# Accurate bounds for derivatives of Legendre polynomials

Let $P_n(x)$ denote the $n$th Legendre polynomial. What bounds can one give for $d_{n,m}(x) = |\frac{d^m}{dt^m}P_n(t)|_{t=x}$ assuming that $|x| \le 1$? Clearly

$$d_{n,m}(x) \le d_{n,m}(1) = \frac{(m+n)!}{2^m m! (n-m)!}$$

works, but this is pessimistic unless is $x$ is very close to 1. For example, if $n = 100, m = 1$, the above bound gives $d_{n,m}(x) \le 5050$, but one can check numerically that $d_{n,m}(x) < 100$ when $|x| < 0.98$ and $d_{n,m}(x) < 10$ when $|x| < 0.5$, so the bound is far from accurate on most of $[-1,1]$.

I'm looking for a simple bound that captures the order of magnitude of the envelope of $d_{n,m}(x)$, ignoring the oscillations (the bound should ideally be a monotonically increasing function of $|x|$, and it should be easily computable like the bound given above, not involving Legendre functions etc.).

A bound of this kind is useful for error analysis when numerically evaluating $P_n(x)$ derivatives thereof. A solution just for $m = 1$ and $m = 2$ would be acceptable.

One difficulty here is that many formulas for Legendre polynomials break down close to $x = \pm 1$. But it might be sufficient to restrict the problem to say $|x| < 1 - C/n$ for some constant $C$ and use the derivative at $x = 1$ for $1 - C/n \le x \le 1$.

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A cheap way to improve the trivial bound is to use Bernstein (if you are far away from the endpoints ) or Markov (if you are close to the endpoints) inequalities for the derivatives of polynomials. In case $m=1,$ Berstein inequality gives
$$\|P'_n\|\le \frac{\|P_n\|}{\sqrt{1-x^2}},$$ while Markov leads to $$\|P'_n\|\le n^2\|P_n\|.$$