Friday, May 12, 2006

Really Rooty Real Roots. Topic: Algebra/Calculus. Level: AIME/Olympiad.

Problem: (Problem-Solving Through Problems - 6.5.5(b)) If $ a_0, a_1, \ldots, a_n $ are real numbers satisfying

$ \frac{a_0}{1}+\frac{a_1}{2}+\cdots+\frac{a_n}{n+1} = 0 $,

show that the equation $ a_0+a_1x+\cdots+a_nx^n = 0 $ as at least one real root.

Solution: Consider integrating this function over the interval $ [0,1] $. Let $ f(x) = a_0+a_1x+\cdots+a_nx^n = 0 $. We have

$ \displaystyle \int_0^1 (a_0+a_1x+\cdots+a_nx^n)dx = \left[\frac{a_0x}{1}+\frac{a_1x^2}{2}+\cdots+\frac{a_nx^{n+1}}{n+1}\right]^1_0 = \frac{a_0}{1}+\frac{a_1}{2}+\cdots+\frac{a_n}{n+1} = 0 $.

If $ f $ had no real roots on the interval $ [0,1] $ it would be strictly positive or negative. Then the integral of $ f $ over $ [0,1] $ would also be strictly positive or negative, respectively. Hence $ f $ must take on both positive and negative values. But since $ f $ is continuous, we know that there must be a $ c \in [0,1] $ such that $ f(c) = 0 $ and therefore $ f $ has a real root. QED.

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Comment: The condition involving the fractions should give away integration; from there, it's not hard to see that we want the interval $ [0,1] $ and this gives us the result immediately.

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Practice Problem: (Problem-Solving Through Problems - 6.5.5(a)) Show that $ 5x^4-4x+1 $ has a root between $ 0 $ and $ 1 $.

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