# Find an explicit second-degree polynomial given a definition of an inner product on a polynomial vector space

Let $$\mathcal{P}_2$$ denote the vector space of of all second-degree polynomials from $$[a,b]$$ to $$\mathbb{R}$$. Define the inner product on $$\mathcal{P}_2$$ to be:

$$\left< p,q \right> = \int_{-1}^{1} p(x)q(x) dx$$

How can I find a unique, explicit polynomial $$q(x) \in \mathcal{P}_2$$ such that for every $$p \in \mathcal{P}_2$$, we have:

$$p(1) = \int_{-1}^{1} p(x)q(x) dx$$

I started by defining $$p(x) = a + bx + cx^2$$ and $$q(x) = \alpha + \beta x + \delta x^2$$. Then I calculated $$p(1)$$, which equals to $$a+b+c$$. From this we get:

$$a+b+c = \int_{-1}^{1} (a + bx + cx^2)(\alpha + \beta x + \delta x^2) dx$$

However, after this, no matter what I try, I cannot get explicit values for $$\alpha, \beta$$, and $$\delta$$. I tried plugging it into WolframAlpha, but each value depends on another value and the values keep changing from one case to another. Moreover, we cannot simply fix values for the constants because the above statement must hold for every polynomial $$p \in \mathcal{P}_2$$.

• Here's a hint: Since you know it must be the case that $\langle p , q\rangle=p(1)$ for $\textbf{every } p$, how about (perhaps carefully) selecting a few different $p$? Commented Nov 19, 2019 at 5:51
• Theo's hint is good. Your approach works, too; multiply out that integrand, do the integration, and you'll get an equation in which the coefficients of $a,b,c$ (which are linear in the unknowns $\alpha,\beta,\delta$) have to be zero: three linear equations in three unknowns. Commented Nov 19, 2019 at 5:56
• Any thoughts about the several comments and answers that have been posted, Curiosity? Commented Nov 20, 2019 at 12:35
• I'm voting to close this question as off-topic because OP has abandoned it. Commented Nov 21, 2019 at 23:06

For $$p(x)=1$$ we get $$1=\int_{-1}^{1} q(x) dx$$.

For $$p(x)=x$$ we get $$1=\int_{-1}^{1} xq(x) dx$$.

For $$p(x)=x^2$$ we get $$1=\int_{-1}^{1} x^2q(x) dx$$.

Can you proceed ?

It is sufficient to define $$q$$ on a basis. A suitable basis is $$e_k(x)= x^k$$ for $$k=0,1,2$$.

Let $$q(x) = q_0+ q_1x+q_2 x^2$$.

Since $$e_k(1) = 1$$ for all $$k$$ we require $$\langle e_k, q \rangle = 1$$ for all $$k$$.

This gives the equations $$2q_0 + {2 \over 3} q_2 = 1, {2 \over 3} q_1 = 1, {2 \over 3} q_0 + {2 \over 5} q_2 = 1$$.

Hence $$q_0 = -{3 \over 4}, q_1 = {3 \over 2}$$, $$q_2 = {15 \over 4 }$$.