I was looking at the question How to convert finite automata to regular expressions? to convert DFA to regex.
The question, I was trying to solve is:
I have got the following equations:
$Q_0=aQ_0 \cup bQ_1 \cup \epsilon$
$Q_1=aQ_1 \cup bQ_1 \cup \epsilon$
When solved, we will get $Q_0=a^*b(a \cup b)^* \cup\ \epsilon$
But my doubt is that, in the DFA starting state is also the final state so, even if we dont give any $b$, it will be accepted, if we give some $a$. But in the regex we have $b$, instead of $b^*$. Why is it so? Is it because,we have that regex $\cup$ $\epsilon$ ?