User:Dingo1729/Sandbox

In mathematics, the three classical Pythagorean means are the arithmetic mean (A), the geometric mean (G), and the harmonic mean (H). They are defined by:


 * $$ A(x_1, \ldots, x_n) = \frac{1}{n}(x_1 + \cdots + x_n) $$
 * $$ G(x_1, \ldots, x_n) = \sqrt[n]{x_1 \cdots x_n} $$
 * $$ H(x_1, \ldots, x_n) = \frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}} $$

Each mean has the following properties:
 * Value preservation: $$ M(x,x, \ldots,x) = x $$
 * First order homogeneity: $$ M(bx_1, \ldots, bx_n) = b M(x_1, \ldots, x_n) $$
 * Invariance under exchange: $$ M(\ldots, x_i, \ldots, x_j, \ldots ) = M(\ldots, x_j, \ldots, x_i, \ldots) $$ for any $$i$$ and $$j$$.
 * Averaging: $$ \min(x_1,\ldots,x_n) \leq M(x_1,\ldots,x_n) \leq \max(x_1,\ldots,x_n)$$

These means were studied with proportions by Pythagoreans and later generations of Greek mathematicians (Thomas Heath, History of Ancient Greek Mathematics) because of their importance in geometry and music.

There is an ordering to these means (if all of the $$ x_i $$ are positive), along with the quadratic mean $$Q=\sqrt{\frac{x_1^2+x_2^2+ \cdots + x_n^2}{n}}$$:


 * $$ \min \leq H \leq G \leq A \leq Q \leq \max $$

with equality holding if and only if the $$ x_i $$ are all equal. This is a generalization of the inequality of arithmetic and geometric means and a special case of an inequality for generalized means. This inequality sequence can be proved for the $$n=2$$ case for the numbers a and b using a sequence of right triangles (x, y, z) with hypotenuse z and the Pythagorean theorem, which states that $$x^2 + y^2 = z^2$$ and implies that $$z > x$$ and $$z > y$$. The right triangles are


 * $$\left(\frac{b-a}{b+a}\sqrt{ab}, \frac{2ab}{a+b}, \sqrt{ab}\right) = \left(\frac{b-a}{b+a}\sqrt{ab}, H(a,b), G(a,b)\right),$$

showing that $$H(a,b) < G(a,b)$$;


 * $$\left(\frac{b-a}{2}, \sqrt{ab}, \frac{a+b}{2}\right) = \left(\frac{b-a}{2}, G(a,b), A(a,b)\right),$$

showing that $$G(a,b) < A(a,b)$$;

and


 * $$\left(\frac{b-a}{2}, \frac{a+b}{2}, \sqrt{\frac{a^2+b^2}{2}}\,\right) = \left(\frac{b-a}{2},A(a,b), Q(a,b)\right),$$

showing that $$A(a,b) < Q(a,b)$$.