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Let $\{x_1,\dots,x_n\}$ be pairwise distinct complex numbers and $\{l_1,\dots,l_n\}$ a vector of natural numbers such that $l_1+l_2+\dots+l_n=N$. Let $$ h_j(x)=\prod_{i\neq j,i=1,\dots, n} (x-x_i)^{-l_i}. $$

Assume $|x_i|\leq 1$ for all $i=1,\dots,n$ and $|x_i-x_j|\geq \delta>0$ for all $i\neq j$.

Question: bound (from above) the successive derivatives of $h_j$ at $x_j$, i.e. $$ |h_j^{(t)}(x_j)|,\qquad t=0,\dots,l_j-1 $$ in terms of $N,n,\delta,t$.

(For motivation, see this MO question).

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1 Answer 1

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Leibniz formula applied to the logarithmic derivative of $h_j$ and an induction on $t\geqslant0$ lead to the conclusion that, for each $t\geqslant0$, there exists some universal polynomial $P_t$ of degree $t$ such that $$ |h^{(t)}_j(x_j)|\leqslant P_t(N)\,\delta^{-N-t}. $$ More precisely, $|h_j(x_j)|\leqslant \delta^{-N}$ and the identity $$ h'_j(x)=h_j(x)\sum_{i\ne j}\frac{\ell_i}{x-x_i}, $$ yields the recursion formula $$ h^{(t+1)}_j(x)=\sum_{k=0}^t{t\choose k}h^{(k)}_j(x)\sum_{i\ne j}\frac{(-1)^{t-k}(t-k)!\ell_i}{(x-x_i)^{t-k+1}}, $$ hence $$ |h^{(t+1)}_j(x_j)|\leqslant\sum_{k=0}^t{t\choose k}|h^{(k)}_j(x_j)|\sum_{i\ne j}\frac{(t-k)!\ell_i}{|x_j-x_i|^{t-k+1}}, $$ implies that $$ |h^{(t+1)}_j(x_j)|\leqslant\sum_{k=0}^t{t\choose k}\frac{P_k(N)}{\delta^{N+k}},\frac{(t-k)!N}{\delta^{t-k+1}}. $$ In other words, one can choose $P_0(N)=1$ and, for every $t\geqslant0$, $$ P_{t+1}(N)=N\sum_{k=0}^t\frac{t!}{k!}P_k(N). $$ This yields $P_t(N)=N(N+1)\cdots(N+t-1)$, and finally, $$ |h^{(t)}_j(x_j)|\leqslant\frac{\Gamma(N+t)}{\Gamma(N)\,\delta^{N+t}}. $$

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Thanks for the great answer! Exactly what I was looking for. –  dima_b Dec 1 '12 at 12:49
    
Hey, I wrote a short note incorporating your proof, based on this MO question. Would you like to be credited in addition to a usual reference? Plz contact me (see my user page) –  dima_b Dec 2 '12 at 16:26
    
The usual reference http://math.stackexchange.com/a/248559 to this MSE page is enough. –  Did Dec 2 '12 at 17:49

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