User john bentin - MathOverflow

Which compositions have these sum-like and product-like properties on the positive reals?

2013-04-01T11:24:12Z

Consider a binary composition $\star:\Bbb R^2_{>0}\rightarrow \Bbb R_{>0}:(x,y)\mapsto x\star y$ with the following properties.

(Commutativity)$\quad x\star y=y\star x\;$for all $x,y\in\Bbb R_{>0};$

(Associativity)$\quad(x\star y)\star z=x\star(y\star z)\;$for all $x,y,z\in\Bbb R_{>0};$

(Continuity)$\quad x\mapsto a\star x\;(x\in\Bbb R_{>0})\;$is continuous, for each $a\in \Bbb R_{>0};$

(Upper monotony)$\quad x\mapsto a\star x\;(x\in\Bbb R_{>0})\;$is strictly increasing, for each $a\in \Bbb R_{>0};$

(Surjectivity)$\quad \star\;$is unbounded above and below in $\Bbb R_{>0}.$

These properties hold for addition and multiplication in $\Bbb R_{>0}.$ Do they hold for any other composition in $\Bbb R_{>0}$? If so, is an example of such a composition known? If not, is the above list redundant---that is, can any of the five conditions be weakened or removed?

Which real scalings of the natural numbers approximately accommodate the unbounded powers of a noninteger?

2013-03-24T18:43:53Z

That is, what are the possible values of a real number $\lambda$ for which there exists a nonintegral real $\alpha >1$ such that, given any $\varepsilon >0,$ all but finitely many powers of $\alpha$ lie within $\varepsilon$ of an integral multiple of $\lambda$? As an example, we may take $\lambda =\sqrt 5$ with $\alpha =\frac 12+{\frac 12} \sqrt5.$

Formally, define$$\begin{align}\Lambda &:=\\\{\lambda &\in \Bbb R_{>0}: (\exists\,\alpha\in \Bbb R_{>1}{\setminus}\Bbb N)(\forall\,\varepsilon\in\Bbb R_{>0})(\exists\,k\in\Bbb N)(\forall\, n\in\Bbb N_{>k})(\exists\,m\in\Bbb N)\;\; |\alpha^n-\lambda m|< \varepsilon\}.\end{align}$$What are the known properties of $\Lambda$? In particular, can $\Lambda$ possess a rational or a transcendental element?

How long can a primal egyptian fraction be, that optimally approaches unity?

2012-11-01T16:40:41Z

Thus, do there exist $n$ distinct primes whose summed reciprocals fall short of $1$ by the reciprocal of their product, for some $n\geqslant6$? I can get as far as $n=5$: $$\dfrac{1}{2}=1-\dfrac{1}{2},$$ $$\dfrac{1}{2}+\dfrac{1}{3}=1-\dfrac{1}{2\cdot3},$$ $$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7}=1-\dfrac{1}{2\cdot3\cdot7},$$ $$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{43}=1-\dfrac{1}{2\cdot3\cdot7\cdot43},$$ $$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{11}+\dfrac{1}{23}+\dfrac{1}{31}=1-\dfrac{1}{2\cdot3\cdot11\cdot23\cdot31}.$$ But I can't see the way beyond that. If there is a fraction for some $n\geqslant6$, then one may naturally ask: is there a bound on such $n$?

Answer by John Bentin for I know that you know...

2012-11-24T08:07:05Z

The article http://plato.stanford.edu/entries/common-knowledge/ is more extensive than wikipedia's. However, it doesn't deal with the issue of infinite information.

Does strict order-preservation of powerset curtail the candidates for violation of CH?

2012-11-19T22:43:36Z

Thus, let $\mathrm{OPP}$ be the axiom that $|A|\lt|B| \Rightarrow |2^A|\lt|2^B|$ for any sets $A$ and $B$; and, for any ordinal $\alpha$, let $\mathrm{CH}_\alpha$ be the hypothesis that $\aleph_\alpha=\frak c$ (so that $\mathrm{CH}_1=\mathrm{CH}$ ) . Define $S$ to be the set of those ordinals $\alpha\in\frak c$ such that $\mathrm{CH}_\alpha$ does not provably (within $\mathrm{ZFC}$ ) violate $\mathrm{ZFC}$ (for example, it is known that $\omega\backslash${ $0$ }$\subseteq S$ and $\omega \notin S$); and let $S'$ be the set of those $\alpha\in\frak c$ such that $\mathrm{CH}_\alpha$ does not provably (within $\mathrm{ZFC}$ ) violate $\mathrm{ZFC}$ $\&$ $\mathrm{OPP}$ . Clearly $S'\subseteq S$. But is $S'= S$ ? Or are any elements of $S$ known to be not in $S'$ ?

My guess is that $\mathrm{OPP}$ can't restrict the possibilities for violations of $\mathrm{CH}$ because the sets it talks about in the consequent---especially $2^B$--- are too big to be relevant; but I'm not sure of my footing here.

What is known about the polynomial factorization of power series?

2012-08-13T10:01:21Z

Some power series factorize; $1+\sum_{n=1}^\infty x^n=\prod_{n=1}^\infty (1+x^{2^n})$ and $1+\sum_{n=1}^\infty x^{2n}/(2n+1)!=\prod_{x=1}^\infty (1+x^2/n^2\pi^2)$ for example; while others do not----in particular, $1+\sum_{n=1}^\infty x^n/n!$. Specifically the question is: what is known about necessary and sufficient conditions on the coefficients of the power series $1+\sum_{n=1}^\infty a_nx^n$ (assuming a positive radius of convergence) for it to factorize as $\prod_{n=1}^\infty (1+ b_nx+c_nx^2)$, where the coefficients $a_n$, $b_n$, and $c_n$ are real constants?

Remarks The Weierstrass factorization theorem doesn't directly answer the question, since here we allow Taylor series of non-entire functions (e.g. the first example above) and exclude non-polynomial factors. The roots of the question are algebraic: loosely stated, under what conditions can the fundamental theorem of algebra be pushed to infinity? A positive radius of convergence is supposed because formal power series with no functional meaning make me uncomfortable, and admitting them might complicate the answer. I previously posted an unanswered version of this question on Math.StackExchange.

Edit Thanks to juan and Douglas Zare for showing my error in the preamble of the question by citing the interesting factorization of the exponential series by Gingold et al. As Aaron Meyerowitz indicates, restriction to linear and quadratic factors leads respectively to quite different situations. Alexandre Eremenko answered the linear case. Allowing polynomial factors of unrestricted degree opens a wide vista. The rather simple power series $1$ includes among its polynomial product representations, valid for $|x| \lt r_0$, such forms as $$\left(1-\dfrac{p(x)}{r} \right) \prod_{n=0}^\infty \left[1+\left(\dfrac{p(x)}{r} \right)^{2^n} \right],$$ where $r_0$ is an arbitrary positive constant, $p(x)$ is an arbitrary polynomial, and $r$ is any real constant strictly exceeding $|p(x)|$ whenever $|x|\leqslant r_0$. Myriad expansions like these may be woven into any polynomial product expansion of any power series. I find this vista daunting and so will stick to the specified quadratic case, which remains unanswered so far.

Answer by John Bentin for Do approximately the same polynomials have approximately the same roots?

2012-08-03T16:18:30Z

For the headline question, no. See, for example, Wilkinson's polynomial: http://en.wikipedia.org/wiki/Wilkinson's_polynomial.

Edit: From the response to this answer, it seems that some expansion is needed. The OP asks two questions. In the the title, and repeated in the text, the question is: Do approximately the same polynomials have approximately the same roots? The answer to this question is no, as Wilkinson shows with his monic degree-20 polynomial, which has well separated roots of moderate size, namely the integers 1 to 20. A decrement of $2^{-23}$ in the coefficient of the degree-19 term (which is $-210$)---a proportional change of less than 1 in 1.7 billion---induces substantial changes in the roots: for example, the root pair $16.5\pm 0.5$ becomes the root pair $16.73024\pm 2.81262\mathrm i$ (to 5 d.p.). The second question asked by the OP is whether the coefficients-to-roots map is continuous, which was already answered (affirmatively) by others.

