Finding a basis from a given dual basis Let $V=\mathbb{R}[x]^{\leq3}$ that is the space of polynomials with degree less or equal 3. 
Now I showed that $B^*=\{\phi_0,\phi_1,\phi_2,\phi_3\}$ is a Basis of $V^{*}$, the dual space of V, where 
$$\phi_i(f) = f(i-1)$$ for $i=0,1,2,3$.
Now I want to find a Basis $B$, such that the dual basis of $B$ is $B^*$.
I know there exists an isomorphism between $V$ and $V^{**}$, which immediately gives us the existence of such a basis, but I'm having a hard time actually "computing it". An example on one of the dual vectors would help tremendously!
 A: Note that your dual basis is given by nothing but the evaluation at the $4$ points $-1,0,1,2$. By very definition of dual basis, you want to find $4$ linearly independent vectors in your base space, such that
$$\phi_i(v_j)=\delta_{ij},$$
In other words, you want to find $4$ linearly independent polynomials of degree smaller equal $3$ such that each vanishes at exactly $3$ of the four evaluation points $-1,0,1,2$ and such that the evaluation at the fourth point is $1$.
Writing the candidate polynomial as
$$f(x)=a(x-b)(x-c)(x-d)$$
makes it clear that for $f$ to vanish for example at $-1,0,1$ it must be of the form
$$f(x)=a(x+1)(x)(x-1)$$
and lastly evaluating it at $2$ gives the value of $a$, namely
$$f(2)=a(2+1)(2)(2-1)=6a\overset{!}{=}1\Rightarrow a=\frac16$$
hence the first basis element is
$$f(x)=\frac16(x+1)(x)(x-1).$$
The same reasoning applies for the remaining $3$.
Of course one shall last argue that these are indeed linearly independent. But if $B^\ast$ is known to be a (dual) basis, then this immediate.
