On Fri, 3 Oct 2003, Scott Dattalo wrote:

> On Fri, 3 Oct 2003, Russell McMahon wrote:
>

> > eg using said calculator
> >
> >     log10(X) ~= (sqrt.sqrt.sqrt.sqrt.sqrt.sqrt.sqrt.sqrt.(X) - 1 ) x 889
>
> This can be re-written as:
>
>    log10(X) ~=  (X^(1/2^11) - 1) * 889
>
> (11 instead of 10 because of a subsequent RM post).
>
> I don't see any 'obvious' analytical way to prove this. However, it might
> be instructive to look at series expansions of log10 and X^Y and see if
> there's some way to correlate the coefficients of the terms in the
> expansions.

Actually, with a little fiddling around it's easy to see the relationship:


              log10(X) = (X^(1/2^11) - 1) * 889

      1/889 * log10(X) = (X^(1/2^11) - 1)

  1/889 * ln(X)/ln(10) = (X^(1/2048) - 1)

    889 * ln(10) = 2.3025 * 889 = 2047  <=== !!

          ln(X) / 2047 = (X^(1/2048) - 1)

       ln(X ^ 1/2047)  = X^(1/2048) - 1

and a series expansion for ln(z) is

  ln(z) = (z-1) - 1/2 * (z-1)^2  + 1/3 * (z-1)^3 -+ ...

 for abs(z-1) <= 1 and z != 0

(see Abramowitz and Stegun eq. 4.1.26)

In our case, z = X^1/2047. For reasonably small values for X, z will be
extremely close to 1. Thus z-1 will be extremely close to 0. Thus all high
order terms in the series expansion can be dropped!

The approximation can be made even more accurate if we use

   2048/ln(10) = 889.43

instead of 889

Scott

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