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  • 8
    $\begingroup$ I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. $\endgroup$
    – Asaf Karagila Mod
    Jul 18 '12 at 8:35
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    $\begingroup$ (+1) For thinking outside the (sand)box. $\endgroup$
    – cardinal
    Jul 18 '12 at 19:40
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    $\begingroup$ At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! $\endgroup$
    – Grace Note StaffMod
    Oct 5 '12 at 14:45
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    $\begingroup$ To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. $\endgroup$
    – leo
    Dec 17 '12 at 18:03
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    $\begingroup$ PSA: Between the creation of this sandbox (in July 2012) and today (December 2015), technology has advanced. Something like StackEdit (or others, it's simply the only one I know) essentially solves all the limitations of this sandbox. You can have multiple concurrent drafts, you don't have to worry about polluting meta's front page, you can leave your draft untouched for days and expect it to still be there, you don't have to explicitly clear up your draft when you're done... Maybe someday we can get rid of this outdated crutch. $\endgroup$ Dec 2 '15 at 14:07

17 Answers 17


With $n+k= {\lceil n \log_23\rceil}<n \log_23+1$ we can use the Rhin bound (page 160: $|µ_1\log 2+µ_2\log 3|\geq H^{-13.3}$ with $H=max(|µ_1|,|µ_2|)$)

$$|(n+k) \log2 - n \log3|>\frac 1{(n+k)^{13.3}}>\frac1{(n \log_23+1)^{13.3}}>\frac 1{n^a}$$ If we choose $a=15$ the above is true for $n>41$ (note that $(\log_23)^{13.3}\sim457$)

Now (growth of $x^a$ vs $a^x$) and with $\mu = e^{\log(2)-.1}=1.80967483607...$,

$$\frac 1{n^a}>\frac{\mu}{(\frac{3}{\mu^{\log_2(3)}})^n}$$ is true for $n>\frac{W_{-1}\Big(\frac{-\log\frac{3}{\mu^{\log_2(3)}}}{a\cdot \mu^\frac{1}{a}}\Big)}{\Big(\frac{-log\frac{3}{\mu^{\log_2(3)}}}{a}\Big)}$ or $n>610$ with chosen $a=15$ ($W$ is the productlog), and we also have (using $x>log(1+x)$) $$\frac{\mu}{(\frac{3}{\mu^{\log_2(3)}})^n}=\frac{\mu^{n\log_2(3)+1}}{3^n}>\frac{\mu^{\lceil n\log_2(3)\rceil}}{3^n}>\log(1+\frac{\mu^{\lceil n\log_2(3)\rceil}}{3^n})$$ So for $n>610$ we have $$|(n+k)\log2 - n \log3|=\log(\frac{2^{n+k}}{3^n})>\log(1+\frac{\mu^{\lceil n\log_2(3)\rceil}}{3^n})$$ or $$\frac{2^{n+k}}{3^n}>1+\frac{\mu^{\lceil n\log_2(3)\rceil}}{3^n}$$ $$2^{\lceil n \log_23\rceil}-3^n>\mu^{\lceil n\log_2(3)\rceil}$$

This leads to ($n>610$):

$$\begin{array}{|c|}\hline 2^{\lceil n \log_23\rceil}-3^n>\mu^{\lceil n\log_2(3)\rceil}\\\hline\end{array}$$ and with manual checking, inequality holds for ${\lceil n \log_23\rceil}>27$

Similarly, ....


The lower bounds of distances of perfect powers $\Delta=\mid a^A-b^B \mid$ is one of those attractive, simple-to-state, but difficult-to-solve problem. For me it occured in the question of cycles in the Collatz-problem, where we ask for the lower bound of distances of perfect powers of $2$ and $3$: $$\Delta(N,S)=|2^S - 3^N|$$ or $$\Lambda(N,S)=|S \cdot \ln2-N \cdot \ln 3| \qquad .$$
What I've found so far is

  • the estimate by J.W.Ellison formulated in the 1970'ies (adressing $\Delta(N,S)$), and
  • the estimate by G.Rhin formulated in 1987 (adressing $\Lambda(N,S)$), as reported in J.Simons & deWeger[2007] on the existence of "m-cycles" in the Collatz-problem.

