+ 0iwX^RIt^RIHtRtR# ]dLi;i ]dLi;i)Nxcfa\S4p\V3Rl\V444p\V, P \^V4P 4pV^,.p\V4F0pVP VR,V,P44K2 \P!^4.p\^V4F<pVP WE,P\V^, 4V, 4K> VUu.uFp\P!V4NK ppV#uupi)atGiven a series f(x) = a[1]*x + a[2]*x**2 + ... + a[n-1]*x**(n - 1), use the Lagrange inversion formula to compute a series g(x) = b[1]*x + b[2]*x**2 + ... + b[n-1]*x**(n - 1) so that f(g(x)) = g(f(x)) = x mod x**n. We must have a[0] = 0, so necessarily b[0] = 0 too. The algorithm is naive and could be improved, but speed isn't an issue here and it's easy to read. c3Z<"TF pSV,\V,,xK" R#5i)Nr).0ias& ]/var/www/html/photoedit/myenv/lib/python3.14/site-packages/scipy/special/_precompute/utils.py %lagrange_inversion..s (x!AaDAIIxs(+) lensumrangerseriesremoveOappendexpandmpmpfcoeff)r nfhhpowerkbrsf r lagrange_inversionr s AA (uQx ((A 1 Q1%%'AdVF 1X vbz!|++-.  A 1a[ AE*1,-AqAA H s D.)mpmathr ImportError sympy.abcrrr r$sA        s))