用不動(dòng)點(diǎn)法求數(shù)列通項(xiàng)

1、用不動(dòng)點(diǎn)法求數(shù)列的通項(xiàng)用不動(dòng)點(diǎn)法求數(shù)列的通項(xiàng)定義：定義：方程xxf?)(的根稱為函數(shù))(xf的不動(dòng)點(diǎn).利用遞推數(shù)列)(xf的不動(dòng)點(diǎn)，可將某些遞推關(guān)系)(1??nnafa所確定的數(shù)列化為等比數(shù)列或較易求通項(xiàng)的數(shù)列，這種方法稱為不動(dòng)點(diǎn)法.定理定理1：若)10()(????aabaxxfp是)(xf的不動(dòng)點(diǎn)，na滿足遞推關(guān)系)1()(1???nafann，則)(1paapann????，即pan?是公比為a的等比數(shù)列.證明：證明：因?yàn)閜是)(

2、xf的不動(dòng)點(diǎn)pbap???appb????由baaann????1得)(11paapbaapannn?????????所以pan?是公比為a的等比數(shù)列.定理定理2：設(shè))00()(??????bcadcdcxbaxxf，na滿足遞推關(guān)系1)(1???nafann，初值條件)(11afa?（1）：若)(xf有兩個(gè)相異的不動(dòng)點(diǎn)qp，則qapakqapannnn????????11（這里qcapcak???）（2）：若)(xf只有唯一不動(dòng)點(diǎn)p，

3、則kpapann?????111（這里dack??2）證明：證明：由xxf?)(得xdcxbaxxf????)(，所以0)(2????bxadcx（1）因?yàn)閝p是不動(dòng)點(diǎn)，所以?????????????0)(0)(22bqadcqbpadcp??????????????qcabqdqpcabpdp，所以qapaqcapcaqcabqdapcabpdaqcapcaqdbaqcapdbapcaqdcabaapdcabaaqapannnnnnn

4、nnnnn??????????????????????????????????????????1111111111)()(令qcapcak???，則qapakqapannnn???????11??????????????221122xaexbxaexb??????????0)2(0)2(21xeabxeab?1121xx0??方程組有唯一解aeb20??例3：已知數(shù)列na中211222Nnaaaannn?????求數(shù)列na的通項(xiàng).其實(shí)不

5、動(dòng)點(diǎn)法除了解決上面所考慮的求數(shù)列通項(xiàng)的幾種情形還可以解決如下問題:例4：已知1011??aa且)1(4162241?????nnnnnaaaaa求數(shù)列na的通項(xiàng).解:作函數(shù)為)1(416)(224????xxxxxf解方程xxf?)(得)(xf的不動(dòng)點(diǎn)為ixixxx3333114321??????.取11???qp作如下代換:423423422422411)11(146414641)1(4161)1(41611?????????????

6、????????????nnnnnnnnnnnnnnnnnnnnaaaaaaaaaaaaaaaaaaaa逐次迭代后得:111141414141)1()1()1()1(???????????nnnnaaaaan已知曲線從點(diǎn)向曲線引斜率為22:20(12)nCxnxyn?????(10)P?nC的切線，切點(diǎn)為(0)nnkk?nl()nnnPxy（１）求數(shù)列的通項(xiàng)公式；nnxy與（２）證明：1352112sin1nnnnnxxxxxxxy??