已知平面上有 $n$ 个横坐标各不相同的点 $\left( {{x}_{i}},{{y}_{i}} \right)$，$i=1,2,\cdots ,n$，设直线 $y=bx+a$，求证：当 $\displaystyle \sum\limits_{i=1}^{n}{{{\left( b{{x}_{i}}+a-{{y}_{i}} \right)}^{2}}}$ 最小时，$\displaystyle b=\dfrac{\sum\limits_{i=1}^{n}{\left( {{x}_{i}}-\overline{x} \right)\left( {{y}_{i}}-\overline{y} \right)}}{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}{{y}_{i}}-n\overline{xy}}}{\sum\limits_{i=1}^{n}{x_{i}^{2}-n{{\overline{x}}^{2}}}}$，$a=\overline{y}-b\overline{x}$．
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【答案】
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【解析】
$\displaystyle \sum\limits_{i=1}^{n}{{{\left(b{{x}_{i}}+a-{{y}_{i}} \right)}^{2}}}={{b}^{2}}\left( \sum{x_{i}^{2}}\right)+n{{a}^{2}}+2ab\left( \sum{{{x}_{i}}} \right)-2b\left(\sum{{{x}_{i}}{{y}_{i}}} \right)-2a\left( \sum{{{y}_{i}}}\right)+\sum{y_{i}^{2}}$
最小时，$\displaystyle b=\dfrac{-a\left( \sum{{{x}_{i}}}\right)+\sum{{{x}_{i}}{{y}_{i}}}}{\sum{x_{i}^{2}}}$，$\displaystyle a=\dfrac{\sum{{{y}_{i}}-b\left({{x}_{i}} \right)}}{n}$，
解得 $\displaystyle b=\dfrac{\sum\limits_{i=1}^{n}{\left({{x}_{i}}-\overline{x} \right)\left( {{y}_{i}}-\overline{y}\right)}}{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x}\right)}^{2}}}}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}{{y}_{i}}-n\overline{xy}}}{\sum\limits_{i=1}^{n}{x_{i}^{2}-n{{\overline{x}}^{2}}}}$，$a=\overline{y}-b\overline{x}$．
最小时，$\displaystyle b=\dfrac{-a\left( \sum{{{x}_{i}}}\right)+\sum{{{x}_{i}}{{y}_{i}}}}{\sum{x_{i}^{2}}}$，$\displaystyle a=\dfrac{\sum{{{y}_{i}}-b\left({{x}_{i}} \right)}}{n}$，
解得 $\displaystyle b=\dfrac{\sum\limits_{i=1}^{n}{\left({{x}_{i}}-\overline{x} \right)\left( {{y}_{i}}-\overline{y}\right)}}{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x}\right)}^{2}}}}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}{{y}_{i}}-n\overline{xy}}}{\sum\limits_{i=1}^{n}{x_{i}^{2}-n{{\overline{x}}^{2}}}}$，$a=\overline{y}-b\overline{x}$．
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