Wikipedia:Reference desk/Archives/Mathematics/2010 November 8

= November 8 =

Fibonacci-like Series
Consider the series $$\sum^{\infin}_{k=2} b^{-F_k}$$, which is the number in base b that has ones in its expansion at indices corresponding with the Fibonacci numbers. Is this number algebraic? Similarly, is the number $$\sum^{\infin}_{k=2} p^{F_k}$$ algebraic in the p-adic numbers? Black Carrot (talk) 01:34, 8 November 2010 (UTC)
 * The answer to the first question is "almost certainly not, but I very much doubt anyone has a proof". It should be the case that all algebraic irrationals are normal numbers to every base.  Basically that's because almost all real numbers are normal, so if a number is not normal then there should be a reason it isn't, and algebraic irrationals don't really know from positional number representations so they shouldn't have such a reason.  Given that there are only countably many algebraic irrationals, and the probability of any of them not being normal should be zero, the probability that there's even a single algebraic irrational that's not normal should also be zero.
 * But of course that's not a proof. --Trovatore (talk) 07:37, 8 November 2010 (UTC)
 * I don't believe in this one. Some rational numbers have a very good reason not to be normal, so why couldn't some algebraic irrational numbers also have a good reason?  On the other hand, I do think that this particular sum is not algebraic.  &#x2013; b_jonas 11:18, 8 November 2010 (UTC)
 * All rational numbers have a very good reason not to be normal. They have repeating expansions in every base. No such reason is apparent for algebraic irrationals.  There probably isn't one, and most likely all algebraic irrationals are in fact normal.
 * This is a case with a huge disconnect between what we basically know (in the sense of things that we'd say we know in, for example, physics) and what we can prove by the rules of mathematics. We basically know that all algebraic irrationals are normal to all bases.  But I am unaware that anyone has managed to prove that even one algebraic irrational is normal to even one base. --Trovatore (talk) 18:38, 8 November 2010 (UTC)
 * So if a number is normal with respect to all digits but 8, but its decimal expansion has no 8s in it, you're saying it's probably transcendental? Michael Hardy (talk) 16:56, 10 November 2010 (UTC)
 * Yes. --Trovatore (talk) 18:53, 10 November 2010 (UTC)
 * You might be interested in the article Liouville number, though I don't believe it applies in this case. I also suspect your first number is not algebraic in any base, but do not have a proof. A plausible generalization might replace F_k with nearestint(e^k) for suitable real e, in light of Binet's formula. 67.158.43.41 (talk) 11:55, 8 November 2010 (UTC)


 * This is not an answer, but still. It is easy to see that the two related sums $$\sum_{k\text{ even}} b^{-F_k}$$ and $$\sum_{k\text{ odd}} b^{-F_k}$$ both have approximation exponent ≥ φ2 > 2, hence they are both transcendental by Roth's theorem. This does not preclude their sum from being algebraic, but it seems quite unlikely.—Emil J. 15:15, 8 November 2010 (UTC)


 * At a first glance, my impression is that for any integer $$\scriptstyle b\geq2$$ the number $$\scriptstyle\beta:=\sum^{\infin}_{k=2} b^{-F_k}$$ is indeed trascendental, and that there should be an elementary proof, arguing on the "base b support" of the number $$\scriptstyle\beta$$ and of its powers. Here by "base b support" of a number x I mean the subset of $$\scriptstyle\Z$$ corresponding to the non-zero digits of x, e.g. for $$\scriptstyle\beta$$, it is the set $$\scriptstyle\mathcal{F}^1$$ of all Fibonacci numbers (negated). Note that $$\scriptstyle\beta^m=\sum_{k\geq 0} c(k,m)b^{-k}$$ where $$\scriptstyle c(k,m)$$ is the number of representations of k as a sum of m Fibonacci numbers, with possible repetitions, and distinguishing the order of summands (in general this is not a base b representation, since the coefficients may exceed b-1). However I think it should be true something like that:
 * i) for any m, c(k,m) is bounded by a constant C(m);
 * ii) for any m, the set $$\scriptstyle\mathcal{F}^{<m}$$ of all natural numbers that have a representation as a sum of less than m Fibonacci numbers has density 0 relative to the set $$\scriptstyle\mathcal{F}^{m}$$ of all natural numbers that have a representation as a sum of exactly m Fibonacci numbers, meaning that the number of elements of $$\scriptstyle\mathcal{F}^{<m}$$ less than x is little oh of the number of elements of $$\scriptstyle\mathcal{F}^{m}$$ less than x, as $$\scriptstyle x\to\infty.$$ Let $$\scriptstyle P(x)=ax^m-Q(x)\in \Z[x]$$ be a polynomial of degree m with integer coefficients and with $$Q(x)$$ of degree less than m.
 * As a consequence of (i), the base b support of $$a\beta^m$$  should be only a sligtly perturbation of the set $$\scriptstyle -\mathcal{F}^{m},$$ and by (ii) it could not be covered by the support of $$Q(\beta)$$, that would make it impossible for $$\beta$$ to be a root of $$P(x)$$. Though I think that the best proof is the one shown by EmilJ via Roth's theorem, I'd be glad to hear any comments or ideas on the above elementary approach. --pm a  23:37, 8 November 2010 (UTC)

