big "mean" math question

Well, my big "mean" math question has to do with the idea of
the "root mean square" (RMS) method of finding averages that Graham
Breed was talking about on the "fat" list...

I'm actually intrigued by this, since I'm not understanding why
squaring everything, adding it all together and then taking the
SQUARE ROOT of the sum is going to lead to an accurate average...

Why is this done this way again?? This is pretty interesting,
actually...

_________ ________ _______
Joseph Pehrson

On 6/8/01 1:40 PM, "jpehrson@rcn.com" <jpehrson@rcn.com> wrote:

> Well, my big "mean" math question has to do with the idea of
> the "root mean square" (RMS) method of finding averages that Graham
> Breed was talking about on the "fat" list...
>
> I'm actually intrigued by this, since I'm not understanding why
> squaring everything, adding it all together and then taking the
> SQUARE ROOT of the sum is going to lead to an accurate average...
>
> Why is this done this way again?? This is pretty interesting,
> actually...

Joseph -

I think there's an easier explanation than using vectors.

It's basically using the Pythagorean triangle;

2 2 2
a + b = c

Try imagining having to take an average of only two variables.
I think it's the easiest way to visualize it.
Imagine an x-y grid.
You have your ideal point,
(by your problem, a JI interval?)
which is somewhere on the grid.
Then you have your test data scattered all over the grid.
If you want to give each variable equal "weight",
effectively you have to give it its own "dimension".
That is, if a 2% error is of equivalent deviance,
no matter which direction it's coming from,
then the error of any one set of two test values
would be equivalent to the distance between
that particular test point and the ideal point.
Which you can calculate by the Pythagorean triangle method.

The same method holds true for 3 dimensional space.
The geometric proof is very short.
Once you prove it true for 3 dimensional space,
given the original quest,
to plot the accuracy of equally relevant data,
you use the square root
of the sum of all the squares of deviances.

I'm in between naps,
so I hope you can follow this wording.

Marc

--- In tuning-math@y..., jpehrson@r... wrote:
> Well, my big "mean" math question has to do with the idea of
> the "root mean square" (RMS) method of finding averages that Graham
> Breed was talking about on the "fat" list...
>
> I'm actually intrigued by this, since I'm not understanding why
> squaring everything, adding it all together and then taking the
> SQUARE ROOT of the sum is going to lead to an accurate average...
>
> Why is this done this way again?? This is pretty interesting,
> actually...

You don't want to take the _straight_ average because it might be
zero just from positives and negative signs canceling out.

The two simplest alternatives are to take the _maximum_ error, or to
take the average of the absolute values of the errors (called MAD,
for Mean Absolute Deviation).

The RMS is known as the Standard Deviation in statistics. It's the
standard measure of error in science and engineering. There are
several reasons for this. Let me give you a rough idea of why it
makes some sense in this context.

Look at the dips in the harmonic entropy curve. Notice how they
are "rounded" at the bottom. Any curve with a round minimum like this
(not getting too technical) approximates a parabola more and more
closely the more you zoom in on the minimum. A parabola is just the
curve representing squared error. So if you sum the squared errors,
you're summing the dissonances, in a sense. And then you have to take
the square root at the end so that the result is comparable with the
units for a _single_ error. For example, in the 3-limit there's only
one interval to evaluate. Let's say it has a 2-cent error. So any
sort of _average_ over this one interval would have to be 2 cents. It
wouldn't make much sense to say the average was 4 cents when there's
only a single 2 cent error, would it? So that's why you have to take
the square root after summing. If you want to get more technical,
check out a statistics book.

On 6/8/01 5:01 PM, "Paul Erlich" <paul@stretch-music.com> wrote:

> The RMS is known as the Standard Deviation in statistics. It's the
> standard measure of error in science and engineering.

THAT'S what standard deviation IS? Ahh..

Thank you Paul. I don't think I ever knew that.
Or if I did, I managed to not retain it...

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

/tuning-math/message/169

Thanks, Paul... this gives me a good overview on this one! It's
pretty interesting...

________ _______ ______
Joseph Pehrson

--- In tuning-math@y..., "Orphon Soul, Inc." <tuning@o...> wrote:
> On 6/8/01 5:01 PM, "Paul Erlich" <paul@s...> wrote:
>
> > The RMS is known as the Standard Deviation in statistics. It's the
> > standard measure of error in science and engineering.
>
> THAT'S what standard deviation IS? Ahh..
>
> Thank you Paul. I don't think I ever knew that.
> Or if I did, I managed to not retain it...

Sorry, I was wrong about that. The Standard Deviation is something
different. It's actually the RMS deviation of a set of measurements
from their collective mean. The RMS we're talking about here, rather,
is the RMS deviation from a pre-determined standard, namely the JI
interval. SO it's somewhat different.

On 6/8/01 5:40 PM, "Paul Erlich" <paul@stretch-music.com> wrote:

> Sorry, I was wrong about that. The Standard Deviation is something
> different. It's actually the RMS deviation of a set of measurements
> from their collective mean.

Right, I remember it had to do with the mean,
just didn't know how it was calculated.
Thanks for clearing that up.

