29
$\begingroup$

I expect this is a classical question, so feel free to point me to classical answers: what is the fastest-growing function $f(t)$ for which we know that $$ |2^t - 3^{t'}| \ge f(\min(t,t')) \;? $$ In particular, do we know that the gaps between powers of 2 and powers of 3 get exponentially large as $t,t'$ increase? Do we know anything like this for any other pair of integers besides 2 and 3?

$\endgroup$
2
  • $\begingroup$ It seems that by Khinchin's theorem, if $\log(3)/\log(2)$ is a typical real number (in some lebesgue measure sense), then $f(x)=3x/(x\log(x)^2)$ would do with at most finitely many exceptions. $\endgroup$ Aug 7, 2017 at 18:10
  • 1
    $\begingroup$ See also this blog-entry of T.Tao terrytao.wordpress.com/2011/08/25/… $\endgroup$ Oct 12, 2019 at 9:05

4 Answers 4

32
$\begingroup$

What you need is the theory of lower bounds for linear forms in logarithms. A good place to start reading about this is the following article by Evertse:

www.math.leidenuniv.nl/~evertse/dio2011-linforms.pdf

In particular, Corollary 1.8 of the article (a Corollary to a famous theorem of Matveev) gives

$$ \lvert 2^a-3^b \rvert \ge \frac{\max(2^a,3^b)}{(e \max(a,b))^{C}} $$ where $C$ is a positive constant (that is easily computable--see the proof and also the statement of Theorem 1.7).

$\endgroup$
4
  • 8
    $\begingroup$ Can you give a rough and ready estimate of $C$ for those who just want to be impressed with this neat result? $\endgroup$ Dec 21, 2012 at 11:05
  • 3
    $\begingroup$ I just composed an answer with a heuristic for the approximants according to the continued fractions of $\log_2(3)$. For that heuristic it seems $C=1.06$ is a good choice allowing only two exceptions, and $C=1.22$ has no exceptions at all up to $b=2^80$ $\endgroup$ Aug 7, 2017 at 17:01
  • 1
    $\begingroup$ The number $C$ computed by the formula in the linked article by inserting values in the formula $C= e*2^{3.5}*30^5*1*\log(3) $ is $C=821013300.694...$ (if I'm not messing up some things). I think there should far smaller values be available meanwhile... $\endgroup$ Sep 7, 2017 at 10:55
  • $\begingroup$ @Siksek - you might like to see my picture for some suggestion for the choice of $C$ in my updated answer above. Can I actually use that picture's suggestion? $\endgroup$ Jun 27, 2019 at 12:25
18
$\begingroup$

I guess you expect $t$ and $t'$ to be integers. In this case, having a small $2^p-3^q$ is related to having a small $\frac{\log 3}{\log 2} - \frac{p}{q}$. So it's Diophantine approximation, and this is very well studied. The first result in Diophantine approximation is that there exists an infinity of rational $p/q$ such that $$ \left|\frac{\log 3}{\log 2} - \frac{p}{q}\right| < \frac{1}{q^2}. $$ In which case it's not hard to compute that $$ \left| 2^p - 3^q \right| = \mathcal{O}\left( \frac{3^q}{q} \right). $$ This is valid of course for all $2$'s and $3$'s.

If now you want lower bounds, then you will need to know a upper bound for the irrationality measure of $\frac{\log 3}{\log 2}$, which is hard to get, but hopefully someone did it. Do you want more details ?

EDIT

Let $\epsilon = p \log 2 - q \log 3$. In particular, $|\epsilon| < \frac{\log 2}{q}$. We compute that $$\left| 2^p - 3^q \right| = 3^q \left| 1 - \exp(\epsilon) \right|$$

As $q\to \infty$, we have $\epsilon \to 0$ so $1-\exp(\epsilon) = \mathcal{O}(\epsilon) = \mathcal{O}(q^{-1})$. Thus $$ \left| 2^p - 3^q \right| = \mathcal{O}\left( \frac{3^q}{q} \right). $$