A lot of interesting responses. Thank you, everyone. @pm a : I've given it some thought, and I can at least justify your first point. To be determined is a function C(m), such that c(k,m)<=C(m) for all k,m. The base case is c(k,1)<=C(1)=1. For other m, we can take permutations to fit inside a constant m! factor, and focus on combinations of summands, which we can also take to be ordered. Let the maximal such in a given sum be called Fn. Holding this fixed reduces the problem to finding the combinations of Fibonacci numbers no greater than Fn which sum to (k-Fn). This is bounded above by c(k-Fn,m-1)<=C(m-1). So, how many such maximal elements can there be for a given k? Let the greatest of these maximal elements be called Fa, and the least be called Fb. The greatest is bounded above by k, and the least is bounded below by k/m. So, Fa<=k<=mFb. Because these values are restricted to Fibonacci numbers, this leaves at most (1+floor(log_phi(m))) possible values of Fn. Putting everything together, we get that c(k,m) <= C(m) = m!C(m-1)(1+log(m)/log(phi)). Can anyone offer suggestions on the second point? It seems to be equivalent to the slightly simpler problem of proving that the density of Fm-1 is zero relative to Fm, since for a given m there are only finitely many classes less than it, each tiny compared to the next. Black Carrot (talk) 14:33, 12 November 2010 (UTC)

Quick query - integral tending to 0 on [0,1]
Hi there,

Could anyone quickly suggest a method I might use to show that $$lim_{n \to \infty} \,\int_0^1 \frac{n \cos{x}}{1+n^2x^{\frac{3}{2}}}=0$$ ? I've spent a long time staring at it and the only possible method I could think of was to say $$|cos(x)|\leq 1,\, \frac{1}{1+n^2x^{\frac{3}{2}}} \leq \frac{1}{1+(nx)^2} $$, but this only gives us a bound of $$ \frac{\pi}{2} $$ via arctan, which is not good enough unfortunately.

I am happy to do all the working myself if anyone could point me in the direction of an effective method! (Limit tests, basic measure theory etc. is fine with me, incidentally).

Thanks! 131.111.1.66 (talk) 15:01, 8 November 2010 (UTC)
 * For $$x\neq0$$, this is just $$\lim_{n\rightarrow\infty}\frac{an}{1+bn^2}$$ which is obviously 0 (since $$b\neq0$$). For $$x=0$$, it's $$\lim_{n\rightarrow\infty}n$$, which is rather not 0.  --Tardis (talk) 15:09, 8 November 2010 (UTC)
 * I suspect that the OP didn't write properly what they actually want, since they talk about integrals, which there aren't any in this expression.—Emil J. 15:18, 8 November 2010 (UTC)
 * Sorry! Corrected now, I'm an idiot :-P 131.111.1.66 (talk) 16:15, 8 November 2010 (UTC)
 * OK, trying again (and assuming a dx in the integral): dropping the cosine factor for simplicity, we have $$\int_0^1f_n(x)\,dx=\int_0^{x_1}f_n(x)\,dx+\int_{x_1}^{x_2}f_n(x)\,dx+\int_{x_2}^1f_n(x)\,dx$$ where $$x_1=n^{-1-\varepsilon_1}$$ and $$x_2=n^{-2/3+\varepsilon_2}$$. As $$n\rightarrow\infty$$, the first term goes to zero because $$f_n(x)\le n$$ so $$f_n(x)x_1\le n^{-\varepsilon_1}$$ which vanishes, the second goes to zero because on that interval $$f_n(x)\le f_n(x_1)=\frac n{1+n^{2+(3/2)(-1-\varepsilon_1)}}=\frac n{1+n^{1/2-3\varepsilon_1/2}}\le n^{1/2+3\varepsilon_1/2}$$ so $$f_n(x)(x_2-x_1)\le f_n(x)x_2\le n^{-1/6+\varepsilon_2+3\varepsilon_1/2}$$ which vanishes for suitably small $$\varepsilon_1,\,\varepsilon_2$$, and the third goes to zero because on that interval $$f_n(x)\le f_n(x_2)=\frac n{1+n^{1+3\varepsilon_2/2}}\le n^{-3\varepsilon_2/2}$$ which vanishes (and the interval length is bounded).  --Tardis (talk) 19:34, 8 November 2010 (UTC)

