contrageometric proportion
Just as one converts the proportion equation
$$\frac{m-x}{y-m}=\frac{x}{y}$$ |
defining the harmonic mean^{} of $x$ and $y$ into the proportion equation
$$\frac{m-x}{y-m}=\frac{y}{x}$$ |
defining their contraharmonic mean (http://planetmath.org/ContraharmonicProportion), one also may convert the proportion equation
$$\frac{m-x}{y-m}=\frac{m}{y}$$ |
defining the geometric mean^{} into a new equation
$\frac{m-x}{y-m}}={\displaystyle \frac{y}{m}$ | (1) |
defining the contrageometric mean $m$ of $x$ and $y$. Thus, the
three positive numbers $x$, $m$, $y$ satisfying (1) are in contrageometric proportion. One integer example is
$1,\mathrm{\hspace{0.17em}4},\mathrm{\hspace{0.17em}6}$.
Solving $m$ from (1) one gets the expression
$m={\displaystyle \frac{x-y+\sqrt{{(x-y)}^{2}+4{y}^{2}}}{2}}=:f(x,y).$ | (2) |
Suppose now that $0\le x\le y$. Using (2) we see that
$$m\ge \frac{x-y+\sqrt{{0}^{2}+4{y}^{2}}}{2}=\frac{x+y}{2}\ge x,$$ |
$${y}^{2}-{m}^{2}=\frac{-{(y-x)}^{2}+(y-x)\sqrt{{(x-y)}^{2}+4{y}^{2}}}{2}=\frac{(y-x)[\sqrt{{(y-x)}^{2}+4{y}^{2}}-(y-x)]}{2}\ge \mathrm{\hspace{0.33em}0},$$ |
accordingly
$x\le f(x,y)\le y.$ | (3) |
Thus the contrageometric mean of $x$ and $y$ also is at least equal to their arithmetic mean^{}. We can also compare $m$ with their quadratic mean by watching the difference
$${\left(\sqrt{\frac{{x}^{2}+{y}^{2}}{2}}\right)}^{2}-{m}^{2}=(y-x)\left(\frac{1}{2}\sqrt{{(y-x)}^{2}+4{y}^{2}}-y\right)\ge \mathrm{\hspace{0.33em}0}.$$ |
So we have
$\frac{x+y}{2}}\le f(x,y)\le \sqrt{{\displaystyle \frac{{x}^{2}+{y}^{2}}{2}}}.$ | (4) |
Cf. this result with the comparison of Pythagorean means (http://planetmath.org/ComparisonOfPythagoreanMeans); there the brown curve is the graph of $f(x,\mathrm{\hspace{0.17em}1})$.
It’s clear that the contrageometric mean (2) is not symmetric with respect to the variables $x$ and $y$, contrary to the other types of means in general. On the other hand, the contrageometric mean is, as other types of means, a first-degree homogeneous function its arguments:
$f(tx,ty)=tf(x,y).$ | (5) |
References
- 1 Mabrouk K. Faradj: http://etd.lsu.edu/docs/available/etd-07082004-091436/unrestricted/Faradj_thesis.pdfWhat mean do you mean? An exposition on means. Louisiana State University (2004).
- 2 Georghe Toader & Silvia Toader: http://rgmia.org/papers/monographs/Grec.pdfGreek means and the arithmetic-geometric mean^{}. RGMIA (2010).
Title | contrageometric proportion |
---|---|
Canonical name | ContrageometricProportion |
Date of creation | 2013-04-19 7:14:46 |
Last modified on | 2013-04-19 7:14:46 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 17 |
Author | pahio (2872) |
Entry type | Definition |
Classification | msc 26E60 |
Classification | msc 11-00 |
Classification | msc 01A20 |
Classification | msc 01A17 |
Related topic | ContraharmonicProportion |
Defines | contrageometric mean |