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The Context

The origin of my question is my own answer to this question, where the continuity of the function $f: [0,1) \to [0,1)$ that only preserves the odd digits of its input value is analyzed.

In this context, I will slightly modify the definition given there, as follows.

The function I am considering now is $f: [0,+\infty) \to [0,+\infty)$ that preserves only the even digits of the input. So if $x$ has decimal expansion

$$x = \sum_{k=-\infty}^{+\infty}a_k 10^k,$$

then

$$f(x)=\sum_{k=-\infty}^{+\infty}a_{2k}10^{k}.$$

So for example

$$f(5\textbf{8}7\textbf{4}1\textbf{2}.7\textbf{8}0\textbf{3}4\textbf{0}5\textbf{1})=842.8301.$$

With a very similar approach to the one given here, it can be shown that $f$ is continuous almost everywhere (the only exception being the numbers whose least significant digit occupies an odd position), and right-continuous everywhere.

Edit. I am assuming, as in the original question, to adopt, in case of ambiguity, the finite version of the number's decimal expansion.

Note the self-similarity

$$f\left(10^{2k} x\right)=10^k f(x), \ \ \forall k\in \Bbb Z.$$

Below an approximate plot of the function $f$ in the range $[0,1]\times[0,1]$.

enter image description here

Red dots represent points actually belonging to the graph of $f$. The yellow line represents the graph of $$s(x) = \sqrt x.$$ By self-similarity, a scaling of $10^{2k}$ of the $x$-axis and of $10^{-k}$ of the $y$-axis, for any $k\in \Bbb Z$, would give an exact replica of the given plot.

Introductory Observation

Aside from the trivial intersections between $f$ and $s$, that is all the points with coordinates

$$\left(10^{2k},10^k\right),$$

there are many other interesting intersections, such as (limiting ourselves to the range shown in the picture)

$$(0.25,0.5),$$ $$(0.36,0.6),$$ $$(0.0121,0.11),$$ and 'trikiest' ones, such as $$(0.5776,0.76),$$ or even $$(0.35295481,0.5941).$$

The Question

Is there any non-terminating decimal (or even irrational) $x$ such that $$y=f(x) = s(x),$$ that is, is there any non-terminating decimal $y$ whose square contains - in the even positioned digits - the digits of the original number $y$?

Edit. I emphasize again that no infinite sequences of $9$'s are allowed, since we are adopting the finite decimal expansion version of the number, if this ambuiguity arises.

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  • $\begingroup$ Perhaps I'm not understanding the question, but why aren't 5, 760, and 5941 solutions? $\endgroup$
    – rogerl
    Sep 17, 2019 at 15:40
  • $\begingroup$ @rogerl you are correct, they are solutions. I only gave some examples of terminating decimals in the range of the picture. Infinite other solutions can be found from them, by self-similarity. $\endgroup$
    – dfnu
    Sep 17, 2019 at 15:42

1 Answer 1

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The answer is on the affirmative.

Consider the sequence \begin{eqnarray} \alpha_0 &=& 1,\\ \alpha_1 &=& 10005,\\ \alpha_2 &=& 1000505,\\ \alpha_3 &=& 10005050005,\\ \alpha_4 &=& 1000505000500000005,\\ \dots & &, \end{eqnarray} where, for $n>1$, $\alpha_{n+1}$ is obtained from $\alpha_n$ by appending a sequence of $2^{n-1}-1$ zeros and then a $5$: $$\alpha_{n+1} = \alpha_n \cdot 10^{2^{n-1}}+5.$$

Let us first show, by induction, that $$f\left(\alpha_n^2\right) = \alpha_n.\tag{1}\label{eq1}$$ Suppose that \eqref{eq1} holds true for a given $n$. Then \begin{eqnarray} f\left(\alpha_{n+1}^2\right) &=& f\left(\left(\alpha_n \cdot 10^{2^{n-1}}+5\right)^2\right)=\\ &=& f\left(\alpha_n^2\cdot 10^{2^n}+\alpha_n\cdot 10^{2^{n-1}+1}+25\right). \end{eqnarray} Note that

  1. By induction and self-similarity of $f$, $$f\left(\alpha^2_n \cdot 10^{2^n}\right) = \alpha_n \cdot 10^{2^{n-1}};$$
  2. For $n\geq 3$, the addition of the second term $\alpha_n\cdot 10^{2^{n-1}+1}$ does not modify any digits of the first term and has $0$'s in every even position;
  3. The first and last term of the sum never interfere.

As a consequence

\begin{eqnarray} f\left(\alpha_{n+1}^2\right) &=& f\left(\alpha_n^2\cdot 10^{2^n} + 25\right) =\\ &=&\alpha_n \cdot 10^{2^{n-1}} + 5=\\ &=& \alpha_{n+1}. \end{eqnarray}

Consider now the sequence \begin{eqnarray} \beta_0 &=& 1,\\ \beta_1 &=& 1.005,\\ \beta_2 &=& 1.00505,\\ \vdots && \vdots\\ \beta_n &=& \alpha_n\cdot 10^{-2^n-2}. \end{eqnarray} The sequence $(\beta_n)$ is monotonic and upper bounded, and thus convergent in $\Bbb R$, and so is the sequence $(\beta_n^2)$.

Let

$$(\beta_n^2) \to \xi.$$

Clearly $\xi$ is a non-terminating decimal. Thus, as shown in the answer to this question, $f(x)$ is continuous in $\xi$, and so is $$ h(x) = f(x) - \sqrt x.$$

We therefore must have

$$\left(h\left(\beta^2_n\right)\right)\to h(\xi).$$

Since, by self-similarity of $h$, for each $n$ $$h\left(\beta_n^2\right) = 0,$$ it must be $$h(\xi) = 0,$$ that is $$f(\xi) = \sqrt \xi.$$


A little update

  • The same reasoning applies to the sequences $(5.0005, 5.000505, 5.0005050005,\dots)$, and $(6.0005, 6.000505, 6.0005050005,\dots)$.
  • In the above mentioned sequences any digit $5$ can be replaced by a $0$, obtaining thus a dense set of intersection points on the right neighborhoods of $1$, $5$, and $6$.
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