Re: [中學] 函數方程

[JohnMash (Paul)]

※ 引述《jurian0101 (Hysterisis)》之銘言：
: 從筆記堆中挖出來的老題目，頗奇妙的
: f(x) + f((x-3)/(x-2)) = x-1
: 試解出f(x)之顯式

   denote u = (x-3)/(x-2)

    (u-3)/(u-2) = [(x-3-3x+6)/(x-2)]/[(x-3-2x+4)/(x-2)]

                = (2x-3)/(x-1)

    denote v = (2x-3)/(x-1)

    (v-3)/(v-2) = [(2x-3-3x+3)/(x-1)]/[(2x-3-2x+2)/(x-1)]

                = x


   f(x) + f(u) = x-1

   f(u) + f(v) = u-1

   f(v) + f(x) = v-1

   2f(x) = x+v-u-1

    f(x) = [x + (2x-3)/(x-1) - (x-3)/(x-2) -1] / 2

         = [x + 2 - 1/(x-1) - 1 + 1/(x-2) -1] / 2

         = [x - 1/(x-1) + 1/(x-2)] / 2


: - - - -
: 有關聯的附題:
: 試解出所有 f(f(f(x)))=x 的函數f

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◆ From: 123.194.224.241
※ 編輯: JohnMash        來自: 123.194.224.241      (08/20 00:35)
推 ouchbgb     :outstanding!
→ jurian0101  :這題關鍵正好是 f(x)=(x-3)/(x-2) 滿足f(f(f(x)))=x
→ ouchbgb     :那麼要怎麼說明所有滿足f(f(f(x)))=x的只有它呢?
→ jurian0101  :嗯，不只有他。像f(x)= (-2x-3)/(x+1)也可以。
→ pcyu16      :f(x)=x  orz..|||
→ jurian0101  :f(x)=zx , z^3=1 怎麼樣
→ ouchbgb     :那麼...您寫的附題, 試解出所有 f(f(f(x)))=x ...