Open ArslanArdavic opened 1 year ago
{\displaystyle H={\frac {n}{{\frac {1}{x{1}}}+{\frac {1}{x{2}}}+\cdots +{\frac {1}{x{n}}}}}={\frac {n}{\sum \limits {i=1}^{n}{\frac {1}{x{i}}}}}=\left({\frac {\sum \limits {i=1}^{n}x_{i}^{-1}}{n}}\right)^{-1}.}
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ArslanArdavic
commented
1 year ago
{\displaystyle H={\frac {n}{{\frac {1}{x{1}}}+{\frac {1}{x{2}}}+\cdots +{\frac {1}{x{n}}}}}={\frac {n}{\sum \limits {i=1}^{n}{\frac {1}{x{i}}}}}=\left({\frac {\sum \limits {i=1}^{n}x_{i}^{-1}}{n}}\right)^{-1}.}