## Wednesday, February 7, 2007

### Solve this operation.

My co-worker and I are working on a large equation and we're having trouble resolving this last piece. Can you figure it out?

Anonymous said...

Are you sure you don't mean "1" to infinity? Because indexing things from "i" to infinity doesn't make any sense. If it's just 1 to infinity, then that's just the geometeric series (a + ar + ar^2 + ar^3 +...) - 1 where a=1 and r=1/2. In this case the sum is 1/(1-r) - 1, or 1/(1/2) - 1 = 2 - 1 = 1.

If "i" was supposed to be some finite number, in which case you just want to find the sum from that point onwards, then you'd take 1 and subtract the partial sum from 1 to i-1 (or even easier, start with 2 and subtract the sum from 0 to i-1). This other sum would just be another geometric series, albeit a finite one. So the answer would be 2 - (1 - (1/2)^i)/(1 - 1/2), or 2 - 2(1 - (1/2)^i).

Kevin said...

I'm assuming that "i" is an imaginary number (i.e. the negative power of 2).

If n=0 we've figured that the solution is a number that approaches 2.

We cannot figure how this operation changes when the low range is "i".

Bond said...

Ummm Kevin.. my head hurts just from looking at it...

LOL

Hope you figure it out

Finally had time during my trip to do a little "walk around the blogs"