f(f(x)f(y))+f(x+y)=f(xy) (1) let y=0 f(f(0)f(x))+f(x)=f(0) (2) From (2), let x=0 f(f(0)^2)=0 (3) f(0) can be either zero or non-zero Case A: f(0)=0, from (2), f(0*f(x))+f(x)=0 => f(x)=0 Case B: f(0) ≠ 0, then f(0)^2 must be 1. Claim: if f(a)=0, then a=1, Assume: there is a ≠ 1 such f(a)=0 from (1), if x+y=xy, then f(f(x)f(y))=0 So f(f(x)f(x/(x-1))=0 for all x ≠ 1 let x=a, f(f(a)f(a/(a-1))=f(0)=0, contradition to f(0) ≠ 0 So f(0)^2 must be 1, and f(0) is either 1 or -1. Case B1: f(0)=1 (4) from (2), f(f(x))=1-f(x) (5) From (1), let y=1, 1+f(x+1)=f(x) f(x+1)=f(x)-1 f(x+N)=f(x)-N for any integer N (6) continue to: f(x)=f(1-f(x)) (7) from (7), guess f(x)=1-x (8) Case B2: f(0)=-1 f(x)=x-1
从(7)到(8), 还需要证明: For any x and y, if f(x)=f(y), then x=y. 谁有简单直接的的证明? 以前的某个版主可能会。
【在 l*******s 的大作中提到】 : f(f(x)f(y))+f(x+y)=f(xy) (1) : let y=0 : f(f(0)f(x))+f(x)=f(0) (2) : From (2), let x=0 : f(f(0)^2)=0 (3) : f(0) can be either zero or non-zero : Case A: f(0)=0, from (2), f(0*f(x))+f(x)=0 => f(x)=0 : Case B: f(0) ≠ 0, then f(0)^2 must be 1. : Claim: if f(a)=0, then a=1, : Assume: there is a ≠ 1 such f(a)=0
【在 l*******s 的大作中提到】 : f(f(x)f(y))+f(x+y)=f(xy) (1) : let y=0 : f(f(0)f(x))+f(x)=f(0) (2) : From (2), let x=0 : f(f(0)^2)=0 (3) : f(0) can be either zero or non-zero : Case A: f(0)=0, from (2), f(0*f(x))+f(x)=0 => f(x)=0 : Case B: f(0) ≠ 0, then f(0)^2 must be 1. : Claim: if f(a)=0, then a=1, : Assume: there is a ≠ 1 such f(a)=0
第一步对的,第二步换元u=f(0)*f(x)就行了 f(u) = a - u/a,( a = f(0) ) 后面都多余的 [在 llaalways (熊大) 的大作中提到:] :f(f(x)f(y))+f(x+y)=f(xy) (1) :let y=0 :f(f(0)f(x))+f(x)=f(0) (2) :From (2), let x=0 :f(f(0)^2)=0 (3) :f(0) can be either zero or non-zero :Case A: f(0)=0, from (2), f(0*f(x))+f(x)=0 => f(x)=0 :Case B: f(0) NotEqual 0, then f(0)^2 must be 1. :Claim: if f(a)=0, then a=1, :Assume: there is a NotEqual 1 such f(a)=0 :..........
l*s
36 楼
你在跟我一开始做的一样。 这只能说明如果 u=f(0)*f(x),then f(u) = a - u/a,( a = f(0) ) 怎么能证明对 任何一个u, 都存在x, 使得 u=f(0)*f(x). 不证明 if f(x)=f(y), then x=y, 是得不到完整的解的。
【在 P****i 的大作中提到】 : 第一步对的,第二步换元u=f(0)*f(x)就行了 : f(u) = a - u/a,( a = f(0) ) : 后面都多余的 : [在 llaalways (熊大) 的大作中提到:] : :f(f(x)f(y))+f(x+y)=f(xy) (1) : :let y=0 : :f(f(0)f(x))+f(x)=f(0) (2) : :From (2), let x=0 : :f(f(0)^2)=0 (3) : :f(0) can be either zero or non-zero
想了一个解抛砖引玉,还有点问题, if f is differentiable f(f(0)f(y)) + f(y) = f(0) for all y then f(f(0)f(x)) - f(f(0)f(y)) = -(f(x)-f(y)) for all x, y d f(f(0).f(y)) / dy * f(0) * f'(y) = - f'(y) so f(0) = 0 or f'(y) = 0 for all y, or d f(f(0).f(y)) / dy *f(0) = -1 for all y from f(0) = 0 or f'(y) = 0 => f(x) = 0 for all x from d f(f(0).f(y)) / dy *f(0) = -1, let C = -1/f(0) - f(y)/C = - Cy + D f(y) = C^2 y - D, C = -1/f(0) take this back to f(f(x)f(y)) + f(x+y) = f(xy), which holds for all x, y, gets C = +/- 1, D = 1 f(y) = y - 1
【在 z*********e 的大作中提到】 : 想了一个解抛砖引玉,还有点问题, : if f is differentiable : f(f(0)f(y)) + f(y) = f(0) for all y : then : f(f(0)f(x)) - f(f(0)f(y)) = -(f(x)-f(y)) for all x, y : d f(f(0).f(y)) / dy * f(0) * f'(y) = - f'(y) : so f(0) = 0 or f'(y) = 0 for all y, or d f(f(0).f(y)) / dy *f(0) = -1 for : all y : from f(0) = 0 or f'(y) = 0 => f(x) = 0 for all x : from d f(f(0).f(y)) / dy *f(0) = -1, let C = -1/f(0)