设f(a) = arctan(a),f'(a) = 1/(1 + a²)
f(a)在(x,y)连续可导,根据拉格朗日中值定理,
| arctanx - arctany | = 1/(1 + c²) * | x - y | < | x - y |,c∈(x,y)
当a = b = 0时arctanx = arctany = 0
∴
| arctanx - arctany | ≤ | x - y |
证明不等式|arctanx-arctany|
设f(a) = arctan(a),f'(a) = 1/(1 + a²)
f(a)在(x,y)连续可导,根据拉格朗日中值定理,
| arctanx - arctany | = 1/(1 + c²) * | x - y | < | x - y |,c∈(x,y)
当a = b = 0时arctanx = arctany = 0
∴
| arctanx - arctany | ≤ | x - y |