Minimal orthogonal projections, and distances in Hilbert spaces. Let $X=L^2\left([-1,1]\right)$ be the inner product space of Lebesgue integrable functions from $[-1,1]$. And denote $\mathcal{P}_{n-1} \subset X$ the subspace of polynomials with a degree not greater than $n-1$. I need to find the distance of $x^n$ from $\mathcal{P}_{n-1}$.

I started by stating that $\mathcal{P}_{n-1} = span \{ p_k \}_{k=0}^{n-1}$ where $p_k$ are the scaled Legendre polynomials, which we know that are complete orthonormal system in $\mathcal{P}_{n-1}$. 
Now $x^n = P_{\mathcal{P}_{n-1}}x^n + \left( I - P_{\mathcal{P}_{n-1}} \ \ \right)x^n$  where $P_\left( \mathcal{P}_{n-1} \  \right)$ is the projection onto $\mathcal{P}_{n-1}$. So basically $\left( I - P_{\mathcal{P}_{n-1}} \ \ \right)x^n$ is the "unscaled" Legendre polynomial, It's norm then is the distance I want to compute. Is this correct thinking? If it is, how do I calculate the norm?
 A: If $\mathcal{H}$ is an inner product space, and $\mathcal{M}$ is a subspace of $\mathcal{H}$, then closest point projection of $f$ onto $\mathcal{M}$ is the same as the orthogonal projection of $f$ onto $\mathcal{M}$. That is, one type of projection exists iff the other does, and, in that case, the two are equal.
In your case, $\mathcal{H}=L^2[-1,1]$, and you want the distance from $f=x^n$ to the subspace $\mathcal{P}_{n-1}$ spanned by $\{1,x,\cdots,x^{n-1}\}$. That distance is $\|f-g\|$ where $g = a_{n-1}x^{n-1}+\cdots + a_0$ is the unique polynomial of order $n-1$ for which $(f-g) \perp\mathcal{P}_{n-1}$. As you noted, the polynomial $f-g$ is a polynomial of order $n$ that is orthogonal to all polynomials of order $n-1$, which makes it a scalar multiple of the $n$-th order Legendre polynomial. The polynomial $f-g$ is a monomial (i.e., highest order coefficient is $1$,) which tells you that
$$
                f-g = \frac{1}{(2n)(2n-1)(\cdots)(n+1)}\frac{d^{n}}{dx^{n}}(x^2-1)^{n} = \frac{n!}{(2n)!}\frac{d^{n}}{dx^{n}}(x^2-1).
$$
The normalized Legendre polynomial of order $n$ is
$$
           P_n(x)= \sqrt{n+1/2}\frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n.
$$
(Normalized means $\|P_n\|=1$.)
So the distance of $x^n$ to $\mathcal{P}_{n-1}$ is
$$
         \|f-g\|= \frac{n!}{(2n)!}\frac{2^n n!}{\sqrt{n+1/2}}
$$
