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ehrenfest
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[SOLVED] roots of a polynomial
Let P(x) be a polynomial of odd degree with real coefficients. Show that the equation P(P(x))=0 has at least as many real roots as the equation P(x) = 0, counted without multiplicities.
By the FTC, P(x) and P(P(x)) factor into complex linear factors.
Please just give me hint.
By the odd degree, we know that both P(x) and P(P(x)) have at least one real root.
By the FTC, P(x) and P(P(x)) factor into complex linear factors.
Oh wait, let \alpha_1,...,\alpha_m be the roots of P(x)=0. Because P(x) has odd degree, we know that p(R) = R. So, we can find distinct \beta_1,...,\beta_n such that P(\beta_i) = \alpha_i. That was easy. I guess I will post it anyway.
Homework Statement
Let P(x) be a polynomial of odd degree with real coefficients. Show that the equation P(P(x))=0 has at least as many real roots as the equation P(x) = 0, counted without multiplicities.
Homework Equations
By the FTC, P(x) and P(P(x)) factor into complex linear factors.
The Attempt at a Solution
Please just give me hint.
By the odd degree, we know that both P(x) and P(P(x)) have at least one real root.
By the FTC, P(x) and P(P(x)) factor into complex linear factors.
Oh wait, let \alpha_1,...,\alpha_m be the roots of P(x)=0. Because P(x) has odd degree, we know that p(R) = R. So, we can find distinct \beta_1,...,\beta_n such that P(\beta_i) = \alpha_i. That was easy. I guess I will post it anyway.
