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Chronos
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Feynman was stumped
« on: May 20th, 2003, 10:19am » |
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(scenario from Surely you're Joking, Mr. Feynman; puzzle by me)
While at Los Alamos, Richard Feynman bragged to his colleagues, "I can work out in sixty seconds the answer to any problem that anybody can state in ten seconds, to 10 percent.". Much to the impressment of all, he delivered on this boast... Until Paul Olum showed up, and asked for the tangent of ten to the hundredth (that's tan(10100)). Feynman, of course, was stumped, since that would require knowing pi to one hundred decimal places, and he admitted defeat.
However, he could have guessed, and thus have had at least a chance, however miniscule, of upholding his reputation. What would his best possible guess have been?
Hint number 1: There are two equally good answers for this one.
Hint number 2: This can be done either approximately or exactly.
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Leo Broukhis
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Re: Feynman was stumped
« Reply #1 on: May 22nd, 2003, 11:33am » |
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The best possible answer would have been 1 or -1, because they are
maxes of abs(atan(0.9*x)-atan(1.1*x)), that's what I can gather by looking at what gnuplot shows me.
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Chronos
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Re: Feynman was stumped
« Reply #2 on: May 25th, 2003, 11:26am » |
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That's almost right, given that 10% is small. But not quite.
Given that the answer is (almost) equal to 1, I suspect that there's a cleverer way to do this than the way I did.
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Leo Broukhis
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Re: Feynman was stumped
« Reply #3 on: May 26th, 2003, 7:42pm » |
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Actually, the function could be
abs(atan(x/0.9)-atan(x/1.1)), or
abs(atan(x/0.9)-atan(x*0.9)), or
abs(atan(x*1.1)-atan(x/1.1)),
depending on what he really meant by "to 10%"
By "a cleverer way" do you mean a quicker way to arrive at the answer "1 or -1", or something else?
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Chronos
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Re: Feynman was stumped
« Reply #4 on: May 29th, 2003, 4:00pm » |
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Hmm... I suppose that one should assume that "within 10%" means that Feynman's answer is between .9x and 1.1x, where x is the correct answer. In retrospect, though, I don't think that's what I assumed. I think that what I actually assumed is that the correct answer is between .9y and 1.1y, where y is Feynman's guess. This doesn't change the approximate answer (-+ 1), but it does change the exact answer, in a manner which I'll have to calculate further to determine.
And whenever the (approximate) answer is something as simple as 1, it seems like there should be a simple way to arrive at that (approximate) answer. But maybe not.
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