Thomae's function

Thomae's function is a real-valued function of a real variable that can be defined as:^[1]: 531 $${\displaystyle f(x)={\begin{cases}{\frac {1}{q}}&{\text{if }}x={\tfrac {p}{q}}\quad (x{\text{ is rational), with }}p\in \mathbb {Z} {\text{ and }}q\in \mathbb {N} {\text{ coprime}}\\0&{\text{if }}x{\text{ is irrational.}}\end{cases}}}$$

Point plot on the interval (0,1). The topmost point in the middle shows f(1/2) = 1/2

It is named after Carl Johannes Thomae, but has many other names: the popcorn function, the raindrop function, the countable cloud function, the modified Dirichlet function, the ruler function (not to be confused with the integer ruler function),^[2] the Riemann function, or the Stars over Babylon (John Horton Conway's name).^[3] Thomae mentioned it as an example for an integrable function with infinitely many discontinuities in an early textbook on Riemann's notion of integration.^[4]

Since every rational number has a unique representation with coprime (also termed relatively prime) ${\displaystyle p\in \mathbb {Z} }$ and ${\displaystyle q\in \mathbb {N} }$, the function is well-defined. Note that ${\displaystyle q=+1}$ is the only number in ${\displaystyle \mathbb {N} }$ that is coprime to ${\displaystyle p=0.}$

It is a modification of the Dirichlet function, which is 1 at rational numbers and 0 elsewhere.

Properties

Proof of periodicity

For all ${\displaystyle x\in \mathbb {R} \setminus \mathbb {Q} ,}$ we also have ${\displaystyle x+n\in \mathbb {R} \setminus \mathbb {Q} }$ and hence ${\displaystyle f(x+n)=f(x)=0,}$ For all ${\displaystyle x\in \mathbb {Q} ,\;}$ there exist ${\displaystyle p\in \mathbb {Z} }$ and ${\displaystyle q\in \mathbb {N} }$ such that ${\displaystyle \;x=p/q,\;}$ and ${\displaystyle \gcd(p,\;q)=1.}$ Consider ${\displaystyle x+n=(p+nq)/q}$. If ${\displaystyle d}$ divides ${\displaystyle p}$ and ${\displaystyle q}$, it divides ${\displaystyle p+nq}$ and ${\displaystyle q}$. Conversely, if ${\displaystyle d}$ divides ${\displaystyle p+nq}$ and ${\displaystyle q}$, it divides ${\displaystyle (p+nq)-nq=p}$ and ${\displaystyle q}$. So ${\displaystyle \gcd(p+nq,q)=\gcd(p,q)=1}$, and ${\displaystyle f(x+n)=1/q=f(x)}$.

Proof of discontinuity at rational numbers

Let ${\displaystyle x_{0}=p/q}$ be an arbitrary rational number, with ${\displaystyle \;p\in \mathbb {Z} ,\;q\in \mathbb {N} ,}$ and ${\displaystyle p}$ and ${\displaystyle q}$ coprime. This establishes ${\displaystyle f(x_{0})=1/q.}$ Let ${\displaystyle \;\alpha \in \mathbb {R} \setminus \mathbb {Q} \;}$ be any irrational number and define ${\displaystyle x_{n}=x_{0}+{\frac {\alpha }{n}}}$ for all ${\displaystyle n\in \mathbb {N} .}$ These ${\displaystyle x_{n}}$ are all irrational, and so ${\displaystyle f(x_{n})=0}$ for all ${\displaystyle n\in \mathbb {N} .}$ This implies ${\displaystyle |x_{0}-x_{n}|={\frac {\alpha }{n}},}$ and ${\displaystyle |f(x_{0})-f(x_{n})|={\frac {1}{q}}.}$ Let ${\displaystyle \;\varepsilon =1/q\;}$, and given ${\displaystyle \delta >0}$ let ${\displaystyle n=1+\left\lceil {\frac {\alpha }{\delta }}\right\rceil .}$ For the corresponding ${\displaystyle \;x_{n}}$ we have $${\displaystyle |f(x_{0})-f(x_{n})|=1/q\geq \varepsilon }$$ and $${\displaystyle |x_{0}-x_{n}|={\frac {\alpha }{n}}={\frac {\alpha }{1+\left\lceil {\frac {\alpha }{\delta }}\right\rceil }}<{\frac {\alpha }{\left\lceil {\frac {\alpha }{\delta }}\right\rceil }}\leq \delta ,}$$ which is exactly the definition of discontinuity of ${\displaystyle f}$ at ${\displaystyle x_{0}}$.