> The RMS we're talking about here, rather,
> is the RMS deviation from a pre-determined standard, namely the JI
> interval. SO it's somewhat different.

Actually I've worked with that myself.
Nice to see other people having the same
intuitions and/or conclusions.

[Joseph Pehrson wrote:]
>Well, my big "mean" math question has to do with the idea of
>the "root mean square" (RMS) method of finding averages that Graham
>Breed was talking about on the "fat" list...

>I'm actually intrigued by this, since I'm not understanding why
>squaring everything, adding it all together and then taking the
>SQUARE ROOT of the sum is going to lead to an accurate average...

>Why is this done this way again?? This is pretty interesting,
>actually...

Is it my imagination, or has nobody already caught the error in this?
Paul E, even you???

Before you take the square root, you divide by the number of values
whose square has been summed. Thus, the RMS of 3 and 4 is:

RMS = sqrt((3^2 + 4^2) / 2)
= sqrt((9 + 16) / 2)
= sqrt(12.5)
=~ 3.54

NOT sqrt(25) = 5!!

JdL

--- In tuning-math@y..., "John A. deLaubenfels" <jdl@a...> wrote:

/tuning-math/message/180

> Is it my imagination, or has nobody already caught the error in
this?
> Paul E, even you???
>
> Before you take the square root, you divide by the number of values
> whose square has been summed. Thus, the RMS of 3 and 4 is:
>
> RMS = sqrt((3^2 + 4^2) / 2)
> = sqrt((9 + 16) / 2)
> = sqrt(12.5)
> =~ 3.54
>
> NOT sqrt(25) = 5!!
>
> JdL

Actually, John... this is interesting because, if I'd known this, I
probably wouldn't have been quite as "mystified" as I was after
Graham's original post. The method you outline immediately above
seems somewhat "averagy" to me... so it would have seemed more
sensible.

Here was Graham's original quote from post 24541:

>Averages are trickier, you do need to consider all intervals then.
>The most popular is the root mean squared (RMS). So you take the
>errors in all intervals, square them all, add them together and
>return the square root.

________ ______ _____
Joseph Pehrson

[Joseph Pehrson wrote:]
>Actually, John... this is interesting because, if I'd known this, I
>probably wouldn't have been quite as "mystified" as I was after
>Graham's original post. The method you outline immediately above
>seems somewhat "averagy" to me... so it would have seemed more
>sensible.

>Here was Graham's original quote from post 24541:

>>Averages are trickier, you do need to consider all intervals then.
>>The most popular is the root mean squared (RMS). So you take the
>>errors in all intervals, square them all, add them together and
>>return the square root.

Right. That'd be the "Root Sum Square", which, as you've surmised,
wouldn't be very "averagy". In fact, I'm not sure what it would be
useful for. I'm sure Graham, and probably all the other people who
responded to your post yesterday, _do_ know the correct definition, but
all had a "brain fart" (which I know a lot about, 'cause I get them all
the time!).

The RMS value will always be less than the largest absolute value which
goes into its calculation (or equal if all input values are the same or
-same). I can see that you were grasping for that in your original
post. So, you have a better math sense than you realized!

JdL

[I wrote:]
>Right. That'd be the "Root Sum Square", which, as you've surmised,
>wouldn't be very "averagy". In fact, I'm not sure what it would be
>useful for.

Oops! Well, maybe it'd be slightly useful for such abstractions as
the length of a hypotenuse of a right triangle. ;->

JdL

--- In tuning-math@y..., "John A. deLaubenfels" <jdl@a...> wrote:

/tuning-math/message/186

> [Joseph Pehrson wrote:]
> >Actually, John... this is interesting because, if I'd known this,
I
> >probably wouldn't have been quite as "mystified" as I was after
> >Graham's original post. The method you outline immediately above
> >seems somewhat "averagy" to me... so it would have seemed more
> >sensible.
>
> >Here was Graham's original quote from post 24541:
>
> >>Averages are trickier, you do need to consider all intervals
then.
> >>The most popular is the root mean squared (RMS). So you take the
> >>errors in all intervals, square them all, add them together and
> >>return the square root.
>
> Right. That'd be the "Root Sum Square", which, as you've surmised,
> wouldn't be very "averagy". In fact, I'm not sure what it would be
> useful for. I'm sure Graham, and probably all the other people who
> responded to your post yesterday, _do_ know the correct definition,
but all had a "brain fart" (which I know a lot about, 'cause I get
them all the time!).
>
> The RMS value will always be less than the largest absolute value
which goes into its calculation (or equal if all input values are the
same or -same). I can see that you were grasping for that in your
original post. So, you have a better math sense than you realized!
>
> JdL

Actually, John... that's pretty funny and, frankly, encouraging.
It's never too late to learn at least *something!* I think part of
the problem was the math "training" I had as a student. Math was
always presented as a "practical" study with ugly and dull-
looking "engineering type" books. No wonder I would read art and
music books instead. The subject was *ruined* for me... or at least
for *my* particular sensibilities...

_______ _____ ________
Joseph Pehrson