Editing Signed-digit representation (section)

===For integers===
Given the digit set <math>\mathcal{D}</math> and function <math>f:\mathcal{D}\rightarrow\mathbb{Z}</math> as defined above, let us define an [[integer]] [[endofunction]] <math>T:\mathbb{Z}\rightarrow\mathbb{Z}</math> as the following:
:<math>T(n) = 
\begin{cases}
\frac{n - f(d_i)}{b} &\text{if } n \equiv i \bmod b, 0 \leq i < b
\end{cases}</math>
If the only [[periodic point]] of <math>T</math> is the [[fixed point (mathematics)|fixed point]] <math>0</math>, then the set of all signed-digit representations of the [[integers]] <math>\mathbb{Z}</math> using <math>\mathcal{D}</math> is given by the [[Kleene plus]] <math>\mathcal{D}^+</math>, the set of all finite [[concatenation|concatenated]] strings of digits <math>d_n \ldots d_0</math> with at least one digit, with <math>n\in\mathbb{N}</math>. Each signed-digit representation <math>m \in \mathcal{D}^+</math> has a [[Valuation (algebra)|valuation]] <math>v_\mathcal{D}:\mathcal{D}^+\rightarrow\mathbb{Z}</math>
:<math>v_\mathcal{D}(m) = \sum_{i=0}^{n}f_\mathcal{D}(d_{i})b^{i}</math>. 
Examples include [[balanced ternary]] with digits <math>\mathcal{D} = \lbrace \bar{1}, 0, 1\rbrace</math>.  

Otherwise, if there exist a non-zero [[periodic point]] of <math>T</math>, then there exist integers that are represented by an infinite number of non-zero digits in <math>\mathcal{D}</math>. Examples include the standard [[decimal numeral system]] with the digit set <math>\operatorname{dec} = \lbrace 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 \rbrace</math>, which requires an [[Radix complement|infinite number of the digit]] <math>9</math> to represent the [[additive inverse]] <math>-1</math>, as <math>T_\operatorname{dec}(-1) = \frac{-1 - 9}{10} = -1</math>, and the positional numeral system with the digit set <math>\mathcal{D} = \lbrace \text{A}, 0, 1\rbrace</math> with <math>f(\text{A}) = -4</math>, which requires an infinite number of the digit <math>\text{A}</math> to represent the number <math>2</math>, as <math>T_\mathcal{D}(2) = \frac{2 - (-4)}{3} = 2</math>.