$n$th derivative of $(x^2-1)^n$

Question

Define $R_n(x)=\dfrac{d^n}{dx^n}(x^2-1)^n$. Show that $R_n(x)$ is orthogonal to $1,x,\ldots,x^{n-1}$ in $L^2([-1,1])$. Also, what is the value of $R_n(1)$?

By definition we have to show that $$\int_{-1}^1R_n(x)x^k=0$$ for $k=0,1,\ldots,n-1$. This looks a lot like integration by parts, so suppose $S_k$ is the $k$-th derivative of $(x^2-1)^n$. Then

$$\int_{-1}^1R_n(x)x^k=x^kS_{n-1}(x)\mid_{-1}^1-k\int_{-1}^1x^{k-1}S_{n-1}(x)$$.

It looks like the second term can be integrated by parts again, but what is the first term?

Also, how would it be possible to compute $R_n(1)$?

Answer 1

By Leibnitz Rule

$$\dfrac{d^n}{dx^n}(x-1)^n(x+1)^n= \sum_{k=0}^n \binom{n}{k} \left( \dfrac{d^k}{dx^k}(x-1)^n \right) \left(\dfrac{d^{n-k}}{dx^{n-k}}(x+1)^n\right)$$

Now, to evaluate $R_n(1)$, just observe that for $0 \leq k \leq n-1$ the term $\left( \dfrac{d^k}{dx^k}(x-1)^k \right)$ still contains an $x-1$. Thus

$$R_n(1)=\left( \dfrac{d^n}{dx^n}(x-1)^n \right) (x+1)^n |_{x=1}=n!2^n$$

P.S. The above also Yields:

$$\dfrac{d^n}{dx^n}(x-1)^n(x+1)^n= \sum_{k=0}^n \binom{n}{k} \left( \dfrac{n!}{(n-k)!}(x-1)^{n-k} \right) \left(\dfrac{n!}{k!}(x+1)^{k}\right) \\= \sum_{k=0}^n n! \binom{n}{k}^2(x-1)^{n-k}(x+1)^{k} $$

Answer 2

This can be differentiated via Lebnitz rule, $$\frac{d^{n-1}}{dx^{n-1}}((x^2 - 1)^{n-1} (x^2 - 1)) = (x^2 - 1)D^{n-1} (x^2 - 1)^{n-1} -2x \binom{n-1}{1} D^{n-2} (x^2 - 1)^{n-1} - 2 \binom{n-1}{3} D^{n-3} (x^2 - 1)^{n-1}$$

The first term is zero. The second term and third term can be evaluated in same way such the order of differentiation decreases and decreases until we get zero so we can argue that $D^k(x^2 - 1)^n$ is zero at $x=\pm 1$ if $k<n$.

To find out, $R_n(1)$, we expand binomially and then differentiate, also using the Rodrigue's formula for Legendre's polynomial, $$P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n}(x^2-1)^n$$ we can evaluate this. as $P_n(1) 2^n n!$