Re: [tuning] Digest Number 1084

Hi Dan,

Yes, that is perfectly clear now, I get it.

So:

n=7 x=1000 + 2,
7000/7000 8002/6999 9004/6998, 10006/6997,...

would show how the undertone series turns into the overtone series when x is large, right.

Then, the undertone series has intervals getting larger as you go up,
overtone series has intervals getting smaller, so if you hit the
boundary between the two, intervals will be aproximately
equal.

n=7
x = sqrt(2)+2

writting r2 as short for sqrt(2).

So starting with (n*x)/(n*x),
adding r2 to numerator each time, subtracting 1 from denom:

gives:
(14+7*r2)/(14+7*r2)
(14+8*r2)/(13+7*r2)
(14+9*r2)/(12+7*r2)
(14+10*r2)/(11+7*r2)
(14+11*r2)/(10+7*r2)
(14+12*r2)/(9+7*r2)
(14+13*r2)/(8+7*r2))
(14+14*r2)/(7+7*r2)

Last note is 2/1
step sizes:
1/1 173.524 cents 171.416 cents 170.176 cents 169.767 cents 170.176 cents 171.416 cents
173.524 cents

You get the x = sqrt(2) + 2 condition by requiring the first and last steps to be equal:
Last step is:
2/ ((14+13*r2)/(8+7*r2))
=
(16+14*r2)/(14+13*r2)

First step is:
(14+8*r2)/(13+7*r2)

and multiplying top and bottom by r2 you get the first step as:
(16+14*r2)/(14+13*r2)
so in this case they are equal.

In general case, for x = r2+2.
Sequence starts with
(2*n+n*r2)/(2*n+n*r2)

first step is
(2*n+(n+1)*r2)/(2*n-1+n*r2)

Last step is
(2*(n+1) + 2*n*r2)/(2*n+(2*n-1)*r2)

and on multiplying first step by r2 you get:
(2*n*r2+(n+1)*2)/((2*n-1)*r2+n*2)
which is the same as the last step.

It's clear this will always work.

Q.e.d. as they say.

I like it, a neat result!

Robert