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Question: Root Manifest For each positive integer n, prove that the equation P_n(x) = x^(n+2) - 2x + 1 = 0 has exactly one root c_n between 0 and 1, and determine lim(n -> +infinity) c_n. Answer: Note that x^(n+2) - 2x + 1 = (x-1)*( -1 + x + x2 + ... + x^n + x^(n+1) ). Hence it suffices to show that Q_n(x) := -1 + x + x2 + ... + x^n + x^(n+1) has a single root between 0 and 1. But this must be true since Q_n(x) is strictly increasing and cts on [0,1], Q_n(0) = -1 < 0, and Q_n(1) = n > 0. Now from (c_n)^(n+1) - 2c_n + 1 = 0, we get c_n = 1/2 + (1/2)(c_n)^(n+2), and c_n > 1/2 for all n. We will now show that c_n --> 1/2 as n goes to infinity. Note that it suffices to show that (c_n)^(n+2) --> 0 as n goes to infinity. Since c_n is a root of Q_n(x) = 0, we have (c_n)^(n+1)
+ (c_n)^n + ... + c_n = 1, and likewise Now if c_(n+1) were greater than or equal to c_n, then (c_(n+1))^k >= (c_n)^k throughout, implying (c_(n+1))^(n+2) + ((c_(n+1))^(n+1) + ... + c_(n+1)) >= (c_(n+1))^(n+2) + ((c_n)^(n+1) + ... + c_n) = (c_(n+1))^(n+2) + 1 > 1, a contradiction. Hence c_(n+1) < c_n. In particular c_n < c_1 for all n > 1. Hence 0 < (c_n)^(n+2) < (c_1)^(n+2). c_1 is between 0 and 1 so taking n -> infinity, we have (c_n)^(n+2) -> 0.
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