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Re: [Phys-L] rms / conic / arithmetic / geometric averages

For some reason JD's post reminds me of an xkcd comic of a while back.

Also I tried looking at a generalization of his 'average' radius truncated cone average problem wherein the trucated cone's sides are not straight lines but are more generally power law curves so that for the generic truncated "cone" it looks like the flared bell of a horn, and the exponent, p, of the power law is a parameter of the horn's shape. So p=1 is an ordinary truncated cone, p=2 corresponds to a horn with a parabolic/quadratic bell flare, and p=3 is, likewise, a cubically flared horn bell, etc. It ends up in such a situation the appropriate formula for the 'mean' radius is
R_avg = √[((r_2^(2 + 1/p) - r_1^(2 + 1/p))/( r_2^(1/p) - r_1^(1/p)))/(2*p+1)]
& where r_1 & r_2 are the radii at each of the ends of the horn. For such a shape the only 2 values of p correspond to instances of JD's formula for his generic average. Those 2 values are p = 1 which gives the N=1 ordinary conic case (i.e. a cheerleader's horn), and p=1/2 which gives the N=0 RMS case. (But because the p=1/2 value is less than 1 the 'horn' is not really flared like horn, but is more shaped like a truncated bowl since the curve is reversed.) Other values of p give an average (above) that don't correspond to JD's N formula because even when the above quotient of differences of powers results in a finite telescoping sum there are many cross terms of products of various powers where the sum of powers on each such cross term is 2. But JD's formula just has 1 kind of cross term with a single power for each radial factor with a coefficient of N. The quotient of differences results in a finite sum when p is either a natural number or a positive half-integer. For other values of p the quotient of differences does not divide out to a finite sum. As an illustration of what I mean consider the p = 2 case. This generates the 5 term finite telescoping sum in the overall square root:
R_avg = √[(r_2^2 + r_2^(3/2)*r_1^(1/2) + r_2*r_1 + r_2^(1/2)*r_1^(3/2 ) + r_1^2))/5]
which does not correspond to JD's N=3 case. Similarly, the p = 5/2 generates a finite 6 term telescoping sum:
R_avg = √[(r_2^2 + r_2^(8/5)*r_1^(2/5) + r_2^(6/5)*r_1^(4/5) + r_2^(4/5)*r_1^(6/5) + r_2^(2/5)*r_1^(8/5 ) + r_1^2))/5],
and the p = 3 case generates a finite 7 term telescoping sum:
R_avg = √[(r_2^2 + r_2^(5/3)*r_1^(1/3) + r_2^(4/3)*r_1^(2/3) + r_2*r_1 + r_2^(2/3)*r_1^(4/3) + r_2^(1/3)*r_1^(5/3 ) + r_1^2))/7],

BTW, in case anyone is interested, the geometric mean of the harmonic mean and the arithmetic mean of any 2 numbers is their geometric mean, i.e. GM(HM(x,y),AM(x,y)) = GM(x,y).

David Bowman

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Subject: [Phys-L] rms / conic / arithmetic / geometric averages

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Hi --

I recently noticed something cute about averages. Maybe everybody but me already knew this, but I was surprised.

Context: Suppose you have a round pot or bucket with uniformly tapering sides. In other words, it is a truncated cone. You know the height, and you know the ID at the bottom and at the top.

We all know the formula for the volume of a cylinder, and this is "almost" a cylinder, so maybe we can use the same formula, using some sort of *average* radius.

What sort of average should we use? RMS? Arithmetic? Geometric????

For a full cone, we know the volume is 1/3 of the volume of a cylinder with the same diameter and radius. So RMS is no good, because that would predict 1/2 of the volume rather than 1/3.
And arithmetic is no good, because it would predict 1/4 of the volume (half of the diameter) instead of 1/3 of the volume.
Geometric is even worse, because it would predict 0 instead of 1/3. We want something that is somewhere between RMS and arithmetic.

The cute thing is that all of the above form a family. You can interpolate between them. The general formula is:
average = √[(Bot2+N*Bot*Top+Top2) / (2+N)]

for all N from 0 to ∞. In particular:

N=0 --> RMS
N=1 --> conic
N=2 --> arithmetic
N=∞ --> geometric

The relationships are graphed here:

The N=1 average is in fact exactly correct for a truncated cone, as you can easily verify in lots of ways. (Grind out the integrals.
Or just subtract one full cone from another.)

Related: Here is an online calculator to figure out the volume of a truncated cone:
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