Elimination, Resultants And Discriminants

Excerpts

Elimination, Resultants And Discriminants
We call $R$ the *resultant* (or *eliminant*) of the two equations.
Elimination, Resultants And Discriminants
Methods of elimination which seem plausible often yield not $R$ itself, but the product of $R$ by an extraneous function of the coefficients.
Elimination, Resultants And Discriminants
Multiply the first equation by $x$ and the second by $x^2$ and $x$ in turn.
Elimination, Resultants And Discriminants
Evidently $D$ is unaltered by the interchange of any two roots.
Elimination, Resultants And Discriminants
The student should employ only methods of elimination (such as those due to Sylvester, Euler, and Bézout) which have been proved to lead to the true resultant.
Elimination, Resultants And Discriminants
Given without proof by Sylvester, *Philosophical Magazine*, 1840, p. 132.

Equations

Elimination, Resultants And Discriminants
x = -\frac{b}{a} = -\frac{d}{c}
If ax+b=0 and cx+d=0 hold for the same x, that common x equals -b/a and also -d/c.
Elimination, Resultants And Discriminants
R \equiv ad - bc = 0
The two linear equations have a common root exactly when R = ad - bc vanishes; R is the resultant (eliminant) of the pair.
Elimination, Resultants And Discriminants
f(x) = a_0x^m + a_1x^{m-1} + \dotsb + a_m
The polynomial f of degree m in x, with coefficients a_0 to a_m.
Elimination, Resultants And Discriminants
g(x) = \;b_0x^n + \;b_1x^{n-1} + \dotsb + \, b_n
The polynomial g of degree n in x, with coefficients b_0 to b_n.
Elimination, Resultants And Discriminants
R(f, g) = a_0^n g(\alpha_1)g(\alpha_2) \dotsm g(\alpha_m)
The resultant of f and g is a_0^n times the product of g evaluated at the roots of f; it vanishes exactly when f and g share a root.
Elimination, Resultants And Discriminants
F = \begin{vmatrix} a_0 & a_1 & a_2 & a_3 & 0 \\ 0 & a_0 & a_1 & a_2 & a_3 \\ b_0 & b_1 & b_2 & 0 & 0 \\ 0 & b_0 & b_1 & b_2 & 0 \\ 0 & 0 & b_0 & b_1 & b_2 \end{vmatrix}
Sylvester's determinant F for f cubic and g quadratic; its vanishing is the condition for a common root.
Elimination, Resultants And Discriminants
b_0^3z^2 + kz + F = 0
Replacing a_3 by a_3 - z turns the determinant into a quadratic in z whose constant term is F; k is an irrelevant middle coefficient.
Elimination, Resultants And Discriminants
F = b_0^3 f(\beta_1) f(\beta_2)
The Sylvester determinant equals b_0^3 times the product of f evaluated at the two roots of g.
Elimination, Resultants And Discriminants
\beta f + \alpha g \equiv 0
Where f and g have no common factor, there is an identity combining f and g with polynomial multipliers alpha and beta, which follows from the vanishing determinant.
Elimination, Resultants And Discriminants
(a_0b_1) = a_0b_1 - a_1b_0
Shorthand: the bracket (a_0b_1) denotes the 2x2 combination a_0b_1 - a_1b_0, and similarly for other index pairs.
Elimination, Resultants And Discriminants
(a_0b_1)(a_2b_3) - (a_0b_2)(a_1b_3) + (a_0b_3)(a_1b_2)
This combination of 2x2 brackets equals zero, by the vanishing of a determinant with two repeated rows.
Elimination, Resultants And Discriminants
F=(a_0b_3)R
The method's determinant F equals the bracket (a_0b_3) times the true resultant R, so the bracket is an extraneous factor.
Elimination, Resultants And Discriminants
f(x) \equiv a_0(x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_m)
The polynomial f factors into linear factors over its roots alpha_1 to alpha_m.
Elimination, Resultants And Discriminants
D = a_0^{2m-2}(\alpha_1 - \alpha_2)^2(\alpha_1 - \alpha_3)^2 \dotsm (\alpha_1 - \alpha_m)^2(\alpha_2 - \alpha_3)^2 \dotsm (\alpha_{m-1} - \alpha_m)^2
The discriminant D of f is a_0^{2m-2} times the product of the squared differences of all pairs of roots; it is a polynomial in the coefficients.
Elimination, Resultants And Discriminants
f'(\alpha_1) = a_0(\alpha_1 - \alpha_2)(\alpha_1 - \alpha_3) \dotsm (\alpha_1 - \alpha_m)
The derivative of f at the root alpha_1 equals a_0 times the product of the differences from alpha_1 to the other roots.
Elimination, Resultants And Discriminants
D = (-1)^{\frac{m(m-1)}{2}} \frac{1}{a_0} R(f, f')
The discriminant of f equals a sign times one over a_0 times the resultant of f and its derivative f'.
Elimination, Resultants And Discriminants
f \equiv (x-c)\alpha
If f and g have a common root c, then f has the factor x - c, with cofactor alpha of degree m-1.

