As Elijah pointed out, you probably are interested only in the rational
solutions, since there an infinity of real solutions, most of which are
likely to be irrational. You can even locate them. Assume you have an
integer b, such that b(b+1) > a (where a is the number you are trying to
reach –
On Tue, 19 Oct 2021, Elijah Stone wrote:
0 = (64*x*x) + (_4096*x) + 4002
Another foolish mistake! It should be '- 4002', not '+ 4002'. (The
answer is still correct; but I did not correctly reproduce my workings.)
--
For in
Thanks Elijah! That's what I was missing.
We could also use J's polynomial solver:
* ] 'x f' =. ; }. p. 4002 _4096 64*
*63.00755956214548803 0.99244043785451341311*
Looks good:
* 4002=64*x*f*
*1*
But only if we rely on J's comparison tolerance.
If we move to rational fractions:
* ] 'x f'
On Tue, 19 Oct 2021, Elijah Stone wrote:
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 :/
Ah, no, I made a foolish mistake. The determin
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*
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 par
Maybe this?
c3=. (%3)^~poss=. 4002x */ >:i.999
n=. poss%(<.c3)*>.c3
1 e. poss=n*(<.n)*>.n
1
On Mon, Oct 18, 2021 at 8:02 PM Raul Miller wrote:
> I think that 64 is too big.
>64^3
> 262144
>
> I don't think this one has a solution, because 4002%15*16 is too
> large, and 4002%16*17 i
I think that 64 is too big.
64^3
262144
I don't think this one has a solution, because 4002%15*16 is too
large, and 4002%16*17 is too small.
I haven't explored gaussian integers here, but my gut feeling is that
there won't be any viable candidates there, either.
I hope this helps,
--
Raul
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
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