Now i remember this. I came across this by wondering what it would be like
if there were decimal points in binary. thus to come up for a series to
devide by some number, say 9 for instance, just do long division in binary
of 9 into 1:
       .0001100011...
1001) 1.0000000000
      -1 001
         1100
        -1001
           10000
           -1001
             110....

which coresponds to 1/9=1/16 + 1/32 + 1/512 + 1/1024...
neat stuff!

------------------------------------------------------------------------------
A member of the PI-100 Club:
3.1415926535897932384626433832795028841971693993751
058209749445923078164062862089986280348253421170679

On Sat, 24 Jul 1999, Dave Bell wrote:

> Bob Drzyzgula <pic@DRZYZGULA.ORG> wrote:
>
> >I was fooling around with this today (avoiding fixing
> >the oven, I'm sure I'll have to pay for that soon),
> >and I found an interesting twist on this series.
>
> >I had this gut feeling that the series Myke listed had
> >popped out of an Euler transformation somehow, and I was
> >fooling around with that. What I noticed, however, was that
> >if the Euler transformation was applied backward of what
> >I had first thought (and I'm not this smart, this is an
> >immediate consequence of a derivaton in R.W. Hamming's
> >book on Numerical Methods, section 12.6, on page 203 of
> >the Dover edition), that you can improve the convergence
> >of such a series while still working exclusively with
> >powers of two. In particular, the series
>
> >  1/2 * ( 1/4 + 1/16 + 1/64 + 1/256 + ... + 1/(4^n) + ... )
>
> >Also converges to 1/3, but at a much quicker rate.
> >I calculated the two series in Excel to demonstrate:
>
> >Index   Myke's Series   Powers of 1/4
> >1       0.5             0.25
> >2       0.25            0.3125
> >3       0.375           0.328125
> >4       0.3125          0.33203125
> >5       0.34375         0.333007813
> >6       0.328125        0.333251953
> >7       0.3359375       0.333312988
> >8       0.33203125      0.333328247
>
> ...  all of which amounts to multiplying the Numerator (N) by the binary
> expansion of 1/the Denominator (1/D). Note that the binary digits of 1/3
> are a repeating fraction, 0.01010101......, and that the positions with
> 1's correspond to the terms that are present in Bob's series. Take the
> same expansion of any fraction, and do likewise. In Myle's original case,
> 1/7 = 0.0010010010......
>
> Dave
>