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The ratio of the unsigned coefficients for the discriminants of $x^n+bx+c$ for $n=2$ to $5$ follow a simple pattern:

$$\left (\frac{2^2}{1^1},\frac{3^3}{2^2},\frac{4^4}{3^3},\frac{5^5}{4^4} \right )=\left ( \frac{4}{1},\frac{27}{4},\frac{256}{27},\frac{3125}{256} \right )$$

corresponding to the discriminants

$$(b^2-4c, -4b^3-27c^2,-27b^4+256c^3,256b^5+3125c^4).$$

Does the pattern for the ratios extend to higher orders? (An online reference would be appreciated.)

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  • $\begingroup$ The general formula for the discriminant curve is given on page 248 of Forsyth's Theory of Differential Equations Part II (Cambridge, 1900) as $\left ( \frac{c}{n-1} \right )^{n-1}-\left ( -\frac{b}{n} \right )^{n}=0$. $\endgroup$ – Tom Copeland Jul 1 '12 at 23:03
  • $\begingroup$ These results (and book about discriminants by Gelfand, Kapranov and Zelevinsky) could interest you too. $\endgroup$ – Raymond Manzoni Feb 24 '14 at 23:41
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    $\begingroup$ See also section 6 of "Discriminating deltas, depressed equations, and generalized Catalan numbers" (tcjpn.wordpress.com/2012/06/13/…) to relate the tangents of the discriminant curve to the equation $x^n + b x + c = 0$. $\endgroup$ – Tom Copeland Aug 1 '16 at 23:33
  • $\begingroup$ See also p. 775 of "Function series, Catalan numbers and random walks on trees" by Bajunaid, Cohen, Colonna, and Singman $\endgroup$ – Tom Copeland Dec 25 '19 at 16:50
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Yes. Sketch: $b$ is a symmetric polynomial of degree $n-1$ in the roots and $c$ is a symmetric polynomial of degree $n$, whereas the entire discriminant is a symmetric polynomial of degree $n(n-1)$. It follows that the discriminant is a linear combination of $b^n$ and $c^{n-1}$, and the coefficients can be determined by setting $b = 0, c = -1$ and then $b = -1, c = 0$ and reducing to the computation of the discriminant of $x^n - 1$.

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  • $\begingroup$ This is perhaps the easier way to see where the factor $n^n$ comes from if you are familiar with the DFT matrix, $S_n$, given in ccrma.stanford.edu/~jos/mdft/Matrix_Formulation_DFT.html . $det (S^{*}_{n}S_n)=n^n$ $\endgroup$ – Tom Copeland Jun 22 '12 at 8:54
  • $\begingroup$ Qiaochu, I noticed you deleted your allusion to the Vandermonde matrix, but I think the corresponding Wikipedia reference helps to make the connection between the discriminant and the DFT matrix (as well as other connections). $\endgroup$ – Tom Copeland Jun 22 '12 at 9:38
  • $\begingroup$ @Tom: well, it wasn't quite the right thing to say. It was a specific Vandermonde matrix rather than a general one. $\endgroup$ – Qiaochu Yuan Jun 22 '12 at 10:46
  • $\begingroup$ See also math.stackexchange.com/questions/103491/… and math.stackexchange.com/questions/18119/… $\endgroup$ – Tom Copeland Jun 30 '12 at 12:20
  • $\begingroup$ Why does the discriminant need to be a linear combination of $b^n$ and $c^{n-1}$, given that they are symmetric polynomials in the roots of those particular degrees? $\endgroup$ – PrimeRibeyeDeal Jan 27 '15 at 23:17
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Use the relation between the disciminant of $f$ and the resultant of $f$ and $f'$. The resultant is easy to calculate since $f'$ is so simple.

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  • $\begingroup$ Thanks, clearly the Sylvester matrix for the resultant has a sub-diagonal related to $n^n$, but then I'm left with understanding where the matrix itself comes from. $\endgroup$ – Tom Copeland Jun 22 '12 at 9:19
  • $\begingroup$ The idea of the resultant of $f$ and $g$ is that it should be zero if and only if $f$ and $g$ have a common (non-constant) factor. That happens if and only if there are polynomials $a$ and $b$ with degree $a$ less than degree $g$, and degree $b$ less than degree $f$, such that $af+bg=0$. And by considering this as a system of linear equations in the coefficients, this happens if and only if the Sylvester matrix is singular. $\endgroup$ – Gerry Myerson Jun 22 '12 at 12:32

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