1、遞uni6B78數列與不uni52D5uni9EDE徐瀝uni6CC9 uni738B繼uni5CB3 uni9673uni6F22冶uni5929uni5730萬uni7269uni5468而復uni59CB, uni5B8F觀精微uni6DF7沌交替, uni6CE2uni6D6A式uni8D77伏uni524Duni9032, uni87BAuni65CB式循環上uni5347, 人uni985Euni793E會亦uni5BD3uni6B64uni7406, 歷uni53F2uni7684經uni9A57值得uni6CE8意。uni8FD1uni671F翻uni95B120世紀 80年代
2、以後中uni570Buni5927uni9678之uni9AD8考試uni984C,uni5C0D遞uni6B78數列uni554Funi984Cuni5C24感興趣,uni9042uni767Cuni6B64文,以uni55BBuni6B64義。uni7279uni5225獻給uni9AD8中uni7684uni540Cuni5B78們。(uni4E00) 線性遞uni6B78uni95DC係uni7684不uni52D5uni9EDEuni4E00. uni4E00uni968E線性uni95DC係 Un = aUn1 + b1. uni9810備uni77E5uni8B58uni9996
3、先考慮常數uni9805 b = 0 uni7684情uni6CC1Un = aUn1 (n 2). (1)我們不固uni5B9Auni521Duni9805 U1, (1) 式中 a 不uni53D60, uni90A3uni9EBC線性uni95DC係 (1) 等價於uni4E00uni65CFuni9996uni9805uni70BA U1, 公uni6BD4uni70BA a uni7684等uni6BD4數列Un|Un = U1an1。uni7528uni6B78uni7D0Duni6CD5uni53EFuni8B49明上述論斷, 茲從略。2. 不uni52D5uni9EDEuni7
4、684uni6982念設線性遞uni6B78uni95DC係 Un = aUn1 + b, uni53C8設 limUn uni7684uni5B58uni5728性已uni7372uni8B49明, uni6C42uni6975uni9650值uni5C31是簡易事情。例: uni521Duni9805 U1 自uni7531, Un = 12Un1 + 1, 已uni77E5 limnUn uni5B58uni5728, uni6C42uni6975uni9650值。解: uni7531 limnUn uni5B58uni5728, 故 limnUn1 uni5B58uni5728; 而兩
5、者uni76F8等。設uni70BA x, uni5728給uni5B9A等式兩uni7AEFuni53D6uni6975uni9650, 於是 x = 12x + 1, limnUn = x = 2, 所uni8B02uni7684不uni52D5uni9EDEuni5C31是方程uni7684根。uni5B9A義: uni5728線性遞推uni95DC係 Un = aUn1 +b 中, a negationslash= 1, uni5982 Un 與 Un1 uni540Cuni53D6uni4E00值, uni5247該值uni53EB做上述uni95DC係uni7684不uni52D5
6、uni9EDE。uni5982不uni52D5uni9EDE記uni70BA f, uni5247 f = b1a。 回uni9867第uni4E00節, 不uni52D5uni9EDE f = 0, uni91CD新表述uni4E00下。不uni52D5uni9EDEuni70BAuni96F6uni7684線性遞uni6B78uni95DC係等價於uni4E00uni65CF等uni6BD4數列U1an18384 數uni5B78傳uni64AD 28uni53771uni671F uni6C1193年3月3. uni4E00般情uni6CC1Un = aUn1 + b (2)(2) 式兩
7、uni7AEFuni5404uni6E1Buni53BB f = b1a, 於是Un f = a(Un1 f)令 Un = Un f, uni5C0D於新uni7684數列 Un = aUn1; uni9019是不uni52D5uni9EDEuni70BAuni96F6uni7684線性遞uni6B78uni95DC係, 故Un = f + (U1 f)an1uni7576 |a| 1, limnUn = f2;uni5982 = 1, 不uni5B58uni5728;| = 1, 不uni5B58uni5728。uni554Funi984C = 1 能成立uni55CE?86 數uni5B7
