1. 设 $f(x) = sqrt{x^2 + c}$. 求 $f(x)$ 之 $n$ 次迭代.
解答:
直接求迭代函数: $$f^{(2)}(x) = sqrt{left(sqrt{x^2 + c} ight)^2 + c} = sqrt{x^2 + 2c},$$ $$f^{(3)}(x) = sqrt{left(sqrt{x^2 + 2c} ight)^2 + c} = sqrt{x^2 + 3c},$$ $$cdots cdots,$$ $$Rightarrow f^{(n)}(x) = sqrt{x^2 + nc}.$$
2. 设 $f(x) = x^2 - 4x + 6$. 求 $f(x)$ 之 $n$ 次迭代.
解答:
直接求迭代函数: $$f(x) = (x - 2)^2 + 2,$$ $$f^{(2)}(x) = (x - 2)^4 + 2,$$ $$f^{(3)}(x) = (x - 2)^8 + 2,$$ $$cdots cdots,$$ $$Rightarrow f^{(n)}(x) = (x - 2)^{2^n} + 2.$$
3. 求函数 $f(x)$, 使 $f^{(3)}(x) = 8x + 7$.
解答:
待定系数法求解:
令 $f(x) = ax+b$, 则有 $$f^{(3)}(x) = a^3x + (a^2 + a + 1)b = 8x + 7,$$ $$Rightarrow egin{cases}a = 2\ b = 1 end{cases} Rightarrow f(x) = 2x + 1.$$
4. 设 $f(x) = displaystyle{x over sqrt{1 + x^2}}$. 求 $f(x)$ 之 $n$ 次迭代.
解答:
直接求迭代函数: $$f^{(2)}(x) = {{x over sqrt{1 + x^2}} over sqrt{1 + left({x over sqrt{1 + x^2}} ight)^2}} = {x over sqrt{1 + 2x^2}},$$ $$f^{(3)}(x) = {{x over sqrt{1 + 2x^2}} over sqrt{1 + left({x over sqrt{1 + 2x^2}} ight)^2}} = {x over sqrt{1 + 3x^2}},$$ $$cdots cdots,$$ $$Rightarrow f^{(n)}(x) = {x over sqrt{1 + nx^2}}.$$
5. 设 $f(x) = displaystyle{x over 1 + ax}$. 求 $f(x)$ 之 $n$ 次迭代.
解答:
直接求迭代函数: $$f^{(2)}(x) = {{x over 1 + ax}over 1 + {ax over 1 + ax}} = {x over 1 + 2ax},$$ $$f^{(3)}(x) = {{x over 1 + 2ax}over 1 + {ax over 1 + 2ax}} = {x over 1 + 3ax},$$ $$cdots cdots,$$ $$f^{(n)}(x) = {x over 1 + nax}.$$
6. 设 $f(x) = 19x + 89$, 则 $f^{(100)}(5)$ 之末位数字是多少?
解答:
由题意得 $$f^{(100)}(x) = 19^{100}x + left(19^{99} + 19^{98} + cdots + 19 + 1 ight)cdot 89$$ $$Rightarrow f^{(100)}(5) = 19^{100}cdot 5 + left(19^{99} + 19^{98} + cdots + 19 + 1 ight)cdot 89$$ $$equiv 5 + left[(-1)^{99} + (-1)^{98} + cdots + (-1) + 1 ight]cdot (-1)$$ $$equiv 5 + 0cdot (-1) equiv 5 ( ext{mod} 10)$$ 即末位数字为 $5$.
7. 设 $f(x) = x + 2sqrt{x} + 1$. 求 $f(x)$ 之 $n$ 次迭代.
解答:
直接求迭代函数: $$f(x) = left(sqrt{x} + 1 ight)^2,$$ $$f^{(2)}(x) = left[sqrt{left(sqrt{x} + 1 ight)^2} + 1 ight]^2 = left(sqrt{x} + 2 ight)^2,$$ $$f^{(3)}(x) = left[sqrt{left(sqrt{x} + 2 ight)^2} + 1 ight]^2 = left(sqrt{x} + 3 ight)^2,$$ $$cdots cdots$$ $$Rightarrow f^{(n)}(x) = left(sqrt{x} + n ight)^2.$$ 构造桥函数: $$g(x) = x+1, phi(x) = sqrt{x}, phi^{-1}(x) = x^2,$$ $$Rightarrow phi^{-1}circ gcirc phi = x^2 circ (x +1)circ sqrt{x} = x^2 circ (sqrt{x} + 1) = x + 2sqrt{x} + 1 = f.$$ $$Rightarrow f^{(n)}(x) = phi^{-1} circ g^{(n)}(x) circ phi = x^2 circ (x + n) circ sqrt{x} = (sqrt{x} + n)^2.$$
8. 设 $f(x) = displaystyle{x^2 over 2(x - 1)}$. 求 $f(x)$ 之 $n$ 次迭代.
解答:
构造桥函数: $$f(x) = {1 over {2over x} - {2over x^2}} = {2 over 1 - left(1 - {2over x} ight)^2},$$ $$Rightarrow g(x) = x^2, phi(x) = 1 - {2over x}, phi^{-1}(x) = {2over 1-x},$$ $$Rightarrow phi^{-1}circ g circ phi = {2over 1-x} circ x^2 circ left(1 - {2over x} ight) = {2over 1-x} circ left(1 - {2over x} ight)^2$$ $$= {2over 1 - left({x - 2 over x} ight)^2} = {2x^2 over x^2 - (x-2)^2} = {x^2 over 2x - 2} = f.$$ $$Rightarrow f^{(n)}(x) = phi^{-1}circ g^{(n)} circ phi = {2over 1-x}circ x^{2^n}circ left(1 - {2over x} ight)$$ $$= {2over 1-x}circ left(1-{2over x} ight)^{2^n} = {2over 1 - left({x - 2 over x} ight)^{2^n}}$$ $$= {2x^{2^n} over x^{2^n} - (x - 2)^{2^n}}.$$