I’m looking to understand the relationship between rectangular product & mean square operations -- or, equivalently, between their root operations, the geometric & quadratic means. Here’s the problem:
Suppose we define an isomorphic mapping CR(a,b) to re-describe argument real pair a,b into their conjugate roots -- i.e., $$CR(a,b) = \left( \sqrt a+\sqrt b, \sqrt a-\sqrt b \right)$$
With the function arguments so ‘re-described’, subtraction becomes re-describable as a product, and addition as a mean square:
$$ \begin{split} \text{Diff}(a,b) = a - b &= \left(\sqrt a+\sqrt b\right)\left(\sqrt a-\sqrt b\right) = \text{Prod}(CR(a,b)) \\ \text{Sum}(a,b) = a + b &= \frac{\left(\sqrt a+\sqrt b\right)^2+\left(\sqrt a-\sqrt b\right)^2}{2} = \text{MeanSq}(CR(a,b)) \end{split} $$
So subtracting a pair of real numbers is always just multiplying their conjugate square roots, and summing them likewise just mean-squaring those same conjugate square roots. 7 is thus not just the difference of 16 and 9, but the product of their conjugate square roots (4+3)(4-3); likewise, 25 is not just their sum, but the mean-square of those conjugate square roots (49+1)/2.
But the sum & difference operations on the left-side of the above equations are inverse operations to one another -- by definition, even. Can the corresponding right-sides of the above equations, multiplying & mean-squaring, be similarly understood as inverse operations of each other??
Since those latter operations, rectangular product & mean-square, are 2-degree area-functions, perhaps the question is best addressed to their 1-degree linear counterparts (i.e., their own square roots), in which case we ask then after the relationship between geometric mean & quadratic mean. And this appears to offer some promise, since among the four Pythagorean means, the other two -- arithmetic & harmonic means -- are, obviously, multiplicative “duals” of one another:
$$ \begin{split} HM(a,b) &= {1 \over AM(\frac1a,\frac1b)} \\ AM(a,b) &= {1 \over HM(\frac1a,\frac1b)} \end{split} $$
…and in that sense they are in "inverse" relationship, with the geometric mean that falls between them being multiplicatively self-dual, $GM(a,b)= {1 \over GM(1/a,1/b)}$.
But can all four Pythagorean means then really be so neatly paired off that the quadratic mean QM be in some way “dual with” or "inverse to" GM? The following diagram in Wikipedia, which relates them as a “central secant” (my term for it) vs. a tangent, looks promising in this regard, but I myself am not quite seeing what the inversion would then be, conceptually:
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Please feel free to suggest other tags appropriate to this question. I am, frankly, not quite sure under what math subfields it falls.
- Image here: https://en.wikipedia.org/wiki/Root_mean_square#/media/File:QM_AM_GM_HM_inequality_visual_proof.svg ( The diagram is drawn from these larger articles: https://en.wikipedia.org/wiki/Pythagorean_means#Inequalities_among_means https://en.wikipedia.org/wiki/Root_mean_square#Relationship_to_other_statistics )