3

For any rational $r=n/d$, define

$$h_s(r) = (nd)^s$$

where $s > 0$ is a free parameter. The intent is for this to be a representation of how "simple" each rational is; simpler rationals are ranked lower.

Now, suppose $S$ is a finitely-generated subgroup of the multiplicative group of strictly positive rationals $\Bbb Q^+$. Then we can use our $h_s$ to define a useful notion of the "size" of the subgroup as follows:

$$h_s(S) = \sum_{r \in S} \frac{1}{h_s(r)}$$

This is guaranteed to converge whenever $s>0$. Furthermore, whenever we have $s \in \Bbb N^+$, it even seems to converge to a rational number. Let's call this the rational associated with $S$ for some value of $s$.

QUESTION: is there some closed-form expression for each sum which gives the rational associated with an arbitrary subgroup, at least for $s=1$? Has this been studied and does it hav ea name?

Here are some examples for $s=1$, ranked from "largest" to "smallest."

For the following examples, I know how to get a closed-form expression:

$$ \begin{align} h_1(\langle 2,3,5,7 \rangle) &= 12 \\ h_1(\langle 2,3,5 \rangle) &= 9 \\ h_1(\langle 2,3,7 \rangle) &= 8 \\ h_1(\langle 2,3,25 \rangle) &= 13/2 = 6.5 \\ h_1(\langle 2,9,5 \rangle) &= 45/8 = 5.625 \\ h_1(\langle 3/2, 7/5 \rangle) &= 126/85 \approx 1.482 \end{align}$$

Here are some examples of subgroups for which I don't have a closed form expression, but which are fairly easy to evaluate numerically, and which can easily be seen to lead to a rational value

$$ \begin{align} h_1(\langle 2,9,5/3 \rangle) &= 5 \\ h_1(\langle 4,3/2,5 \rangle) &= 37/8 \approx 4.625 \\ h_1(\langle 6,10 \rangle) &= 584/315 \approx 1.854 \\ h_1(\langle 3/2,5/2 \rangle) &= 584/315 \approx 1.854 \\ h_1(\langle 6,15/2 \rangle) &= 3459/2090 \approx 1.655 \end{align}$$

These sums are easy to play with, since they converge relatively quickly, at least for small-rank subgroups. Taking them to 1000 terms can be done in a few seconds, and for all of these you get a relatively short continued fraction that suddenly "almost terminates" with some enormous number in the billions or so.

The first list is easy to get in closed form because each subgroup has a basis of pairwise-coprime rationals, which makes things pretty easy to get. To see this, first, note that we have the following for our $h_s(r)$:

  • For any rational $r \neq 1$ and $s>0$, we have $h_s(r) > 1$.
  • For any rational $r$ and any integer $z$, we have $h_s(r^z) = h_s(r)^{|z|}$
  • For any two coprime rationals $r_1, r_2$, we have $h_s(r_1 \cdot r_2) = h_s(r_1) \cdot h_s(r_2)$.

Then if we have $S = \langle r_1, r_2, \ldots, r_n \rangle$, and all the $r_i$ are coprime, then we can write any rational $r$ as ${r_1}^{a_1} \cdot {r_2}^{a_2} \cdot \ldots {r_n}^{a_n}$, so that we have $h_s(r) = h_s(r_1)^{|a_1|} \cdot h_s(r_2)^{|a_2|} \cdot \ldots \cdot h_s(r_n)^{|a_n|}$. As a result, we can write our summation as

$$h_s(\langle r_1, r_2, \ldots, r_n \rangle) = \sum_{a_1, a_2, \ldots, a_n \in \Bbb Z^n} \left( \frac{1}{{h_s(r_1)}^{|a_1|}} \cdot \frac{1}{{h_s(r_2)}^{|a_2|}} \cdot \ldots \cdot \frac{1}{{h_s(r_n)}^{|a_n|}}\right)$$

which we can factor into a product of sums:

