"Tim Little" <tim@[EMAIL PROTECTED]
> wrote in message
news:slrng2nbs2.lul.tim@[EMAIL PROTECTED]
> On 2008-05-14, Jim Dars <jim-dars@[EMAIL PROTECTED]
> wrote:
> > Even with today's computers I suspect the calculation would take a
> > long time.
>
> Approximations to y can be made accurate to as many figures as you can
> handle, very rapidly. The only problem with making it exact is the
> difficulty of representing the answer, which has more than 10^300 bits
> and so a naive representation cannot work.
>
>
> > I wonder how accurate the estimate might be?
>
> It's just looking at harmonic numbers H_n, and tons is known about
> them. In particular,
> |H_n - (ln n + gamma + 1/(2n))| < 1/n^2 for all n.
>
> Since you're just interested in differences of H_y and H_g, the gamma
> term in the approximation drops out entirely, so for y' = g/e and
> y = floor(y'),
> H_g - H_y = ln g - ln y + 1/(2g) - 1/(2y) + O(y^-2),
> which reduces to
> H_g - H_y = 1 + (frac(y') + 1/(2e) - 1/2) / y + O(y^-2).
>
> So the solution y = floor(g/e) is exact if
> frac(g/e) > (e - 1) / 2e + some number of magnitude 10^(-10^100),
> otherwise we need to let y = ceil(g/e). At most the approximation can
> be out by 1.
>
> There might be some clever theoretical argument that can determine
> whether in this particular case the floor or the ceiling operation
> should be taken without having to compute the first 10^100 or so
> decimal digits of e, but I suspect there isn't.
>
>
> - Tim
Hi Tim, Chip,
Well, I know how to go about it now. I suspect mathematical would get
one
pretty close.
Thanks for the input.
Best wishes, Jim
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