8、8傳uni64AD 28uni53771uni671F uni6C1193年3月3. 不uni52D5uni9EDEuni91CDuni5408uni5982f1 = f2 = f,uni63DB言之,方程cx2+(da)xb = 0uni7684uni5224uni5225式 = (da)2+4bc = 0,uni6B64時 f = ad2c 把分式線性uni95DC係 Un = aUn1+bcUn1+d 改uni5BEBuni70BAuni6A19uni6E96形1Un f =1Un1 f + k (8)uni9019個改uni5BEB是uni5408uni7406uni7684, uni
9、56E0uni70BA Un, Un1 uni53EA要有uni4E00個uni53D6值 f, uni53E6uni4E00個也uni53D6值 f; (8) uni548C (3) 有uni76F8uni540Cuni7684不uni52D5uni9EDE, uni5728 (3) 中, 令 Un1 = dc, uni5247 Un = , 代入 (8) 得uni5230:k = 1dc + f= 1dc +ad2c= 2ca + d.故有 1Unf = 1Un1f + 2ca+d, 於是 1Unf = 1U1f + (n1) 2ca+d。uni901Auni9805uni7531uni6
10、B64而uni5B9A。4. uni5C0F結uni76F8uni7570不uni52D5uni9EDE Unf1Unf2 = (acf1acf2)n1U1f1U2f2.uni91CDuni5408不uni52D5uni9EDE 1Unf = 1Un1f + 2ca+d.5. uni9032uni4E00uni6B65uni7684例uni984Cuni554Funi984C: 給uni5B9Auni4E00uni5BE6數列Un, 其中 U0 uni70BAuni6B63整數, 且Un+1 = 2Un + 1Un + 2n 0,1,2,3,.(1) uni8B49明不論 n uni70BA任
11、何自uni7136數, uni5B58uni5728uni4E00個整數 a0, uni6EFFuni8DB3 Un = anU0+(an1)(an1)U0+an, uni6C42出整數列uni7684uni901Auni9805公式;(2) uni6C42 limnUn。略解: 不uni52D5uni9EDE f1 = 1 f2 = 1。uni901Auni9805: Un f1Un f2= (acf1acf2)nU0 1U0 + 1;Un 1Un + 1 = (2112 + 11)nU0 1U0 + 1 =13n U0 1U0 + 1.先解 (2): limnUn = 1 = f1.再解
12、(1) 從給uni5B9Auni95DC係式 Un = anU0+(an1)(an1)U0+an 得:Un 1Un + 1 =12an 1 U0 1U0 + 1.於是 2an 1 = 3n, an = 12(3n + 1), 故 an uni70BA整數。遞uni6B78數列與不uni52D5uni9EDE 876. 習uni984C(a) 已uni77E5 U1 = 5, Un = 3Un1+12Un1+2, uni6C42 Un, limUn。(b) 已uni77E5 U1 = 2, Un = 5Un1+22Un1+1, uni6C42 Un, limUn。uni5982uni679Cun
13、i6975uni9650uni5B58uni5728, 你能有什uni9EBC結論?(c) uni8A73解第 5 節uni7684uni554Funi984C。uni4E09. uni4E8Cuni968E線性遞uni6B78uni95DC係 Un = aUn1 + bUn21. uni9810備uni77E5uni8B58我們希uni671Buni7528uni4E00uni968Euni7684結uni679C來解uni6C7Auni4E8Cuni968Euni7684uni554Funi984Cuni5F15uni7406: uni5982 r 是方程 r2 ar b = 0 uni7
14、684根, uni5247uni4E8Cuni968E線性遞推uni95DC係Un = aUn1 + bUn2 (n 3), (1)等價於 Un rUn1 = (ar)(Un1 rUn2) (n 3). (2)uni8B49: Un rUn1 = (ar)Un1 + bUn2= (ar)Un1 + (r2 ar)Un2= (ar)(Un1 rUn2).uni53CD之, uni7531 (2) uni53EF以推出 (1)。2. uni76F8uni7570uni4E8C根設 、 uni70BA方程 r2 ar b = 0 uni7684uni4E8C個根 ( negationslash= )
