Abacus

I adjusted the indices. I let n=0 be for the first cycle after the first bounce, so the nth term for a bounce height is 2^-(n+1)= 2^-1 * 2^-n = 1/2*2^-n.

r^(n/2) = r^(1/2) * r^n

So sum(n = 0 -> inf) ar^(n/2) = r^(1/2)*sum(n = 0 -> inf) ar^n

At least I think so.

Lobstrosity

Nononono, r^(n/2) = (r^n)^(1/2) = (r^(1/2))^n

r^(1/2) * r^n = r^(n + 1/2)

the two are very different

Abacus

Shoot. You're right. I guess 2.61 is not the answer.

Abacus

Forget this.

Let's try it again.

r^(n/2) = (r^(1/2))^n

Let rp = r^(1/2) (rp is short for r prime)

So you're right Lobstrosity. r would need to be non-negative for rp < 1, which is fine.

So r^(n/2) = rp^n

sum(n = 0 -> inf) ar^(n/2) = sum(n = 0 -> inf) a*rp^n
= a/(1-rp)

*This* is actually what I did. See anything wrong?

Lobstrosity

Damn, why didn't I think of that? That looks fine, so your sum is simply going to be some constant (2/sqrt(5)) times 1/(1 - sqrt(1/2)) and then you simply need to subtract off the extra initial half-bounce you've unnecessarily added in (1/sqrt(5)). Ok, so I think my answer would be 2.61 s. I guess you were right all along, although it would seem that you might have gotten lucky somewhere along the way? I wonder why breaking it up into two series didn't work...perhaps I just messed up the calculation.

For completeness, the exact answer seems to be:

(3 + 4 * sqrt(1/2)) / sqrt(5) ~ 2.61

which is remarkably close to what I calculated originally and which is what I should have calculated originally had I remembered to distribute the four. Damn careless errors!

Anyway, thank you for the very interesting problem, Abacus! It's like Zeno's paradox updated for today's modern ball-bouncing society.

Abacus

When I first worked it out, I did the series in r^(n/2) in my head correctly without writing out the steps. When I tried to work out the steps explicitly in one of my posts on here, I initially did it wrong. Go figure.

I guess I am lucky.

Lobstrosity

Nah, if you did it right the first time because you knew what you were doing, that's not luck. I can see messing it up when trying to be detailed. It seems a bit far-fetched that you could make such a drastic error as r^(n/2) = r^n*r^(1/2) and still get the right answer. It appears we both had the right idea, yours was just infinitely more elegant.

Abacus

For what it's worth, here's the general formula I worked out:

T = sqrt(2*h/g)*(1+2*r^0.5/(1-r^0.5))

where in my example, h=1, g=10, and r=0.5

And that concludes today's brain teaser. Thanks for playing.

Abacus

Yeah, I'm a bit embarrased by that. I generally pride myself on being pretty good at algebra. That's a beginner's error. But I guess I'm only human, afterall.

Abacus

Actually, come to think of it, the ball must come to a rest. Yes, in this ideal case, it bounces an infinite number of times, but we know that all of those bounces must occur before 2.61 seconds. After 2.61 seconds, there is no more bouncing.

Wrap your mind around that.