Proof of continuity at irrational arguments

Since ${\displaystyle f}$ is periodic with period ${\displaystyle 1}$ and ${\displaystyle 0\in \mathbb {Q} ,}$ it suffices to check all irrational points in ${\displaystyle I=(0,1).\;}$ Assume now ${\displaystyle \varepsilon >0,\;i\in \mathbb {N} }$ and ${\displaystyle x_{0}\in I\setminus \mathbb {Q} .}$ According to the Archimedean property of the reals, there exists ${\displaystyle r\in \mathbb {N} }$ with ${\displaystyle 1/r<\varepsilon ,}$ and there exist ${\displaystyle \;k_{i}\in \mathbb {N} ,}$ such that for ${\displaystyle i=1,\ldots ,r}$ we have ${\displaystyle 0<{\frac {k_{i}}{i}}<x_{0}<{\frac {k_{i}+1}{i}}.}$ The minimal distance of ${\displaystyle x_{0}}$ to its i-th lower and upper bounds equals $${\displaystyle d_{i}:=\min \left\{\left|x_{0}-{\frac {k_{i}}{i}}\right|,\;\left|x_{0}-{\frac {k_{i}+1}{i}}\right|\right\}.}$$ We define ${\displaystyle \delta }$ as the minimum of all the finitely many ${\displaystyle d_{i}.}$ $${\displaystyle \delta$$ :=\min _{1\leq i\leq r}\{d_{i}\},\;} so that for all ${\displaystyle i=1,\dots ,r,}$ ${\displaystyle |x_{0}-k_{i}/i|\geq \delta }$ and ${\displaystyle |x_{0}-(k_{i}+1)/i|\geq \delta .}$ This is to say, all these rational numbers ${\displaystyle k_{i}/i,\;(k_{i}+1)/i,\;}$ are outside the ${\displaystyle \delta }$-neighborhood of ${\displaystyle x_{0}.}$ Now let ${\displaystyle x\in \mathbb {Q} \cap (x_{0}-\delta ,x_{0}+\delta )}$ with the unique representation ${\displaystyle x=p/q}$ where ${\displaystyle p,q\in \mathbb {N} }$ are coprime. Then, necessarily, ${\displaystyle q>r,\;}$ and therefore, $${\displaystyle f(x)=1/q<1/r<\varepsilon .}$$ Likewise, for all irrational ${\displaystyle x\in I,\;f(x)=0=f(x_{0}),\;}$ and thus, if ${\displaystyle \varepsilon >0}$ then any choice of (sufficiently small) ${\displaystyle \delta >0}$ gives $${\displaystyle |x-x_{0}|<\delta \implies |f(x_{0})-f(x)|=f(x)<\varepsilon .}$$ Therefore, ${\displaystyle f}$ is continuous on ${\displaystyle \mathbb {R} \setminus \mathbb {Q} .}$

Proof of being nowhere differentiable

For rational numbers, this follows from non-continuity. For irrational numbers: For any sequence of irrational numbers ${\displaystyle (a_{n})_{n=1}^{\infty }}$ with ${\displaystyle a_{n}\neq x_{0}}$ for all ${\displaystyle n\in \mathbb {N} _{+}}$ that converges to the irrational point ${\displaystyle x_{0}}$, the sequence ${\displaystyle (f(a_{n}))_{n=1}^{\infty }}$ is identically ${\displaystyle 0}$, and so ${\displaystyle \lim _{n\to \infty }\left|{\frac {f(a_{n})-f(x_{0})}{a_{n}-x_{0}}}\right|=0}$. On the other hand, consider the sequence of rational numbers ${\displaystyle (b_{n})_{n=1}^{\infty }}$ with ${\displaystyle b_{n}=\lfloor nx_{0}\rfloor /n}$, where ${\displaystyle \lfloor nx_{0}\rfloor }$ denotes the floor of ${\displaystyle nx_{0}}$. Since ${\displaystyle nx_{0}-1<\lfloor nx_{0}\rfloor \leq nx_{0}}$, the sequence ${\displaystyle (b_{n})_{n=1}^{\infty }}$ converges to ${\displaystyle x_{0}}$ using the Squeeze theorem. Also, ${\displaystyle |b_{n}-x_{0}|=|\lfloor nx_{0}\rfloor /n-x_{0}|=|\lfloor nx_{0}\rfloor -nx_{0}|/n\leq 1/n}$ for all ${\displaystyle n}$. Thus for all ${\displaystyle n}$, ${\displaystyle \left|{\frac {f(b_{n})-f(x_{0})}{b_{n}-x_{0}}}\right|\geq {\frac {1/n-0}{1/n}}=1}$. Therefore we obtain ${\displaystyle \liminf _{n\to \infty }\left|{\frac {f(b_{n})-f(x_{0})}{b_{n}-x_{0}}}\right|\geq 1\neq 0}$ and so ${\displaystyle f}$ is not differentiable at any irrational number ${\displaystyle x_{0}}$.

