 Finite observations do not certify Richardson extrapolation accuracy

Richardson extrapolation cancels a specified leading term in an asymptotic expansion, but an unquantified remainder does not certify a numerical error at two finite parameter values. For any positive distinct parameters $h_1,h_2$ and prescribed observations $v_1,v_2$, every candidate limit $L$ can be made consistent with a model $E(h)=L-c{h}^2-d{h}^3$: choose $c,d$ to solve $c{h}_i^2+d{h}_i^3=L-v_i$ for $i=1,2$. The coefficient matrix has nonzero determinant $h_1^2{h}_2^2(h_2-h_1)$. Thus the observations alone cannot identify $L$, even though every model has exactly the advertised $O(h^3)$ remainder in Big O notation. Quantitative remainder bounds or additional information are needed for an error certificate.