find the taylor series of f(x) = 1/(x-1)^3 centered at a = 0

Question

series written in summation form Σ n=0 to infinity

Answer 1

Recall that the Taylor series for a function f(x) at x=a:

f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)²/2! + f'''(a)(x-a)³/3! + ... + fⁿ(a)(x-a)ⁿ/n!

Since you don't specify where to stop, I'll stop at the 3rd derivative.

At a=0:

f(a) = 1/(a-1)³ = 1/(0-1)³ = -1

f'(a) = -3/(a-1)⁴ = -3/(0-1)⁴ = -3

f''(a) = 12/(a-1)⁵ = 12/(0-1)⁵ = -12

f'''(a) = -60/(a-1)⁶ = -60/(0-1)⁶ = -60 <-- Again, we can stop here, but you could go further

f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)²/2! + f'''(a)(x-a)³/3! + ... + fⁿ(a)(x-a)ⁿ/n!

1/(x-1)³ ≈ f(0) + f'(0)(x-0) + f''(0)(x-0)²/2! + f'''(0)(x-0)³/3!

1/(x-1)³ ≈ -1 + (-3x) + (-12x²/2!) + (-60x³/3!)

1/(x-1)³ ≈ -1 - 3x - 6x² - 10x³ centered at a=0

Hope this helped!