Is there a good comparative study of the Banach integral?

2012-05-20T19:54:11Z

The Banach integral is elegant in its definition, and I am intrigued as to why it is so rarely seen. Is it in practice difficult to calculate from the definition? And are there any other problems with it? I would also be interested to see examples of functions that are Banach-integrable, but not Lebesgue-integrable, and functions (if any) whose Banach and Lebesgue integrals over some interval are defined but different. Added in response to comments: The Banach integral is defined in The Encyclopaedic Dictionary of Mathematics. Like the familiar Riemann integral, it's an integral for real functions over a real interval, and not applied to general Banach spaces (except perhaps in some generalization that I know nothing about).

Answer by John Bentin for Fraktur symbols for Lie algebras: $\mathfrak{g}$, etc.

2012-02-06T09:53:42Z

Fraktur was the standard font for cardinal numbers, for example, in writing the continuum hypothesis as $\aleph_1=\frak c$.

Which squares succeed a factorial?

2011-11-09T08:49:52Z

One notices that $5^2=4!+1$, $11^2=5!+1$, and $71^2=7!+1$. Are there others?

Answer by John Bentin for The concept of Duality

2011-08-26T08:32:13Z

Finite-dimensional linear spaces. A particular feature in this case is that the (algebraic) dual of a finite-dimensional vector space, namely the space of linear maps from the vector space into the base field, is isomorphic to the original space (since it is of the same dimensionality) but not canonically so. In contrast, the bi-dual (the dual of the dual) is canonically isomorphic to the original space, and so may be identified with it.

Can Sierpinski's anisotropic bicolouring of the plane, assuming the continuum hypothesis (CH), be extended to three dimensions?

2011-01-02T21:02:41Z

Sierpinski showed that, on the assumption of CH (in fact, equivalently to it), each point in the plane can be coloured (say) black or white so that every section of the plane parallel to the $x$ axis is "almost" white $-$ in the sense that all but countably many points of it are white $-$ while every section parallel to the $y$ axis is almost black. Equivalently, given CH, every line through the origin can be almost white, while every circle centred on the origin is almost black. Is there a corresponding result for three-dimensional space? For example, assuming CH, can we bicolour space so that every plane through the origin is almost white, while every sphere centred at the origin is almost black?

What bounds the ratio of summed partial harmonic means to a sum?

2011-07-01T10:00:39Z

For positive real $x_1$ , $x_2$ ,..., define their $k$th partial harmonic mean as $h_k = k/(1/x_1 +\cdots+1/x_k)$ for $k = 1, 2, ...,$ and let

$\alpha=\sup_{x_1,x_2,... \geqslant0}\: \lim_{n\rightarrow\infty}\dfrac{h_1+\cdots+h_n}{x_1+\cdots+x_n}.$

What is this bound, and for which $x_1$ , $x_2$ ,... is it attained? All I can do is show $\alpha \geqslant 2$ by taking $x_k = 1/k\;\;$ ($k = 1, 2, ...$).

How fast can the base-bumping function in Goodstein's theorem grow?

2010-08-13T19:22:06Z

In the usual presentation of Goodstein's theorem, the base is bumped up by the "add 1" function. Does the theorem still hold when we replace this function by a fast-growing one (e.g. Ackermann or busy beaver)? How far can we push this? For example, let's define $g_0(n)$ to be the number of Goodstein iterations needed to reach 0 when we start with base 2 and seed $n$ (so that $g_0(0)$ = 0). Then we can build a hierarchy of functions by defining $g_{k+1}(n)$ as the number of Goodstein iterations needed to reach 0 with seed $n$ and base-bumping function $g_k$ ($k$ = 0, 1, ...), continuing through the ordinals by diagonalization at each limit ordinal. Surely it's got to break down when we go past $\epsilon_0$, if not long before that!

Is the following statement a correct formulation of the (much doubted) P = NP conjecture?

2011-03-14T19:31:01Z

"Call a Turing machine $A$ a d-machine if, for some polynomial $p(\cdot)$, when $A$ starts with any input string, say of length $n$, in its alphabet on its (otherwise blank) tape, it will halt in a number of steps bounded by $p(n)$ with its tape either blank or bearing just a string of length $n$. Then each d-machine $A$ corresponds to a d-machine $B$ which, when started with any input string in its alphabet, will halt with a blank tape, if $A$ halts with a blank tape for every input the length of $B$'s input, and otherwise will halt with an output tape that, input to $A$, will lead to $A$ halting with a non-blank tape."

Interpretation: Input for $A$ codes candidate "solutions" while blank/non-blank output indicates just refutation/verification. For $B$, input marks only length while output codes an $A$-verifiable solution.

How aspherical can a Gömböc be?

2011-02-23T10:44:12Z

A Gömböc is a homogeneous massive convex solid that can rest on a horizontal plane in just two positions of equilibrium under gravity: one stable and the other unstable. How small a proportion of the volume bounded by its circumscribing sphere can a Gömböc occupy?

What sums of equal powers of consecutive natural numbers are powers of the same order?

2011-01-24T21:42:03Z

Trivially $n^1=n^1$, and everyone knows that $3^2+4^2=5^2$. Denis Serre quoted $3^3+4^3+5^3=6^3$ in a recent MathOverflow question (which prompted this one). Are any other examples known?

Answer by John Bentin for Webpages for specialized communities

2011-02-10T07:26:35Z

Foundations of mathematics: http://cs.nyu.edu/pipermail/fom/

Could the cluster set of a rational sequence be of intermediate cardinality (given not-CH)?

2011-02-05T18:23:34Z

It's easy to construct a sequence of rational numbers whose set of cluster (or accumulation or limit) points is of any finite cardinality, or is countably infinite, or has the cardinality of the continuum. Thus there appears no obvious barrier to forming a rational sequence with a cluster set of any cardinality up to $\mathfrak{c}$. Now suppose that we reject the continuum hypothesis. Is it consistent with $\mathrm{ZFC}$ that there could be an oracular sequence $(a_0 , a_1 , \dots ) \in \{0, 1\}^\mathbb{N}$ such that the cluster set of the sequence $(n_{2i - 1}/n_{2i} : i = 1, 2, \dots)$, where $n_0$ , $n_1$ , ... are the elements of $\{n \in \mathbb{N} : a_n = 1\}$ in natural order, has cardinality strictly between $\omega$ and $\mathfrak{c}$ ?

What patterns have been measured in the graph of the number of two-prime-sum representations of even numbers?

2011-02-01T09:22:47Z

There are remarkable patterns of density in the graph http://upload.wikimedia.org/wikipedia/commons/7/7c/Goldbach-1000000.png plotting the number of representations of even numbers up to a million as the sum of two primes. Has anyone measured these patterns? The referenced graph was uploaded in 2006. Given the enormous data bases of primes, it would seem to be a matter of simple checking of sums to generate the graph, with a complexity that grows not much more than linearly with the size of the even number. Thus, with the growth of computer power, some advance on a million might have been achieved over the past few years. Has this been done?

Answer by John Bentin for Why is "abelian" infrequently capitalized?

2011-02-03T22:12:38Z

A simple rule for this is that, if the adjective has a precise mathematical meaning, it should be lower-cased: for example, abelian (= commutative) group, euclidean metric, or gaussian (= normal) distribution. If the meaning is vaguer, referring to the method, style, or approach generally associated with the originator, then capitalizing is appropriate: for example, Bayesian statistics. If the word does not carry an adjectival ending---e.g. Hilbert (space)---it should always be capitalized.

Answer by John Bentin for Lower bound on the curvature of the curves on $M$

2011-01-10T22:37:54Z

No. For example, on a 2-sphere in $\mathbb{R}^3$, you can draw a circle as small as you like. (Answer withdrawn: this of course answers the question with the bounding in the other direction)

Answer by John Bentin for Principal curvatures and curvature directions

2010-12-20T17:15:25Z

In general, the concept of "eigenvector" is slightly misleading. The fundamental concept is eigenspace. However, eigenspaces of dimension greater than 1 are generally considered pathological and rather a nuisance. Mostly we would rather have one-dimensional eigenspaces, and for these any convenient nonzero element---an eigenvector--- will serve as a representative. It would be pedantically correct but encumbering to insist on talking about one-dimensional eigenspaces rather than eigenvectors. The arbitrariness of eigenvectors becomes clearer when we really have to deal with a multidimensional eigenspace.

Answer by John Bentin for What are some examples of journals that will accept undergraduate student research?

2010-12-13T08:54:27Z

You might first want to read the article by Arthur White & Robin Wilson: "The hunting group", The Mathematical Gazette 79, no. 484, March 1995, which is about the group theory of change ringing.

What polynomials biject from $\mathbb{N}^{2}$ to $\mathbb{N}$?

2010-11-07T15:42:10Z

Perhaps there are none with integral coefficients; so let us admit rational coefficients. The map $(x, y) \mapsto x + \frac{1}{2}(x + y)(x + y + 1)$ is well known, and swapping $x$ and $y$ in the formula yields another, so we have two for starters.

Answer by John Bentin for Suggestions for good notation

2010-10-21T11:15:56Z

Using $(a, b, ... )$ is handy to denote a column vector, which is the transpose of the row vector $[a, b, ... ]$, especially in linear text. Correspondingly, all displayed matrices should be written with brackets, not parentheses. This notation agrees with the usual identification of coordinates with column vectors.

Answer by John Bentin for Generalizing a problem to make it easier

2010-09-26T21:58:44Z

Here's an example in planar euclidean geometry. Consider an equilateral triangle of side $a$ and a general point in the plane distant $b$, $c$, and $d$ from the respective vertices. Then

$3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2$.

This is an an awful slog to get by planar trigonometry. Even harder to do by trig in three dimensions is the corresponding result for the regular tetrahedron. However, it's easy to get the $(n - 1)$-dimensional result for a regular $(n - 1)$-dimensional simplex of side $d_0$, with vertex distances $d_1$ ,..., $d_n$ :

$n(d_0^4 + ... + d_n^4) = (d_0^2 + ... + d_n^2)^2$.

You can do this by embedding the euclidean $(n - 1)$-dimensional space as the hyperplane of points $(x_1 ,..., x_n)$ in euclidean $n$-space such that $x_1 + ... +x_n = d_0/\sqrt2$. The vertices of the simplex can then be represented as the points $(d_0/\sqrt2)(1, 0 ,..., 0)$, ... , $(d_0/\sqrt2)(0 ,..., 0, 1)$ in the hyperplane, and the result drops out in a few lines.

A sequential optimizing task

2010-07-08T10:13:50Z

Find distinct positive real numbers $x_1$ , $x_2$ , ... of least supremum such that, for each positive integer $n$, any two of 0, $x_1$ , $x_2$ ,..., $x_n$ differ by $1/n$ or more.

Note that the hurdle term $1/n$ is optimal in the sense that any replacement for it would need to stay below a constant multiple of it to allow a finite supremum. By a nonconstructive proof, there is a unique solution minimal with respect to the lexicographic ordering of real sequences; so a constructed solution (while eluding me) doesn't seem impossible. Although I haven't seen this problem anywhere, it looks too simple not to have been posed before. Any pointers would be welcome.

