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#11 | ||
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So sum(n = 0 -> inf) ar^(n/2) = r^(1/2)*sum(n = 0 -> inf) ar^n At least I think so. |
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#12 | |
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r^(1/2) * r^n = r^(n + 1/2) the two are very different |
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#13 | |
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#14 | |
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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? |
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#15 |
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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. |
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#16 | |
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I guess I am lucky. |
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#17 |
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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.
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#18 |
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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. |
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#19 | |
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#20 | |
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Wrap your mind around that. |
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