I have a sequence $\lambda_0,\lambda_1,\ldots,$ which is defined recursively as

$$ \lambda_0 = \frac{1}{2},$$


$$\lambda_{k+1} = \max_{\lambda\in [1,b]} \left(\frac{1}{2\lambda}\prod_{0\leq j\leq k}\left(\frac{\lambda-\lambda_j}{\lambda+\lambda_j}\right)^2\right), \qquad k\geq 0,$$

where $b>1$. I would like to find a relatively sharp upper bound for $\lambda_k$. Numerically, it seems that $\lambda_k$ is bounded by something like,

$$\lambda_k \leq \mathcal{O}\left(e^{-8k/\log(b)}\right). $$

For instance, the following very simple argument gives an upper bound that is very weak and I'm seeking techniques to do better:

Note that we have, for $k\geq 0$,

$$ \lambda_{k+1} \leq \max_{\lambda\in[1,b]} \left(\frac{1}{2\lambda}\prod_{0\leq j\leq k-1}\left(\frac{\lambda-\lambda_j}{\lambda+\lambda_j}\right)^2\right)\max_{\lambda\in[1,b]}\left(\frac{\lambda - \lambda_k}{\lambda+\lambda_k}\right)^2\leq \lambda_k \left(\frac{b-1}{b+1}\right)^2$$

and hence,

$$\lambda_k \leq \left(\frac{b-1}{b+1}\right)^{2k}\lambda_0 =\frac{1}{2}\left(\frac{b-1}{b+1}\right)^{2k},\qquad k\geq0.$$

I've already asked this on the Math Stack exchange, but it didn't receive any response:


Thank you in advance.

  • $\begingroup$ I think the usual name for this kind of definition is "recursive", not "implicit". An implicit definition would be of the form $f(\lambda_n)=0$, or $f(\lambda_n, \lambda_{n+1})=0$, etc. Each term in your sequence is defined quite explicitly as a function of the previous terms. Or did I miss something? $\endgroup$ – Goldstern Apr 28 '13 at 23:19
  • $\begingroup$ Yes, I also think the word "recursive" is better than "implicit". Changed. Thanks. I was thinking that the maximum made it "implicit" since the maximum probably does not have a closed form. $\endgroup$ – alext87 Apr 29 '13 at 6:09
  • $\begingroup$ How about using the estimate for $\lambda_k$ that you obtained, when estimating the second $\max$? Currently, for estimates that $\max$ you bound $\lambda_k$ by $1$. It is the bootstrapping method popularized by Münchausen :-) $\endgroup$ – Boris Bukh Apr 29 '13 at 14:58

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