Hi,


I have a question about Wynn's epsilon algorithm for extrapolation of sequences. Say I have a list of *N* sequences, with each sequence being of length *M*. The goal is to evaluate the extrapolated value --- *S[i]*, *i* = {*1, 2, ... N*} --- for each of the N sequences, which should give the dependence *S[i]* as a function of *i*. 


The concern is this: the Wynn extrapolation on each sequence, after say *k* iterations (with *k < M*), gives us *M - k* extrapolated values and it is not clear which of these *M - k* values one should choose as the final *S[i]*. It may so happen that for the sequence *i*, it is the *j* -th value (*j <= M - k*) that converges whereas for the sequence *i'* it is the *j'*-th value, with *j' != j*. By "converges", I mean the result *S[i]* must be a smooth curve between *i* and *i'* without sharp awkward jumps. My question is: is there any theorem for this extrapolation technique which states that if it is the *j*-th value that converges for a given sequence, it must only be the *j*-th value that also converges for a closely related (in terms of arising from a given unknown function) sequence?


Note that it is neither always helpful nor possible to always choose *k* such that *k = M - 1* because only every even-numbered iteration gives a converged list of *M - k* values.


Thanks,
VKV



Valid to use all Wynn-extrapolated values?
There really isn't a theorem you can use, since most of the applicable theory is for sequences with **known** asymptotic behavior. For "in the wild" sequences, you can do no better than to check that the results of the $\varepsilon$-algorithm remain sensible as you proceed. As a general rule however, for the recursion


$$\varepsilon\_{k+1}^{(n)}=\varepsilon\_{k-1}^{(n+1)}+\frac1{\varepsilon\_{k}^{(n+1)}-\varepsilon\_{k}^{(n)}}$$


the so-called "diagonal approximations" (borrowing from the theory of Padé approximants) $\varepsilon\_m^{(0)}$ or $\varepsilon\_m^{(1)}$, depending on which diagonal of the table you're already in, would be the most accurate approximations for the limit of your sequence. Still, for these algorithms, the human eye is a better judge of convergence than the computer, so have your Wynn routine print out the entire array if it can.


For more practical advice, have a look at [E.J. Weniger's survey paper](http://arxiv.org/abs/math/0306302).


