Almost-integers

You'll find something interesting if you take the golden ratio, $$\phi = \frac{1 + \sqrt{5}}{2}$$, to high powers:

$\phi^6 = 17.944$ $\phi^{10} = 122.992$ $\phi^{15} = 1364.0007$

It seems as if large exponents of $$\phi$$ come arbitrarily close to integer values. In fact, $$\phi$$ isn't the only number which does this. For example, $$(2+ \sqrt{3})^{5} = 723.999$$. This is no coincidence.

These numbers - known as Pisot-Vijayaraghavan numbers - produce "almost-integers" when raised to high powers. I'll give a quick sketch of why this is, as well as an easy way to find other numbers with this property:

We can take a number $$q = a + \sqrt{b}$$, where $$a$$ and $$b$$ are integers, and consider how it behaves under an operation called Galois conjugation. Essentially, this is just a more general version of what you may already know as complex conjugation. To find the Galois conjugate of $$q$$, simply change the sign of the square root term: $$q* = a - \sqrt{b}$$. This is equivalent to setting $$q$$ to be the root of the polynomial $$x^2 - 2ax + a^2 - b = 0$$ and finding the second root, or $$q*$$.

A few properties of Galois conjugation can be easily deduced, and are already implicit in the quadratic equation above:

$q + q* = 2a$ $qq* = a^2 - b$ $q^2 + q*^2 = 2a^2 + 2b$ $q^n + q*^n = \sum \binom{n}{k} a^{n-2k}b^{2k}$

The last equality is important here. Notice especially that these four equations all produce integers results on the righthand side. Now, whenever $$|q*| < 1$$, the second term of $$q^n + q*^n$$ becomes arbitrarily small for increasing $$n$$. This immediately implies that

$q^n + q*^n \approx q^n \approx \sum \binom{n}{k} a^{n-2k}b^{2k} \qquad n \gg 1$

Thus, any number $$q$$ whose Galois conjugate satisfies $$|q*| < 1$$ will produce near-integer values when raised to high powers. Here are some examples of such numbers:

$(2 + \sqrt{6})^7 = 34528.0037$ $(4 + \sqrt{21})^9 = 252672968.008$ $(1 + \sqrt{3})^{15} = 3526400.0093$

*It's worth noting that in the case that $$q$$ has a denominator of $$t$$, as is the case with $$\phi$$ where $$t=2$$, then we also need the further restriction that $$t^2 | qq*$$, or that $$a^2 -b$$ is divisible by $$t^2$$ so that $$qq*$$ remains an integer.

3 weeks, 2 days ago

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@Levi Adam Walker Yes, this is amazing right???!! I got to know about these numbers quite recently I must confess........But, check out "Heegner Number "...........Wikipedia has a page on it too...........when it comes to "Close" approximations, I prefer to quote them........:)

- 3 weeks, 1 day ago

Yeah, they're much cooler, but also much harder to explain :p

- 3 weeks, 1 day ago

Does this also work with cuberoots? I suppose $$q*$$ has to be defined differently, right?

- 2 weeks, 4 days ago

It won't generally work with this construction, since $$q^n + q*^n$$ must give an integer and cube roots tend to act erratically when raised to arbitrary powers.

- 2 weeks, 4 days ago

Ah, right. Thanks!

- 2 weeks, 4 days ago