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Topic: hard (Read 712 times) |
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pex
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Re: hard
« Reply #1 on: Aug 23rd, 2007, 9:26am » |
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Hard? Not at all.
More generally, let us prove for any positive integer n that
sqrt(2)sin[(1/2)arcsin{sqrt(1/n2 + (n-1)/n3 + (n-1)/n4)}] = 1/n.
We divide by sqrt(2), take the arcsin, multiply by 2, and simplify the expressions of which arcsins are taken:
arcsin(sqrt((2n2-1)/n4)) = 2*arcsin(sqrt(1/(2n2))).
Recall sin(2x) = 2sin(x)cos(x). Thus,
sqrt((2n2-1)/n4) = 2 * sqrt(1/(2n2)) * sqrt(1-1/(2n2)),
since we are working in the first quadrant. Squaring,
(2n2-1)/n4 = 4 * 1/(2n2) * (2n2-1)/(2n2),
which is easily verified. Plugging in n = 2006 gives the desired result.
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| « Last Edit: Aug 23rd, 2007, 9:30am by pex » |
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Sameer
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Pie = pi * e
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Re: hard
« Reply #2 on: Aug 23rd, 2007, 9:38am » |
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tony is incorrigible 
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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pex
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Re: hard
« Reply #3 on: Aug 23rd, 2007, 9:48am » |
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on Aug 23rd, 2007, 9:38am, Sameer wrote:| tony is incorrigible |
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Well, this is an old thread. The first request for more descriptive titles I could find (Aryabhatta's) dates from almost a full day after this puzzle was posted...
Edit: No! I found an earlier one here - made by yourself, which is probably not that much of a coincidence. So yes, he is (or at least was) incorrigible...
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| « Last Edit: Aug 23rd, 2007, 10:16am by pex » |
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Barukh
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Re: hard
« Reply #4 on: Aug 23rd, 2007, 11:48am » |
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Nicely done, pex! 
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