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Let's call $\Bbb{R[x]}_4$ a linear space of all polynomials of 4-th degree with real coefficients. We are given functionals $\phi_j \in (\Bbb{R[x]}_4)^*$:

  • $\phi_0(p)=p(-1)$
  • $\phi_1(p)=p(0)+p(1)$
  • $\phi_2(p)=p(0)-p(1)$
  • $\phi_3(p)=p(2)$
  • $\phi_4(p)=p(-2)$

The problems asks us to prove that given functionals are basis for $(\Bbb{R[x]}_4)^*$. Then we are to find coefficients $a_0,a_1,a_2,a_3,a_4 \in \Bbb{R}$ such that if given functionals are dual basis to polynomials $p_0,p_1,p_2,p_3,p_4$ and $p(t)=t$ then $p=\sum_{i=0}^4a_ip_i$.

When it comes to the 1st part of this problem I thought of picking functionals $\beta_0=p(0)$, $\beta_1=p(1)$, $\beta_2=p(-1)$, $\beta_3=p(2)$, $\beta_4=p(-2)$, and these functionals must be dual basis for Lagrange basis constructed from them. Then you can show that every $\beta$ function can be written as a linear combination of $\phi$ functions, so their span is equal, and as we know that $\beta$ functions must be basis of $(\Bbb{R[x]}_4)^*$ so the same must be true for $\phi$ functions.

When it comes to the 2nd part of the problem I thought of computing the Lagrange basis for $\beta$ functions and then do some manipulations to it to get the Lagrange basis of $\phi$ functions and from that point I would compute the coordinates of $p$ in that basis which would give the answer. But this solution seems too complex and too hard to compute. Is there a simpler one? Also, are my earlier arguements correct?

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  • $\begingroup$ do you already experimented on ${\Bbb{R}}[x]_2$? $\endgroup$ – janmarqz Jan 6 '15 at 2:39
  • $\begingroup$ what do you mean? $\endgroup$ – qiubit Jan 6 '15 at 2:57
  • $\begingroup$ that you are trying to solve for ${\Bbb{R}}[x]_4$, try for ${\Bbb{R}}[x]_2$ or/and ${\Bbb{R}}[x]_3$ $\endgroup$ – janmarqz Jan 6 '15 at 3:03
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Firstly, you want to show that the functionals are linearly independent. Suppose $ \phi_c = \sum_i c_i \phi_i = 0 $, then $ \phi_c( x (x-1) (x-2) (x+2) ) = 0 $ but also $ \phi_c( x (x-1)(x_2)(x+2) ) = c_0 $, by directly checking explicitly what it evaluates for the other basis functionals. So $ c_0 = 0 $. Similarly conclude that the other $ c_i = 0 $.

Secondly, you want to show that they span the vector space. Can you appeal to the theorem that the dual space has the same dimension?

Finally, if $ t = \sum_i a_i p_i(t) $, then $ \phi_j(t) = a_j $ by the fact that $ \phi_j (p_i) = \delta_{ij} $. So just evaluate the functionals on $ t $.

You might also try to calculate explicitly the dual basis for practice. For example $ p_0 = \frac{-1}{6} x(x-1)(x-2)(x+2) $ evaluates to zero at the right places, and is scaled so that $ \phi_0(p_0) = 1 $. Then you can check explicitly if the $ a_i $ are correct.

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