HMMT 十一月 2015 · 冲刺赛 · 第 26 题
HMMT November 2015 — Guts Round — Problem 26
题目详情
英文原题
- [ 13 ] Let f : R → R be a continuous function satisfying f ( xy ) = f ( x ) + f ( y ) + 1 for all positive realsx, y . If f (2) = 0, compute f (2015).
解析
英文解析
- [ 13 ] Let f : R → R be a continuous function satisfying f ( xy ) = f ( x ) + f ( y ) + 1 for all positive realsx, y . If f (2) = 0, compute f (2015).
Proposed by: Alexander Katz
Answer: log 2015 − 1
Let g ( x ) = f ( x ) + 1 . Substituting g into the functional equation, we get that 2
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 setxh ( 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 areal number c . Going all the way back to g , we can get that g ( x ) = c log x. Since g (2) = 1 , c = .1
log 2
log 2015
Therefore, g (2015) = c log 2015 = = log 2015 .
log 22
Finally, f (2015) = g (2015) − 1 = log 2015 − 1 . 2