What is the expansion of the regular expression $(a^*b^*)^*$

Consider the regular expression $(a^b^)^*$

Does this mean that there are zero or more occurences of the term $(a^b^)$ before the term is expanded or that there are zero of more occurences of the term $(a^b^)$ after the term itself has been expanded?

Here is an example of what I'm try to say

zero or more occurences of $(a^b^)$ BEFORE expanding $(a^b^)$
$(a^b^)^ = a^b^a^b^a^b^* = aababbab$

or

Zero or more occurences of $(a^b^)$ AFTER expanding $(a^b^)$
$(a^b^)^ = (aab)^ = aabaabaab$

Expand the outer starred expression first and then expand the terms. If $r$ is a regular expression, then $r^$ denotes the language consisting of zero or more copies of $r$, concatenated together, so $r^$ could take the form $\epsilon$ (the empty string, resulting from no copies of $r$) or $r$ or $rr$ or $rrr$, and so on.

In your example, with $r=a^b^$, then $r$ denotes all the strings consisting of zero or more $a$'s, followed by zero or more $b$'s, and so $(a^b^)^$ will denote all possible concatenations of terms $(a^b^)$. For example, one possible concatenation would be $(a^b^)(a^b^)(a^b^*)$ and then after expanding each term, we might have $(abb)(aab)(bbb)=aabaabbbb$.

By choosing to expand the inner term and then concatenating the results you will likely miss the general form. If, for example, we chose a particular string in $r$ and then concatenated them, we might replace $r=a^b^$ with $abb$. Then, if we concatenated just those terms three times, we'd have $(abb)(abb)(abb)$. While this is in the language $(a^b^)^*$, it is of a special form that doesn't capture the full generality of strings in the language.

While it's not directly related to the question you asked, it's worth noting that $(a^b^)^=(a\mid b)^$ (some authors would write $(a\mid b)$ as $(a\cup b)$ or $(a+ b)$). In other words, $(a^b^)^*$ denotes the language consisting of all possible strings over $a$ and $b$. Can you see why?

StackExchange Computer Science Q#64530, answer score: 5