高中數(shù)學(xué)：第二章：函數(shù)-函數(shù)迭代(競賽精講)

1、word2.3 函數(shù)迭代知識提要年45 5 x x ii4x (x 1)，i 5i1i44 xf(x) (x1) (x4)4且x5504x f(x) (x4)4514x f(f(x)( ) (x4)42254x f(f(f(x)( ) (x4)43354x f(f(f(f(x)( ) (x4)44454x f(f(f(f(f(x)( ) (x4)45554xx5 4 5 | x55D f :D是xD ，記 f (x) x ， f (x) f (x) ， f (x) f (f (x) ，一個函數(shù)，對(0)(1)(2)f(x) f (f (x)nN f (x)為 f(x)的nn為 f (x)的(n

2、1)(n)(n)(n) f(x)(n)n-1- / 4word xcf (x) xnc(n)f(x)f (x) a x f(x)；(n)nxxf (x)(n)f (x) x f (x) xan； f(x)；a(n)1ax1nax1anb a1（則 f (x) a x f(x)b；(n)n1af (x)(n)例題講解例 f(x) f (x)(10)1024x1023 f(x) f f (k)f (11)n例 2 k f (k) f (k)1n1n1（nN）N5（C）（D） f 12f (x) f f (x) nN a a 99例 f (x)1，nf 2x1n 11nnn b f(x)bbf(x)

3、a(x)1a 1abbf (x)a (x)1a 1a2-2- / 4wordbbf (x)a (x)1a 1a3bbf (x)a (x)1a 1a(n)nbb xf(x) xax f(x)的不1ax f(x)x f (x)(n)00 f(x)的 n 19x 93例 f(x)2 f (x)(n) (x) (x)得 f (x)(g( (x) f(x)11(x)和g(x) f(x)和g(x) (x)(g( (x)(g ( (x)(n)f (x)(n) f(x)和g(x) f (x)11 2x 1f (x)(n)例 f (x)2例 f (x)x2 x1 f (x)(n)課后練習(xí)x1x1 f(x)A， ff(xB） RABBA A B5A B-3- / 4wordn3(n1000)(n1000)(n)f N f）f(n7)f997 998999（D）10002(x)ff(x4x3 f(x) f(xa)|x2|x2|ff(a3a a,b f6 f(x)fb，有 ，則5設(shè) y是定義在 上的函數(shù)，且對于任意實(shí)數(shù)R| f(1999)|(n)k k3函數(shù) f， nf 3F(n) f(f(f( f(n) )令，求證：10fFf f(5) f Ffff1 k ,n（nna0.50.5 ，a 0.5ak11k2n2999ka k（kk 屆(x) x af (x) f (x)