Exercise 3.F.8

Proof. _(a)_ Suppose there are $a_1,\dots ,a_m\in 𝔽$ such that

It is easy to see that, after expanding all terms, the only one containing $x^m$ is $a_m{x}^m$, therefore $a_m=0$, since there is no other term containing $x^m$ to cancel it. Thus

The same argument can be used to show that all $a$’s are zero. Hence $1,x-5,\dots ,{(x-5)}^m$ is a linearly independent list of length $m+1$. But , therefore it also a basis of .

_(b)_ Define ${\phi }_0,\dots ,{\phi }_m$, where ${\phi }_j(p)=\frac{p^{(j)}(5)}{j!}$. By Exercise 7, this is the dual basis (just swap $x$ for $x-5$). □