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Prove that $L:P_5(\Bbb R) \mapsto \Bbb R^5$ defined by $L(f(x))=(f(-2),f(-1),f(0),f(1),f(2))$ is an isomorphism, then show that for any $f(x)$ there exists unique $a_1,a_2$ such that $$\int_{-1}^1 f(x)dx=2(1-a_1-a_2)f(0)+a_1(f(1)+f(-1))+a_2(f(2)+f(-2))$$

I have shown that the linear map is an isomorphism, but I'm stuck at the second part. Suppose $L^{-1}$ is the inverse map, $L_0,\dots,L_4$ are linear combination of the values $(a,b,c,d,e)\in \Bbb R^5$, then we have

$$f(x)=L^{-1}(a,b,c,d,e)=L_0(a,b,c,d,e)+L_1(a,b,c,d,e)x+L_2(a,b,c,d,e)x^2+L_3(a,b,c,d,e)x^3+L_4(a,b,c,d,e)x^4$$

$$\int_{-1}^1f(x)dx=2L_0+\frac 23 L_2 +\frac 25 L_4$$

How do I complete the proof?

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  • $\begingroup$ I would avoid finding the inverse of $L$. However, you can use Lagrange interpolation to get $$f(x)=\sum\limits_{k=-2}^2\,f(k)\,P_k(x)\,.$$ Then, the coefficients will be the integrals of the Lagrange polynomials $P_k(x)$. The uniqueness part is quite obvious. You only need to deal with $k\geq 0$ due to symmtry. $\endgroup$ – Batominovski Aug 25 '16 at 3:26
  • $\begingroup$ Well, just realized. Lagrange interpolation is the inverse. $\endgroup$ – Batominovski Aug 25 '16 at 3:33
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Using your equation where you write $f(x) = L^{-1}(a,b,c,d,e)$, we get: $$\int_{-1}^1f(x)dx = 2(1-a_1-a_2)f(0)+a_1(f(1)+f(-1))+a_2(f(2)+f(-2))$$$$ = 2(1-a_1-a_2)L_0+a_1(2L_0+2L_2+2L_4)+a_2(2L_0+8L_2+32L_4) $$$$= 2L_0+(2a_1+8a_2)L_2+(2a_1+32a_2)L_4$$

You also have the equation you've already obtained, $$2L_0+\frac{2}{3}L_2+\frac{2}{5}L_4.$$

All that's left is to equate coefficients, leaving you to solve a system of two equations with two variables.

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  • $\begingroup$ So are you suggesting that I should explicitly find out $L^{-1}$. I am not sure but since this is from a proof based course, I thought there would be a clever algebraic method. $\endgroup$ – lEm Aug 25 '16 at 7:49

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