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IMC / 1994 / Problems / Day 2, P9

IMC 1994 · Day 2 · P9

real analysisworth 14 pts

Let ff be a real-valued function with n+1n+1 derivatives at each point of R\mathbb{R}. Show that for each pair of real numbers aa, bb, a<ba < b, such that ln⁡(f(b)+f′(b)+⋯+f(n)(b)f(a)+f′(a)+⋯+f(n)(a))=b−a\ln \left( \frac{f(b) + f'(b) + \cdots + f^{(n)}(b)} {f(a) + f'(a) + \cdots + f^{(n)}(a)} \right) = b - a there is a number cc in the open interval (a,b)(a,b) for which f(n+1)(c)=f(c).f^{(n+1)}(c) = f(c). Note that ln⁡\ln denotes the natural logarithm.

Solution (official)

Set g(x)=(f(x)+f′(x)+⋯+f(n)(x))e−xg(x) = \left( f(x) + f'(x) + \cdots + f^{(n)}(x) \right) e^{-x}. From the assumption one get g(a)=g(b)g(a) = g(b). Then there exists c∈(a,b)c \in (a,b) such that g′(c)=0g'(c) = 0. Replacing in the last equality g′(x)=(f(n+1)(x)−f(x))e−xg'(x) = \left( f^{(n+1)}(x) - f(x) \right) e^{-x} we finish the proof.

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