Dave was questioning the lack of alternatives, so let's look at the

standard Diophantine approximation ones. In the 7-limit, the

n^(-1/3) exponent puts the solutions into a cube of side n^(-1/3),

and hence of volume 1/n. This gives a density of solutions

proportional to 1/n, and since the integral of 1/n is unbounded, an

infinity of solutions may be expected.

In general, if f(n)>0 is such that its integral is unbounded, then

for d irrational numbers xi,

max f(n)^(-1/d) |round(n*xi) - n*xi| < c

"almost always" has an infinite number of solutions. This isn't as

tight a theorem as when we use exponents, but in practice it works

for our problem.

Most obviously, we could use an exponent less than 1/3, so that using

fourth roots instead will still give an infinite number of soutions--

a fact which already is obvious without the above. We could even use

1/ln(x), and get solutions in droves, like prime numbers. On the

other hand, we could fade just a little faster by using 1/(n ln n),

which makes the high end die out more, but probably not quite enough

to kill off an infinite list of solutions.

I don't see any advantages here, but there it is.

--- In tuning-math@y..., "genewardsmith" <genewardsmith@j...> wrote:

> Dave was questioning the lack of alternatives, so let's look at the

> standard Diophantine approximation ones.

Why not look outside Diophantine approximation alternatives?

> In general, if f(n)>0 is such that its integral is unbounded, then

> for d irrational numbers xi,

> max f(n)^(-1/d) |round(n*xi) - n*xi| < c

> "almost always" has an infinite number of solutions. This isn't as

> tight a theorem as when we use exponents, but in practice it works

> for our problem.

...

> I don't see any advantages here, but there it is.

There probably aren't any advantages here.

But why does badness have to be of the form

f(n)* |round(n*xi) - n*xi|

at all?

Why not

f(n) + |round(n*xi) - n*xi|

or

f(n) * g(|round(n*xi) - n*xi|)

?