Republic of Mathematics blog

Some stubborn problems associated with the number 3/2

Posted by: Gary Ernest Davis on: December 4, 2010

Fractional parts of powers of \frac{3}{2}

The powers of \frac{3}{2} increase exponentially:

However, if we keep only the fractional part of (\frac{3}{2})^n – that is, (\frac{3}{2})^n-Floor[(\frac{3}{2})^n] – quite a different picture results:

The first few of these numbers are 1/2, 1/4, 3/8, 1/16, 19/32, 25/64, 11/128, 161/256, 227/512, 681/1024, 1019/2048, 3057/4096.

Despite much study, there are things about this sequence of fractional numbers that are not known.

Equidistribution

If we take an interval of numbers such as [0.2, 0.3] and for each n\geq 1 count how many times (\frac{3}{2})^n lies between 0.2 \textrm{ and } 0.3, will the ratio of the number of times to n approach the length of this interval – namely, 0.1 – as n increases?

For example, in the second picture above, it is true that for exactly 101 values of n we have 0.2\leq (\frac{3}{2})^n\leq 0.3, and \frac{101}{1000} =0.101\approx 0.1.

This property, when true for any interval [a,b]\textrm{ with } 0\leq a < b\leq 1 is known as equidistribution of the sequence of numbers (\frac{3}{2})^n.

Back in 1914 Hardy & Littlewood  proved that for “most” numbers x the numbers x^n-Floor[x^n] are equidistributed in the interval [0,1].

(Hardy, G. H. and Littlewood, J. E. “Some Problems of Diophantine Approximation.” Acta Math. 37, 193-239, 1914)

Godfrey Hardy and John Littlewood

It is not known if x=\frac{3}{2} is one of those numbers.

Density

An equidistributed sequence of numbers 0\leq x_n\leq 1 must land in every interval [a,b] \textrm{ with } 0\leq a < b\leq 1 because the x_n must land in [a,b] the right number of times.

The property of landing in each interval [a,b] at least once is called density of the sequence x_n.

It is not even known if the sequence (\frac{3}{2})^n is dense, let alone whether it is equidistributed.

A powerful, but stubborn, inequality

It is widely believed, but still not known, if the fractional part of (\frac{3}{2})^n, namely (\frac{3}{2})^n-Floor[(\frac{3}{2})^n], satisfies the inequality:

(\frac{3}{2})^n-Floor[(\frac{3}{2})^n] \leq 1-(\frac{3}{4})^n for all n\geq 1………………..(*)

If this inequality is true then a famous problem of number theory – Waring’s problem – is known to be true.

The inequality (*) seems innocent enough, yet no-one to date has been able to prove it is true.

If (*) is true and we plot F(n):=1-(\frac{3}{4})^n -((\frac{3}{2})^n-Floor[(\frac{3}{2})^n]) versus n we should never get a negative value.

A plot of the first 10,000 of these numbers F(n) looks as follows:

Of course it’s hard to see from this plot how close to 0 the values of F(n) get.

There are values of n for which F(n) is smaller than F(i) for all previous i.

The first few of these values of n and the corresponding values of F(n) are:

Values of n for which F(n) is less than any previous F(i) F(n)
5 0.168945
14 0.0529218
46 0.0317909
58 0.0297681
105 0.0144072
157 0.00725365
163 0.00449855
455 0.00314395
1060 0.00110873
1256 0.000564205
2677 0.000363945
8093 0.0000809868
28277 0.0000604566
33327 0.0000549506
49304 0.00000736405

x

Is this a “good” problem?

The inequality (*) is intriguing, and if you can prove it you will undoubtedly become (mathematically) famous.

But it is it a good problem to work on?

A graduate student should not work on this problem because the chances of getting anywhere new are very slim.

A professional mathematician might play with it, in the hope of having a few ideas.

School students might also play with it in the hope of getting a sense of what the problem means and why it is hard: there are very few such hard problems in mathematics, with such strong implications, that are so easily stated.

A good starting point for school students to experiment is with the the fractional part of the powers of \frac{1+\sqrt{5}}{2}.

1 Response to "Some stubborn problems associated with the number 3/2"

You asked on Twitter: Is there an x* > 0 such that 1-(3/4)^x-frac[(3/2)^x ] > 0 for all x > x* ? I think there is no such x*, but I don’t think I can rigorously prove why… This is assuming that x can be any real number, and is not restricted to integers.

My reason has to do with the nice continuous nature of f(x) = (3/2)^x. All that frac[(3/2)^x] does is translate a section of f(x) = (3/2)^x down.

For x from negative infinity to 0, you translate down 0 units. For x from 0 to [log (2)/log (3/2)], you translate down 1 unit. Since f(x) = (3/2)^x is continuous everywhere, then (3/2)^x gets as close as you want to 2 as x approaches [log (2)/log (3/2)]. And, it gets there faster than 1-(3/4)^x gets close to 1, since the slope of 1-(3/4)^x is decreasing, whereas the slope of (3/2)^x is increasing.

The same thing happens on every interval, and the intervals get smaller and smaller as x gets bigger, so it becomes even more frequent that frac[(3/2)^x ] “crosses” 1 – (3/4)^x

Can anyone help me formalize that?

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