Answer by John Bentin for A sequential optimizing task

2010-09-04T09:38:24Z

Belated thanks to Kevin O'Bryant for his pointer to discrepancy theory. This led me eventually to a source where the problem is solved: See Theorem 6.7 in Harald Niederreiter's book Random Number Generation and Quasi Monte Carlo Methods (SIAM 1992). The logarithmic sequence described by Tracy Hall is due to Rusza and is indeed optimal.

Answer by John Bentin for A sequential optimizing task

2010-07-21T06:55:09Z

I'll now put forward my candidate solution to the problem. It clearly satisfies the hurdle condition, but I can't prove its optimality. To get a handle on the algorithm, let's represent $x_1$ , $x_2$ , … as hotel guests numbered accordingly. Recall Hilbert's hotel, where the unfortunate guests were ever being shunted from their room to a higher-numbered room. The hotel in this case, rather than having a countably infinite number of discrete rooms, is a continuum of “rooms”, represented by the points of a closed bounded real interval; [0 , 1.4] is big enough, as it turns out. The “guests”, numbered 1, 2, … , are movable tags, each assigned to a rational point in the interval. Unlike Hotel Hilbert, which is always full, this hotel starts empty apart from the proprietor who resides permanently at 0, and the guests arrive one by one in the order 1, 2, … . The proprietor operates the strict rule that, when there are a total of $m$ guests in the hotel, there must be a space of at least $1/m$ between the residents (including himself), for $m$ = 1, 2, … . Guest 1 is assigned to the point 1. Guest 2 is placed at 1/2. When guest 3 arrives, she is put at 1/3, while guests 1 and 2 are moved up by a distance 2/3 – 2/(3 + 1) = 1/6. When guest 4 comes, he is allotted to 1/2 + 1/6 + 1/4 = 11/12. The general rule is as follows. For $k$ = 1, 2, … , after $2^k$ guests have been accommodated, the next $2^k$ are put consecutively into the $2^k$ spaces between the proprietor and the first $2^k$ guests, in left-to-right order: An odd-numbered arrival, say guest $2n$ – $1$, is assigned to the point below her right-hand neighbour (guest $n$) that is 1/($2n$ – $1$) above her left-hand neighbour, while all the guests to her right are moved up by a distance 2/($2n$ – 1) – $1/n$; when the next, even-numbered, guest arrives (guest $2n$), he can just go to the midpoint of the next space up, at a distance 1/$2n$ above his left-hand neighbour (guest $n$), and mercifully no resident has to move until the next (odd-numbered) guest arrives. The result is that the guests, now identified with their limiting room positions, are each the sum of two parts: The first part is a sum of a finite number of distinct fractions of the form $1/n$, while the second is an infinite series whose terms are of the form 1/$n$($2n$ – 1), where $n$ is a positive integer. Generally an infinite number of terms of the latter type are absent; only in the case of $x_1$ is the series free of gaps, and then the first “sum” has only one term. Thus $x_1$ = 1 + ∑{1/$j$($2j$ – 1) : $j$ = 2, 3, …} = ln 4, and this is also the supremum of the sequence.

An alternative characterization, suggested by Gerhard Paseman, is as follows: For j from 2^(k-1)+1 to 2^k, you will arrange to place guest (2j-1) to the left of guest j and guest 2j to the right of guest j. Since space to the left of guest j has previously been guaranteed to be 1/j from his/her lefthand neighbor, guest (2j-1) needs more since it needs 1/(2j-1) space to his/her left and right. So add the difference delta_j = (2/(2j-1) – 1/j) to the left of guest j and shift guest j and every guest on the right of guest j by this difference delta_j. Since guest 2j does not need more than 1/j = (2/2j) space, no such adjustment is needed for guest 2j. This gives guest 1 infinitely many adjustments; 1 ends up at place 1 + sum(j > 1) delta_j = 1 + sum(j > 1) {2 [ 1/(2j-1) - 1/2j ] } = 2 ln(2).

