Article 6YR2G Powers of 3 + √2

Powers of 3 + √2

by
John
from John D. Cook on (#6YR2G)

Yesterday's post looked at the distribution of powers ofx mod 1. For almost allx > 1 the distribution is uniform in the limit. But there are exceptions, and the post raised the question of whether 3 + 2 is an exception.

A plot made itlook like 3 + 2 is an exception, but that turned out to be a numerical problem.

power_mod_n1.png

A higher precision calculation showed that the zeros on the right end of the plot were erroneous.

power_mod_n2.png

So this raises the question of how to calculate (3 + 2)n accurately for large n. The way I created the second plot was to use bc to numerically calculate the powers of 3 + 2. In this post, I'll look at using Mathematica to calculate the powers symbolically.

For all positive integers n,

(3 + 2)n = an + bn2

where an and bn are positive integers. We want to compute thea andb values.

If you ask Mathematica to compute (3 + 2)n it will simply echo the expression. But if you use the Expand function it will give you want you want. For example

 Expand[(3 + Sqrt[2])^10]

returns

 1404491 + 993054 2

We can use the Coefficient function to split a + b 2 into a and b.

 parts[n_] := Module[{x = (3 + Sqrt[2])^n}, {Coefficient[x, Sqrt[2], 0], Coefficient[x, Sqrt[2], 1]}]

Now parts[10] returns the pair {1404491, 993054}.

Here's something interesting. If we set

(3 + 2)n = an + bn2

as above, then the two halves of the expression on the right are asymptotically equal. That is, asn goes to infinity, the ratio

an / bn2

converges to 1.

We can see this by defining

 ratio[n_] := Module[ {a = Part[ parts[n], 1], b = Part[parts[n], 2]}, N[a / (b Sqrt[2])]]

and evaluating ratio at increasing values of n. ratio[12] returns 1.00001 and ratio[13] returns 1, not that the ratio is exactly 1, but it is as close to 1 as a floating point number can represent.

This seems to be true more generally, as we can investigate with the following function.

 ratio2[p_, q_, r_, n_] := Module[{x = (p + q Sqrt[r])^n}, N[Coefficient[x, Sqrt[r], 0]/(Coefficient[x, Sqrt[r], 1] Sqrt[r])]]

When r is a prime and

(p+ qr)n = an + bnr

then it seems that the ratio an / bn r converges to 1 asn goes to infinity. For example, ratio2[3, 5, 11, 40] returns 1, meaning that the two halves of the expression for (3 + 511)n are asymptotically equal.

I don't know whether the suggested result is true, or how to prove it if it is true. Feels like a result from algebraic number theory, which is not something I know much about.

Update: An anonymous person on X suggested a clever and simple proof. Observe that

3sqrt2proof.svg

In this form it's clear that the ratio an / bn 2 converges to 1, and the proof can be generalized to cover more.

The post Powers of 3 + 2 first appeared on John D. Cook.
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