frac{1}{2}\biggl(-\frac{4B_2}{3}-\frac{B_3^2}{2}-\frac{(12B_0+B_2^2+3B_1B_3)}{3\digamma}\\ -\frac{\digamma}{3} -\frac{i(-8B_1+4B_2B_3+B_3^3)}{4\ell}\biggr)^{1/2},\label{sol-3} \end{multline} \begin{multline} \omega_{4}=-i \frac{B_3}{4}-\frac{1}{2}\ell +\frac{1}{2}\biggl(-\frac{4B_2}{3}-\frac{B_3^2}{2}-\frac{(12B_0+B_2^2+3B_1B_3)}{3\digamma}\\ -\frac{\digamma}{3} -\frac{i(-8B_1+4B_2B_3+B_3^3)}{4\ell}\biggr)^{1/2},\label{sol-4} \end{multline} where \begin{multline} \digamma=\frac{1}{2^{1/3}}\Biggl(-27 B_1^2-72B_0 B_2+2B_2^3+9B_1B_2B_3-27B_0B_3^2\\ +\biggl\{-4(12B_0+B_2^2+3B_1B_3)^3+(-27B_1^2-72B_0B_2+2B_2^3+9B_1B_2B_3-27B_0B_3^2)^2\biggr\}^{1/2}\Biggr)^{1/3} \end{multline} and \begin{equation} \ell=\biggl(-\frac{2B_2}{3}-\frac{B_3^2}{4}+\frac{(12B_0+B_2^2+3B_1B_3)}{3\digamma}+\frac{\digamma}{3}\biggr)^{1/2}. \end{equation}

All the four solutions of this equation contain the term $\digamma$. The $\digamma$ can be written as $\digamma\approx(B_2^3)^{1/3}$, because the term $(2B_2^3)$ is the most dominant in the parenthesis of $\digamma$. Since $B_2$ is negative, we can also write write: $\digamma\approx|B_2|(-1)^{1/3}$. The term $(-1)^{1/3}$ has 3 distinct roots which are: $(-1)$, $\Bigl(\frac{1}{2}+i\frac{\sqrt{3}}{2}\Bigr)$ and $\Bigl(\frac{1}{2}-i\frac{\sqrt{3}}{2}\Bigr)$.

So, my question is: since all the four solutions of this fourth-degree equation contain $(-1)^{1/3}$, does each solution have 3 different values corresponding to distinct roots of $(-1)^{1/3}$. For example: $\omega_1$ have 3 values corresponding to each root of $(-1)^{1/3}$. In that way, is it correct to say, this fourth-degree equation will have 12 solutions in total. Please see below all the four roots:

I have solved the fourth degree equation with Mathematica, and given numerical values to $B$ coefficients. It is found that by default, mathematica choses the root: $\Bigl(\frac{1}{2}+i\frac{\sqrt{3}}{2}\Bigr)$ for $(-1)^{1/3}$. But mathematically, other roots are also valid. Which one root should I chose?

