泰勒展开与牛顿法解析

1、写出以下每个函数在x = 0处的二阶泰勒展开式：(a) x²；(b) x³；(c) x⁴；(d) cos(x)。

函数 $ f(x) $ 在 $ x = a $ 处的二阶泰勒展开式为  
$$
f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)(x - a)^2}{2!}
$$

本题中 $ a = 0 $，则展开式为  
$$
f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2}
$$

### (a) 对于 $ f(x) = x^2 $  
- $ f(0) = 0 $  
- $ f'(x) = 2x \Rightarrow f'(0) = 0 $  
- $ f''(x) = 2 \Rightarrow f''(0) = 2 $  

二阶泰勒展开式为  
$$
f(x) = 0 + 0 \cdot x + \frac{2x^2}{2} = x^2
$$

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### (b) 对于 $ f(x) = x^3 $  
- $ f(0) = 0 $  
- $ f'(x) = 3x^2 \Rightarrow f'(0) = 0 $  
- $ f''(x) = 6x \Rightarrow f''(0) = 0 $  

二阶泰勒展开式为  
$$
f(x) = 0 + 0 \cdot x + \frac{0 \cdot x^2}{2} = 0
$$

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### (c) 对于 $ f(x) = x^4 $  
- $ f(0) = 0 $  
- $ f'(x) = 4x^3 \Rightarrow f'(0) = 0 $  
- $ f''(x) = 12x^2 \Rightarrow f''(0) = 0 $  

二阶泰勒展开式为  
$$
f(x) = 0 + 0 \cdot x + \frac{0 \cdot x^2}{2} = 0
$$

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### (d) 对于 $ f(x) = \cos(x) $  
- $ f(0) = 1 $  
- $ f'(x) = -\sin(x) \Rightarrow f'(0) = 0 $  
- $ f''(x) = -\co