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Has anyone come across a graph like this?

enter image description here

The black circles represent rationals in $(0,1)$ and their heights are roughly proportional to the reciprocal of the square of their lowest terms denominator. The red lines are drawn by eye on the pattern of the black dots.

This came from trying to create a probability distribution on the rationals where $$\Pr\left(X = \frac{a}{b}\right) = \frac{\zeta(k)}{\zeta(k-1) - \zeta(k) } \left(\frac{1}{b}\right)^k$$ where $0 \lt a \lt b$ with $a$ and $b$ coprime and where $k \gt 2$.

The red lines look somewhat like the left half of the Stern-Brocot Tree except that points with different denominators are at different heights.

Henry
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    This is a very nice looking graph. – NoChance Sep 01 '12 at 23:32
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    I'd say that the red lines are the left half of the Stern-Brocot tree. The heights used in a diagram that represents the tree are not an intrinsic feature of the tree itself. – hmakholm left over Monica Sep 01 '12 at 23:39
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    It seems interesting and highly symmetric like a gaussian or laplace distribution. Meanwhile what is $\zeta$ – Seyhmus Güngören Sep 02 '12 at 00:05
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    I think you mean the heights are roughly proportional to the reciprocal of the square of the denominator. The function $f(p/q)=1/q$, $f(x)=0$ for $x$ irrational is the standard example of a function continuous at $x$ if and only if $x$ is irrational. You are (roughly) squaring that function. – Gerry Myerson Sep 02 '12 at 06:04
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    @Seyhmus: $\zeta(k)$ is the Riemann zeta function – Henry Sep 02 '12 at 07:48
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    @Gerry: Fair point - I will edit – Henry Sep 02 '12 at 07:49
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    There is somewhat similar graph here http://blogs.scientificamerican.com/critical-opalescence/2012/08/06/how-do-you-count-parallel-universes-you-cant-just-go-1-2-3/ – NoChance Sep 09 '12 at 08:47
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    "their heights are roughly proportional to the reciprocal of the square of their lowest terms denominator" Substitute "exactly" for "roughly" and you've got the locations of the centers of the Ford circles. See my answer below. – Michael Hardy Jan 01 '15 at 20:52
  • Nice graph. What value of $k$ did you use to make the graph? Also, how did you obtain the normalizing constant? – user551774 Aug 28 '21 at 01:22
  • @apelt001 $k$ marginally over $2$ and since there is no marked vertical scale there is no need to normalise to draw the graph. $\Pr\left(X = \frac{a}{b}\right) = \frac{\zeta(k)}{\zeta(k-1) - \zeta(k) } \left(\frac{1}{b}\right)^k$ indicates what it would be – Henry Aug 28 '21 at 01:56
  • @Henry yes it’s just interesting that the normalizing constant is what it is, and I’m not sure how that works, because the lebesgue integral of such a density function between $0$ and $1$ would evaluate to $0$. – user551774 Aug 28 '21 at 04:36
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    @apelt001 It is not a density function but a probability mass function, which sums to $1$ over $q\in \mathbb Q \cap (0,1)$, for $k>2$ – Henry Aug 28 '21 at 11:46
  • @Henry Oh I see, thanks for explaining my mistake. I got my own answer to the normalizing constant: $$\sum_{a,b:\ (a,b)=1 \ 0<\frac{a}{b}< 1} = \sum_{b=2}^\infty\sum_{(a,b)=1} (\frac{1}{b})^k = \sum_{b=2}^\infty \frac{\varphi(b)}{b^k}=\frac{\zeta(k-1)}{\zeta(k)}-1.$$ – user551774 Aug 28 '21 at 16:08
  • @apelt001 dividing by $\frac{\zeta(k-1)}{\zeta(k)}-1$ is the same as multiplying by $\frac{\zeta(k)}{\zeta(k-1) - \zeta(k) }$ – Henry Aug 28 '21 at 16:16
  • @Henry I know. I was just stating that that's the answer I wanted thanks. – user551774 Aug 28 '21 at 16:17

2 Answers2

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As Gerry mentions, this is (roughly) the square of Thomae's function, also known as the popcorn function, the raindrop function, the countable cloud function, the modified Dirichlet function, the ruler function, the Riemann function, or the Stars over Babylon...

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It is surprising that no one has mentioned Ford circles. The heights of the centers of Ford circles are exactly proportional to the reciprocals of the denominators.

Let us suppose that the content of the section on total area of Ford circles is correct. Then if a point is randomly chosen in the interior of one of the Ford circles corresponding to a rational number in the interval $[0,1]$, the probability of its being in any region proportional to the region's area, then the probability that it is in the circle corresponding to a particular rational number with denominator $k$ is $$ \frac{\zeta(4)}{\zeta(3)} \left(\frac 1 k\right)^4. $$

Michael Hardy
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