15、。uni4E8Cuni968E線性遞uni6B78uni95DC係 (1) uni5C31是:Un Un1 = (a)(Un1 Un2),或 Un Un1 = (Un1 Un2).遞推之 Un Un1 = n2(U2 U1). (3)uni540Cuni7406 Un Un1 = n2(U2 U1). (4)解 (3) 與 (4) 兩式, uni5247有uni901Auni9805公式Un = U2n1 n1 + bU1n2 n2 (5)88 數uni5B78傳uni64AD 28uni53771uni671F uni6C1193年3月例: U1 = 0, U2 = 1, Un = 12Un
16、1 + 12Un2 (n 3), uni6C42 limUn。解: uni7279徵方程 r2 12r 12 = 0, = 1, = 12, 於是Un = 1(12)n11(12) =231 + (1)n 12n1 23.3. uni91CD根把 (5) 式改uni5BEB成Un = U2(n2 + n3 + n3) + bU1(n3 + n4 + n4).令 = , uni5247 Un = U2(n1)n2 + bU1(n2)n2。例: U1 = 1, U2 = 2, Un = 2Un1 Un2 (n 3), uni6C42 Un。解: = 1, Un = 2(n1)1n2 11(n2)1
17、n3 = n。4. uni6590uni6CE2uni90A3契數列 Fn = Fn1 + Fn2 (n 3)uni901A常uni53D6 F1 = F2 = 1, uni5247uni5B83有uni5982下性uni8CEA:(i) uni901Auni9805 Fn = F2n1 n1 F1n2 n2= n1 n1 +n2 n2 .uni6CE8意uni5230 、 uni70BA方程 r2 r 1 = 0 uni7684根, 故 1 + = 2, 1 + = 2。於是: Fn = n1(+1)n2(1) = nn 。uni7531於 = 1+52 , =152 , 故 Fn =15(
18、n n)。(ii) n uni9805uni7684uni548Csummationtextnk=1 Fk = Fn+2 1 = 15(n+2 n+2)1。(iii) 平方uni548Csummationtextnk=1 F2k = FnFn+1。5. lucas uni95DC於uni6590uni6CE2uni90A3契數列uni7684uni7814究(i) F2 + F4 + F6 + F2n = F2n+1 1。(ii) F1 F2 + F3 F4 + F2n1 F2n = F2n1 + 1。(iii) lucas and Catalan F2n1 + F2n = F2n1; F2n
19、+1 F2n1 = F2n 。(iv) FnFn+1 Fn1Fn2 = F2n1。(v) F3n + F3n+1 F3n1 = F3n。(vi) C0n + C1n1 + C2n2 + = Fn+1。遞uni6B78數列與不uni52D5uni9EDE 896. (uni5927uni9678uni5730uni5340) uni9AD8考uni984Cuni7684背景介詔1981年uni7684uni9644uni52A0uni984C, uni539Funi984Cuni5982下:已uni77E5以 AB uni70BAuni76F4徑uni7684uni534Auni5713, 有u
20、ni4E00個內接uni6B63方形 CDEF, 其邊uni9577uni70BA 1 (uni5982uni5716), 設 AC=a,BC=b, 作數列 U1 = a b, U2 = a2 ab + b2, U3 = a3 a2b + ab2 b3,.,Uk =ak ak1b + ak2b2 + (1)kbk, uni6C42uni8B49: Un = Un1 + Un2 (n 3)。uni8B49明: a + (b) = 1, a(b) = 1, uni5247 a, (b) uni70BA方程 r2 r 1 = 0 根U1 = 1, U2 = 2, U3= 3,.,Uk = ak+1
21、(b)k+1a(b)於是: Un1+Un2 = an (b)na(b) +an1 (b)n1a(b)= an1(1+a)(b)n11+(b)a(b)= an+1 (b)n+1a(b)= Un.A BCDEF11252uni5BE6uni969B上 Fibonacci 數列: F1 = F2 = 1, Fn = Fn1 +Fn2 (n 3), 具uni9AD4是 1, 1, 2, 3, 5,., n2n2 , n1n1 , nn , n+1n+1 ,. 而我們uni7684數列 Un = Fn+1。1982年uni7684uni9644uni52A0uni984C: 已uni77E5數列 a1,
