<s>[INST] Solving for coefficients in polynomial I have the following equation:
$$8\eta_c-6x\eta_c+\frac{x^3}{2}\eta_c=2.5\eta_c+C_{01}x^0+C_{11}x+C_{21}x^2+C_{31}x^3$$
If I solve for these coefficients, I get:
$$C_{01}=5.5\eta_c$$
$$C_{11}=-6\eta_c $$
$$C_{21}=0$$
$$C_{31}=1/2\eta_c$$
If I change my equation from (x) to (x-1) like this, why do I get that the coefficients are the same? Wouldn't this imply that $x = x-1$?
My derivation with $x-1$ is below.
$$\eta_c\left(8-6(x-1)+\frac{(x-1)^3}{2}\right)=2.5\eta_c+C_{01}(x-1)^0+C_{11}(x-1)^1+C_{21}(x-1)^2+C_{31}(x-1)^3$$
When I expand the LHS:
$$\eta_c\left(8-6x+6+\frac{x^3}{2}-\frac{3x^2}{2}+\frac{3x}{2}-\frac{1}{2}\right)=2.5\eta_c+C_{01}+C_{11}(x-1)+C_{21}(x-1)^2+C_{31}(x-1)^3$$
Collecting terms on the LHS:
$$\eta_c\left(13.5-4.5x+\frac{x^3}{2}-\frac{3x^2}{2}\right)=2.5\eta_c+C_{01}+C_{11}(x-1)+C_{21}(x-1)^2+C_{31}(x-1)^3$$
Now if I expand the RHS:
$$\eta_c\left(13.5-4.5x+\frac{x^3}{2}-\frac{3x^2}{2}\right)=2.5\eta_c+C_{01}+C_{11}x-C_{11}+C_{21}x^2-2C_{21}x+C_{21}+C_{31}x^3-3C_{31}x^2+3C_{31}x-C_{31}$$
Now collecting terms on the RHS:
$$\eta_c\left(13.5-4.5x-\frac{3x^2}{2}+\frac{x^3}{2}\right)=(2.5\eta_c+C_{01}-C_{11}+C_{21}-C_{31})+x(C_{11}-2C_{21}+3C_{31})+x^2(C_{21}-3C_{31})+x^3(C_{31})$$
Then I solve for all coefficients and get the same as before.
 [/INST] Rearranging the original equation gives
$$(C_{01}-5.5\eta_c)+(C_{11}+6\eta_c)x+C_{21}x^2+(C_{31}-\frac{\eta_c}{2})x^3=0$$
Now if we replace $x$ with any function $f(x)$ for example $f(x)=x-1$ we get
$$(C_{01}-5.5\eta_c)+(C_{11}+6\eta_c)f(x)+C_{21}f^2(x)+(C_{31}-\frac{\eta_c}{2})f^3(x)=0$$
Now this equation is always true when
$$C_{01}=5.5\eta_c$$
$$C_{11}=-6\eta_c$$
$$C_{21}=0$$
$$C_{31}=\frac{\eta_c}{2}$$
because both sides of the equation will then be equal to zero;
$$(5.5\eta_c-5.5\eta_c)+(-6\eta_c+6\eta_c)f(x)+(0)f^2(x)+(\frac{\eta_c}{2}-\frac{\eta_c}{2})f^3(x)=0$$
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