Can anybody help me with this problem please?
How do I find the taylor series of f(x) = (27 - x)^(1/3) centered at 0? Thanks
F(x) = f(0) + f(0)' * x / 1! +f(0)'' * x^2 / 2! + f(0)''' * x^3 /3! + .... (1)
now
f(0) = 3
f(x)' = -1/3 *( 27 - x ) ^ ( -2/3) ,,, --> f(0)' = - 1 / 27
f(x)'' = 2/9 *( 27 - x ) ^ ( -5 /3) ,,, --> f(0)'' = 2 / 2187
f(x)''' = -10/27 *( 27 - x ) ^ ( -8/3) ,,, --> f(0)''' = 10 / 177147
.....................
now by (1) we get
f(x)= 3 - (1/27) x + 2/2187 x^2 / 2! - 10/ 177147 x^3 / 3! + ....
f(x)= 3 - (1/27) x + 1/2187 x^2 - 5 / 531441 x^3 + ....
rem:- at taylor series , when x centered at 0 , it`s becam`s Maclaurin series