Expand 1+2x+3x2 +4x3 in powers of x+1
f(x) = 1+2x+3x2 +4x3
Taylor series expansion of f(x) in terms of (x-a)
f(x) = f(a) + (x-a) f'(a) + {(x-a)2 f''(a) }/2! + {(x-a)3 f'''(a) }/3! + {(x-a)4 f''''(a) }/4! + ---- Eqn 1
we need to expand f(x) in terms of (x+1) then a = -1
f(x) = 1+2x+3x2 + 4x3 then f(-1) = 1 - 2 + 3 - 4 = -2
f'(x) = 2 + 6x + 12x2 then f'(-1) = 2 - 6 + 12 = 8
f''(x) = 6 + 24x then f''(-1) = 6 - 24 = -18
f'''(x) = 24 then f'''(-1) = 24
f''''(x) = 0 then f''''(-1) = 0
Substituting this values in Eqn 1
f(x) = -2 + (x+1)8 - (x+1)2(18)/2! + (x+1)3(24)/3! + 0
f(x) = - 2 + (x+1)8 - 9(x+1)2 + 4(x+1)3
