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Find a basis $B$ for $P_2$

$[p]_B = \begin{bmatrix}p(0)\\p(1)\\p(2)\end{bmatrix}$

and its coordinates to a second degree polynomial

The solutions says:

$p(x) = p(0)e_1(x) + p(1)e_2(x) + p(2)e_3(x)$, where

$e_1(x): e_1(0) = 1,\ e_1(1) = 0,\ e_1(2) = 0$

$e_2(x): e_2(0) = 0,\ e_2(1) = 1,\ e_2(2) = 0$

$e_3(x): e_3(0) = 0,\ e_3(1) = 0,\ e_3(2) = 1.$

But I don't understand the solution. Can someone please explain?

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  • $\begingroup$ Please show your work $\endgroup$ – PackSciences May 24 at 19:01
  • $\begingroup$ What this basis is doing is interpolating each polynomial $p$. Therefore, it should consists of the Lagrange polynomials for the points $0,1,2$. That would be $b_1(x)=\frac{(x-1)(x-2)}{(0-1)(0-2)}, b_2(x)=\frac{x(x-2)}{(1-0)(1-2)},$ and $b_2(x)=\frac{x(x-1)}{(2-0)(2-1)}$. $\endgroup$ – logarithm May 24 at 19:19

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