Comment by John Bentin

2013-05-23T13:53:25Z

@Douglas Zare: Good editing! However, there is still room for small improvements in the punctuation (which I don't have the rep points to do). (1) In line 1, commas are needed after "Wikipedia" and "small", with the comma after "good" preferably deleted. (2) In line 3, a comma should follow "edits". (3) The comma splice in line 6 needs attention, perhaps by replacing "click" with "by clicking".

Comment by John Bentin

2013-05-18T17:42:07Z

Your latter cube is indeed a cube, but the first-mentioned "cube" is not. Such a rectangular body is called a cuboid.

Comment by John Bentin

2013-05-05T20:05:54Z

Do you mean the symmetric matrix associated with the quadratic form?

Comment by John Bentin

2013-04-30T06:57:43Z

You talk about "subsets" of $S$, but then you have discarded the multiset structure of $S$. Do you mean sub-multisets of $S$?

Comment by John Bentin

2013-04-02T08:50:31Z

This is a beautifully clear answer to the question I should have asked. It all seems so simple now! Just one query: In the second line of the proof of lemma 2, should the initial term of the inequalities be $y$, not $x$?

Comment by John Bentin

2013-04-01T17:28:30Z

@Todd: Yes, your example works well with, say, $f(x)=x^2$. I would happily take this as the "accepted answer".

Comment by John Bentin

2013-04-01T17:13:24Z

@Gerry, @Gerald: I think that Gerry's composition satisfies all the conditions except upper monotony. Take, for example, $a=1/\mathrm e$.

Comment by John Bentin

2013-03-26T08:45:28Z

Thank you very much for this answer. Just a few queries concerning your 4th paragraph: (1) I can't read "$\sum_1^{\infty}\|\frac{1][\sqrt{5}} \Phi^n\| \lt \infty$"; (2) I guess the equality sign in "$\|x=y\| \le \|x\|+\|y\|$" should be a plus sign; (3) By "$\lambda$ is any real $\lambda=\frac{a}{\sqrt{5}}+b+c\Phi$" do you mean "$\lambda$ is any real of the form $\lambda=\frac{a}{\sqrt{5}}+b+c\Phi$ with integral $a$, $b$, and $c$"?

Comment by John Bentin

2013-03-25T14:19:59Z

@Douglas: Thank you for all your work on this. However, it seems to me that your answer addresses a question different from the one I posed, namely with the order of quantification of $\alpha$ and $\varepsilon$ inverted. In my question, the order is expressed as $(\exists\,\alpha\in \Bbb R_{>1}{\setminus}\Bbb N)(\forall\,\varepsilon\in\Bbb R_{>0})$, while your $\alpha$ is dependent on a previously chosen $\epsilon$.

Comment by John Bentin

2013-03-24T23:00:37Z

Sorry to be so dense, but I can't see how a method to construct a number whose powers, modulo $1$, are uniformly distributed over $(0, 1)$ can be applied to finding one whose power sequence converges, modulo $1$, to $0$.

Comment by John Bentin

2013-03-24T20:37:08Z

Thank you. Could you please give me a reference (or suitable search terms) for "the usual construction of non-integral real numbers greater than 1 whose powers are close to integers"?

Comment by John Bentin

2013-01-07T13:49:36Z

+1, especially for "We should always, always distinguish a function $f$ from its value $f(x)$ at $x$". But I don't get your next point: If we define (say) $y$ := $x^2+1$, then what is wrong with saying that (the variable) $y$ depends on (the variable) $x$?

Comment by John Bentin

2013-01-01T09:06:38Z

Thanks, Aryeh. I was confusing erf with a cumulative distribution function.

Comment by John Bentin

2013-01-01T08:30:20Z

Yes. But the function $f$ here doesn't tend to zero at both ends of the interval $[0, \infty)$, because $f(0)=\mathrm{erf}\, 0-\mathrm{tanh}\,0=\frac{1}{2}-0=\frac{1}{2}$.

Comment by John Bentin

2012-12-31T21:20:50Z

Sorry, I can only see that, on your supposition, $f'$ has $2$ (or more) positive zeros. But anyway that's enough for the rest of the argument to work.