A calculation to first order gave 2 n (2 - 2) B 1 2 n x ~ - - ---------------------- 4 8 n (2 n - 1) E 2 n - 2 as a root of B_2n(x). This bottoms out at B_2n(x) ~ .8e-20 for n=25. But to first order, n 1 4 - 2 x ~ - - ---------- 4 2 n 2 %pi 4 or just 1 1 x ~ - - --------. 4 n 2 %pi 4 But these are vastly inferior, bottoming out at ~3e-10 for n=12. But they are almost exactly mutual negatives, so that their average, n 1 4 - 1 x ~ - - ---------- 4 2 n 2 %pi 4 manages to bottom out at ~ .5e-14 for n=19. So there was something fortuitously 2nd order about that initial calculation. In any case, we have the mild paradox that for large n, B_2n(1/4) -> oo, even as a root of B_2n(x) -> 1/4. --rwg PS, It seems to me that floating mantissas should display with the point to the right of the first digit if it's <5 (or maybe sqrt 10), and to the left if it's greater. This leaves the best exponent if the mantissa is discarded.