when will math end?
what is the most likely avenue for demise of the entire field of mathematics as we know it today and when is it most likely to take place? My impression is that it must be empirical evidence (not proof, of course =) ) that any axiomatic system is inconsistent, and it seems to me this is most likely to take place in the next 150 years. I would place a 90% chance on this happening.  But this is just my impression.  What are the facts, referenced ideas on this subject?  Thank you. 84.153.236.235 (talk) 18:42, 8 November 2010 (UTC)
 * Take a look at Gödel's incompleteness theorems, and the references linked at our consistency article to learn about formal mathematical consistency. At the reference desk, we will not speculate about future developments in mathematics; that means we won't delve into guessing about "the demise" of mathematics (whatever that even means).  Nimur (talk) 18:51, 8 November 2010 (UTC)
 * My guess would be because life has ended and the timescale would be some thousands maybe millions of millions of years. Hopefully evolution and technical fixes would have made people better at it by then. Dmcq (talk) 19:33, 8 November 2010 (UTC)
 * Agreed with Dmcq, the demise of humanity is by far the most likely scenario for math ending. But I think it's very likely we will destroy ourselves this century.
 * A discovery like what you describe, if indeed it takes place, may change some views in philosophy of mathematics, but it will not change how mainstream math is practiced.
 * I'd say that the probability that humanity as we know it will continue, but math as we know it will end this millennium, is about 0.00000001% (though I am in no way qualified to make such estimates). -- Meni Rosenfeld (talk) 20:00, 8 November 2010 (UTC)
 * I'd nitpick on Dmcq's and Meni Rosenfeld's answer on account of the fact that the study of mathematics has only been able to pick up due to the luxury of humanity's ability to focus on the mental exercises of math, sciences, and the arts, instead of the constant need to survive a harsh environment. It would not take the complete demise of humanity to end math as we know it; only the end of modern civilizations. --COVIZAPIBETEFOKY (talk) 21:31, 8 November 2010 (UTC)
 * Depending on what you mean by "as we know it", the "field of mathematics as we know it today" could end abruptly if a strong AI were developed which ordinary humans couldn't keep up with, effectively ending humanity's direct participation in mathematical discovery. Math would certainly exist and continue after that, but not really "as we know it". Staecker (talk) 00:53, 9 November 2010 (UTC)
 * If humans can get along with each other and the planet well enough, I dont think math will ever end. I recall Max Born once saying "physics, as we know it, will end in 2 weeks" only to find out the explosive development of new stuff aft that. I think, as new theories are developed to answer current problems, new problems arise from the new theories at a rate thats faster than the rate that current problems are being solved. Money is tight (talk) 14:26, 9 November 2010 (UTC)

Every time a theorem is proved, a part of mathematics dies. Count Iblis (talk) 23:59, 9 November 2010 (UTC)

I'm inclined to say that it won't even within Staecker's not-too-unlikely scenario. Human mathematics could potentially even leap forward with such a strong AI, as there would likely be a need to try to keep up the best we could from a variety of different motivations. As far as the demise of humanity is concerned, while the possibility of WMDs eliminating humans entirely is a posibility nobody can attach a reasonable probability to, a more likely scenario for the demise of mathematics would be a permanently legally enforced ludditism toward human abstract thought founded on the premise that it could only do harm once AI reigns superior. I think this is unlikely though, even if humanity goes through a prolonged degeneracy.Julzes (talk) 15:17, 13 November 2010 (UTC)