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pcbouhid
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Even and odd
« on: Jan 26th, 2006, 10:43am » |
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Let [z] mean the greatest integer less than or equal to z.
Find a positive real number X, such that [X^n] is an even number whenever n is even, and [X^n] is an odd number whenever n is odd.
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towr
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Re: Even and odd
« Reply #1 on: Jan 26th, 2006, 11:06am » |
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I think you could solve any appropriate recurrence where
f(0) is even, f(1) is odd, and f(n) = a f(n-1) + b f(n-2) with a even and b odd
If you select the parameters with care carefull, you'll have f(n) = c1 l1n + c2 l2n with one of the terms being negligible.
I'll pick 5+sqrt(26) if you want a definite answer.
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| « Last Edit: Jan 26th, 2006, 11:14am by towr » |
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towr
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Re: Even and odd
« Reply #2 on: Jan 26th, 2006, 3:09pm » |
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I'm not so sure about the number I gave now. Seems I overlooked a few things.
Maybe I'll try again later.
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Eigenray
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Re: Even and odd
« Reply #3 on: Jan 26th, 2006, 7:23pm » |
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I remember thinking about this very problem before (was it on this site?). I'm not sure, but I don't think I ever got a solution that way -- it seems like it always does the opposite of what you want it to do. If you do get it to work out, please share.
But one can prove such an X exists. Start with the equation 3 < X < 4. Inductively, suppose n>1 and we are given the equation
3n < kn < Xn < kn+1. (*n)
Now, if f(x) = x(n+1)/n, then f(kn+1)-f(kn) > f'(kn) > kn1/n > 3,
so the interval [f(kn), f(kn+1)] must contain 3 consecutive integers. Let kn+1 be the smallest integer in that interval of the same parity as n+1, namely,
kn+1 = 2 ceil[f(kn)/2], if n+1 is even,
= 2 ceil[(f(kn)-1)/2] + 1, if n+1 is odd.
It follows that [kn+1, kn+1+1] is contained in [f(kn), f(kn+1)], i.e.,
(3n)1/n < kn1/n < kn+11/(n+1) < (kn+1+1)1/(n+1) < (kn+1)1/n, (**)
which means that the equation
3n+1 < kn+1 < Xn+1 < kn+1+1 (*n+1)
implies (*n).
Now define X=3.21124734... to be the limit of the sequence kn1/n. Equation (**) shows this limit exists, and that for all n, (*n) is satisfied:
[Xn] = kn == n mod 2.
[Editted typo in (*n)]
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| « Last Edit: Jan 29th, 2006, 5:54pm by Eigenray » |
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towr
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Re: Even and odd
« Reply #4 on: Jan 29th, 2006, 11:47am » |
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on Jan 26th, 2006, 7:23pm, Eigenray wrote:| I'm not sure, but I don't think I ever got a solution that way -- it seems like it always does the opposite of what you want it to do. If you do get it to work out, please share. |
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It turns out to be impossible that way. 
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pcbouhid
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Re: Even and odd
« Reply #5 on: Jan 30th, 2006, 6:48am » |
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The solution I have:
One example is ½(3+√17). How do we get this?
Consider the binomial equation x²-ax-b=0, with a odd, b even, 0 < b < a.
Let u be the positive root, and v be the negative root.
Create the sequence S as S(n) = u^n + v^n.
We have that S(1) = a and S(2) = a² + 2b
Notice that S(1) is odd and S(2) is odd. Now note that a = u+v and b = -uv, giving S(n) = a*S(n-1) + b*S(n-2).
This implies that S(n) is odd for all n > 0.
Next, by the nature of how a and b were restricted, we see that:
-1 < v < 0, because v=-b/u, and b < a < u.
Now, [u^n] = [S(n) – v^n],
so [u^n] = S(n) when n is odd and
[u^n] = S(n)-1 when n is even.
Therefore, u has the desired property.
As an example, pick a=3, b=2, u = ½(3+√17)
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Barukh
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Re: Even and odd
« Reply #6 on: Jan 30th, 2006, 9:02am » |
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Excellent! Straight from the Book! 
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towr
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Re: Even and odd
« Reply #7 on: Jan 30th, 2006, 9:11am » |
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Quote:| This implies that S(n) is odd for all n > 0. |
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Doh.. no wonder it went totally wrong for me. And it's so obvious when you say it..
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