The lower bounds from those estimates are -in the view of an amateur- astonishingly far away from the actual lowest distances -as checked for $N$ of about $1$ million digits. (Of course, the only relevant $N$ are selected according to the convergents of the continued fraction of $\log_2(3)$.)

Often discussions of this topic state the general form of estimation-formulae, but don't give explicite values for the constants involved - for instance in the well-visited blog of T. Tao on the "separation of powers of 2 and 3". Naive functional estimates for such lower bounds based on heuristics would suggest small values for such constants, for instance while G.Rhin gives something like $1/457/N^{13.3}$ a naiive fitting suggests something like $.../N^1$ or $.../N^{1.01}$ instead, valid for $N$ in my mentioned range of search.
A recent guess, using $.../N/\ln(N)$ instead, leaving aside any $1+\varepsilon$-exponent at all seems to be a really good estimate for the lower bounds. However, the famous estimate for the number of primes below some number $n$ as value of the integrallogarithm $Li()$ as proposed by the young K.F.Gauss has later been proven to be not really a lower bound, but where the first counterexample could only be upper-limited by something like $10^{10^{10^{34}}}$(WP:Skewes-Number) (modern computation reduced that upper bound to $1.39 \cdot 10^{316}$, see there).

Actual computations with large exponents diverge extremely far from that given estimates so I once tried to find an own algebraical pathway to such an lower-bound estimate. My current attempt provides a lower bound which is much nearer to the empirical values than the two other estimates - unfortunately this is not based on an analytical argument but only on heuristic finding.
Here I intend to create a "biglist" of immediately usable such lower bounds and invite other readers to add information and / or alternatives to the two established estimates and to try to improve my proposed estimate with analytical arguments.

While Ellison immediately looks at $\Delta(N,S) = |2^S - 3^N|$ and gives $$\Delta(N,S) \gt \exp(S \cdot (\ln 2 - 0.1)) \qquad S\gt 27 \tag {\text{EL}}$$

G. Rhin looks at the difference of logarithms $\Lambda(N,S)=| S \ln2 - N \ln 3 |$ and gives $$\Lambda(N,S) \gt {1\over 457 N ^{13.3} }| \tag {\text{RH}}$$ I've looked as well at $\Lambda(N)$ and propose $$\Lambda(N,S) \gt {1\over 10 N \ln N } \tag {\text{HE}}$$ To compare that bounds, we must norm for the data of $\Delta()$ or for that of $\Lambda()$. The best adaption is surely dependent on the question that one has - either using a formula involving the exponential expression or one involving the logarithmic expression. So the discussion of this should be done in the single answers of the big-list where we reference some question in MSE.
To trigger the interest, here some pictures, how the three estimates behave in contrast to the empirical values. Here I adapted the Ellison-estimate to a $\Lambda()$-version.

image1We see, that the empirical $\Lambda()$ jitter between $0$ and $1/2 \cdot \ln 2$ (blue dots), the values at $N=\{2,5,12,53,...\}$ show very small distances, and even decreasing towards zero. The idea is to have a continuous function $f(N)$ which can supply a lower bound, such that we can say $\Lambda(N,S) \gt f(N)$ .
The red curve for the Ellison-estimate is below of the empirical values only for $N \gt 17$ while the green curve for the Rhin-estimate is so near the zero-line, that we barely see it. My estimate shown by the brown curve is always below the $\Lambda()$ .
To see a bit more detail, and to get aware of the basic characteristics of the estimates, a picture in log()-scalings is more appropriate. Picture 2: x,y logarithmic scaling image2 Here we see the characteristics better. All three methods of estimates work for $N$ towards $=200$. Ellison took a much different shape compared with Rhin's, and while Rhin's lower bound is unfortunate small over the whole range, the Ellison's is initally too large, then better than Rhin's but from about $N \approx 700$ it decreases much faster than Rhin's. So, for the larger $N$, Rhin's lower bound should be preferred, but for the moderate values in $N$ Ellison's estimate gives a simpler formulation and a better lower bound.
But the "LBLam"-function has the best characteristic; it is nearly parallel to a lower hullcurve, and also is valid from the smallest $N=2$ on. That this property holds forever, that means the formula is analytically useful for all $N$, has not been proved, but empirical heuristic show that it holds for all $N$ up to $1$ million decimal digits and is always tight to the most extreme small values of $\Lambda()$...
A last picture, for the aesthetical impression is the following. I rescaled the values of $\Lambda(N,S)$ to the open interval $-1[...]1$ by $w(N,S)=4 \cdot \Lambda(N,S) / \ln 2 -1$ and then show the values $Y(N,S)=\tanh^{-1}(w(N,S))$ and the accordingly rescaled values for the three lower-bound-estimates. image3The slightly jittering effect in the Ellison-curve is because its function uses the somehow jittering values $S$ instead that of $N$ for its argument.


