15
$\begingroup$

This was asked earlier at MSE.

Let $A = \{a_0, a_1, a_2, \dotsc\}$ denote a weakly decreasing sequence of positive terms whose sum converges. Next introduce plus minus signs in every possible way, to form the collection $L(A) = \{\sum \limits_{j=0}^\infty \epsilon_j a_j : \epsilon_j = \pm1\}$ of values of the signed sums. All of these series converge since they converge absolutely.

For example, let $a_j = 1/j!$, giving the usual series for $e = 2.718\ldots$. Then one can check that $L(A)$ has the following general properties: it is a proper closed uncountable subset of the interval $[-e,+e]$.

Questions: (1) Is it possible to be more specific and to say exactly what the set $L(A)$ actually is (apart from just recapitulating the definition)?

(2) Can one recover the sequence from the values of its signed sums? Or does there exist some $B = \{b_0, b_1, b_2, \dotsc\}$ satisfying (as above) $b_j \geq b_{j+1} > 0$ and such that $L(B) = L(A)$?

Thanks

$\endgroup$
  • 2
    $\begingroup$ I think that it is generally regarded as bad practice to switch out of math mode for things like a$_j$ (as opposed to $a_j$). I have edited accordingly. If you really want upright letters, then I believe that the accepted convention is to use \mathrm, like $\mathrm a_j$ ($\mathrm a_j$). $\endgroup$ – LSpice Jun 28 '18 at 0:25
  • $\begingroup$ Another (equivalent) way to do this: the set of all sums $\sum_{j\in A} a_j$, where $A$ ranges over all subsets of $\mathbb N$. This one is an affine image of yours and vice versa. $\endgroup$ – Gerald Edgar Jun 28 '18 at 0:44
19
$\begingroup$

If $a_{n+1} \ge a_n/2$ for all $n$, then $L(A) = [-s,s]$ where $s = \sum_n a_n$.

On the other hand, if $a_{n+1} < a_n/2$ for all $n$, then $L(A)$ is a nowhere dense Cantor-type set, and you can recover $A$ from $L(A)$.

EDIT: I don't have references, but these are not hard to prove. As Gerald Edgar remarks, it's equivalent to consider sums $S(A) = \sum_{j \in A} a_j$ for $A \subseteq \mathbb N$.

In the case $a_{n+1} \ge a_n/2$, so $\sum_{j > n} a_j \ge a_n$, you can use a greedy approach: to make a sum $s$, where $0 \le S(\mathbb N)$, include $n$ in $A$ iff $S(A \cap [0\ldots n-1]) + a_n \le s$, and note that by induction $$S(A \cap [0\ldots n]) + S([n+1 \ldots \infty)) \ge s \ge S(A \cap [0\ldots n])$$

In the case $a_{n+1} < a_n/2$, consider the intervals $J_n(A) = [S(A), S(A \cup [n+1\ldots \infty)]$ for $A \subseteq [1,\ldots, n]$, so that $S(B) \in J_n(A)$ if $A \subseteq B \subseteq A \cup [n+1\ldots\infty)$, and note that $J_{n+1}(A)$ and $J_{n+1}(A \cup \{n+1\})$ are disjoint closed subintervals of $J_n(A)$.

$\endgroup$
  • 4
    $\begingroup$ A reference here would be great. $\endgroup$ – Winther Jun 28 '18 at 10:39
12
$\begingroup$

Let $A$ be the sequence $1/2,1/4,1/8,1/16,1/32\dots$, and let $B$ be the sequence $1/4,1/4,1/4,1/8,1/16,1/32,\dots$. Then $L(B)=L(A)$.

$\endgroup$
0
$\begingroup$

A fun remark :

If you consider the natural measure $\mu$ on $L(A)$ defined as follow : $L_N(A)=\{\sum_{j=0}^{N-1}\epsilon_j a_j : \epsilon_j=\pm 1 \}$ and $$\mu_N=\frac{1}{2^N}\sum_{\epsilon\in \{\pm 1 \}^N} \delta_{\sum_{j=0}^{N-1}\epsilon_j a_j} = \frac{1}{2^N}\sum_{x\in L_N(A)}\delta_x \text{ (if all the sums are disjoint)}$$ and $\mu=\lim_{N\rightarrow} \mu_N$. Then we have $$\int e^{i\lambda x}d\mu(x) = \prod\cos(\lambda a_j)$$ and the sequence $a_i$ is related to the zeros $\lambda_{i,k}$ of the Fourier transform of $\mu$: $$\lambda_{j,k} = \frac{1}{a_j}(\frac{\pi}{2}+k\pi) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.