Steve Hardy wrote: > > > From: James Musselman > > > > Scott Dattalo wrote: > > > > > 1/10 = 0.00011001100110011001100110011001100 ..... (base 2) > > > > > > This can be rewritten to emphasize the optimization: > > > 1/10 = 2 * (3/16 + 3/256 + 3/4096 + ... + 3/(2^(4*m)) ) > > > > Isn't the above line wrong? Isn't 2 * 3/16 greater than 1/10? Or am I > missing > > something? > > James > > > > Scott meant the initial coefficient to be 1/2 not 2. Scott > must have had an April 1 calculator since he also claimed that > > > > 1/10 = 6* [1/ (1 -1/16) - 1] = 6 ( 16/15 - 1) = 1/10 (The check is good!) The Derivative of r cubed divided by three. With egg dripping off his face he confesses, "I inadvertantly propogated an error." 8^( If you're really interested, here are the corrections for Rev 1.0001: This line is correct: 1/10 = 0.00011001100110011001100110011001100 ..... (base 2) This line was incorrect: > 1/10 = 2 * (3/16 + 3/256 + 3/4096 + ... + 3/(2^(4*m)) ) and should have been 1/10 = (3/16 + 3/256 + 3/4096 + ... + 3/(2^(4*m)) ) / 2 The summation is incorrect: > i->infinity > ___ > \ > 1/10 = 6*/__ 2^(-4*i) > i=1 And should have been i->infinity ___ 3 \ 1/10 = --- * /__ 2^(-4*i) 2 i=1 The **check** was incorrect: > 1/10 = 6* [1/ (1 -1/16) - 1] = 6 ( 16/15 - 1) = 1/10 (The check is good!) and should have been 1/10 = 3* [1/ (1 -1/16) - 1]/2 = 3 ( 16/15 - 1)/2 = 1/10 (The check is good!) ((I think)) Sorry folks... I guess people do read this stuff. So, as Steve points out, I inadvertantly was multiplying by two when I should have been dividing by two (thanks Steve). However the theory is sound. BTW Thomas Coonan wrote: (in Response to Andy's posting) > > So, what's your general problem-solving technique for this class of > problem? Are you applying some basic number theory tricks about rational > numbers, or is this trial-n-error? Yes. Scott PS The answer is: r * dr * r. Get it? Hardy-har-har. Sorry Steve, it's something I stole from the Simpsons.