Note that (x-2)*(x-1)*x is equal to 6*3!x .
So the problem is reduced to identifying binomial coefficients.
Den 17:02 mandag den 10. august 2015 skrev 'Pascal Jasmin' via Programming
<programm...@jsoftware.com>:
To answer why the last digit in p. result identifies whether it has integer
root, no matter how many consecutive integers are used.
for 3 digits, the root will be x^3. For 4, x^4, and the result of p. will
identify x in that last location.
----- Original Message -----
From: Roger Hui <rogerhui.can...@gmail.com>
To: Programming forum <programm...@jsoftware.com>
Cc:
Sent: Monday, August 10, 2015 10:52 AM
Subject: Re: [Jprogramming] Detecting special products
p. orders the roots by decreasing magnitude, and within that (for roots
with equal magnitude) the ordering used by \: . For example:
p. p. < 3 1 4 1 5 9
┌─┬───────────┐
│1│9 5 4 3 1 1│
└─┴───────────┘
p. p. <3 1 4 1 5 9 3j4 3j_4 _3j4 _3j_4
┌─┬───────────────────────────────┐
│1│9 5 3j4 3j_4 _3j4 _3j_4 4 3 1 1│
└─┴───────────────────────────────┘
If I may be permitted to show off a bit:
p. p. <1+i.20x
┌─┬──────────────────────────────────────────────────┐
│1│20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1│
└─┴──────────────────────────────────────────────────┘
https://en.wikipedia.org/wiki/Wilkinson%27s_polynomial
On Mon, Aug 10, 2015 at 6:54 AM, Don Guinn <dongu...@gmail.com> wrote:
> Thanks. I was being a little dense in forgetting that it is solving for x,
> not for all values. My thinking gets a little muddled with grandkids
> around. I was just surprised at the neatness of the solution.
>
> I played with products of consecutive numbers greater than 3 (are those
> still special products?) and p. still works until p. becomes unstable due
> to the order of the polynomial. I'm curious as to why the desired value for
> x is always the last root. Just the luck of the draw or is there some
> reason as to why it is always last. Do real values always come last? For
> the products of even consecutive numbers the number of real roots found by
> p. must be even. But the last one still seems to be the correct one.
>
> Take of two different product of 4 sequential numbers. Their gcd many cases
> is s special product. Those that aren't are a multiple of a special
> product. This seems to be the case that the gcd for the products of n
> consecutive integers is the a product or a multiple of n-1 and n-2 ...
> consecutive integers. The multiples seem to be in one consecutive group.
>
> Now that the grandkids are gone maybe I can play with this more with a
> clearer mind.
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