$$h_s(\langle r_1, r_2, \ldots, r_n \rangle) = \left( \sum_{a_1 \in \Bbb Z} \frac{1}{{h_s(r_1)}^{|a_1|}} \right) \left( \sum_{a_2 \in \Bbb Z} \frac{1}{{h_s(r_2)}^{|a_2|}} \right) \ldots \left( \sum_{a_n \in \Bbb Z} \frac{1}{{h_s(r_n)}^{|a_n|}} \right)$$

which, as long as $s>0$, converges via some simple power series manipulations to

$$h_s(\langle r_1, r_2, \ldots, r_n \rangle) = \left( \frac{h_s(r_1)+1}{h_s(r_1)-1} \right) \left( \frac{h_s(r_2)+1}{h_s(r_2)-1} \right) \ldots \left( \frac{h_s(r_n)+1}{h_s(r_n)-1} \right)$$

Any subgroup generated by prime numbers will be of the above form. So since this expression converges for all such subgroups, we know it will converge for any multiplicative subgroup of $\Bbb Q^+$.

Mike Battaglia
  • 8,582

2 Answers2

2

For $S$ a finite index subgroup of $T = \{2,3,\dots,p\}$ for some prime number $p$,

$$\sum_{r\in S} \frac{1}{h_1(r)} = \sum_{r\in T} \frac{[r\in S]}{h_1(r)} = \frac{1}{[T : S]}\sum_{r\in T} \sum_{\chi : T/S \to \mathbb{C}^*} \frac{\chi(r)}{h_1(r)} $$

where the second sum ranges over characters of $T/S$ interpreted as functions from $T$ to $\mathbb{C}$. But since such characters are completely multiplicative, and $h_1$ is multiplicative, this equals

$$\frac{1}{[T : S]} \sum_\chi \prod_{q}\left( \sum_{r\in \langle q\rangle} \frac{\chi(r)}{h_1(r)} \right)=\frac{1}{[T : S]} \sum_\chi \prod_{q}\left( \frac{q + \chi(q)}{q - \chi(q)} \right)$$

where the products range over primes $2,3,\dots,p$. If $n := [T : S]$, the latter is an element of $\mathbb{Q}[e^{2\pi i / n}]$ fixed by the Galois group of $\mathbb{Q}[e^{2\pi i / n}] / \mathbb{Q}$, hence rational.

Praveen Venkataramana
  • 21
1

I'm almost sure that $h_{s}(S)$ are rational for $s\in \mathbb{Z}_{>0}$, and here's a computation for $h_{1}(\langle 2, 9, 5/3\rangle) = 5$.

Each $r\in S = \langle 2, 9, 5/3\rangle$ has a form of $$ r = 2^{a} 3^{2b} \left(\frac{5}{3}\right)^{c} = 2^{a} 3^{2b-c}5^{c} $$ for $a, b, c\in \mathbb{Z}$, and $$ h_{1}(S) = \sum_{a, b, c\in \mathbb{Z}} \frac{1}{2^{|a|}3^{|2b-c|}5^{|c|}} \\ = \left(\frac{2 + 1}{2 - 1}\right) \sum_{b, c\in \mathbb{Z}} \frac{1}{3^{|2b-c|} 5^{|c|}} \\ = 3 \sum_{c\in \mathbb{Z}} \frac{1}{5^{|c|}} \sum_{b\in \mathbb{Z}} \frac{1}{3^{|2b-c|}} $$ Now the sum over $b$ depends on the parity of $c$: $$ \sum_{b\in \mathbb{Z}} \frac{1}{3^{|2b-c|}} = \begin{cases} \sum_{b\in \mathbb{Z}} \frac{1}{3^{2|b|}} = \frac{5}{4} & 2|c \\ \sum_{b\in \mathbb{Z}}\frac{1}{3^{|2b-1|}} = \frac{3}{4} & 2\nmid c \end{cases} $$ so $$ h_{1}(S) = 3\left( \frac{5}{4} \left( 1 + \frac{2}{5^{2}} + \cdots\right) + \frac{3}{4} \left( \frac{2}{5} + \frac{2}{5^{3}} + \cdots\right)\right) = 5. $$

For general $S$, I think some linear algebra may help to show that $h_{1}(S)$ are rational.

Seewoo Lee
  • 15,670