Final answer:
To prove that (φ₁,φ₂,φ₃) is a basis for P₂(R)′, we need to show that it spans P₂(R)′ and is linearly independent. To find a basis (p₁,p₂,p₃) of P₂(R) whose dual is (φ₁,φ₂,φ₃), we can choose p₁(x) = 1, p₂(x) = x, and p₃(x) = x² as a basis.
Step-by-step explanation:
To prove that (φ₁,φ₂,φ₃) is a basis for P₂(R)′, we need to show that it spans P₂(R)′ and is linearly independent.
To show that it spans P₂(R)′, we need to show that any polynomial in P₂(R) can be written as a linear combination of φ₁,φ₂, and φ₃. Let's consider any polynomial p(x) ∈ P₂(R). We can express p(x) as a linear combination of φ₁,φ₂, and φ₃ as follows:
p(x) = a₁φ₁(x) + a₂φ₂(x) + a₃φ₃(x), where a₁, a₂, and a₃ are coefficients.
To show that (φ₁,φ₂,φ₃) is linearly independent, we need to prove that the only solution to the equation a₁φ₁(x) + a₂φ₂(x) + a₃φ₃(x) = O is a₁ = a₂ = a₃ = 0. Let's assume that a₁φ₁(x) + a₂φ₂(x) + a₃φ₃(x) = O. We can differentiate both sides of this equation to get (a₁φ₁)'(x) + (a₂φ₂)'(x) + (a₃φ₃)'(x) = O. Expanding and simplifying this equation, we get (a₁φ₁)'(x) + (a₂φ₂)'(x) + (a₃φ₃)'(x) = a₁φ'₁(x) + a₂φ'₂(x) + a₃φ'₃(x) = O. Since φ₁(x), φ₂(x), and φ₃(x) are linearly independent, a₁φ'₁(x) + a₂φ'₂(x) + a₃φ'₃(x) = O only if a₁ = a₂ = a₃ = 0.
Hence, (φ₁,φ₂,φ₃) is a basis for P₂(R)′. To find a basis (p₁,p₂,p₃) of P₂(R) whose dual is (φ₁,φ₂,φ₃), we need to find polynomials p₁,p₂, and p₃ such that their duals are φ₁,φ₂, and φ₃, respectively. We can choose p₁(x) = 1, p₂(x) = x, and p₃(x) = x² as a basis. The duals of these polynomials are φ₁(p₁) = φ₁(1) = 0, φ₂(p₂) = φ₂(x) = x(2-1) = x, and φ₃(p₃) = φ₃(x²) = x²(3-1) = 2x².