The content in today's blog is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.
Lemma 1: Depressed Quartic Equation
If:
x4 + bx3 + cx2 + dx + e = 0
Then there exists y,p,q,r such that:
y4 + py2 + qy + r = 0
where:
y = x + (b/4)
p = c - 6(b/4)2
q = d - (b/2)c + (b/2)3
r = e - (b/4)d + (b/4)2c - 3(b/4)4
Proof:
(1) Let y = x + (b/4) so that x = y - (b/4)
(2) We note that (see here for details on the Binomial Theorem if needed):
(u - v)2 = u2 - 2uv + v2
(u - v)3 = u3 - 3u2v + 3uv2 -v3
(u - v)4 = u4 - 4u3v + 6u2v2 -4uv3 + v4
(3) This then gives us:
(y - [b/4])4 = y4 - by3 + (3/8)b2y2 -b3y/16 + (b/4)4
b(y - [b/4])3 = by3 - (3/4)b2y2 + (3/16)b3y - b4/64
c(y - [b/4])2 = cy2 - bcy/2 + c(b/4)2
d(y - [b/4]) = dy - bd/4
(4) Putting this all together gives us:
x4 + bx3 + cx2 + dx + e =
= y4 + (3/8)b2y2 -b3y/16 + (b/4)4 - (3/4)b2y2 + (3/16)b3y - b4/64 + cy2 - bcy/2 + c(b/4)2 + dy - bd/4 +e =
= y4 + [c-(3/8)b2]y2 + [ (1/8)b3 -bc/2 + d]y + [e - bd/4 + c(b/4)2 - 3b4/256 ]
QED
Lemma 2: Solution for y4 + py2 + r = 0
If y4 + py2 + r = 0
Then:
y = ± √ (1/2)[-p ± √p2 - 4r
Proof:
(1) y4 + py2 + r = 0
(2) (y2)2 + p(y2) + r = 0
(3) Using the quadratic equation (see here):
y2 = (1/2)[-p ± √p2 - 4r
(4) Solving for y2, we get:
y = ± √ (1/2)[-p ± √p2 - 4r
QED
Lemma 3:
If:
y4 + py2 + qy + r = 0 and q ≠ 0
Then there exists u ≠ 0 such that:
(y2 + p/2 + u)2 = [√2uy - q/(2√2u)]2
Proof:
(1) y4 + py2 + qy + r = 0
(2) y4 + py2 + (p/2)2 = -qy -r + (p/2)2
(3) (y2 + p/2)2 = -qy -r + (p/2)2
(4) Let u be the solution of the following equation:
8u3 + 8pu2 + (2p2 -8r)u - q2 = 0.
We know that this solution exists from the general cubic equation (see here)
(5) We know that u ≠ 0 since if u = 0, then q = 0 but q ≠ 0.
(6) So, we can rearrange our equation in step #4 to:
8u3 + 8pu2 + (2p2 -8r)u = q2
And after dividing both sides by 8u, we get:
u2 + pu + (p/2)2 - r = q2/8u.
(7) Now, we know that:
(y2 + p/2 + u)2 = -qy -r + (p/2)2 + 2uy2 + pu + u2
(8) So, using step #6, we have:
-qy -r + (p/2)2 + 2uy2 + pu + u2 =
= -qy + 2uy2 + q2/8u =
= [√2uy - q/(2√2u)]2
QED
Thereom: General Quartic Equation
If:
ax4 + bx3 + cx2 + dx + e = 0
p = (c/a) - 6([b/a]/4)2
q = (d/a) - ([b/a]/2)c + ([b/a]/2)3
r = (e/a) - ([b/a]/4)d + ([b/a]/4)2c - 3([b/a]/4)4
s = ([b/a]/4)
u = the solution to the cubic equation: 8u3 + 8pu2 + (2p2 -8r)u - q2 = 0.
Then:
if q = 0,
x = ± √ (1/2)[-p ± √p2 - 4r - s
if q ≠ 0, then:
x = ± (1/2) [√2uy + √2u - 2p - 4u + 2q/(√2u)] - s
or
x = ± (1/2) [-√2uy + √2u - 2p - 4u - 2q/(√2u)] - s
Proof:
(1) ax4 + bx3 + cx2 + dx + e = 0
(2) If we divide all sides by a, we get:
x4 + (b/a)x3 + (c/a)x2 + (d/a)x + (e/a) = 0
(3) Using Lemma 1 above, we get:
y4 + py2 + qy + r = 0
where:
y = x + ([b/a]/4)
p = (c/a) - 6([b/a]/4)2
q = (d/a) - ([b/a]/2)c + ([b/a]/2)3
r = (e/a) - ([b/a]/4)d + ([b/a]/4)2c - 3([b/a]/4)4
(4) If q=0, then using Lemma 2 above, we have:
y = ± √ (1/2)[-p ± √p2 - 4r
(5) If q ≠ 0, then using Lemma 3 above, we have:
(y2 + p/2 + u)2 = [√2uy - q/(2√2u)]2
where
u is the solution to:
8u3 + 8pu2 + (2p2 -8r)u - q2 = 0.
We can find this solution using the cubic equation (see here).
(6) This then gives us:
y2 + p/2 + u = ± √2uy - q/(2√2u).
(7) This then gives us four solutions:
(a) Case I: y2 + p/2 + u = + √2uy - q/(2√2u).
In this case:
y2 - √2uy + [p/2 + u + q/(2√2u)] = 0
Using the quadratic equation:
y = ± (1/2) [√2uy + √2u - 2p - 4u - 2q/(√2u)]
(b) Case I: y2 + p/2 + u = - √2uy + q/(2√2u).
In this case:
y2 + √2uy + [p/2 + u - q/(2√2u)] = 0
Using the quadratic equation:
y = ± (1/2) [-√2uy + √2u - 2p - 4u + 2q/(√2u)]
(8) Since x = y - ([b/a]/4), we have the following solutions:
(a) If q=0, then:
x = ± √ (1/2)[-p ± √p2 - 4r - ([b/a]/4)
(b) If q ≠ 0, then:
x = ± (1/2) [√2uy + √2u - 2p - 4u - 2q/(√2u)] - ([b/a]/4)
or
x = ± (1/2) [-√2uy + √2u - 2p - 4u + 2q/(√2u)] - ([b/a]/4)
QED
References
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations
, World Scientific, 2002