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Icarus
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Powers approximating integers.
« on: Mar 12th, 2006, 3:29pm » |
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The golden ratio  = 1.6180..., has an unusual property: (n --> 0) mod 1 as n -->  .
Which real numbers have this property?
Of course, the Integers themselves do, and almost as trivially, all x with |x| < 1 satisfy the condition.
What other numbers have approximately integral powers?
[ I have only a partial answer for this. ]
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| « Last Edit: Mar 12th, 2006, 3:31pm by Icarus » |
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towr
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Re: Powers approximating integers.
« Reply #1 on: Mar 13th, 2006, 12:37am » |
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There's a large range of numbers that can easily be identified.
Namely if you have x and x', both roots of a the same quadratic equation, such that |x'| < 1 and xn + x'n is integer. Of which we can generate a good number from second order recurrences. (Because, obviously x'n goes to 0, so xn must converge to an integer)
One might ask whether all numbers fall under this, but that's probably not the case. I'd imagine you could have a third order recurrence that gives (nonzero) y, y', and y'' such that the latter two are smaller in magnitude than 1, and yn + y'n + y''n being integer. (And similarly for higher order recurrences)
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Icarus
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Re: Powers approximating integers.
« Reply #2 on: Mar 13th, 2006, 3:40pm » |
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That is, effectively, my partial answer. Algebraic integers all of whose conjugates have magnitude < 1 should satisfy the property.
I'm curious if there is another approach that might shine more light on the matter though. And what about transcendental numbers? Can any of them satisfy the property? I'm not even examined the question of whether or not any rationals might work.
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Barukh
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Re: Powers approximating integers.
« Reply #3 on: Mar 14th, 2006, 12:22am » |
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Icarus, I think the following recent paper may partially answer one of your questions.
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| « Last Edit: Mar 14th, 2006, 8:28am by Barukh » |
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Icarus
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Re: Powers approximating integers.
« Reply #4 on: Mar 14th, 2006, 5:31am » |
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Possibly - but I'm too chintzy to spring for the $30 it would cost to find out for sure.
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"Pi goes on and on and on ...
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I wonder: Which is larger
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towr
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Re: Powers approximating integers.
« Reply #5 on: Mar 14th, 2006, 5:54am » |
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You need to pay to see it?
I'll send you a copy
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Icarus
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Re: Powers approximating integers.
« Reply #6 on: Mar 14th, 2006, 3:06pm » |
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The link takes me to a log-in screen with a second panel telling me that if I don't have a log-in, I can purchase the article for $30. There is a link to the abstract, so I can read that, but the article itself requires payment.
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Eigenray
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Re: Powers approximating integers.
« Reply #7 on: Mar 15th, 2006, 9:51pm » |
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on Mar 13th, 2006, 3:40pm, Icarus wrote:| That is, effectively, my partial answer. Algebraic integers all of whose conjugates have magnitude < 1 should satisfy the property. |
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These are known as Pisot Numbers, and they indeed satisfy the property. For, the sum of the n-th powers of all conjugates of any algebraic integer x is always an integer: it can be expressed as a polynomial (with coefficients in Z) in the elementary symmetric polynomials (cf. e.g. Newton's Identities), which are the coefficients of the minimal polynomial of x, and therefore integers.
As a partial converse, these are the only such algebraic x>1. Let ||x|| denote the distance from x to the nearest integer. I follow an argument in Algebraic Numbers and Harmonic Analysis, by Yves Meyer.
Theorem: If x>1 is algebraic, and ||xk|| -> 0, then x is Pisot: it is an algebraic integer, whose other conjugates are all < 1 in magnitude.
Proof (sketch): Write
xk = pk + rk,
where pk is an integer, and rk -> 0. If
P(t) = cn tn + cn-1tn-1 + . . . + c0 = cn(t-x1)(t-x2)...(t-xn)
is the minimal polynomial of x=x1 (note the {xi} are distinct, Q being perfect), with integer coefficients, then xk*P(x) = 0 gives
cnrk+n + ... + c0rk = -cnpk+n - ... - c0pk.
Since the LHS goes to 0, but the RHS is an integer, it follows that the above quantity is identically 0 for k>K sufficiently large. That is, {rk} satisfies a linear recurrence with characteristic polynomial P, and therefore we can write
rk = a1x1k + . . . + an xnk, k>K,
for some complex numbers ai. It is left as an exercise to the reader to show that
rk -> 0 implies that ai=0 whenever |xi| >1.
In particular, a1=0. Renumbering if necessary, we may also assume, for some 2<m<n, that
rk = a2x2k + ... + amxmk for k>K,
where a2,...,am are non-zero, and hence |xi|<1 for 2<i<m. Now, the power series
[sum]k>K pktk = [sum] (xk-rk)tk
= xKtK/(1-xt) - a2x2KtK/(1-x2t) - . . . - amxmKtK/(1-xmt)
= F(t)/G(t),
where G(t) = (1-tx1)...(1-txm), and F is a polynomial relatively prime to G. Since F/G has integer coefficients, and G(0)=1, it follows from a "well-known result" (Meyer calls it Fatou's lemma) that in fact G has integer coefficients. Then tmG(1/t) is a monic integral polynomial of degree m, with x as a root. It follows that m=n, and that x is Pisot.
Quote:| I'm curious if there is another approach that might shine more light on the matter though. And what about transcendental numbers? Can any of them satisfy the property? |
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As far as I know, this is unknown, but conjectured to be "no". What is known is the following (again from Meyer):
If x>1, and c>0 is such that [sum] ||c xn||2 is finite, then x is Pisot.
Quote:| I'm not even examined the question of whether or not any rationals might work. |
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The case of x rational is covered by the above Theorem (and has come up before), but the proof can be simplified. For, if x > 1 is rational, then arguing as above, writing xk = pk + rk, and taking n=1, we find simply rk = a xk for k sufficiently large. Since this goes to 0, either a=0, in which case x is an integer, or |x|<1.
Exercise: For any integers a>2, and n>1, there is a Pisot number x satisfying xn(x-a)=1. Hence a is a limit point of the set of Pisots.
In fact, it can be shown that the set of Pisots is closed. Also of note: if K is a number field of degree n, discriminant d, with r real embeddings and 2s complex ones, then any real interval of length T contains
2r-1[pi]s/|d| T + o(T)
Pisot numbers of degree n, lying in K.
But as to what Pisot numbers have to do with harmonic analysis, I really don't know enough to say.
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