Im following the book Number Systems and Foundations of Anaysis by Elliot Mendelson.
First I need to put the definitions the author are using before asking the question itself.
Definition ans concepts:
Let $\mathfrak{D} = (D,+,\times)$ be an integral domain, by a polynomial over $\mathfrak{D}$ we mean any function $f:\mathbb{N}\rightarrow D$ such that exists some $n_0$ in $\mathbb{N}$ for which $(\forall n)(n \ge n_0 \Rightarrow f(n) = 0$)
Let $\mathfrak{D}[x]$ denote the set of all polynomials over $\mathfrak{D}$, by the zero polynomial □ we mean the constant function on $\mathbb{N}$ which is always equal to 0.
The operations of addition and multiplication of polynomials is defined as follows, consider $f,g \in \mathfrak{D}[x]$.
By $f+g$ we mean the polynomial $h$, such that $h(n) = f(n) + g(n)$, $\forall n \in \mathbb{N}.$ and by $f \times g$ we mean the polynomial $p$ such that $p(n) = \sum_{i}^nf(i).g(n-1)$, $\forall n \in \mathbb{N}$.
From here the author shows that $(\mathfrak{D}[x], +, \times)$ is an integral domain too, then its said that this previous definition is often called a polynomial form, and that polynomials is usually treated in another way, as functions, then he provided a corresponding function as follows:
Let $u$ be an element of the domain $D$, we shall define a corresponding function $E_u:\mathfrak{D}[x]\rightarrow D$.
Intuitively, if $g$ is a polynomial which is written in the traditional form as $a_nx^n+...+a_1x+a_0$, then $E_u(g) = a_nu^n+...+a_1u+a_0$, that is the result of "substituting" $u$ for $x$ and "evaluating".
First let $E_u($□$)=0$. Second assume that we have defined $E_u(h)$ for all polynomials $h$ of degree $\lt n$, and assume that a polynomial $g$ has degree $n$. Then if $a_n = g(n)$ and $h = g - a_nx^n$, then $h$ is either □ or a polynomial of degree $\lt n$. Then, let $E_u(g) = E_u(h) + a_nu^n$.
Now, by the polynomial function associated with a polynomial $g$ we mean the function $ĝ:D\rightarrow D$ defined by $$ĝ:D\rightarrow D = E_u(g)$$
The exercise: Prove that $E_u(f+g) = E_u(f)+E_u(g)$ and $E_u(fg) = E_u(f).E_u(g)$
I tried first the addition just substituting $x$ for $u$ in then "evaluating",
First let $m = max(deg(g), deg(g))$, and as $(f+g)(n) = f(n)+g(n)$ for all $n$, we got $$E_u(f+g) = \sum_{i=0}^m(f+g)(i).u^i$$ $$= \sum_{i=0}^m[(f(i)+g(i)]u^i$$ $$= \sum_{i=0}^m f(i)u^i +g(i)u^i$$ $$= \sum_{i=0}^m f(i)u^i + \sum_{i=0}^m g(i)u^i$$ $$= E_u(f) + E_u(g)$$
I dont managed to got something in my attempt for the multiplication, I want to know if Im going through the right path or Im supposed to used the recursive definition of $E_u$ in this proof, any kind of tips or explanation on difference of polynomial forms and functions and their evaluations will be appreciated too.