Let the function f(x)=x⁴-2x²+8 be given.
On 3/28/2025 8:33 AM, Richard Hachel wrote:
Let the function f(x)=x⁴-2x²+8 be given.
Define '⋅'
⟨0,1⟩⋅⟨0,1⟩ = ⟨-1,0⟩
⟨1,0⟩⋅⟨0,1⟩ = ⟨0,1⟩
⟨0,1⟩⋅⟨1,0⟩ = ⟨0,1⟩
⟨1,0⟩⋅⟨1,0⟩ = ⟨1,0⟩
f(⟨x,y⟩) = ⟨x,y⟩⁴-2⟨x,y⟩²+8⟨1,0⟩
It follows that
f(⟨x,y⟩) = ⟨0,0⟩ ⇔
⟨x,y⟩ ∈ {⟨a,b⟩,⟨-a,b⟩,⟨a,-b⟩,⟨-a,-b⟩}
f(⟨x,y⟩) = ⟨x-a,y-b⟩⋅⟨x+a,y-b⟩⋅⟨x-a,y+b⟩⋅⟨x+a,y+b⟩
a = 2³ᐟ⁴cos(½tan⁻¹(√7))
b = 2³ᐟ⁴sin(½tan⁻¹(√7))
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