Problems

Exercise Page144

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Exercise Page147

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Exercise Page150

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Exercise Page152

Exercise Page152, problem 1, p. 152
$x^2-y^2=9$, $xy = 5y$.
Printed answer:
- $y^2(16 - y^2)$; $y=0$, $x=±3$; $y=±4$, $x=+5$.
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- solve: FLAG-PARSE [{x: 3, y: 0}, {x: -3, y: 0}, {x: 5, y: 4}, {x: 5, y: -4}]
Exercise Page152, problem 2, p. 152
$x^2 + y^2 = 25$, $x^2 + 3(c-1)x + c(y^2 - 25) = 0$.
Printed answer:
- $(c-1)^2(y^2 - 25)(y^2 - 16)$. If $c\neq 1$, $y=±5$, $x=0$; $y=±4$, $x=+3$.
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- solve: FLAG-PARSE [{x: 0, y: 5}, {x: 0, y: -5}, {x: 3, y: 4}, {x: 3, y: -4}]
Exercise Page152, problem 3, p. 152
When $x^2 + ax + b = 0$ has a double root, what $3$-rowed determinant is zero?
Printed answer:
- $\begin{vmatrix} 1 & a & b \\ 2 & a & 0 \\ 0 & 2 & a \end{vmatrix} = 4b-a^2$.
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Exercise Page152, problem 4, p. 152
Find the roots of $x^6 + 3x^4 + 32x^3 + 67x^2 + 32x + 65 = 0$ by §79.
Printed answer:
- $2±3i$, $-2±i$, $±i$. $\vphantom{\begin{vmatrix}1\\ 1\\ 1\end{vmatrix}}$
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- solve: the printed answer does not match the problem [2 + 3*I, 2 - 3*I, -2 + I, -2 - I, I, -I]

Exercise Page153b

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Exercise Page153

Exercise Page153, problem 1, p. 153
Find the equation whose roots are the abscissas of the points of intersection of two general conics.
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Exercise Page153, problem 10, p. 153
Prove that the equation whose roots are the $n(n-1)$ differences $x_j-x_k$ of the roots of $f(x)=0$ may be obtained by eliminating $x$ between the latter and $f(x+y)=0$ and deleting from the eliminant the factor $y^n$ (arising from $y = x_j - x_j = 0$). The equation free of this factor may be obtained by eliminating $x$ between $f(x)=0$ and % f(x+y) - f(x)/y = f’(x) + f”(x)y1·2 + + f^(n)(x)y^n-11·2n = 0. This eliminant involves only even powers of $y$, so that if we set $y^2 = z$ we obtain an equation in $z$ having as its roots the squares of the differences of the roots of $f(x)=0$. % (Lagrange *Résolution des équations*, 1798, §8.)
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Exercise Page153, problem 11, p. 153
Compute by Ex. 10 the $z$-equation when $f(x) = x^3 + px + q$.
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- See Ex. 15, p. 134.
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Exercise Page153, problem 2, p. 153
Find a necessary and sufficient condition that f(x) x^4 + px^3 + qx^2 + rx + s = 0 shall have one root the negative of another root. When this condition is satisfied, what are the quadratic factors of $f(x)$? Apply to Ex. 4, §74. Hint: add and subtract $f(x)$ and $f(-x)$.
Printed answer:
- $pqr - p^2s - r^2 = 0$, $x^2 + r/p$, $x^2 + px + ps/r$.
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Exercise Page153, problem 3, p. 153
Solve $f(x) \equiv x^4 - 6x^3 + 13x^2 - 14x + 6 = 0$, given that two roots $\alpha$ and $\beta$ are such that $2\alpha + \beta = 5$. Hint: $f(x)$ and $f(5-2x)$ have a common factor.
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- $1$, $3$, $1± i$.
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- solve: the printed answer does not match the problem [1, 3, 1+I, 1-I]
Exercise Page153, problem 4, p. 153
Solve $x^3 + px + q = 0$ by eliminating $x$ between it and $x^2 + vx + w = y$ by the greatest common divisor process, and choosing $v$ and $w$ so that in the resulting cubic equation for $y$ the coefficients of $y$ and $y^2$ are zero. The next to the last step of the elimination %% -----File: 160.png---Folio 154------- gives $x$ as a rational function of $y$. (Tschirnhausen, *Acta Erudit.*, Lipsiae, II, 1683, p. 204.)
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Exercise Page153, problem 5, p. 153
Find the preceding $y$-cubic as follows. Multiply $x^2 + vx + w = y$ by $x$ and replace $x^3$ by $-px-q$; then multiply the resulting quadratic equation in $x$ by $x$ and replace $x^3$ by its value. The determinant of the coefficients of $x^2$, $x$, $1$ must vanish.
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Exercise Page153, problem 6, p. 153
Eliminate $y$ between $y^3 = v$, $x = ry + sy^2$, and get x^3 - 3rsvx - (r^3v + s^3v^2) = 0. Take $s=1$ and choose %[** PP: Typo chose] $r$ and $v$ so that this equation shall be identical with $x^3 + px + q = 0$, and hence solve the latter. (Euler, 1764.)
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Exercise Page153, problem 7, p. 153
Eliminate $y$ between $y^3 = v$, $x = f + ey + y^2$ and get vmatrix 1 & e & f-x e & f-x & v f-x & v & ev vmatrix =0. This cubic equation in $x$ may be identified with the general cubic equation by choice of $e$, $f$, $v$. % [** PP: , -> .] Hence solve the latter.
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Exercise Page153, problem 8, p. 153
Determine $r$, $s$ and $v$ so that the resultant of y^3 = v, y = x+ry+s shall be identical with $x^3 + px + q = 0$. (Bézout, 1762.)
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Exercise Page153, problem 9, p. 153
Show that the reduction of a cubic equation in $x$ to the form $y^3 = v$ by the substitution x = r + sy1 + y is not essentially different from the method of Ex. 7. [Multiply the numerator and denominator of $x$ by $1 - y + y^2$.]
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