We equip $\mathbb{R}_{n}[X]$ with the inner product defined by $$\langle P, Q \rangle = \int_{-1}^{1} P(x)Q(x)\,dx$$ For $j \in \mathbb{N}$, we define the polynomial $$P_{j}(X) = \frac{1}{2^{j} j!} \frac{d^{j}}{dX^{j}}\left[(X^{2}-1)^{j}\right]$$ By convention, $P_{0} = 1$. The subspace of $\mathbb{R}_{n}[X]$ formed by even polynomials is denoted $\Pi_{n}$, and that of odd polynomials is denoted $J_{n}$. (a) Show that the family $\left(P_{j}\right)_{0 \leqslant j \leqslant n}$ is a basis of $\mathbb{R}_{n}[X]$. (b) Deduce that the family $\left(P_{2j}\right)_{0 \leqslant j \leqslant \frac{n}{2}}$ is a basis of $\Pi_{n}$, while the family $\left(P_{2j+1}\right)_{0 \leqslant j \leqslant \frac{n-1}{2}}$ is a basis of $J_{n}$.
We equip $\mathbb{R}_{n}[X]$ with the inner product defined by
$$\langle P, Q \rangle = \int_{-1}^{1} P(x)Q(x)\,dx$$
For $j \in \mathbb{N}$, we define the polynomial
$$P_{j}(X) = \frac{1}{2^{j} j!} \frac{d^{j}}{dX^{j}}\left[(X^{2}-1)^{j}\right]$$
By convention, $P_{0} = 1$. The subspace of $\mathbb{R}_{n}[X]$ formed by even polynomials is denoted $\Pi_{n}$, and that of odd polynomials is denoted $J_{n}$.
(a) Show that the family $\left(P_{j}\right)_{0 \leqslant j \leqslant n}$ is a basis of $\mathbb{R}_{n}[X]$.
(b) Deduce that the family $\left(P_{2j}\right)_{0 \leqslant j \leqslant \frac{n}{2}}$ is a basis of $\Pi_{n}$, while the family $\left(P_{2j+1}\right)_{0 \leqslant j \leqslant \frac{n-1}{2}}$ is a basis of $J_{n}$.