Example of a power of 3 which is close to a power of 2

I'm looking for a power of 3 close to a power of 2.

Let's say, what is $(n,m)$ such that $$\left|\frac{2^n}{3^m}-1\right| = \min\left \{\left|\frac{2^i}{3^j}-1 \right|, 1\leq i,j\leq 20\right\} \quad ?$$

Why?

The idea is to understand the intervals between musical notes. Any link on this subject is welcomed (I don't know music theory but I'm interested).

• $\dfrac{2^1}{3^{20}}$ sounds good? (no pun intended :) But if you want a power of $3$ which is close to a power of $2$ maybe you wanted $$\min\left \{|3^j-2^i|, 1\leq i,j\leq 20\right\}$$ – Raffaele Oct 20 '17 at 7:48
• Thanks, I will edit!!! – Colas Oct 20 '17 at 7:51
• @Raffaele I just understood your good (intended!) pun. Another way to see the issue is to say that's about two superpowers that should be equilibrated. – Jean Marie Oct 20 '17 at 10:37

You want $2^i \approx 3^j$ with $i,j \in \mathbb N$

(preferably $1 \leq i,j \leq 20$).

Take the logarithm (in any base) of both sides: $$i \log(2) \approx j \log(3).$$

Thus the issue is now to obtain a good rational approximation

$$\underbrace{\dfrac{\log(3)}{\log(2)}}_{N}\approx \dfrac{i}{j}$$

Such rational approximations are obtained as the so-called "convergents" of the continued fraction expansion of number $N$ (https://en.wikipedia.org/wiki/Continued_fraction):

$$N=a_0+\tfrac{1}{a_1+\tfrac{1}{{a_2+\tfrac{1}{\cdots}}}}$$

with $(a_0; a_1, a_3, \cdots )=(1;1,1,2,2,3,1,5,2,23,2,2,1,55,1,...).$

with successive convergents:

$$i/j = 3/2, \ 5/3, \ 8/5, \ 11/7, \ 19/12, \ 46/29, \ 65/41, \ 84/53, ...$$

$$317/200,\ 401/253, \ 485/306, \ 569/359, \ 1054/665...$$

(that can be found in (https://oeis.org/A254351/internal)).

Good convergents are obtained by (in plain terms) "stopping just before a big $a_k$".

If the "big $a_k$" (!) is 3, one gets :

$$\dfrac{19}{12} \approx 1.5833 \ \ \ \text{of} \ \ \ N = 1.5850...$$

Thus a good tradeoff is to take $i=19$ and $j=12$.

This is connected to the "Pythagorician comma". (https://en.wikipedia.org/wiki/Pythagorean_comma).

Very related : (http://www.math.uwaterloo.ca/~mrubinst/tuning/12.html)

• I know the definition of continued fractions, but I never thought it had a link to rational approximation. – Colas Oct 20 '17 at 8:00
• That makes senses since there are 12 musical notes. – Colas Oct 20 '17 at 8:15
• see as well (math.stackexchange.com/q/11669) – Jean Marie Oct 20 '17 at 8:50
• Related (andrewduncan.net/cmt) – Jean Marie Oct 20 '17 at 10:49
• Each term in the continued fraction ($a_n$) corresponds to a convergent. You are giving the "semiconvergents" when you include $\frac53$ and $\frac{11}7$. For their denominators, these are not as close to the number being approximated as the convergents. – robjohn Oct 20 '17 at 14:11

Lemma: Let $\alpha, \beta$ be linearly independent over rationals $\mathbb{Q}$; i.e. $\dfrac{\alpha}{\beta} \notin \mathbb{Q}$.
Then the group generated by $\alpha, \beta$ is dense in $\mathbb{Q}$;
i.e. $\{ i\alpha+j\beta \ | \ i,j \in \mathbb{Z} \} = \{ i\alpha-j\beta \ | \ i,j \in \mathbb{Z} \}$ is dense in $\mathbb{Q}$.

Now let $\alpha:=\log2$ and $\beta:=\log3$, note that $\dfrac{\beta}{\alpha}=\dfrac{\log3}{\log2}=\log_23 \notin \mathbb{Q}$;
so if we let $i, j$ varies arbitrary in $\mathbb{N}$,
then the phrase $i\alpha-j\beta$ will become as small as we want;

i.e. in more mathematical discipline we can say:
for every $0 < \delta \in \mathbb{R}$ there exist $i, j \in \mathbb{N}$ such that $0 < i\alpha-j\beta < \delta$.

Note that: $$10^{i\alpha-j\beta}= \dfrac{10^{i\alpha}}{10^{j\beta}}= \dfrac{(10^{\alpha})^{i}}{(10^{\beta})^{j}}= \dfrac{2^i}{3^j}$$

So we can conclude that:
for every $0 < \epsilon \in \mathbb{R}$ there exist $i, j \in \mathbb{N}$ such that $1 < \dfrac{2^i}{3^j} < 1+ \epsilon$.

