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In the context of physics, I stumbled over the following problem: I have $N$ equations, all are quadratic in a single scalar, real variable $x$: \begin{eqnarray} 0 &= A_1x^2 + B_1x + C_1 \\ &\dots\\ 0 &= A_Nx^2 + B_Nx + C_N \end{eqnarray} Also the coefficients $A_n,B_n,C_n$ are real scalars. For a solution $x$ to exist, it is necessary that the coefficients $A_n,B_n,C_n$ satisfy compatibility conditions. For small $N$ I can obtain them numerically from Gröbner bases by eliminating $x$. For example, for $N=2$, I obtain \begin{equation} 0 = C_1C_1 A_2A_2 + A_1A_1 C_2C_2 + B_1B_1 C_2A_2 + C_1A_1 B_2B_2 - C_1B_1 A_2B_2 - C_1A_1 C_2A_2 - C_1A_1 A_2C_2 - B_1A_1 B_2C_2 \end{equation} For higher $N$ there are similar other quartic compatibility equations, but I obtain also cubic ones and others of order five. The number of these conditions grows rapidly with increasing $N$, and I would like to understand their structure.

I would be grateful if someone could help me to understand the structure of these compatibility conditions, or point me to the relevant literature. I did not find a publication on this, and I cannot estimate whether this is standard textbook stuff or not known at all.

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    $\begingroup$ The quartic expression you've given in your example is the resultant of the polynomials $A_1x^2 + B_1x + C_1$ and $A_2x^2 + B_2x + C_2$. In general, given two single-variable polynomials $f$ and $g$, the resultant is a polynomial in the coefficients of $f$ and $g$ that vanishes if and only if $f$ and $g$ have a common zero. I might suggest taking a look at the multivariate version (the Macaulay resultant) and its variants as a starting point. $\endgroup$
    – John Doyle
    Jan 16 at 22:07
  • $\begingroup$ Thank you, John Doyle, for your quick help. I had a look at both types of resultants in the links to Wikipedia you proposed. They apply well to the case of two equations, but unfortunately I need to extend to N. The Macaulay resultant needs homogeneous polynomials, which can be achieved by adding another variable, but still I found that theory only for two equations. Do you have another idea where I could look for, say N>2 quadratic equations? $\endgroup$
    – Michael Schindler
    Jan 17 at 18:10
  • $\begingroup$ Hi @Michael, this Math Stack Exchange discussion includes a reference that discusses several single-variable polynomials. Perhaps this could be useful. $\endgroup$
    – John Doyle
    Jan 18 at 3:04
  • $\begingroup$ This is a very useful link. In particular, the described method by Kronecker generates quartic and cubic conditions. The latter are the determinant of the coefficients in three quadratic equations. This explains why I also observe both of them. $\endgroup$
    – Michael Schindler
    Jan 18 at 16:15

1 Answer 1

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First, we can assume that homogeneous pairs $(A_i : B_i)$ are pairwise distinct; otherwise equal pairs correspond to either redundant (proportional) or inconsistent equations.

Then for each pair of equations with indices $i\ne j$, if they have a common root, then it is expressed uniquely as $$\frac{A_jC_i-A_iC_j}{A_iB_j-A_jB_i}.$$ For an overall common solution to exist, it is therefore necessary that the above expression is constant for all consecutive pairs of equations: $$\frac{A_2C_1-A_1C_2}{A_1B_2-A_2B_1} = \frac{A_3C_2-A_2C_3}{A_2B_3-A_3B_2} = \dots = \frac{A_NC_{N-1}-A_{N-1}C_N}{A_{N-1}B_N-A_NB_{N-1}}.$$ To make it sufficient we also need to plug in this constant into one of the equations (say, first one) and verify that it evaluates to zero: $$A_1\big(\tfrac{A_2C_1-A_1C_2}{A_1B_2-A_2B_1})^2 + B_1\big(\tfrac{A_2C_1-A_1C_2}{A_1B_2-A_2B_1}) + C_1=0.$$

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  • $\begingroup$ Thank you, Max Alekseyev for your answer. Your argument is straightforward and clear. If I take the (consecutive) pairs of equations, I can expand them to polynomials of degree 4 which evaluate to zero. That makes them similar to what I has before. I would expect two sets of equations which encode the same ideal. A close look (and an implementation in sage) however shows me that these compatibility ideals are different. I included also the ideals from John Doyle's answer and found them also to be different. Am I wrong expecting the same compatibility ideals? $\endgroup$
    – Michael Schindler
    Jan 24 at 18:38

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