腾讯课堂目标2017高中数学联赛基础班-2作业题解答-6

Posted 2020-08-15 赵胤数学课

 

课程链接：目标2017高中数学联赛基础班-2（赵胤授课）

 

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: $$f^{(2)}(x) = \sqrt{\left(\sqrt{x^2 + c}\right)^2 + c} = \sqrt{x^2 + 2c},$$ $$f^{(3)}(x) = \sqrt{\left(\sqrt{x^2 + 2c}\right)^2 + c} = \sqrt{x^2 + 3c},$$ $$\cdots \cdots,$$ $$\Rightarrow f^{(n)}(x) = \sqrt{x^2 + nc}.$$

 

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 $$f^{(3)}(x) = a^3x + (a^2 + a + 1)b = 8x + 7,$$ $$\Rightarrow \begin{cases}a = 2\\ b = 1 \end{cases} \Rightarrow f(x) = 2x + 1.$$

 

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: $$f^{(2)}(x) = {{x \over \sqrt{1 + x^2}} \over \sqrt{1 + \left({x \over \sqrt{1 + x^2}}\right)^2}} = {x \over \sqrt{1 + 2x^2}},$$ $$f^{(3)}(x) = {{x \over \sqrt{1 + 2x^2}} \over \sqrt{1 + \left({x \over \sqrt{1 + 2x^2}}\right)^2}} = {x \over \sqrt{1 + 3x^2}},$$ $$\cdots \cdots,$$ $$\Rightarrow f^{(n)}(x) = {x \over \sqrt{1 + nx^2}}.$$

 

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 $$f^{(100)}(x) = 19^{100}x + \left(19^{99} + 19^{98} + \cdots + 19 + 1\right)\cdot 89$$ $$\Rightarrow f^{(100)}(5) = 19^{100}\cdot 5 + \left(19^{99} + 19^{98} + \cdots + 19 + 1\right)\cdot 89$$ $$\equiv 5 + \left[(-1)^{99} + (-1)^{98} + \cdots + (-1) + 1\right]\cdot (-1)$$ $$\equiv 5 + 0\cdot (-1) \equiv 5\ (\text{mod}\ 10)$$ 即末位数字为 $5$.

 

 

7. 设 $f(x) = x + 2\sqrt{x} + 1$. 求 $f(x)$ 之 $n$ 次迭代.

解答:

直接求迭代函数: $$f(x) = \left(\sqrt{x} + 1\right)^2,$$ $$f^{(2)}(x) = \left[\sqrt{\left(\sqrt{x} + 1\right)^2} + 1\right]^2 = \left(\sqrt{x} + 2\right)^2,$$ $$f^{(3)}(x) = \left[\sqrt{\left(\sqrt{x} + 2\right)^2} + 1\right]^2 = \left(\sqrt{x} + 3\right)^2,$$ $$\cdots \cdots$$ $$\Rightarrow f^{(n)}(x) = \left(\sqrt{x} + n\right)^2.$$ 构造桥函数: $$g(x) = x+1,\ \phi(x) = \sqrt{x},\ \phi^{-1}(x) = x^2,$$ $$\Rightarrow \phi^{-1}\circ g\circ \phi = x^2 \circ (x +1)\circ \sqrt{x} = x^2 \circ (\sqrt{x} + 1) = x + 2\sqrt{x} + 1 = f.$$ $$\Rightarrow f^{(n)}(x) = \phi^{-1} \circ g^{(n)}(x) \circ \phi = x^2 \circ (x + n) \circ \sqrt{x} = (\sqrt{x} + n)^2.$$

 

 

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: $$f(x) = {1 \over {2\over x} - {2\over x^2}} = {2 \over 1 - \left(1 - {2\over x}\right)^2},$$ $$\Rightarrow g(x) = x^2,\ \phi(x) = 1 - {2\over x},\ \phi^{-1}(x) = {2\over 1-x},$$ $$\Rightarrow \phi^{-1}\circ g \circ \phi = {2\over 1-x} \circ x^2 \circ \left(1 - {2\over x}\right) = {2\over 1-x} \circ \left(1 - {2\over x}\right)^2$$ $$= {2\over 1 - \left({x - 2 \over x}\right)^2} = {2x^2 \over x^2 - (x-2)^2} = {x^2 \over 2x - 2} = f.$$ $$\Rightarrow f^{(n)}(x) = \phi^{-1}\circ g^{(n)} \circ \phi = {2\over 1-x}\circ x^{2^n}\circ \left(1 - {2\over x}\right)$$ $$= {2\over 1-x}\circ \left(1-{2\over x}\right)^{2^n} = {2\over 1 - \left({x - 2 \over x}\right)^{2^n}}$$ $$= {2x^{2^n} \over x^{2^n} - (x - 2)^{2^n}}.$$