Wikipedia on Collatz Conjecture in Binary


After 3x+1... For certain x:

With all "1" digits (or if it ends in just 3 "1"s), result in ending in 10.

With ending in "011", it ends in 00010.

With ending in "01", it ends in 00.

With ending "101", it ends in 1000.

At most, 3x+1 can increase an odd number number by ~3x, and must result in a division by 2... So can increase at most ~3/2...

I guess something like this, but more focused...

Something all "1"s leads to something where there must be a 0 in it, which leads to something with more 0s, on and on, until it hits 1?

A.k.a if 3x +1 for 2^n - 2^y 2^z -1, for all n, y, and z, except z can be 0, result in something of the same form except -2^a, or more terms for any "a".

A counter-argument would be something all 1s except for one "0", when 3x + 1 = all ones... Sorta. Then you have to prove that that strings of 1s doesn't turn into more zeros...

... A.k.a, is there any number ((2^n - 1) - 1) / 3

A.k.a any odd number (2^n - 2) / 3... No... Even / Odd = Even. There must be a 2 in the divisors...

So, there are no strings of "1", for the Collatz, that preserve 1111...


Let $g$ be a continuous function on $\mathbb R$ with the following property:

  • $0\le g\le 1$,
  • there are positive sequences $(a_n),(x_n),(y_n),(b_n)$ all converging to zero and $b_{n-1}<a_n<x_n<y_n<b_n<a_{n+1}$ so that $g =1$ in $[x_n, y_n]$, $g=0$ outside $[a_n, b_n]$ and $$\sum b_n-a_n <+\infty.$$

Define $f(x) =\int_0^x g(s)ds$. Then $$\frac{f(y_n)-f(x_n)}{y_n-x_n}=1$$ But $f'(0)=0$: to see this, note that for any $\epsilon >0$


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The objective is to find $f(x)$ such that $$f(h(x)) = j(f(x)) \tag 1$$ where $h(x)$ and $j(x)$ are known functions. In order to get rid of the $h(x)$ inside, you want to somehow transform this into the equation $$g(t+1) = j(g(t)) \tag 2$$ This reveals a system of equations: $$\begin{align}f(h(x)) &= g(t+1)\tag 3\\ f(x) &= g(t) \tag 4\end{align}$$

$x$ is now a function of $t$ that we must find. In order to find $x(t)$, you can plug in $t+1$ to $(4)$ to get $$f(x(t+1)) = g(t+1)$$ which must be equal to $f(h(x(t))$. That is, you have that $$f(x(t+1)) = f(h(x(t))$$

Assuming $f$ is one-to-one, you can apply the inverse to get $$x(t+1) = h(x(t)) \tag 5$$

Note how $(2)$ and $(5)$ are pretty much the same. So the same techniques for solving $(5)$ can be applied to $(2)$. The question is, how is $(5)$ solved?

Starting from a fixed point, $x(t_0) = c_0$, you can find that $x(t_0+1) = h(c_0)$, $x(t_0+2) = h(h(c_0)), \cdots,$ and $$x(t_0+m) = \underbrace{h(h(...h}_{h \text{ applied } m \text{ times}}(c_0)...)) \tag 6$$

With $g(t_1) = c_1$

$$g(t_1+n) = \underbrace{j(j(...j}_{j \text{ applied } n \text{ times}}(c_1)...)) \tag 7$$

If you assume everything's sufficiently nice, you can say $t = t_0+m$ to find an equation for $x$ in terms of $t, t_0, c_0$ and $t = t_1+n$ to find an equation for $g$ in terms of $t, t_1, c_1$.