$\endgroup$
5
  • $\begingroup$ Apologies, but what does "This is valid of course for all $2$'s and $3$'s" mean? $\endgroup$ Dec 29, 2018 at 14:54
  • 1
    $\begingroup$ @samerivertwice I take it to mean that you can replace $2$ and $3$ with any two integers that aren't rational powers of each other, and the "same" result holds. Certainly any two primes, or relatively prime numbers. If there's a common factor, it can be pulled out and fed to the big-O. $\endgroup$ Jun 5, 2021 at 14:34
  • $\begingroup$ From $$ \left|\frac{\log 3}{\log 2} - \frac{p}{q}\right| < \frac{1}{q^2} $$ I get to $$ max(\frac{3^q}{2^p}, \frac{2^p}{3^q}) < 2^{\frac{1}{q}} $$ , but I didn't find the way to your formula with big-O notation. Can you add details how to get there? $\endgroup$
    – Daniel S.
    Aug 5, 2022 at 12:26
  • 1
    $\begingroup$ @DanielS. I added some details $\endgroup$
    – Lierre
    Aug 8, 2022 at 7:19
  • $\begingroup$ I'm currently proposing the installation of a "BigList" on known such bounds, see meta.mathoverflow.net/q/5423/7710 $\endgroup$ Aug 16, 2022 at 8:43
13
$\begingroup$

I was a little hesitant to post the following thing after the very thorough answers and references, yet it contains a concrete inequality, and may be of interest as a first elementary approach towards the full complexity of the problem.

The idea is that if $2^t$ and $3^ {t'}$ are too close to each other, then $2^{t+1}$ is close to $2\cdot3^{t'}$, hence it is roughy in the middle between $3^ {t'}$ and $3^ {t'+1}$, and therefore far from any power of $3$. To make this into a more quantitative form: assume that $t$ and $t'$ satisfy $$|2^t -3^ {t'}| < \frac{1}{5} 2^t\, .$$ Then it follows plainly

$$ 3^ {t'} + \frac{1}{5} 2^t < 2^{t+1} < 3^ {t'+1} - \frac{2}{5} 2^t \, .$$ Therefore the closest power of $3$ to $2^{t+1}$ is either $3^ {t'}$ or $3^ {t'+1}$, in any case not closer than $ \frac{1}{5} 2^{t+1}$. This tell us that the inequality $$\min _ {t'\in\mathbb{N}} |2^t -3^ {t'}| > \frac{1}{5} 2^t$$ holds for at least one out of two consecutive integers $t$ and $t+1$. So at least half of the powers of $2$, in a density sense, have a distance from the powers of three of at least one fifth of their size.

$\endgroup$
4
$\begingroup$

Just to satisfy the curiosity of @FelixGoldberg and other cursory readers. Here is a heuristic which pointed me to try to use $C=1.06$ for an example.

We look at the distances

$$\left|1-{3^b\over2^a}\right| \overset{???}\ge { 1\over (e \cdot a)^C} $$

with $a \gt b$ and $2^a \gt 3^b$ (fixing the $\max()$-terms.

In the table $w=\log_2\left|1-{3^b\over2^a}\right| $ and $u=-\log_2 (e a)$. The quotient $w/u$ should give an impression of the missing factor $C$, and in this table for all except $2$ cases ( idx=15,idx=21 ) a value of $C=1.06$ suffices to make the inequality true.

The table reports the cases according to the continued fraction of $ß=\log_23)$ so only the best possible approximants (with $2^a \gt 3^b$) are displayed (the convergents, each second of them)

