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The functional is given by $E:V\rightarrow R$ by $E(p(x))=p(3)$ i.e that is evaluation at $3$ where $V$ is the vector space of all polynomials with real coefficients and the given basis is $\{1,x,x^{2},x^{3},x^{4},x^{5}\}$ . I have to express it as the linear combination of the dual basis.

Now if $\{E_{0},E_{1},E_{2},E_{3},E_{4},E_{5} \}$ is the dual basis then $E$ can be expressed as $$E=\sum_{i=0}^{5} E(x^{i})E_{i} $$ Then by evaluation of $E$ at the basis vectors I get $$E=E_{0}+3E_{1}+9E_{2}+27E_{3}+81E_{4}+243E_{5}$$ and $E_{i}$ is a function that takes all basis elements except for $x^{i}$ to $0$ .Writing this much is not enough. I need to know how exactly do the $E_{i}$'s look $?$ How to proceed$?$

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By definition of the dual basis, you have $E_i(x^j)=\delta_{ij}$ ($1$ if $i=j$, $0$ otherwise). Then, for a polynomial $P$, $E_i(P)=E_i(\sum_{j=0}^5a_jx^j)=\sum_{j=0}^5a_jE_i(x^j)=a_i$.

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The dual basis consists of the coefficient functions: If $p = \sum a_nx^n$, then $E_n(p) = a_n$.

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