用生成函数求解下列递归方程 f(n)=2f(n-1)+1 n>1 f(1)=2 n=1

递归方程:

\[\begin{cases} f(n)=2f(n-1)+1 &(n>1)&\\ f(1)=2 &(n=1)& \end{cases} \]

构造生成函数求解

\[\begin{array}{lcl} G(x)=2x^1+5x^2+11x^3+23x^4+\cdots\\\\ 2x\cdot G(x)=\; +4x^2+10x^3+22x^4+\cdots\\\\ (1-2x)G(x)=2x+x^2+x^3+x^4+\cdots\\\\ \qquad\qquad\qquad =x+(x+x^2+x^3+x^4+x^5+\cdots)\\\\ \qquad\qquad\qquad=x+\frac{1}{1-x}-1=x+\frac{x}{1-x}\\\\ G(x)=\frac{x}{1-2x}+\frac{x}{(1-2x)(1-x)}\\\\ \qquad\;\, =\frac{x}{1-2x}+\frac{1}{1-2x}-\frac{1}{1-x}\\\\ \qquad\;\,=(2^0x+2^1x^2+2^2x^3+\cdots)+(2^1x^1+2^2x^2+2^3x^3+\cdots)\\\\ \qquad\qquad-(1+x+x^2+x^3+x^4+\cdots)\\\\ G(x)=(2^1+2^0-1)x+(2^2+2^1-1)x^2\cdots+(2^n+2^{n-1}-1)x^n+\cdots\\\\ \Rightarrow f(n)=2^n+2^{n-1}-1 \end{array} \]

posted @ 2020-10-07 12:32  lvjiuluan  阅读(1592)  评论(0编辑  收藏  举报