  idx   b     a     log2(b)  log2(a)      w       u         w/u      1.06*u
----------------------------------------------------------------------------
   3      1      2  0.E-201  1.00000  -2.00000  -2.44270  0.818768  -2.58926
   5      5      8  2.32193  3.00000  -4.29956  -4.44270  0.967782  -4.70926
   7     41     65  5.35755  6.02237  -6.45514  -7.46506  0.864714  -7.91297
   9    306    485  8.25739  8.92184  -9.93479  -10.3645  0.958537  -10.9864
  11  15601  24727  13.9294  14.5938  -15.7461  -16.0365  0.981894  -16.9987
  13  79335  SSSSS  16.2757  16.9401  -18.0579  -18.3828  0.982323  -19.4858
  15  NNNNN  SSSSS  17.5397  18.2042  -23.8860  -19.6469   1.21576  -20.8257
  17  NNNNN  SSSSS  23.3620  24.0265  -26.2877  -25.4692   1.03214  -26.9973
  19  NNNNN  SSSSS  27.3572  28.0217  -29.0580  -29.4644  0.986209  -31.2322
  21  NNNNN  SSSSS  28.5666  29.2311  -33.1373  -30.6738   1.08031  -32.5142
  23  NNNNN  SSSSS  32.6169  33.2814  -36.5236  -34.7241   1.05182  -36.8075
  25  NNNNN  SSSSS  37.0009  37.6654  -40.0173  -39.1081   1.02325  -41.4546
  27  NNNNN  SSSSS  42.2986  42.9630  -43.7861  -44.4057  0.986046  -47.0701
  29  NNNNN  SSSSS  43.3957  44.0601  -46.1400  -45.5028   1.01400  -48.2330
  31  NNNNN  SSSSS  48.6152  49.2797  -49.4134  -50.7224  0.974193  -53.7657
  33  NNNNN  SSSSS  52.3527  53.0172  -53.0620  -54.4599  0.974331  -57.7275
  35  NNNNN  SSSSS  56.8562  57.5206  -58.9521  -58.9633  0.999810  -62.5011
  37  NNNNN  SSSSS  58.4640  59.1284  -62.5155  -60.5711   1.03210  -64.2054
  39  NNNNN  SSSSS  62.0089  62.6734  -65.0073  -64.1161   1.01390  -67.9630
  41  NNNNN  SSSSS  64.5731  65.2376  -66.4207  -66.6803  0.996108  -70.6811
  43  NNNNN  SSSSS  66.0744  66.7389  -67.9931  -68.1815  0.997236  -72.2724
  45  NNNNN  SSSSS  67.4786  68.1431  -73.2504  -69.5858   1.05266  -73.7609
  47  NNNNN  SSSSS  74.7217  75.3861  -81.0514  -76.8288   1.05496  -81.4386
  49  NNNNN  SSSSS  80.5354  81.1999  -82.0659  -82.6426  0.993022  -87.6011


The Pari/GP script is

fmt(200,8) \\ internal precision 200 dec digits, user-procedure
{e=exp(1);l3=log(3);l2=log(2);ld3 = l3/l2;
cf = contfrac(ld3);
cvgts= mkContFracConvergents(cf,50) ; \\ user-procedure
listlogs=vectorv(50);ix=0;
forstep(i=3,50,2,
          a=cvgts[1,i];      \\ ===> a > b  and also 2^a > 3^b
          b=cvgts[2,i];   
          ix++; listlogs[ix]=[i,
                 if(b<100 000,b,'NNNNN),      if(a<100 000,a,'SSSSS),
                 log(b)/l2,                   log(a)/l2,
                 w=log((1.0-2^(ld3*b-a)))/l2, u=-log(e*a)/l2,
                 w/u      ,                   1.06*u];
        );
 listlogs=Mat(VE(listlogs,ix))}

Remark: a bit more introduction and tables and graphs for $b \to 10^{10800} \approx 2^{36000} $ can be found at my pages . Note, that I use $N$ for what we use $b$ here, and $S$ for what we use $a$ here, thus discussing $2^S-3^N$.


update A better visualization of the properties of selecting some constant $C=1+\epsilon$ using up to $b =10^{1000}$ taken from the convergents of the continued fraction of $\log(3)/\log(2)$
I show how empirically the values of $C(b)$ were when $a,b$ are inserted in the formula and $C(b)$ is computed. The image shows, that the empirical $C(b)$ are except in two cases smaller than $C=1.06$ and moreover, that possibly we can choose any $C=1+\epsilon$ and getting only finitely many cases where not $C(b) \le C$
Legend: In the picture I used my standard-notation $N$ for $b$ here and $S$ for $a$ here.

image

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.