Taylor Series

[summergrl]

Find the Taylor series for f(x) centered at the given value of a.

f(x)=x^3, a=-1

I got (-1(x+1)^k)/k!, is this right?

 

[skeeter]

the taylor series for x³ is finite and has a max order of 3 since f⁽⁴⁾(x) and beyond = 0.

f(x) = x³ ....... f(-1) = -1
f'(x) = 3x² ....... f'(-1) = 3
f"(x) = 6x ..........f"(-1) = -6
f'''(x) = 6 ...........f'''(-1) = 6
f⁽⁴⁾(x) = 0 ......... f⁽⁴⁾(-1) = 0

f(x) = f(-1) + f'(-1)(x+1) + f"(-1)(x+1)²/2! + f'''(-1)(x+1)³/3!

f(x) = -1 + 3(x+1) - 3(x+1)² + (x+1)³