• Thomae's function ${\displaystyle f}$ is bounded and maps all real numbers to the unit interval:${\displaystyle f:\mathbb {R} \to [0,1].}$
• ${\displaystyle f}$ is periodic with period ${\displaystyle 1:\;f(x+n)=f(x)}$ for all integers n and all real x.
• ${\displaystyle f}$ is discontinuous at every rational number, so its points of discontinuity are dense within the real numbers.
• ${\displaystyle f}$ is continuous at every irrational number, so its points of continuity are dense within the real numbers.
• ${\displaystyle f}$ is nowhere differentiable.
• ${\displaystyle f}$ has a strict local maximum at each rational number.^{[citation needed]}
  See the proofs for continuity and discontinuity above for the construction of appropriate neighbourhoods, where ${\displaystyle f}$ has maxima.
• ${\displaystyle f}$ is Riemann integrable on any interval and the integral evaluates to ${\displaystyle 0}$ over any set.
  The Lebesgue criterion for integrability states that a bounded function is Riemann integrable if and only if the set of all discontinuities has measure zero.^[5] Every countable subset of the real numbers - such as the rational numbers - has measure zero, so the above discussion shows that Thomae's function is Riemann integrable on any interval. The function's integral is equal to ${\displaystyle 0}$ over any set because the function is equal to zero almost everywhere.
• If ${\displaystyle G=\{\,(x,f(x)):x\in (0,1)\,\}\subset \mathbb {R} ^{2}}$ is the graph of the restriction of ${\displaystyle f}$ to ${\displaystyle (0,1)}$, then the box-counting dimension of ${\displaystyle G}$ is ${\displaystyle 4/3}$.^[6]

Related probability distributions

Empirical probability distributions related to Thomae's function appear in DNA sequencing.^[7] The human genome is diploid, having two strands per chromosome. When sequenced, small pieces ("reads") are generated: for each spot on the genome, an integer number of reads overlap with it. Their ratio is a rational number, and typically distributed similarly to Thomae's function.

If pairs of positive integers ${\displaystyle m,n}$ are sampled from a distribution ${\displaystyle f(n,m)}$ and used to generate ratios ${\displaystyle q=n/(n+m)}$, this gives rise to a distribution ${\displaystyle g(q)}$ on the rational numbers. If the integers are independent the distribution can be viewed as a convolution over the rational numbers, ${\textstyle g(a/(a+b))=\sum _{t=1}^{\infty }f(ta)f(tb)}$. Closed form solutions exist for power-law distributions with a cut-off. If ${\displaystyle f(k)=k^{-\alpha }e^{-\beta k}/\mathrm {Li} _{\alpha }(e^{-\beta })}$ (where ${\displaystyle \mathrm {Li} _{\alpha }}$ is the polylogarithm function) then ${\displaystyle g(a/(a+b))=(ab)^{-\alpha }\mathrm {Li} _{2\alpha }(e^{-(a+b)\beta })/\mathrm {Li} _{\alpha }^{2}(e^{-\beta })}$. In the case of uniform distributions on the set ${\displaystyle \{1,2,\ldots ,L\}}$ ${\displaystyle g(a/(a+b))=(1/L^{2})\lfloor L/\max(a,b)\rfloor }$, which is very similar to Thomae's function.^[7]

The ruler function

Main article: Ruler function

For integers, the exponent of the highest power of 2 dividing ${\displaystyle n}$ gives 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, ... (sequence A007814 in the OEIS). If 1 is added, or if the 0s are removed, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, ... (sequence A001511 in the OEIS). The values resemble tick-marks on a 1/16th graduated ruler, hence the name. These values correspond to the restriction of the Thomae function to the dyadic rationals: those rational numbers whose denominators are powers of 2.

Related functions

A natural follow-up question one might ask is if there is a function which is continuous on the rational numbers and discontinuous on the irrational numbers. This turns out to be impossible. The set of discontinuities of any function must be an F_σ set. If such a function existed, then the irrationals would be an F_σ set. The irrationals would then be the countable union of closed sets ${\textstyle \bigcup _{i=0}^{\infty }C_{i}}$, but since the irrationals do not contain an interval, neither can any of the ${\displaystyle C_{i}}$. Therefore, each of the ${\displaystyle C_{i}}$ would be nowhere dense, and the irrationals would be a meager set. It would follow that the real numbers, being the union of the irrationals and the rationals (which, as a countable set, is evidently meager), would also be a meager set. This would contradict the Baire category theorem: because the reals form a complete metric space, they form a Baire space, which cannot be meager in itself.

A variant of Thomae's function can be used to show that any F_σ subset of the real numbers can be the set of discontinuities of a function. If ${\textstyle A=\bigcup _{n=1}^{\infty }F_{n}}$ is a countable union of closed sets ${\displaystyle F_{n}}$, define $${\displaystyle f_{A}(x)={\begin{cases}{\frac {1}{n}}&{\text{if }}x{\text{ is rational and }}n{\text{ is minimal so that }}x\in F_{n}\\-{\frac {1}{n}}&{\text{if }}x{\text{ is irrational and }}n{\text{ is minimal so that }}x\in F_{n}\\0&{\text{if }}x\notin A\end{cases}}}$$

Then a similar argument as for Thomae's function shows that ${\displaystyle f_{A}}$ has A as its set of discontinuities.

See also

• Blumberg theorem
• Cantor function
• Dirichlet function
• Euclid's orchard – Thomae's function can be interpreted as a perspective drawing of Euclid's orchard
• Volterra's function