Manpreet Singh
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  • I am not sure about your point, but I would like to say that (assume you are talking about the 4th root formula that contains certain term) $B_2^{1/3}$ indeed have three roots, but did you check that whether no matter which of the three values of $B_2^{1/3}$ you plug in, you get the same set of four roots of the equation? – JetfiRex Feb 23 '22 at 01:07
  • Hi, JetfiRex I plugged each of the three roots of $(B_2^3)^{1/3}$ into all the four roots of the polynomial, it was found that all the 12 roots were distinct, and all of them satisfy the polynomial equation. – Manpreet Singh Feb 23 '22 at 02:21
  • I am not sure that condition is correct since I have not read the formula (it may be clearer to include that in your problem statement), but I really doubt this... It is really weird to have $(B_2^3)^{1/3}$ instead of $B_2$... Also for did you put all $B_2^{1/3}$ in the formula same? Take an example as the quadratic formula $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, no matter the square root take $+$ or $-$, they are the same two roots... – JetfiRex Feb 23 '22 at 02:34
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    @ManpreetSingh "all the 12 roots were distinct, and all of them satisfy the polynomial equation" $;-;$ That's certainly not possible. You must have made a mistake along the way, but it's hard to guess what/where since you did not include those calculations in the question. – dxiv Feb 23 '22 at 06:15
  • I have now added more details to the question. @dxiv – Manpreet Singh Feb 23 '22 at 06:26
  • Hi @JetfiRex, I have now added more details to the question. I hope you have more insight into my question. – Manpreet Singh Feb 23 '22 at 06:29
  • @ManpreetSingh It does not matter which cube root you choose for $\digamma$ but it must be the same for all roots. With different choices, you'll get the same four roots but in a different order. Same goes for $\ell$, changing the square root $\ell \mapsto -\ell$ just swaps $\omega_{1,2}$ with $\omega_{3,4}$. – dxiv Feb 23 '22 at 06:34
  • I agree with @dxiv. I also think no matter which cube root you choose it is just some permutation. "All the 12 roots were distinct, and all of them satisfy the polynomial equation" may be caused by miscalculation. I can't tell if you are changing the cube root, how the roots permutes. But as you said "12 roots are different" could you show your counterexample (like, how you choose four numbers, what is $\digamma$ and $l$, how you get 12 roots, etc.)? (I personally suspect but don't know why that changing the cube root will cause a 4-cycle permutation) – JetfiRex Feb 23 '22 at 07:00
  • @dxiv Yes, ideally, with different choices, I should get the same four roots but in a different. In order to verify this, I assigned each root of $(-1)^{1/3}$ to solutions $\omega_{1,2,3,4}$, and labeled them respectively $\omega_{1a}$, $\omega_{1b}$, $\omega_{1c}$, up to $\omega_{4a}$, $\omega_{4b}$, $\omega_{4c}$ (total 12 solutions). – Manpreet Singh Feb 23 '22 at 07:11
  • @dxiv There should be four distinct roots but in different orders. In order to verify this last statement, I found unique differences between these roots, e.g., $\omega_{1a}-\omega_{4b}$ or $\omega_{2a}-\omega_{4c}$. There are 53 such unique differences. If the difference between any two roots is zero, then these roots should be equal. For example, if $\omega_{1a}-\omega_{4b}=0$, then $\omega_{1a}=\omega_{4b}$. But, I found that all the 53 differences are non-zero, which means all the 12 roots are distinct. – Manpreet Singh Feb 23 '22 at 07:11
  • Also, I would suggest a way. Please try $x^4-1$, use your formula to calculate, choose different cube roots and see how it promutes. This will give you some intuition. – JetfiRex Feb 23 '22 at 07:16
  • @ManpreetSingh "found that all the 53 differences are non-zero" $;-;$ Again, that's certainly not possible. Of course, this applies to the actual roots. If you use approximations such as $\digamma\approx|B_2|(-1)^{1/3}$ then it is possible to get $12$ different values, but those are not the roots, just $12$ approximations of the same $4$ roots. – dxiv Feb 23 '22 at 07:22
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    This polynomial could have at most four roots, by the fundamental theorem of algebra. Try to search for "\( a z^4+z^3+z^2+dz+e=0\) counting roots" on SearchOnMath. You can find this thread on how to count roots of a quartic equation, for instance. – José Claudinei Ferreira Feb 23 '22 at 01:16
  • @dxiv, On further analysis of the numerical values of the 12 roots with mathematica, it is found that due to numerical accuracy issues with mathematica, I was getting different roots. But, on some more manipulations of the equations, I got only 4 unique solutions of this fourth degree equation. Now, for different roots of $(-1)^{1/3}$, I am getting same four solutions in different orders. ... Thanks for the help. – Manpreet Singh Feb 24 '22 at 00:08
  • @JetfiRex thanks for the help. Problem is now solved. – Manpreet Singh Feb 24 '22 at 00:09

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$$\omega^4+iB_3\omega^3+B_2\omega^2+iB_1\omega+B_0=0$$ Let $\omega=ix$ $$x^4+B_3x^3-B_2x^2-B_1x+B_0=0$$ With the given values for the $B_n$, there two real roots and two complex conjugate roots.

Now, use the same method as in your post.

Claude Leibovici
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