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Topic: Square and Sum the Odd Tans (Read 469 times) |
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TenaliRaman
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Square and Sum the Odd Tans
« on: Dec 3rd, 2003, 11:33am » |
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Prove tan2(1o) + tan2(3o) + tan2(5o) + .... + tan2(89o) is an integer.
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« Last Edit: Dec 3rd, 2003, 11:35am by TenaliRaman » |
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Eigenray
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Re: Square and Sum the Odd Tans
« Reply #1 on: Dec 3rd, 2003, 1:51pm » |
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You could explicitly construct a polynomial whose roots are the 45 terms in the sum, but I don't know if I'd call that "easy," so maybe there's a simpler way.
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TenaliRaman
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Re: Square and Sum the Odd Tans
« Reply #2 on: Dec 3rd, 2003, 10:11pm » |
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you have the right approach and there is an easier way to go about. Hint: (quite vague)ask De Moivre for help
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Eigenray
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Re: Square and Sum the Odd Tans
« Reply #3 on: Dec 4th, 2003, 11:55pm » |
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If t = (2n+1)o, then 0 = cos(90t) = Re[ ei90t ] = Re[ (cos t + isin t)90 ] = [sum]k=045 (90C2k)(-1)kcos90-2kt sin2kt. Dividing by cos90t shows that x = tan2t is a root of the polynomial f(x) = [sum]k=045 (90C2k)(-1)kxk = 0. Since f can have at most 45 roots, its roots are exactly {tan2(2n+1)o, n=0..44}, which therefore sum to 90C2 = 4005, an integer.
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TenaliRaman
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Re: Square and Sum the Odd Tans
« Reply #4 on: Dec 5th, 2003, 5:36am » |
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Awesome!!!
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Barukh
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Re: Square and Sum the Odd Tans
« Reply #5 on: Dec 5th, 2003, 6:35am » |
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I agree with TenaliRaman, it's absolutely brilliant! Eigenray, could you please write a bit about how did you arrive at this solution? TenaliRaman, do you think this problem belongs to easy section?
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TenaliRaman
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Re: Square and Sum the Odd Tans
« Reply #6 on: Dec 5th, 2003, 11:31am » |
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I think it does belong to easy section, doesn't it? i mean if i could solve it in one and a half days, it should be easy.
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