[clean-list] Re: Re: 1.Floating Point Support (clean-list digest)

[Richard A. O'Keefe]

	On Mon, 20 Nov 2000, Siegfried Gonzi wrote:
	> per def.:     lim(n->infinity) (1 + 1/n)^n == e
	> 
	> with the old 68040 Motorola they were able to use n>1.0e15, but with the 64bit and
	> n>1.0e15 the above formula delivers 1.
	
RT Happe wrote:
	In fact, the (...)^n converge to e but slowly.
	Taking that limit doesn't make for an effective algorithm.
	
Think outside the box.  It is clear that

	lim(n->infinity) (1+1/n)**n
    ==
	lim(k->infinity) (1+2**(-k))**(2**k)

and that *does* converge quickly enough to be useful.
Using 64-bit IEEE arithmetic, Sun pow() function, it only takes
53 iterates to get relative error better than 2E-16.
Simply plugging in the largest value of k you can gets you quite
a good answer.