Characteristic of a field and basis of a dual space. Let $K$ be a field. $E=K_n[x]=\{p(x)\in K[x]|\text{degree}(p(x))\leq n\}$
For a $p(x)\in E$ we define $p^{(0)}(x)=p(x)$ and for every positive integer $i$, $p^{(i)}(x)=(p^{(i-1)})'(x)$. For $i\in \{0,...,n\}$ consider $w_i:E\longrightarrow K$ given by $w_i(p(x))=p^{(i)}(1)$. Prove that $B=(w_0,...,w_n)$ is basis of $E^*$ if and only if $K$ has characteristic $0$ or characteristic $p>n$.
I can't solve this problem because I don't even understand the problem statement.
First of all, what is $p^{(i)}(x)$? Of course I understand that $p(x)=a_0+a_1x+...+a_nx^n$ so $p^{(3)}(x)$ would be $((a_0+a_1x+...+a_nx^n)^2)'$?
Second of all, what is the characteristic of $K$?
Thank you for your time and excuse me if I made any grammar mistake, my english is not too good.
 A: $p^{(3)}$ would be $((a_0+a_1x+...+a_nx^n)^2)'$.
No, you have to build it up so
$$
(a_0+a_1x+...+a_nx^n)^{(3)}=
(a_0+a_1x+...+a_nx^n)^{(2)}{'}=
(a_0+a_1x+...+a_nx^n)^{(1)}{''}=
(a_0+a_1x+...+a_nx^n)^{(0)}{'''}=
(a_0+a_1x+...+a_nx^n){'''}
$$
and you now just have to differentiate the polynomial three times.
The characteristic of a field $\mathbb{F}$ is just the least positive $k$ such that $k\cdot 1=0$; or if there are none, $0$. In fact such a $k$ will be a prime number $p$. 
That answers your questions, good luck with the calculation to show the $w_i$ are linearly independent exactly when the characteristic is $0$ or greater than $n$.
A: The notation means that $p^{(i)}$ is the $i$-th (formal) derivative of $p$. Thus $w_i(p)$ associates to $p$ the value of its $i$-th derivative at $1$.
If the characteristic of $K$ is either $0$ or is greater than $n$, then the Taylor expansion
$$
p(x)=p(1)+\frac{p'(1)}{1!}(x-1)+\dots+\frac{p^{(n)}(1)}{n!}(x-1)^n
$$
holds for every polynomial $p$ of degree less than or equal to $n$ (proof?).
In particular, since the set of polynomials
$$
\mathscr{B}=\left\{1,x-1,\frac{1}{2!}(x-1)^2,\dots,\frac{1}{n!}(x-1)^n\right\}
$$
is a basis for $E$ (proof?), we see that $\{w_0,w_1,\dots,w_n\}$ is the  dual basis for $\mathscr{B}$. Indeed,
$$
w_i\left(\tfrac{1}{j!}(x-1)^j\right)=
\begin{cases}
1 & \text{if $j=i$} \\[4px]
0 & \text{if $j\ne i$}
\end{cases}
$$
If the characteristic of $K$ is a prime $q$ less than or equal to $n$, then $w_n(x^i)=0$ for every $i=0,1,\dots,n$, so $w_n=0$ and the set $\{w_0,\dots,w_n\}$ is not linearly independent.
A: You can take derivatives formally. 
Every finite field with $p^n$ elements has characteristic $p$ which is the smallest nonzero integer $k$ such that $1+1+\cdots+1=k\times 1=0.$ So you'd have modulo $k$ operations.
$R$ and $C$ have characteristic zero, sums of the additive identity are never zero.