22、 a2,.,an . uni548C數列 b1, b2,., bn . 其中 a1 = p,b1 = q, an = qan+1, bn = qan1 + rbn1 (n 2) (p、 q、 r 是已uni77E5常數, 且 q negationslash= 0,p r 0)。(1) uni7528 p、 q、 r、 n 表uni793A bn, 並uni7528數uni5B78uni6B78uni7D0Duni6CD5uni8B49明;(2) uni6C42 limn bna2n+b2n。略uni8B49: an = a1qn1 = pqn1,bn rbn1 = qan1 = an = pq
23、n1,bn1 rbn2 = pqn2, ()bn rbn1bn1 rbn2 = q, bn rbn1 = q(bn1 rbn2),遞推之, 終有 bn rbn1 = qn2(b2 rb1). (1)uni53C8 bn qbn1 = r(bn1 qbn2), uni5247bn qbn1 = rn2(b2 qb1). (2)uni7531 (1) uni548C (2) 解得 bn = pq(qn1 rn1)q r qrn1.90 數uni5B78傳uni64AD 28uni53771uni671F uni6C1193年3月從 (1) 式uni770B出, uni672Cuni984Cuni5
24、BE6uni8CEA是uni4E8Cuni6B21遞推uni554Funi984C:bn = (q+r)bn1 qrbn1 (uni9019uni88E1 a = q+r, b = qr, b1 = p, b2 = q(p+r),q、 r 是uni7279徵方程 x2 (q + r)x + qr = 0 uni7684根, 故bn = b2qn1 rn1q r + b1(qr)qn2 rn2q r .把 b1 = q, b2 = q(p + r) 代入上式, uni5373得 bn = qp(qn1 rn1)q r qrn1,與uni524Duni540C, 下略。(uni4E8C) 不uni
25、52D5uni9EDE與uni975E線性遞推數列uni9023續幾年uni7684uni5927uni9678uni9AD8考uni90FD考了遞推數列, 1984年uni9996uni6B21考了uni975E線性遞推數列,uni7531線性遞推uni5230uni975E線性遞推, uni7121論從uni96E3度uni548C方uni6CD5上uni90FD是uni4E00uni6B21uni98DB躍,uni73FE以 1984、 1986 兩年uni7406科uni9AD8考第uni516Buni984Cuni70BA例,說明uni975E線性遞推數列uni7684背景、構uni
26、9020方uni6CD5uni548C解uni9019uni985Euni554Funi984Cuni7684uni51FD數方uni6CD5。uni4E00. 牛uni9813公式與uni54C8萊迭代1984年uni7406科uni9AD8考第uni516Buni984C, 下簡稱 84 uni516B設 2, 給uni5B9A數列 xn, 其中 x1 = , xn+1 = x2n2(xn1), (n = 1,2,.)。uni6C42uni8B49:(1) xn 2, 且 xn+1xn 3, uni90A3uni9EBCuni7576 n lg 3lg 43時必有 xn+1 0) uni7
27、684根, uni5229uni7528 (I) 式uni53EF得 xn+1 = 12(xn + axn), uni5247 limnxn = a。再回uni5230試uni984Cxn+1 = x2n2(xn 1) = xn x2n 2xn2(xn 1).不妨設f(x) = x22x, uni5247 f(x) = 2(x1), 所以uni9019個數列uni53EF以uni770B成是uni7528牛uni9813切線uni6CD5uni6C42方程 x2 2x = 0 uni7684uni4E00個根。但uni5C0Duni4E00個uni9AD8中uni751F來說uni986Fun
28、i7136uni9084不能uni7406解uni6B64式。不uni904Euni53EF採uni53D6數形結uni5408uni7684方uni6CD5,先作出uni51FD數f(x) = x(x2)uni7684uni5716uni8C61,uni5728f(x) = x(x2)上uni53D6uni4E00uni9EDEPn(xn,x2n2xn),其中 xn 2, uni904E Pn 作 f(x) uni7684切線, 其方程uni70BA y (x2n 2xn) = 2(xn 1)(xxn), 令切線與 x 軸uni7684交uni9EDE坐uni6A19uni70BA (xn+
