证明不等式|arctanx-arctany|

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证明不等式|arctanx-arctany|
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证明不等式|arctanx-arctany|
证明不等式|arctanx-arctany|

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 |

令f(x)=arctanx
则(arctana-arctanb)/(a-b)=f'(c)=1+1/(1+c^2)<=1 (不妨设aa且c上式添个绝对值仍是小于1
所以|arctanx-arctany|<=|x-y|
你应该大学了吧 别告诉我还是初中