Ah--that is much clearer!
I am not sure of a good iterative solution, but there is a fairly simple
analytic solution.
Say we would like to find a solution with a floor of 64; that is, where 64
= <.x.
Then we have:
4002 = x * 64 * (x - 64)
= (64*x*x) + (_4096*x)
which means that:
0 = (64*x*x) + (_4096*x) + 4002
Which is a quadratic equation. We may attack it as normal, deriving a
solution as:
] x=.(2*64) %~ 4096 + %: (_4096^2) - 4*64*_4002
64.9626
x * (<.x) * (x - <.x)
4002
However, I think you asked for a rational answer; this solution is not
rational because the determinant is not an integer square.
The solution with a floor of k has a determinant of (k^4) + 4*k. Are
there any values for k in range such that that value is a perfect square?
1 e. (= <.) %: (^&4 + 4&*) >:i.150
0
It appears not :/
-E
On Tue, 19 Oct 2021, Skip Cave wrote:
I made a mistake in the equation in my first post..
The three terms that are multiplied are
1. x
2. floor of x = (<.x)
3. fractional part of x = (x - <.x) This is what I got wrong in my first
post.
I can get close by manual trial & error:
*x=.64.962573478*
Floor of x = <.x = 64
Fractional part of x = (x -<.x ) = 0.962573478
* x: x * (<.x) * (x - <.x)*
*213746821r53410*
Close, but no cigar.
* 4002 = x * (<.x) * (x - <.x)*
*0*
A closer look - as a decimal fraction:
* x * (<.x) * (x-<.x)*
*4002.0000187231198652*
Yep. Not close enough.
How to design an iterative solution? There should be multiple solutions
with (<.x) = 63, 64, 65. 66 ... with the fraction
*(x-<.x) *getting smaller & smaller.
Skip Cave
Cave Consulting LLC
On Mon, Oct 18, 2021 at 6:25 PM Skip Cave <[email protected]> wrote:
How to solve this problem?
4002x = n * (<.n) * (>.n)
What is n, where n is a rational fraction greater than 1, and the answer
is a rational fraction? There are likely many answers, so find some answers
near 64. The result in J should be a 1:
4002x = n * (<.n) * (>.n)
1
Skip
Skip Cave
Cave Consulting LLC
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