Aleph Null

Here I will describe a strategy or technique to solve some finite state automata problems. This strategy was first taught to us by my Theory of Computation Professor. The term 'Function Automata' is given to me since we use function-like structures in this.     Suppose you are given a language that is a regular set \(L\). So there exists a \(DFA\) for \(L\), \(L_D=(Q,\Sigma,\delta, q_1, F)\) where \(Q\) is the set of states, \(\Sigma\) is the set of alphabets, \(\delta\) is the transition function, \(S\) is the starting state and \(f\) is the set of final states. Now let \(Q=\{q_1,q_2,\dots,q_n\) where \(n\in\mathbb{N}\). Now we will do a kind of subset construction but every new state will have \(n\) states of \(Q\) but they can repeat. So $$(\underbrace{q_1,q_1,\dots,q_1}_{n\text{ times}})$$ is a state of the new automata. Now, what is the meaning of these new states? Let \(f=(q_{k_1},q_{k_2},\dots,q_{k_n})\) be a new state where \(q_{k_i}\)'s are not necessary to be diff

Equicontinuity : A collection \(\mathcal{F}\) of real-valued functions on a metric space \(X\) is equicontinuous at the point \(x\in X\) provided for every \(\epsilon>0,\ \exists\) \(\delta>0\) such that \(\forall\ f\in\mathcal{F}\) and \(y\in X\) $$d(x,y)<\delta \implies |f(x)-f(y)|<\epsilon$$The collection \(\mathcal{F}\) is said to be equicontinuous on \(X\) provided it is equicontinuous at every point in \(X\) Uniform Equicontinuity :  A equicontinuous collection \(\mathcal{F}\) of real-valued functions on a metric space \(X\) is uniformly equicontinuous if for every \(\epsilon>0,\ \exists\) \(\delta>0\) such that \(\forall\ f\in\mathcal{F}\) and \(x,y\in X\) $$d(x,y)<\delta \implies |f(x)-f(y)|<\epsilon$$ Just like we know that a continuous function on a compact set is uniformly continuous here we are showing that for a collection of functions the same \(\epsilon-\delta\) pair works. Now coming to the proof, since \(\mathcal{F}\) is equicontinuous for ever

 Let \((a_n)\) be a sequence of real numbers. \(\sum\limits_{n=0}^{\infty}a_n\) be a series. Let $$s_n=\sum_{k=0}^{n}a_k$$Then the sequence \((a_n)\) is Cesaro Summable  with Cesaro Sum \(s\in \mathbb{R}\) if $$\lim_{n\to\infty}\frac{\sigma_n}{n+1}=\lim_{n\to\infty}\frac{1}{n+1}\sum_{k=0}^n s_n=s$$Let $$f(x)=\sum\limits_{k=0}^{\infty}a_nx^n$$ be power series. Then the sequence is Abel Summable  if the power series \(f(x)\) converges with a radius of convergence \(|x|<1\).  You can see that if \(s_n\to L\) as \(n\to \infty\) then \(\sigma_n\to s\) as \(n\to \infty\) i.e. convergence of the series implies Cesaro Summability We will prove Abel Summability is much stronger than Cesaro summability i.e. if a series is Cesaro Summable then it is Abel Summable.  So assume \(a_n\) is Cesaro summable. Hence $$\lim_{n\to\infty}\frac{\sigma_n}{n+1}=\lim_{n\to\infty}\frac{1}{n+1}\sum_{k=0}^n s_n=L$$Hence the sequence \(\left( \frac{\sigma_n}{n+1}\right)\) is Abel summable. Therefore $$f(x)=\sum_