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For any cubic equation, $ax^{3}+bx^{2}+cx+d=0$, we know there is always a real root if $a,b,c,d$ are all real. Suppose that $a,b,c,d$ are continuous and real function with respect of $i\in \mathbb{R}$, formulated as $a(i),b(i),c(i),d(i)$, is there the continuous real root, $x(i)$, of the cubic equation? If there is, I wonder whether there is a closed formula to present it.

Lee White
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    It seems unlikely to me. Think about the critical points. For a general cubic, varying $d$ continuously leads to transitions from three roots to two roots (one double) to one root and you can't pick the highest root or the lowest root to follow as there will be a transition in which that root disappears. And the middle root disappears in both transitions. But my thoughts on general grounds like these have been wrong before. – Mark Bennet Oct 08 '22 at 17:14
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    At least if you pick $a_0, b_0, c_0, d_0, x_0$ for which $3a_0x_0^2+2b_0x_0+c_0\neq 0$ and $a_0x_0^3+b_0x_0^2+c_0x_0+d_0=0$ you have that locally $x$ is a smooth function of $a,b,c,d$ (by the implicit function theorem), but this doesn't quite answer the question. – Mor A. Oct 08 '22 at 17:24

1 Answers1

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No, this is not possible and Mark Bennet's argument in the comments is correct. Suppose that the roots begin at the positions $(-1 - i, -1 + i, 1)$, so that our continuous real root must be $1$. Next have the complex roots travel towards the real line until they meet there, forming a double root at $-1$. Our continuous real root must still be $1$. Then have the double root separate into two real roots, the middle one of which travels from $-1$ to $1$. Our continuous real root must still be $1$. Now we have a double root at $1$ which can separate into a pair of complex roots, so that we end up with the roots $(-1, 1 + i, 1 - i)$. At the point at which our double root at $1$ travels away from the real line the only real root is $-1$ but our continuous real root is stuck at $1$.

However, the possibility of the roots colliding is the only obstacle, and away from the discriminant locus, where all the roots must be distinct, this is possible. The complement of the discriminant locus has two connected components; in one of them there's a unique real root which varies continuously (even smoothly as Mor A. says in the comments), and in another one there are three distinct real roots, all three of which vary continuously (again, even smoothly as Mor A. says in the comments).

Qiaochu Yuan
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  • Thanks for your answer. Do you mean one possible condition for the continuous (even smooth) of the real root is the discriminant is continuous and at most one root? – Lee White Oct 09 '22 at 03:23
  • @Lee: no, I mean the condition is that the discriminant is always nonzero, so there are never any repeated roots. – Qiaochu Yuan Oct 09 '22 at 03:47
  • @Yuan: I can understand that if the discriminant is always non-zero, the real root is continuous. However, I also think that if there is only one zero for the discriminant, the root is still continuous. – Lee White Oct 09 '22 at 04:29
  • @Yuan: Denote the discriminant as $\Delta(i)$. Suppose the only zero is at $i=I\in \mathbb{R}$. Thus $\Delta(i)>0$ when $i<I$ and $\Delta(i)<0$ when $i>I$ (or $\Delta(i)<0$ when $i<I$ and $\Delta(i)>0$ when $i>I$). When $\Delta(i)<0$ is continuous, three real roots are continuous, and when $\Delta(i)>0$, one real root is continuous. Thus I think that by picking one of the real roots when $\Delta(i)<0$ and the real root when $\Delta(i)>0$, we can get a continuous function on $\mathbb{R}$. – Lee White Oct 09 '22 at 04:30
  • @LeeWhite: I haven't thought about it carefully but that sounds right. The counterexample above requires the discriminant to be zero twice. – Qiaochu Yuan Oct 09 '22 at 16:07