www.piclist.com/techref/method/math.htm?key=divide

Steve Hardy wrote:

{Quote hidden}

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.

See also: www.piclist.com/techref/method/math.htm?key=divide