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Consider the set of points $x_j=j/t,\ j\in\{0,1,\dots t\}$ (so we have equally spaced points on the unit interval). The lagrange basis polynomials are $$L_j(x)=\prod_{0\le m\le t,\ m\ne j}\frac{x-x_m}{x_j-x_m}=\prod_{0\le m\le t,\ m\ne j}\frac{x-\frac{m}{t}}{\frac{j}{t}-\frac{m}{t}}=\prod_{0\le m\le t,\ m\ne j}\frac{tx-m}{j-m}$$

If I calculated the double derivative of this correctly, we should be getting $$L_j''(x)=\sum_{\ell\ne j}\frac{t}{j-\ell}\left(\sum_{m\ne (j,\ell)}\frac{t}{j-m}\prod_{k\ne(j,\ell,m)}\frac{tx-k}{j-k}\right)$$ I suppressed the lower and upper limits, they are still from $0$ to $t$. I would like a good upper bound on $L_j''(1)$. My gut feeling tells me this is $\mathcal{O}(\text{poly}(t))$ but the naive bounds are all exponential.

Can this be shown to be upper bounded by a polynomial in $t$?


For context, I need to understand this term for a particular approximation scheme of polynomials given random but controlled evaluations. I was previously stuck on an approach (and had asked on mse), but if we can show this then we will be done!

HackR
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  • As it is a polynomial, you can use Cauchy's formula to express the derivative as a contour integral: does this yield a satisfying bound for you? – Nicolas Nov 16 '24 at 08:02
  • I haven't checked it, but let me try it when I get time. Thanks for the idea! – HackR Nov 16 '24 at 08:03

1 Answers1

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To avoid sums or products over the empty set, we assume that $t\ge 3$.

Suppose first that $j\ne t$. Then $$L''_j(1)=\frac t{j-t}\cdot\sum_{m=0,\, m\ne j}^{t-1} \frac t{j-m} \prod_{k=0,\, k\ne j,m}^{t-1} \frac{t-k}{j-k}=$$ $$ \frac {t^2}{j-t}\cdot\left(\prod_{k=0,\,k\ne j}^{t-1} \frac {t-k}{j-k}\right)\cdot \sum_{m=0,\, m\ne j}^{t-1} \frac 1{t-m}=$$ $$ \frac {t^2}{j-t}\cdot \frac{t!}{(t-j)j!(-1)^{t-1-j}(t-1-j)!} \cdot \left(H(t)-\frac 1{t-j}\right)= $$ $$ \frac {t^2}{j-t}\cdot (-1)^{t-1-j}{t\choose j} \cdot \left(H(t)-\frac 1{t-j}\right), $$

where $H(t)=\sum_{k=1}^t \frac 1{k}\sim \ln t$ is the partial sum of the harmonic series. We can estimate ${t\choose j}$ using bounds and asymptotic formulas for binomial coefficients. So, unfortunately, the upper bound can be exponential, because if $t=2j$ then ${t\choose j}\sim \frac{2^t}{\sqrt{\pi j}}$.

If you still need the value of $L''_t(1)$ then I expect that I can evaluate and estimate it similarly to the above.

Alex Ravsky
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    Ah I understand. Thank you so much! This actually made me realise a fundamental flaw in this aproach in solving my own problem. Thanks a bunch. – HackR Nov 20 '24 at 14:49