ó \è”h¸ãó•SSjrSrg)écór•UR5nURXU-5nUR5nX0- $)a5 Takes as input a polynomial expression and the variable used to construct it and returns the difference between function's value when the input is incremented to 1 and the original function value. If you want an increment other than one supply it as a third argument. Examples ======== >>> from sympy.abc import x, y, z >>> from sympy.series.kauers import finite_diff >>> finite_diff(x**2, x) 2*x + 1 >>> finite_diff(y**3 + 2*y**2 + 3*y + 4, y) 3*y**2 + 7*y + 6 >>> finite_diff(x**2 + 3*x + 8, x, 2) 4*x + 10 >>> finite_diff(z**3 + 8*z, z, 3) 9*z**2 + 27*z + 51 )ÚexpandÚsubs)Ú expressionÚvariableÚ incrementÚ expression2s ÚK/var/www/auris/envauris/lib/python3.13/site-packages/sympy/series/kauers.pyÚ finite_diffr s<€ð*×"Ñ"Ó$€JØ—/‘/ (°yÑ,@ÓA€KØ×$Ñ$Ó&€KØ Ñ #Ð#ócóz•URnURHnURUSUSS-5nM U$)ae Takes as input a Sum instance and returns the difference between the sum with the upper index incremented by 1 and the original sum. For example, if S(n) is a sum, then finite_diff_kauers will return S(n + 1) - S(n). Examples ======== >>> from sympy.series.kauers import finite_diff_kauers >>> from sympy import Sum >>> from sympy.abc import x, y, m, n, k >>> finite_diff_kauers(Sum(k, (k, 1, n))) n + 1 >>> finite_diff_kauers(Sum(1/k, (k, 1, n))) 1/(n + 1) >>> finite_diff_kauers(Sum((x*y**2), (x, 1, n), (y, 1, m))) (m + 1)**2*(n + 1) >>> finite_diff_kauers(Sum((x*y), (x, 1, m), (y, 1, n))) (m + 1)*(n + 1) ééÿÿÿÿr)ÚfunctionÚlimitsr)ÚsumrÚls r Úfinite_diff_kauersrs<€ð*|‰|€HØ ZŒZˆØ—=‘=  1¡ q¨¡v°¡zÓ2Šñà €Or N)r)r r©r r Úrsðô$ó4r