<p>I now know that <span class="math-container">$K \langle A \rangle$</span> is the set of all formal power series with finite support for some alphabet <span class="math-container">$A$</span> and field <span class="math-container">$K$</span>. Now I have another question.</p>
<p>Later this expression comes up <span class="math-container">$\lambda \mu (K\langle A \rangle) = K^{1 \times n}$</span>, where <span class="math-container">$\lambda \in K^{1 \times n}$</span> and <span class="math-container">$\mu$</span> a function from <span class="math-container">$A^{*}$</span> to <span class="math-container">$K^{n \times n}$</span>. Shouldn‘t it make more sense if <span class="math-container">$K \langle A \rangle$</span> is a set of words, because <span class="math-container">$\mu$</span> can’t work with a set of polynomials? Or can anyone explain what set <span class="math-container">$\mu (K\langle A \rangle)$</span> specifies?</p>
<p>Here the paper: <a href="https://www.sciencedirect.com/science/article/abs/pii/S0020019097001622" rel="nofollow noreferrer">https://www.sciencedirect.com/science/article/abs/pii/S0020019097001622</a></p>
<p>Here another one which uses the same notation page 111 (and if you can’t access the first one): <a href="https://www.sciencedirect.com/science/article/pii/S030439750000298X" rel="nofollow noreferrer">https://www.sciencedirect.com/science/article/pii/S030439750000298X</a></p>

CC BY-SA 4.0
2022-10-22T10:28:10.493Z
Isn't a formal power series with finite support the same thing as a polynomial? In the noncommutative case such polynomials would correspond to linear combinations of finite words.
Andrej Bauer
CC BY-SA 4.0
2022-10-22T12:56:50.070Z
I don't have time to look into those papers, but could this simply be the lifting of a function $f : A \rightarrow B$ to powersets $f(S) = \{f(a) \in B| a \in S\}$? This is usually done without comment.
Martin Berger
CC BY-SA 4.0
2022-10-22T13:39:04.537Z
This would be my first idea too. But if $f \in K \langle A \rangle$ then f is a function. Obviously $\mu$ can’t work with functions, only words or a set of words. Thx for both of your comments.
Tim567
CC BY-SA 4.0
2022-10-23T11:11:45.533Z
Perhaps it's extended from $A^*$ to $K\langle A\rangle$ by $K$-linearity?
Emil Jeřábek
CC BY-SA 4.0
2022-10-24T02:50:41.190Z
Yes, this is similar to the notation as in Berstel & Reutenauer's book that I linked on the other question. I agree with Emil's guess.
Joshua Grochow
CC BY-SA 4.0
2022-10-24T11:00:01.063Z
Sorry, I am still new to this topic. Emil or Joshua, would it be possible, if you can give me a simple example.
Tim567
CC BY-SA 4.0
2022-10-25T03:00:51.157Z
$\mu(a)=\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}, \mu(b) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. Then $\mu(a+b) = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}, \mu(ab+2b) = \begin{bmatrix} 0 & 3 \\ 4 & 0 \end{bmatrix}$. People on this question and the other have recommended excellent books that are available online, you should really check them out.
Joshua Grochow
CC BY-SA 4.0
2022-10-25T07:50:03.193Z
Oh, thx you very much. Now it makes all sense to me. 

I will check the books out :).
Tim567
<automata-theory><definitions>
Definition in some old paper about formal power series (automata theory) Part 2
Tim567
# Definition in some old paper about formal power series (automata theory) Part 2

(Asked by: username_0 on 2022-10-22)

I now know that $K \langle A \rangle$ is the set of all formal power series with finite support for some alphabet $A$ and field $K$. Now I have another question.

Later this expression comes up $\lambda \mu (K\langle A \rangle) = K^{1 \times n}$, where $\lambda \in K^{1 \times n}$ and $\mu$ a function from $A^{\*}$ to $K^{n \times n}$. Shouldn‘t it make more sense if $K \langle A \rangle$ is a set of words, because $\mu$ can’t work with a set of polynomials? Or can anyone explain what set $\mu (K\langle A \rangle)$ specifies?

Here the paper: <https://www.sciencedirect.com/science/article/abs/pii/S0020019097001622>

Here another one which uses the same notation page 111 (and if you can’t access the first one): <https://www.sciencedirect.com/science/article/pii/S030439750000298X>

**Tags:** automata-theory, definitions
> **Comments:**
> **username_1**: Isn't a formal power series with finite support the same thing as a polynomial? In the noncommutative case such polynomials would correspond to linear combinations of finite words.  
> **username_2**: I don't have time to look into those papers, but could this simply be the lifting of a function $f : A \rightarrow B$ to powersets $f(S) = \{f(a) \in B| a \in S\}$? This is usually done without comment.  
> **username_0**: This would be my first idea too. But if $f \in K \langle A \rangle$ then f is a function. Obviously $\mu$ can’t work with functions, only words or a set of words. Thx for both of your comments.  
> **username_3**: Perhaps it's extended from $A^*$ to $K\langle A\rangle$ by $K$-linearity?  
> **username_4**: Yes, this is similar to the notation as in Berstel & Reutenauer's book that I linked on the other question. I agree with Emil's guess.  
> **username_0**: Sorry, I am still new to this topic. Emil or Joshua, would it be possible, if you can give me a simple example.  
> **username_4**: $\mu(a)=\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}, \mu(b) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. Then $\mu(a+b) = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}, \mu(ab+2b) = \begin{bmatrix} 0 & 3 \\ 4 & 0 \end{bmatrix}$. People on this question and the other have recommended excellent books that are available online, you should really check them out.  
> **username_0**: Oh, thx you very much. Now it makes all sense to me.   I will check the books out :).  

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## 0 answers