29、1,0), uni5247得 xn+1 xn = x2n2xn2(xn1), uni5373 xn+1 = x2n2(xn1)。再採uni53D6uni8B8Auni63DB n = xn 1, 把uni539F數列uni5316成 n+1 = 12(n + 1n) uni7684模式來解。 uni9019uni5C31是有些uni96DCuni8A8C上採uni53D6n+1 = 12(n + 1n) 解uni984Cuni7684數uni5B78背景。再uni770B 1986年uni7406科uni9AD8考第uni516Buni984C (下簡稱 86 uni516B)。已uni77E
30、5 x1 0, x1 negationslash= 1, 且 xn+1 = xn(x2n+3)3x2n+1, 試uni8B49: 數列 xn 或者uni5C0D任意自uni7136數 n uni90FDuni6EFFuni8DB3 xn xn+1。uni5BE6uni969B上uni53EF以uni770B成是uni6C42uni975E線性方程 f(x) = 0 uni7684根uni7684uni54C8萊 (uni56E0計算uni54C8萊慧星軌uni9053而聞uni540D於世) 迭代uni6CD5 (uni95DC於uni54C8萊迭代uni53EFuni53C3uni770B
31、1986年第uni516Buni671F數uni5B78uni901Auni5831 p.3942)。其公式uni70BA:xn+1 = xn f(xn)f(xn)f(xn)f(xn)2f(xn) (II)uni53D6 f(x) = x2 1, uni5247 f(x) = 2x, f(x) = 2, 代入 (II) 式, uni5247有xn+1 = xn x2n 12xn 2(x2n1)2(2xn) = xn 2xn(x2n 1)3x2n + 1 . xn+1 = xn(x2n + 3)3x2n + 1 .92 數uni5B78傳uni64AD 28uni53771uni671F uni
32、6C1193年3月uni9019uni5C31是試uni984Cuni7684形式、牛uni9813迭代uni6CD5收uni6582是uni4E8Cuni968Euni7684, 而uni54C8萊迭代收uni6582是uni4E09uni968Euni7684, 所以uni5728uni4E00uni5B9Auni689D件下uni8D77著uni52A0uni901F收uni6582uni7684作uni7528。以上分析說明, uni9019兩uni9053試uni984C有著uni6DF1uni523Buni7684數uni5B78背景, uni540C時也uni770Buni523
33、0, uni53EA要uni9078uni53D6適uni5408uni4E00uni5B9Auni689D件uni7684f(x), uni5C31能得uni5230uni975E線性遞推數列, uni7279uni5225uni7576 f(x) 是某uni4E00uni591Auni9805式uni7684時候, 便得uni5230uni9AD8考試uni984Cuni985E型。下uni9762uni8209幾個uni5229uni7528上述方uni6CD5構uni9020uni975E線性遞推數列uni7684例uni5B50。1. uni53D6 f(x) = x(x4), 代入
34、 (I) 按 84 uni516B 編uni984C設 4, 給uni5B9A數列 xn, 其中 x1 = , xn+1 = x2n2(xn2), (n = 1,2,.), uni6C42uni8B49(1) xn 4, 且 xn+1xn 5, uni90A3uni9EBCuni7576 n lg 5lg 65時必有 xn+1 2, 給uni5B9A數列xn, 其中 x1 = , xn+1 = x3n3x2n6xn+4, (n = 1,2,.), uni6C42uni8B49(1) xn 2, 且 xn+1xn 3, uni90A3uni9EBCuni7576 n lg 3lg 43時必有 x
35、n+1 x0, uni5247uni7531數uni5B78uni6B78uni7D0Duni6CD5uni53EFuni8B49xn+1 = F(xn) F(xn1) = xn, uni5373 xn 遞uni589E; 若 F(x0) = x1 3 時 xn 很快趨uni5411於 2。.0 22 xyy=xx = 1y= 12(x+1). . . .至於試uni984C 84 uni516B, 86 uni516B, 能uni901Auni904Euni6C42uni901Auni9805uni7684方uni6CD5來解主要根據是下列uni547Duni984C。uni547Duni9