Then from $(4)$, you got $$f(x(t)) = g(t) \to f(x(x^{-1}(u))) = f(u) = g(x^{-1}(u)) \tag 8$$

With $h(x) = 10x$ and $j(x) = x+1$, $$x(t_0+m) = 10^m c_0 \to x(t) = 10^{t-t_0}c_0 = 10^{t-C_0}$$ and $$g(t_1+n) = n+c_1 \to g(t) = t-t_1+c_1 = t-C_1$$

(we're gonna ignore that this doesn't cover negative $x$)

Using $(8)$, $$f(10^{t-C_0}) = t-C_1 \to f(10^{\log_{10}(u)+C_0-C_0}) = \log_{10}(u)+C_0-C_1$$ Which yields $$f(u) = \log_{10}(u) + C$$


(Deutsch: MathJax: LaTeX Basic Tutorial und Referenz)

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I'm not yet much experienced with analysis of generalized Pell-equation, but part of a specific question is to find all solutions (and possibly a parametric expression) for a couple of generalized Pell-equations.
For my first subproblem I look at the set of integer solutions for $$ a^2 + b^2 = 2 c^2 \tag 1 $$ Of course, if I've found some solution $(a,b,c)$ then I've even found $(\pm a,\pm b,\pm c)$ because we deal with squares.

Brute force: Just to have an initial list of solutions, I just did a twofold-loop over positive integers $a$ and $b$ and checked, whether $CC=(a^2+b^2)/2$ is a square number, and if, then $c=\sqrt {CC}$ and $(\pm a,\pm b,\pm c)$ are solutions.

Generating solutions by matrix-multiplication It is surely well known, that Pell-solutions have the property, that from one handful of solutions one can find infinitely many of them, just by detection some recursion scheme, in fact to find repeated matrix-multiplication on the initial vector $[a,b,c]$.

Fiddling with this gave the following scheme, which looks as if it would generate all solutions.


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$b=3a+1$ \ on odd a then the Syracuse-equivalent transformation is $C(b)= \{b\}_2 \cdot 3 +1 $ , then $C°^2(b)= \{\{b\}_2 \cdot 3 +1\}_2\cdot 3+1 $ , and so on.

Then given that $b$ is $4$-rooted, all $ 4^kb$ are $4$-rooted. Next, using $b_k = 4^kb$ then $ (b_k-1)$
$$ \begin{bmatrix}1&5\end{bmatrix} \cdot \begin{bmatrix} 1&1 \\0 &4 \end{bmatrix}$$ $$ \begin{bmatrix}1&5\end{bmatrix} \cdot \begin{bmatrix} 1&-1/3 \\0 &2/3 \end{bmatrix}$$

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@Zach Siegel

Thanks for your anwer. With all due respect, I do not think that is appropriate.

At first, please allow me to use $X^-$ to replace $X^+$ because we are discussing $Var(X|X<a)$. In fact, $X^-$ in the answer is a function of $X$, not the variable after truncation we are concerned. Note that $E(X|X<a)$ and $Var(X|X<a)$ depend on the truncated pdf function $f(x|X<a)=\frac{f(x)I(x<a)}{\int_{-\infty}^a \ \ f(x)dx}$. However, $E(X^+)=\int xI(x<a)f(x)dx$ and $Var(X^+)=\int (xI(x<a)-E(X^+))^2f(x)dx$ still rely on the distribution of $X$.

Therefore, $Var(X^-)$ is not equivalent to $Var(X|X<a)$, and the given inequality may need more conditions.


The post was closed as a duplicate, but it is not. In the duplicate target, it is assumed that $M$ is a metric space with a metric topology. In this post I am asking for a general topological space. The only answer in the duplicate used the metric in an essentially way, that I don't think that can generalized here.

I was trying to solve the following problem.

If every bijection $f:M\to M$ is a homeomorphism, then every subset of $M$ is clopen.

I tried to prove the equivalent statement,

if every bijection is homeomorphism, then every singleton is open (as if every singleton is open, then every subset is union of open sets and therefore open, hence all subsets are clopen, and converse holds trivially).

I tried to argue by contradiction. Assume that every bijection is homeomorphism and there exists some singleton that is not open, say $\{a\}$. Then, we can construct a bijection that maps every element to itself, but maps this singleton to some singleton which is open, say {b}. Then, this function is not a homeomorphism and so it's not true that every bijection is homeomorphism, contradiction. However, this argument assumes that at least some singleton is open, and I cannot see what goes wrong if none of the singletons were open. Did I even go in the right direction, and if so what could go wrong if none of the singletons were open?


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