Norms on polynomials Let $p_N (x) = d^N /dx^N ((x^2 -1)^N)$ for $N=0,1,2....$ 
Consider these polynomials as elements of the space $C[−1, 1]$ with
the norm $||.||_2$
Show that the inner product of $p_N$ and $p_M = 0$ if $N$ $\ne$$ M$
Find the norm $||p_N||_2$
I have been to to use IBP for the first part, but I'm unsure where to begin. I know how to find the innerproduct normally, but not of something like this. I also am confused by the second part, despite knowing that to find that norm, you do Pythagoras on the polynomial.
 A: Note that $p_N$ has  antiderivatives that vanish both at $-1$ and $1$. Assume (without loss of generality) that $N>M$;  we can iterate integration by parts to get
$$
\langle p_N,p_M\rangle=(-1)^N\int_{-1}^1(x^2-1)^N\,\frac{d^{M+N}}{dx^{M+N}}(x^2-1)^M\,dx=0,
$$
where the derivative is zero as the degree of $(x^2-1)^M$ is less than $N+M$. 
When $N=M$, we need $$\frac{d^{2N}}{dx^{2N}}(x^2-1)^N.$$ The $2N$-derivative of a monic polynomial of degree $2N$ is the constant $(2N)!$. Thus (using Wolfram Alpha)
$$
\langle p_N,p_N\rangle = (-1)^N\int_{-1}^1(x^2-1)^N\,(2N)!\,dx=(-1)^N\,(2N)! \frac{\sqrt\pi(-1)^N\Gamma(N+1)}{\Gamma(N+\tfrac32)}=\frac{\sqrt\pi\,(2N)!\,N!}{\Gamma(N+\tfrac32)}
$$
A: Your $p_n$-s are the Legendre polynomials. I think you can do it by showing that $p_n$ satisfies the following differential equation:
$$\frac{\mathrm{d}}{\mathrm{d}x}\left[(1-x^2)\frac{\mathrm{d}p_n(x)}{\mathrm{d}x}\right]+n(n+1)p_n(x)=0$$
And then use the differential equation to show orthogonality: let $n \neq m$. Then we have that
$$\frac{\mathrm{d}}{\mathrm{d}x}\left[(1-x^2)\frac{\mathrm{d}p_n(x)}{\mathrm{d}x}\right]+n(n+1)p_n(x)=0$$
$$\frac{\mathrm{d}}{\mathrm{d}x}\left[(1-x^2)\frac{\mathrm{d}p_m(x)}{\mathrm{d}x}\right]+m(m+1)p_m(x)=0$$
Multiplying the first one by $p_m$ and the second one by $p_n$, we get that:
$$p_m(x)\frac{\mathrm{d}}{\mathrm{d}x}\left[(1-x^2)\frac{\mathrm{d}p_n(x)}{\mathrm{d}x}\right]+n(n+1)p_n(x)p_m(x)=0$$
$$p_n(x)\frac{\mathrm{d}}{\mathrm{d}x}\left[(1-x^2)\frac{\mathrm{d}p_m(x)}{\mathrm{d}x}\right]+m(m+1)p_m(x)p_n(x)=0$$
And subtract them:
$$p_m(x)p_n(x)[n(n+1)-m(m+1)]+p_m(x)\frac{\mathrm{d}}{\mathrm{d}x}\left[(1-x^2)\frac{\mathrm{d}p_n(x)}{\mathrm{d}x}\right]-p_n(x)\frac{\mathrm{d}}{\mathrm{d}x}\left[(1-x^2)\frac{\mathrm{d}p_m(x)}{\mathrm{d}x}\right]=0$$
Which can be rearranged:
$$p_m(x)p_n(x)[n(n+1)-m(m+1)]+[(1-x^2)(p_n'(x)p_m(x)-p_m'(x)p_n(x))]'=0$$
And integrating it from $-1$ to $1$:
$$\int_{-1}^{1}p_m(x)p_n(x)[n(n+1)-m(m+1)] \mathrm{d}x=0$$
