What is the intuition behind root mean square?

Physics news on Phys.org
Essentially the root mean square is a distance. If you were to calculate the distance the point (x, y, z) is from (0,0,0) you would calculate \sqrt{x^2+ y^2+ z^2}. The "root mean square" is really an "average distance", thinking of each value in the set as a "dimension".

Of course, that's not the only way to define an "average". Also used is to, not "inverse" every sign since that would mean changing positive to negative, take the arithmetic average of the absolute values:
\frac{|a_1|+ |a_2|+ \cdot\cdot\cdot+ |a_n|}{n}
 
Thanks but why the distance needs to divide by n(in this case n=3)?
ie
\frac{\sqrt{(x^2+ y^2+ z^2)}{n}

Seems like it is the average distance of each dimension in the n-dimension to the original point. Am I right?
If yes, may I know what is the significant value to define it as such?

>>not "inverse" every sign since that would mean changing positive to negative
Sorry typo, it should be taking absolute value as you mentioned:)
 
Last edited:
jack1234 said:
Thanks but why the distance needs to divide by n(in this case n=3)?
ie
\frac{\sqrt{(x^2+ y^2+ z^2)}}{n}

Seems like it is the average distance of each dimension in the n-dimension to the original point. Am I right?
If yes, may I know what is the significant value to define it as such?
When you measure an "average" of a number of things you are, basically, measuring 'how large' they are. A distance measure is a natural analogy to use.

>>not "inverse" every sign since that would mean changing positive to negative
Sorry typo, it should be taking absolute value as you mentioned:)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top