What is the fourth Taylor Polynomial of f(x)=(1-x)^2/3?

Answer 1

you need to say around which point....
let us say around x=0:

f(0) = 1
f'(x)= (2/3) (1-x)^{-1/3} ( -1) = -2/3(1-x)^{1/3}
f'(0) = -2/3
f''(x) = (-2/3)(-1/3) (1-x)^{-4/3} ( -1) = -2/9( 1-x)^{4/3}
f''(0) = -2/9
f'''(x) = (-2/9) (-4/3) ( 1-x) ^{-7/3} (-1) = -8/27(1-x)^{7/3}
f'''(0) = -8/27

so the taylor polynomial is:
1 + (-2/3) x + (-2/9) x^2 /2! + ( -8/27) x^3 / 3!
= 1- 2x/3 -x^2 /9 - 4 x^3 /81 .

Answer 2

The fourth polynomial is for n=3. So:
T3(x) = SUM(k=0,n)[(-1^k)(x-1)^k]

It is a very simple expansion.

Answer 3

Do you know what you are asking? If yes why can't you just do it?