# Let P_2 is real polynomial space with the highest degree is 2. Specify the x^2"-coordinate " of the base " "{x^2+x,x+1,x^2+1}" in "P_2 ?

Apr 9, 2017

$\left(\frac{1}{2} , - \frac{1}{2} , \frac{1}{2}\right)$

#### Explanation:

I think you are asking for the coordinates of ${x}^{2}$ using the given base.

We find:

$\frac{1}{2} \left({x}^{2} + x\right) - \frac{1}{2} \left(x + 1\right) + \frac{1}{2} \left({x}^{2} + 1\right) = {x}^{2}$

So the coordinates are: $\left(\frac{1}{2} , - \frac{1}{2} , \frac{1}{2}\right)$

Apr 9, 2017

${x}^{2} = \frac{1}{2} \left({x}^{2} + x\right) - \frac{1}{2} \left(x + 1\right) + \frac{1}{2} \left({x}^{2} + 1\right)$

#### Explanation:

We have that

${x}^{2} = \alpha \left({x}^{2} + x\right) + \beta \left(x + 1\right) + \gamma \left({x}^{2} + 1\right)$

so

${x}^{2} = \left(\alpha + \gamma\right) {x}^{2} + \left(\alpha + \beta\right) x + \left(\beta + \gamma\right)$

and we need

$\left\{\begin{matrix}\alpha + \gamma = 1 \\ \alpha + \beta = 0 \\ \beta + \gamma = 0\end{matrix}\right.$

solving we have

$\alpha = \frac{1}{2} , \beta = - \frac{1}{2} , \gamma = \frac{1}{2}$ and finally

${x}^{2} = \frac{1}{2} \left({x}^{2} + x\right) - \frac{1}{2} \left(x + 1\right) + \frac{1}{2} \left({x}^{2} + 1\right)$