Poly math geniuses--Jules needs your help!

[KrazyHorse]

Duh. Did you think nobody else was getting the pun?

[KrazyHorse]

I did enjoy how you got UR, though...

[JohnT]

Originally posted by KrazyHorse
Duh. Did you think nobody else was getting the pun?

Perhaps I misunderstood what you were saying/to whom you were saying it to in the post above mine. My apologies.

[KrazyHorse]

I was answering the original question. I skimmed over the UR/you banter...

[JohnT]

Oh, I just ignored all that math stuff in my attempt to fill this thread with some tasty spam.

[Urban Ranger]

Originally posted by JohnT


The fact that you felt you had to specify what field they are knowledgable in automatically disqualifies them from Polymath status. Good going, UR!

Yeah, but an expert at polymath is not what you think it meant. So there

[Jules]

Originally posted by JohnT
Oh, I just ignored all that math stuff in my attempt to fill this thread with some tasty spam.

I thank you for keeping the thread bumped at least.

*Jules drops the 50-ton weight on JohnT*


But unfortunately I'm no closer to solving the problem.

[KrazyHorse]

Use ****ing induction like I told you to.

Assume it's true for the t case and use to prove for t+1

The only trickey thing to realize is that your inductive hypothesis for the stayer has to take into account that the movers he's seen started life at 2^(t-1) and beyond

After that it's simply a matter of the union of two sets.

You've covered proofs by induction, right?

[Urban Ranger]

I think he is having problems with the non-trivial case.

[KrazyHorse]

Which non-trivial case? There is none. The trading histories are obvious for all t due to the nice property that the sum of 2^n from n=0 to t is equal to 2^(t+1)-1

If the number of steps moved was not 2^(t-1) but was instead some other power then it would be slightly trickier. If it was a more complicated function of t then it would border on the unsolvable.

[Urban Ranger]

2^(t-1) is not the number of steps moved, but the step size at t.

[Jules]

Originally posted by KrazyHorse
Use ****ing induction like I told you to.

Assume it's true for the t case and use to prove for t+1

The only trickey thing to realize is that your inductive hypothesis for the stayer has to take into account that the movers he's seen started life at 2^(t-1) and beyond

After that it's simply a matter of the union of two sets.

You've covered proofs by induction, right?

Yeah, I know I have to use proof by induction, but it's been a looooong time since I've done one of those.

And, yeah, the more I stared at a particular numerical example, the more I think I see how to generalize it. Thanks.

[Jules]

Perhaps it is as simple as KrazyHorse suggested.

1) Assuming it is true for the t case, then mover j's set Qt is all movers and stayers indexed j who started life at {0, 1, 2, ..., 2^(t-1) - 1}.

2) But also at time t, there is another mover j at TP (2^t - 1) who began life at TP 2^(t-1). His set Qt contains all movers and stayers indexed j who started life at {2^(t-1), ..., 2^t - 1}, because we are assuming that the rule holds for the t case.

3) Then in t+1, our original mover j's set Qt+1 is just
{0, 1, 2, ..., 2^(t-1) - 1} U {2^(t-1), ..., 2^t - 1} = {0, 1, 2, ..., 2^t - 1}.

So it holds for the t+1 case. QED?


EDIT: I think the key thing to recognize in going from 2) to 3) is that the set Qt for mover j who begins life at TP 2^(t-1) is identical to the set Qt for stayer j at TP 2^t - 1.

[MrWhereItsAt]

Design an experiment to test it.

[Jules]

Do I get to use Poly posters as guinea pigs?



Ok, taking applications for movers and stayers here in this thread! What shall we use as the consumption good? How about beer?! Now all we need is a really, really, really long straight line!