So ignoring some mathematical disciplines we can say:
if we let $i, j$ varies arbitrary in $\mathbb{N}$,
then the phrase $\dfrac{2^i}{3^j}$ will become as close to $1$ as we want.

We can restase some similar facts about $-\delta < i\alpha-j\beta < 0$ and $-\delta < i\alpha-j\beta < \delta$;
also for $1- \epsilon < \dfrac{2^i}{3^j} < 1$ and $1- \epsilon < \dfrac{2^i}{3^j} < 1+ \epsilon$; but it does not matters.

I have seen this problem before:

unsolved problems in number theory by Richard K. Guy;
$F$(None of the above);
$F23$(Small differences between powers of $2$ and $3$.) :

Problem $1$ of Littlewood's book asks how small $3^n — 2^m$ can be in comparison with $2^m$.
He gives as an example $$\dfrac{3^{12}} {2^{19}} = 1+ \dfrac{ 7153 } {524288} \approx 1+ \dfrac{ 1 } { 73 } \qquad \text{ (the ratio of } D^{\sharp} \text{ to } E^{\flat}).$$ [Where $\sharp$ and $\flat$ stands for dièse and Bémol.]

The first few convergents to the continued fraction (see $F20$) $$log_23= 1+\tfrac{1} {1+\tfrac{1} {1+\tfrac{1} {2+\tfrac{1} {2+\tfrac{1} {3+\tfrac{1} {1+\tfrac{1} {\cdots} } } } } } }$$

for $\log 3$ to the base $2$ are: $$\dfrac{ 1}{ 1}, \dfrac{ 2}{ 1}, \dfrac{ 3}{ 2}, \dfrac{ 8}{ 5}, \dfrac{19}{12}, \dfrac{65}{41}, \dfrac{84}{53}, \dots$$ so Victor Meally observed that
the $\color{Blue}{\text{octave}}$ may conveniently be partitioned into $12$, $41$ or $53$ intervals, and that the system of temperament with $53$ degrees is due to Nicolaus Mercator ($1620-1687$; not Gerhardus, $1512-1594$, of map projection fame).

Ellison used the Gel'fond-Baker method to show that

$$|2^x-3^y| > 2^x e^{\dfrac{-x}{10}} \qquad \text{for} \qquad x > 27$$

and Tijdeman used it to show that there is a $c \geq 1$ such that $2^x - 3^y > \dfrac{2^x}{x^c}$.

Croft asks the corresponding question for $n! — 2^m$. $\ \ \cdots$

That was the exact text from the book, note that the pair $(m,n)$ has been used in the reversed order in that book versus this question has been asked in this site!

At the end of section $F23$; there are four refrences as follows:

• F. Beukers, Fractional parts of powers of rationals, Math. Proc. Cambridge Philos. Soc., $90(1981) 13-20$; MR $83$g:$10028$.

• A. K. Dubitskas, A lower bound on the value of $||(3/2)^k||$ (Russian), Uspekhi Mat. Nauk, $45(1990) 153-154$; translated in Russian Math. Surveys, $45(1990) 163-164$; MR $91$k:$11058$.

• W. J. Ellison, Recipes for solving diophantine problems by Baker's method, S6m. Thiorie Nombms, $1970-71$, Exp. No. $11$, C.N.R.S. Talence, $1971$.

• R. Tijdeman, On integers with many small factors, Compositio Math., $26 (1973) 319-330$.

• I have found an article by Kurt Mahler with the same name as the first one here.

Also in page $14$ of this paper of Michel Waldschmidt; you can find something.

But I don't know how this problem arise from (music);
and from where the upper bound 20 comes from.

The continued fraction for $\frac{\log(3)}{\log(2)}$ is $$1;1,1,2,2,3,1,5,2,23,2,2,1,1,55,1,4,3,1,1,\dots$$ The convergents are $$\frac11,\frac21,\frac32,\frac85,\frac{19}{12},\frac{65}{41},\frac{84}{53},\frac{485}{306},\frac{1054}{665},\frac{24727}{15601},\frac{50508}{31867},\frac{125743}{79335},\frac{176251}{111202},\frac{301994}{190537},\frac{16785921}{10590737},\frac{17087915}{10781274},\frac{85137581}{53715833}, \frac{272500658}{171928773},\frac{357638239}{225644606}, \frac{630138897}{397573379},\cdots$$ This gives a number of "close" approximations. For example, the best with exponents between $1$ and $20$, is $$\left|\frac{2^{19}}{3^{12}}-1\right|=0.01345963145485576$$ This gives the approximation $2^{19/12}=2.9966$. Since their are $12$ half-steps to the octave, this approximation is very useful.

• A certain number of convergents are missing in your list : $5/3$, $11/7, 46/29...$ – Jean Marie Oct 20 '17 at 10:10
• @JeanMarie: are you talking about convergents, or semiconvergents, which are not as close as convergents for their denominators? – robjohn Oct 20 '17 at 14:03