= Solution
This is the <Gram-Schmidt process> applied degree by degree. Inductively, subtracting projections onto $p_0,\ldots,p_n$ makes $p_{n+1}$ orthogonal to every earlier polynomial. Since every subtracted polynomial has degree at most $n$, the leading coefficient of the monic $q_{n+1}$ remains one, so $p_{n+1}$ is a real <monic orthogonal polynomial> of degree exactly $n+1$.
Now take $q_{n+1}=xp_n$. For $k\leq n-2$, symmetry of the <inner product> gives
$$
\langle xp_n,p_k\rangle=\langle p_n,xp_k\rangle=0,
$$
because $xp_k$ has degree at most $n-1$. Only the $p_n$ and $p_{n-1}$ projections remain, yielding the <three-term recurrence for monic orthogonal polynomials>
$$
p_{n+1}=(x-\alpha_n)p_n-\beta_np_{n-1},
$$
where
$$
\boxed{\alpha_n=\frac{\langle xp_n,p_n\rangle}{\langle p_n,p_n\rangle},
\qquad
\beta_n=\frac{\langle p_n,p_n\rangle}{\langle p_{n-1},p_{n-1}\rangle}.}
$$
The second identity uses $xp_{n-1}=p_n$ plus lower-degree terms.
For the symmetric Legendre inner product, parity gives $\alpha_n=0$. Starting with $p_0=1$, one obtains $p_1=x$, then
$$
\beta_1=\frac{\int_{-1}^1x^2\,dx}{\int_{-1}^1 1\,dx}=\frac13,
\qquad
\beta_2=\frac{\int_{-1}^1(x^2-1/3)^2\,dx}{\int_{-1}^1x^2\,dx}=\frac4{15}.
$$
Thus the first four monic <Legendre polynomials> are
$$
\boxed{p_0=1,\qquad p_1=x,\qquad
p_2=x^2-\frac13,\qquad
p_3=x^3-\frac35x.}
$$
Back to article page
