RMS and the Pythagorean Theorem

[kotreny]

Today I was thinking about the root mean square, and I figured out a definite relationship with the Pythagorean theorem. Specifically, the root mean square of the legs of a right triangle is equal to the "average leg," i.e. the leg of a square with the hypotenuse as it's diagonal. It appears to me that this is a fairly interesting and important connection, certainly applying to distance on Cartesian coordinates and maybe explaining the usefulness of the RMS. However, when I googled it, nothing came up! I'm trying to see if this relationship has any meaning whatsoever, which I believe it should, and if so, what it means. I'm pretty sure my math isn't wrong.

 

[mathador]

Hi Kotreny,

It's a nice observation. The geometric meaning is illustrated in the figure (I modified the figure found at http://mathworld.wolfram.com/PythagoreanMeans.html).

Mathador

 

[kotreny]

Thanks for the attachment, but there seems to be something wrong. I tried substituting a=4 and b=3. The RMS is then 5/\sqrt{2}. But the line labeled RMS in the figure should be of length 7/\sqrt{2}, right? It looks like the diagonal of a square with side A--or the arithmetic mean, which would be 7/2--and the diagonal is always equal to side*\sqrt{2}. Correct me if I'm wrong. Thanks again!

 

[kotreny]

Today I was thinking about the root mean square, and I figured out a definite relationship with the Pythagorean theorem. Specifically, the root mean square of the legs of a right triangle is equal to the "average leg," i.e. the leg of a square with the hypotenuse as it's diagonal.

I should clarify exactly what I mean.

Let's say you have a right triangle with legs a and b and hypotenuse c.
The Pythagorean Theorem says that a² + b² = c².
Now, the root mean square of the two legs is \sqrt{(a^2 + b^2)/2}. But wait! Combine the two equations to get,

RMS of a and b = \sqrt{c^2/2} = c /\sqrt{2}

Now imagine a 45-45-90 triangle with legs equal to c /\sqrt{2}. The length of the hypotenuse would then be
c /\sqrt{2} * \sqrt{2}, which is equal to c. The conclusion is that the RMS of legs a and b gives you the leg of a 45-45-90 triangle with the same hypotenuse c. A little extra work shows that it applies to 3 or more dimensions too.

I'll bet this is used to find average vector components, or something, though they probably don't take the time to mention the connection with the RMS. I dunno.

Does standard deviation have something to do with this?

 

[kotreny]

http://en.wikipedia.org/wiki/Standard_deviation#Geometric_interpretation"

This is essentially what I'm talking about, although worded differently. Seems strange that nowhere else mentions it; you'd think this is an important fact!

So can standard deviation really, formally be visualized like this? If you take a data set, can each data point's deviation be considered as inhabiting its own "dimension"? Does it have mathematical significance at all?

I'd love to get some answers, opinions, and especially corrections! Any feedback would be appreciated. Please comment, and thanks very much!