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From: Jan Burse <janburse@fastmail.fm>
Newsgroups: comp.lang.java.programmer
Subject: Re: higher precision doubles
Date: Tue, 09 Aug 2011 01:08:25 +0200
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BGB schrieb:
> division is mostly calculating the reciprocal of the second value, and
> multiplying the first by this (can be done using Newton's Method).

I don't know exactly what you understand by
this algorithm. But if I understand this as
calculating the reciprocal first adn then
multiplying, you will have as a result:

    o(x*o(1/y))

Instead of

    o(x/y).

you will usually get less accurate results
since by the reciprocal-math option method
two approximations will be involved.

This is just a guess from the hip. But here
you see a simple excel verifying the claim:

	1	2	3	4	5	6	7	8
1	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00
2	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00
3	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00
4	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00
5	0.00	0.00	0.00	0.00	0.00	-0.01	-0.01	0.00
6	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00
7	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00
8	0.00	0.00	0.01	0.00	0.00	-0.01	0.00	0.00

Used formula:
=ROUND($G6/H$5;2)-ROUND($G6*ROUND(1/H$5;3);2)

	1	2	3	4	5	6	7	8
1	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00
2	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00
3	0.00	0.00	0.01	0.00	0.00	0.01	0.00	0.00
4	0.00	0.00	0.00	0.00	0.00	0.00	0.01	0.00
5	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00
6	0.00	0.00	0.01	0.00	0.00	0.01	0.00	0.00
7	0.00	0.00	0.00	0.00	0.00	0.00	0.01	0.00
8	0.00	0.00	0.00	0.00	0.00	0.01	0.01	0.00

Used formula:
=FLOOR($G6/H$5;2)-FLOOR($G6*FLOOR(1/H$5;3);2)

Errors are also seen for larger numbers of digits.

Bye