Verify language that i found of this automata

Question

Sorry for my bad english.

Automata:

My answer: $$L = \{0^n, 0^n1X \mid n = 1,2,3\dots\}$$

X is any strings that is not empty.

Answer 1

You have not accounted for the strings which begin with $1$. In your notation: $$ \{1^n0^{m+1}|m,n\in \mathbb N\}\cup\{1^n0^{m+1}1X|m,n\in \mathbb N,X\text{ non-empty}\} $$ where $\mathbb N=\{0,1,2.\dots\}$. In regex notation, $$ (1^*00^*|1^*00^*1(0|1)(0|1)^*) $$

Answer 2

The automaton accepts any string containing at least one $0.$ In the notation I am used to, I would write $\{0,1\}^*0\{0,1\}^*$ as a regular expression for the language. Here $\{0,1\}^*$ means any string of $0$'s and $1$'s, possibly empty.

In the notation you are using the answer would be something like $$ L=\{1^n0, 1^n0X| n = 0,1,2,\dots\}$$ That is, a (possibly empty) string of $1$'s, followed by a $0$, followed by anything. If you have a notation (say Y) for a string that may be empty, then you can shorten it to $$ L=\{1^n0Y| n = 0,1,2,\dots\}$$

Answer 3

A short regular expression for the language accepted by this DFA is $$ 1^*0^+(\varepsilon \cup 1\{0,1\}^*). $$