递推公式求通项 |
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不动点求通项
1、一阶线性递推式: a n + 1 = c a n + d , c ≠ 0 , c ≠ 1 a_{n+1}=ca_n+d,c\neq 0,c\neq 1 an+1=can+d,c=0,c=1,已知 a 1 a_1 a1的值 特征方程: f ( x ) = a x + b f(x)=ax+b f(x)=ax+b,令 f ( x ) = x f(x)=x f(x)=x,解得特征根为 x 0 x_0 x0,则可得: a n − x 0 = c ( a n − 1 − x 0 ) = c n − 1 ( a 1 − x 0 ) a_n-x_0=c(a_{n-1}-x_0)=c^{n-1}(a_1-x_0) an−x0=c(an−1−x0)=cn−1(a1−x0) a n = c n − 1 ( a 1 − x 0 ) + x 0 a_n=c^{n-1}(a_1-x_0)+x_0 an=cn−1(a1−x0)+x0 当 x 0 = a 1 x_0= a_1 x0=a1时, a n = x 0 a_n=x_0 an=x0 当 x 0 ≠ a 1 x_0\neq a_1 x0=a1时, a n = c n − 1 ( a 1 − x 0 ) + x 0 a_n=c^{n-1}(a_1-x_0)+x_0 an=cn−1(a1−x0)+x0 2、二阶线性递推: a n + 2 = p a n + 1 + q a n a_{n+2}=pa_{n+1}+qa_{n} an+2=pan+1+qan,已知 a 1 , a 2 a_1,a_2 a1,a2,可以联立求得A、B 特征方程: x 2 = p x + q x^2=px+q x2=px+q,设有特征根 x 1 , x 2 x_1,x_2 x1,x2, 当 x 1 ≠ x 2 x_1\neq x_2 x1=x2时, a n = A x 1 n − 1 + B x 2 n − 1 a_n=Ax_1^{n-1}+Bx_2^{n-1} an=Ax1n−1+Bx2n−1 当 x 1 = x 2 x_1=x_2 x1=x2时, a n = ( A + B n ) x 1 n − 1 a_n=(A+Bn)x_1^{n-1} an=(A+Bn)x1n−1 3、分式递推式: a n + 1 = a a n + b c a n + d a_{n+1}=\frac {aa_{n}+b}{ca_n+d} an+1=can+daan+b, r ≠ 0 , a d ≠ b c , a 1 ≠ − d c r\neq 0,ad\neq bc,a_1\neq -\frac dc r=0,ad=bc,a1=−cd,已知 a 1 a_1 a1的值 特征方程: x = a x + b c x + d x=\frac {ax+b}{cx+d} x=cx+dax+b,设 x 1 , x 2 x_1,x_2 x1,x2是两个特征根, 当 x 1 ≠ x 2 x_1\neq x_2 x1=x2时, a n − x 1 a n − x 2 = a − x 1 c a − x 2 c × a n − 1 − x 1 a n − 1 − x 2 \frac {a_n-x_1}{a_n-x_2}=\frac {a-x_1c}{a-x_2c}\times \frac{a_{n-1}-x_1}{a_{n-1}-x_2} an−x2an−x1=a−x2ca−x1c×an−1−x2an−1−x1 当 x 1 = x 2 x_1= x_2 x1=x2时, 1 a n − x 1 = 1 a n − 1 − x 1 + 2 c a + d \frac 1{a_n-x_1}=\frac 1{a_{n-1}-x_1}+\frac {2c}{a+d} an−x11=an−1−x11+a+d2c 型如: a n + 1 = a n 2 + b a n + d a_{n+1}=\frac {a_n^2+b}{a_n+d} an+1=an+dan2+b 例:已知数列 { a n a_n an}, a n + 1 = a n 2 + 2 2 a n , a 1 = 2 a_{n+1}=\frac {a_n^2+2}{2a_n},a_1=2 an+1=2anan2+2,a1=2,求通项 解: f ( x ) = x 2 + 2 2 x = x , x 1 = 2 , x 2 = − 2 f(x)=\frac {x^2+2}{2x}=x,x_1=\sqrt 2,x_2=-\sqrt 2 f(x)=2xx2+2=x,x1=2 ,x2=−2 a n + 1 − 2 a n + 1 + 2 = ( a n − 2 a n + 2 ) 2 = ( a 1 − 2 a 1 + 2 ) 2 n − 1 \frac {a_{n+1}-2}{a_{n+1}+2}=(\frac {a_{n}-2}{a_{n}+2})^2=(\frac {a_1-2}{a_1+2})^{2^{n-1}} an+1+2an+1−2=(an+2an−2)2=(a1+2a1−2)2n−1 解得: a n = 2 × ( 2 + 2 ) 2 n − 1 + ( 2 − 2 ) 2 n − 1 ( 2 + 2 ) 2 n − 1 − ( 2 − 2 ) 2 n − 1 a_n=\sqrt 2\times \frac{ (2+\sqrt 2)^{2^{n-1} } +(2-\sqrt 2)^{2^{n-1}} }{ (2+\sqrt 2)^{2^{n-1} } -(2-\sqrt 2)^{2^{n-1}} } an=2 ×(2+2 )2n−1−(2−2 )2n−1(2+2 )2n−1+(2−2 )2n−1 |
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