HMMT 十一月 2015 · 冲刺赛 · 第 26 题

HMMT November 2015 — Guts Round — Problem 26

[ 13 ] Let f : R → R be a continuous function satisfying f ( xy ) = f ( x ) + f ( y ) + 1 for all positive reals x, y . If f (2) = 0, compute f (2015).

解析

[ 13 ] Let f : R → R be a continuous function satisfying f ( xy ) = f ( x ) + f ( y ) + 1 for all positive reals x, y . If f (2) = 0, compute f (2015). Proposed by: Alexander Katz Answer: log 2015 − 1 2 Let g ( x ) = f ( x ) + 1 . Substituting g into the functional equation, we get that g ( xy ) − 1 = g ( x ) − 1 + g ( y ) − 1 + 1 g ( xy ) = g ( x ) + g ( y ) . ′ ′ x y + Also, g (2) = 1 . Now substitute x = e , y = e , which is possible because x, y ∈ R . Then set x h ( x ) = g ( e ) . This gives us that ′ ′ ′ ′ x + y x y ′ ′ ′ ′ g ( e ) = g ( e ) + g ( e ) = ⇒ h ( x + y ) = h ( x ) + h ( y ) ′ ′ for al x , y ∈ R . Also h is continuous. Therefore, by Cauchy’s functional equation, h ( x ) = cx for a 1 real number c . Going all the way back to g , we can get that g ( x ) = c log x. Since g (2) = 1 , c = . log 2 log 2015 Therefore, g (2015) = c log 2015 = = log 2015 . 2 log 2 Finally, f (2015) = g (2015) − 1 = log 2015 − 1 . 2