36、84Cuni4E00: 設 f(x) = x2 + px + q uni7684兩uni5BE6根uni70BA 、 , f(x) = 2x + p, 代入 (I) 式得xn+1 = xn x2n + pxn + q2xn + p =x2n q2xn + p,uni5247有xn xn = (xn1 xn1 )2 = (xn2 xn2 )22 = = (x1 x1 )2n1.uni547Duni984Cuni4E8C: 設 f(x) = x2 +px+q uni7684兩個uni5BE6數根uni70BA、 , f(x) = 2x+p, f(x) = 2, 代入 (II) 式得xn+1 = x
37、n x2n + pxn + q(2xn + p) 2(x2n+xn+p)2(2xn+p) =x3n 3qxn pq3x2n + 3pxn + (p2 q)uni5247有 xn xn = (xn1 xn1 )3 = (xn1 xn1 )32 = = (x1 x1 )3n1.uni9650於uni7BC7uni5E45, uni8B49明略。94 數uni5B78傳uni64AD 28uni53771uni671F uni6C1193年3月uni70BAuni6C42 86 uni516B uni7684uni901Auni9805, uni986Funi7136 = 1, = 1 是方程 x
38、= (x) uni5373 x = x(x2+3)3x2+1 uni7684兩個根。uni5728遞推數列 xn+1 = xn(x2n+3)3x2n+1uni7684兩uni7AEF分uni5225uni6E1Buni53BB uni548C , 兩式uni76F8uni6BD4, 得xn+1 1xn+1 + 1 = (xn 1xn + 1)3, 遞推之終有 xn+1 1xn+1 + 1 = (x1 1x1 + 1)3n, 解出xn+1 = (x1 + 1)3n + (x1 1)3n(x1 + 1)3n (x1 1)3n, 或 xn =(x1 + 1)3n1 + (x1 1)3n1(x1 +
39、1)3n1 (x1 1)3n1借uni52A9於uni901Auni9805, uni5C07給我們解uni6C7A感興趣uni7684uni554Funi984C提供uni6975uni5927uni7684方便,uni7531於uni984C設中xn 是uni6B63uni9805級數,uni5373uni5C0D任意自uni7136數 n, 恆有 xn 0, 於是uni5229uni7528uni6BD4值xn+1xn =(x1 + 1)3n12 (x1 + 1)3n1(x1 1)3n1 + (x1 1)3n12(x1 + 1)3n12 + (x1 + 1)3n1(x1 1)3n1 +
40、(x1 1)3n12.不uni96E3uni770B出, uni5C0D任意自uni7136數 n:(i) uni7576 0 1 時, 恆有 xn xn+1。uni9019uni88E1, 我們uni5C0D試uni984C中uni7684uni554Funi984C感興趣, uni9032uni4E00uni6B65uni7684結uni679C不再贅述。uni70BA了熟悉上uni9762uni7684方uni6CD5, 下列習uni984C供uni53C3考。1. 設 32, 給uni5B9A數列xn, 其中 x1 = , xn+1 = 2xn + 3 (n = 1,2,.), uni
41、6C42uni8B49(1) uni7576 1 (n 2), 且 xn 3 92(23)n1;(2) uni7576 3 4, uni90A3uni9EBCuni7576 n lg 4lg 113時必有 xn+1 4。2. 設 2 2, 給uni5B9A數列 xn, 其中 x0 = , xn+1 = 2xn (n = 1,2,.), uni6C42uni8B49:(1) uni7576 2 x0 1, uni7528數uni5B78uni6B78uni7D0Duni6CD5uni8B49明, 數列 x2n 遞uni589E, 且有 1 12n x2n 1; 數列 x2n+1 遞uni6E1B, 且有 1 x2n+1 1 + 12n;(2) uni7576 1 x0 2 時, uni8B49明 xn 遞uni6E1B; x2n+1 遞uni589E, 並uni8B49明 xn uni7684uni6975uni9650是 1。uni672C文作者任職於中uni570Buni7121uni932B市教育uni7814究中心
