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Version 1 |
| Three positive numbers $x$, $m$, $y$ are in {\em contraharmonic proportion}, if the ratio of the difference of the second and the first number to the difference of the third and the second number is equal the ratio of the third and the first number, i.e. if |
Three positive numbers $x$, $m$, $y$ are in {\em contraharmonic proportion}, if the ratio of the difference of the second and the first number to the difference of the third and the second number is equal the ratio of the third and the first number, i.e. if |
| \begin{align} |
\begin{align} |
| \frac{m-x}{y-m} = \frac{y}{x}. |
\frac{m-x}{y-m} = \frac{y}{x}. |
| \end{align} |
\end{align} |
| The middle number $m$ is then called the {\em contraharmonic mean} of the first and the last number. |
The middle number $m$ is then called the {\em contraharmonic mean} of the first and the last number. |
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The contraharmonic proportion has very probably been known in the proportion doctrine of the Pythagoreans, since they have in a manner \PMlinkescapetext{similar} to (1) described the classical Babylonian means:
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The contraharmonic proportion has very probably been known in the proportion doctrine of the Pythagoreans, since they have in a manner similar to (1) described the classical Babylonian means:
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| $$\frac{m-x}{y-m} = \frac{x}{x} \qquad (\mbox{arithmetic mean }m)$$ |
$$\frac{m-x}{y-m} = \frac{x}{x} \qquad (\mbox{arithmetic mean }m)$$ |
| $$\frac{m-x}{y-m} = \frac{x}{m} \qquad (\mbox{geometric mean }m)$$ |
$$\frac{m-x}{y-m} = \frac{x}{m} \qquad (\mbox{geometric mean }m)$$ |
| $$\frac{m-x}{y-m} = \frac{x}{y} \qquad (\mbox{harmonic mean }m)$$ |
$$\frac{m-x}{y-m} = \frac{x}{y} \qquad (\mbox{harmonic mean }m)$$ |
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The contraharmonic mean $m$ is indeed between $x$ and $y$. Indeed, if we solve it from (1), we get |
| The contraharmonic mean $m$ is between $x$ and $y$. Indeed, if we solve it from (1), we get |
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| $$m = \frac{x^2+y^2}{x+y},$$ |
$$m = \frac{x^2+y^2}{x+y},$$ |
| and if we assume that\, $x \leqq y$, we see that |
and if we assume that\, $x \leqq y$, we see that |
| $$x = \frac{x^2+xy}{x+y} \leqq \frac{x^2+y^2}{x+y} \leqq \frac{xy+y^2}{x+y} = y.$$ |
$$x = \frac{x^2+xy}{x+y} \leqq \frac{x^2+y^2}{x+y} \leqq \frac{xy+y^2}{x+y} = y.$$ |
| The contraharmonic mean $c$ is the greatest of all the mentioned means, |
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| $$x \leqq h \leqq g \leqq a \leqq c \leqq y$$ |
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| where $a$ is the arithmetic mean, $g$ the geometric mean and $h$ the harmonic mean. It is easy to see that |
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| $$\frac{c+h}{2} = a \quad\mbox{and}\quad \sqrt{ah} = g.$$ |
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| \begin{thebibliography}{8} |
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| \bibitem{DD}{\sc Diderot \& d'Alembert}: {\em Encyclop\'edie}.\, Electronic version in |
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| \PMlinkexternal{L'Encyclop\'edie de Diderot et d'Alembert}{http://obliques.free.fr/encyclotexte/}. |
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| \bibitem{HH}{\sc Horst Hischer}: ``Viertausend Jahre Mittelwertbildung''. --- {\em mathematica didactica} \textbf{25} (2002). |
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| \end{thebibliography} |
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