Let’s discover the beauties of Thomae’s function also named the popcorn function, the raindrop function or the modified Dirichlet function.
Thomae’s function is a real-valued function defined as:
\[f:
\left|\begin{array}{lrl}
\mathbb{R} & \longrightarrow & \mathbb{R} \\
x & \longmapsto & 0 \text{ if } x \in \mathbb{R} \setminus \mathbb{Q} \\
\frac{p}{q} & \longmapsto & \frac{1}{q} \text{ if } \frac{p}{q} \text{ in lowest terms and } q > 0
\end{array}\right.\]
\(f\) is periodic with period \(1\)
This is easy to prove as for \(x \in \mathbb{R} \setminus \mathbb{Q}\) we also have \(x+1 \in \mathbb{R} \setminus \mathbb{Q}\) and therefore \(f(x+1)=f(x)=0\). While for \(y=\frac{p}{q} \in \mathbb{Q}\) in lowest terms, \(y+1=\frac{p+q}{q}\) is also in lowest terms, hence \(f(y+1)=f(y)=\frac{1}{q}\). Continue reading A function continuous at all irrationals and discontinuous at all rationals→
From the mean value theorem, a real function whose derivative is strictly positive at every point of an interval is strictly increasing. In particular, a continuously differentiable function \(f\) defined in a non-degenerate interval \(I\) with a strictly positive derivative at a point \(a\) of the interval is strictly increasing near that point. For the proof, we just have to notice that as \(f^\prime\) is continuous and \(f^\prime(a) > 0\), \(f^\prime\) is strictly positive within an interval \(J \subset I\) containing \(a\). By the mean value theorem, \(f\) is strictly increasing on \(J\).
We now suppose that \(f\) is differentiable on an interval \(I\) containing \(0\) with \(f^\prime(0)>0\). For \(x>0\) sufficiently close to zero we have \(\displaystyle \frac{f(x)-f(0)}{x-0} > \frac{f^\prime(0)}{2}>0\), hence \(f(x)>f(0)\). But that doesn’t imply that \(f\) is strictly increasing in a neighborhood of zero. Let’s prove it with a counterexample. Continue reading A function whose derivative at 0 is one but which is not increasing near 0→
In that article, I gave examples of real valued functions defined on \((0,+\infty)\) that converge to zero and whose derivatives diverge. But those functions were not monotonic. Here I give an example of a decreasing real valued function \(g\) converging to zero at \(+\infty\) and whose derivative is unbounded.
We first consider the polynomial map:
\[P(x)=(1+2x)(1-x)^2=1-3x^2+2x^3\] on the segment \(I=[0,1]\). \(P\) derivative equals \(P^\prime(x)=-6x(1-x)\). Therefore \(P\) is decreasing on \(I\). Moreover we have \(P(0)=1\), \(P(1)=P^\prime(0)=P^\prime(1)=0\) and \(P^\prime(1/2)=-3/2\). Continue reading A decreasing function converging to zero whose derivative diverges (part2)→
In this article, I consider real valued functions \(f\) defined on \((0,+\infty)\) that converge to zero, i.e.:
\[\lim\limits_{x \to +\infty} f(x) = 0\] If \(f\) is differentiable what can be the behavior of its derivative as \(x\) approaches \(+\infty\)?
Let’s consider a first example:
\[\begin{array}{l|rcl}
f_1 : & (0,+\infty) & \longrightarrow & \mathbb{R} \\
& x & \longmapsto & \frac{1}{x} \end{array}\] \(f_1\) derivative is \(f_1^\prime(x)=-\frac{1}{x^2}\) and we also have \(\lim\limits_{x \to +\infty} f_1^\prime(x) = 0\). Let’s consider more sophisticated cases! Continue reading Differentiable functions converging to zero whose derivatives diverge (part1)→
In that article, I provide basic counterexamples on sequences convergence. I follow on here with some additional and more advanced examples.
If \((u_n)\) converges then \((\vert u_n \vert )\) converge?
This is true and the proof is based on the reverse triangle inequality: \(\bigl| \vert x \vert – \vert y \vert \bigr| \le \vert x – y \vert\). However the converse doesn’t hold. For example, the sequence \(u_n=(-1)^n\) is such that \(\lim \vert u_n \vert = 1\) while \((u_n)\) diverges.
If for all \(p \in \mathbb{N}\) \(\lim\limits_{n \to +\infty} (u_{n+p} – u_n)=0\) then \((u_n)\) converges?
The assertion is wrong. A simple counterexample is \(u_n= \ln(n+1)\). It is well known that \((u_n)\) diverges. However for any \(p \in \mathbb{N}\) we have \(\lim\limits_{n \to +\infty} (u_{n+p} – u_n) =\ln(1+\frac{p}{n+1})=0\).
The converse proposition is true. Assume that \((u_n)\) is a converging sequence with limit \(l\) and \(p \ge 0\) is any integer. We have \(\vert u_{n+p}-u_n \vert = \vert (u_{n+p}-l)-(u_n-l) \vert \le \vert u_{n+p}-l \vert – \vert u_n-l \vert\) and both terms of the right hand side of the inequality are converging to zero. Continue reading Counterexamples on real sequences (part 2)→
We build here a continuous function of one real variable whose derivative exists on \(\mathbb{R} \setminus \mathbb{Q}\) and doesn’t have a left or rightderivative on each point of \(\mathbb{Q}\).
We build here a continuous function of one real variable whose derivative exists except at \(0\) and is bounded on \(\mathbb{R^*}\).
We start with the even and piecewise linear function \(g\) defined on \([0,+\infty)\) with following values:
\[g(x)=
\left\{
\begin{array}{ll}
0 & \mbox{if } x =0\\
0 & \mbox{if } x \in \{\frac{k}{4^n};(k,n) \in \{1,2,4\} \times \mathbb{N^*}\}\\
1 & \mbox{if } x \in \{\frac{3}{4^n};n \in \mathbb{N^*}\}\\
\end{array}
\right.
\] The picture below gives an idea of the graph of \(g\) for positive values.Continue reading A differentiable function except at one point with a bounded derivative→
Recall that a function of bounded variation, also known as a BV-function, is a real-valued function whose total variation is bounded (finite).
Being more formal, the total variation of a real-valued function \(f\), defined on an interval \([a,b] \subset \mathbb{R}\) is the quantity:
\[V_a^b(f) = \sup\limits_{P \in \mathcal{P}} \sum_{i=0}^{n_P-1} \left\vert f(x_{i+1}) – f(x_i) \right\vert\] where the supremum is taken over the set \(\mathcal{P}\) of all partitions of the interval considered. Continue reading A continuous function which is not of bounded variation→
Let \(f_1(x) = |x|\) for \(| x | \le \frac{1}{2}\), and let \(f_1\) be defined for other values of \(x\) by periodic continuation with period \(1\). \(f_1\) graph looks like following picture:
